136 - Algebra.an.approach.via.module.theory.pdf 5d501

34

Chapter 1. Groups

It is easy to check that o(a) = n and that i3a/3 = a-'. If the vertices of P are numbered n, 1, 2, ... , n -1 counterclockwise starting at (1, 0), then Den is identified as a subgroup of S by

a,-4(12... n) (1 n - 1)(2 n - 2)

(In - 1)(2 n - 2)

((n - 1)/2 (n + 1)/2) (n/2 - 1 (n/2) + 1)

when n is odd, when n is even.

Thus, we have arrived at a concrete representation of the dihedral group that was described by means of generators and relations in Example 2.8 (13).

(5.12) Examples. (1)

If X is the rectangle in R2 with vertices (0, 1), (0, 0), (2, 0), and (2, 1) labelled from 1 to 4 in the given order, then the symmetry group of X is the subgroup

H = {e, (13)(24), (12)(34), (14)(23)} of S4, which is isomorphic to the Klein 4-group. (2) DB °-° S3 since D6 is generated as a subgroup of S3 by the permutations

a= (123) andp=(23). (3) D8 is a (nonnormal) subgroup of S4 of order 8. If a = (1234) and a = (13) then Ds = {e, a, a2, a3, Q, aQ, a2Q, a3,3}.

There are two other subgroups of S4 conjugate to Ds (exercise).

1.6 Direct and Semidirect Products (6.1) Definition. If N and H are groups the (external) direct product of N and H, denoted N x H, is the cartesian product set N x H with the multiplication defined componentwise, i.e.,

(n, h)(n', h') = (nn', hh'). It is easy to check that N x H is a group with this multiplication. Associated to N x H there are some natural homomorphisms

aN : N x H -. N

7TH:NxH-+H

((n, h) - n) ((n,h)i--ih)

LN:N-4NxH (n-*(n,e))

tH:H-.NxH (hs(e,h)).

1.6 Direct and Semidirect Products

35

The homomorphisms 7rN and nH are called the natural projections while LN and tH are known as the natural injections. The word canonical is used interchangeably with natural when referring to projections or injections. Note the following relationships among these homomorphisms Ker(7rH) = Im(LN) Ker(7rN) = Im(tH) 7rH o tH = 1H

7rNOLN=IN

(1c refers to the identity homomorphism of the group G). In particular, N x H contains a normal subgroup N = Im(LN) = Ker(7H) = N and a normal subgroup H = Im(tH) = Ker(7rN) 25 H

such that N fl H = {(e, e)} is the identity in N x H and N x H = NH. Having made this observation, we make the following definition.

(6.2) Definition. Let G be a group with subgroups N and H such that N f1 H = {e}

and NH=G.

(1) If N and H are both normal, then we say that G is the internal direct

(2)

product of N and H. If N is normal (but not necessarily H), then we say that G is the semidirect product of N and H.

The relationship between internal and external direct products is given by the following result. We have already observed that N x H is the internal

direct product of N and H, which are subgroups of N x H isomorphic to N and H respectively. (6.3) Proposition. If G is the internal direct product of subgroups N and

H, then G N x H.

Proof. Let a E G. Then a = nh for some n E N, h E H. Suppose we may also write a = n1h1 for some n1 E N, h1 E H. Then nh = n1h1 so that n-lnl = hhi 1 E N n H = {e}. Therefore, n = n1 and h = h1 so that the factorization a = nh is unique. Define f : G N x H by f (a) = (n, h) where a = nh. This function is well defined by the previous paragraph, which also shows that f is a one-toone correspondence. It remains to check that f is a group homomorphism.

36

Chapter 1. Groups

Suppose that a = nh and b = n1h1. Then ab = nhnlhl. We claim that hn1 = n1h for all n1 E N and h E H. Indeed, (hnlh-')nj 1 E N since N is normal in G and h(nIh-'ni 1) E H since H is normal. But N n H = {e}, so hnlh-Ini' E N n H = {e}, and thus, hn1 = n1 h. Therefore, f(ab) = f(nhnlhi) = f(nnlhhl) = (nn1, hh1) = (n, h)(n1, h1) = f(a)f(b) so that f is a group homomorphism, and, hence, a group isomorphism since it is a one-to-one correspondence. 0 (6.4) Examples. (1) Recall that if G is a group then the center of G, denoted Z(G), is the

set of elements that commute with all elements of G. It is a normal subgroup of G. Now, if N and H are groups, then it is an easy exercise

(do it) to show that Z(N x H) = Z(N) x Z(H). As a consequence, one obtains the fact that the product of abelian groups is abelian. (2) The group Z2 X Z2 is isomorphic to the Klein 4-group. Therefore, the two nonisomorphic groups of order 4 are Z4 and Z2 X Z2(3) All the hypotheses in the definition of internal direct product are necessary for the validity of Proposition 6.3. For example, let G = S3, N = A3, and H = ((12)). Then N oG but H is not a normal subgroup of G. It is true that G = NH and N n H = {e}, but G 9t N x H since G is not abelian, but N x H is abelian. (4) In the previous example S3 is the semidirect product of N = A3 and

H = ((12)). (6.5) Lemma. If G is the semidirect product of N and H then every a E G can be written uniquely as a = nh where n E N and h E H.

Proof. By hypothesis, G = NH, so existence of the factorization is clear. Suppose a = n1h1 = n2h2. Then n2'n1 = h2hi 1 E Nf1H = {e}. Therefore, 0 n1=n2 and h1=h2. According to this lemma, G is set theoretically the cartesian product set N x H, but the group structures are different. If G is the semidirect product of N and H, then the second isomorphism theorem (Theorem 3.12) shows that

H = H/(H n N)

(HN)/N = (NH)/N = GIN.

Thus, H is determined once we have N. A natural question is then, given all groups C such that G is the semidirect product groups N and H, H. As one answer to N and H of subgroups N and H where N this problem, we will present a construction showing how to produce all semidirect products. We start with the following definition:

1.6 Direct and Semidirect Products

37

(6.6) Definition. Let N and H be groups. An extension of N by H is a group G such that

(1) G contains N as a normal subgroup. (2) G/N H.

The first isomorphism theorem shows that for G to be an extension of N by H means that there is an exact sequence of groups and group homomorphisms

1-+N- -+ G--w-+ H-1. a

In this sequence, 1 = {e} and exactness means that it is surjective, B is injective, and Ker(7r) = Im(B). The extension G of N by H is a split extension if there is a homomor-

phism a : H - G such that 7r o a = 1H. In this case we say that the above sequence is a split exact sequence. The relationship between semidirect products and extensions is given by the following result:

(6.7) Proposition. G is a semidirect product of N and H if and only if G is a split extension of N by H. Proof. Suppose G is a semidirect product of N and H with N a G. Define 7r : G --+ H by a(a) = h where a = nh. Lemma 6.5 shows that it is well defined. To see that 7r is a homomorphism, note that hln2hi 1 = nz E N whenever hl, n2 E N (because N a G). Thus,

ir(nlhln2h2) = ir(nlnihlh2) = h1h2 = 7r(nlhl)x(n2h2), so it is a homomorphism. It is clear that Im(ir) = H and Ker(ir) = N. Let a : H -' G be the inclusion map, i.e., a(h) = h. Then it o a(h) = h for all h E H, so the extension determined by 7r is split.

Conversely, assume that G is a split extension of N by H with 7r : G - H and a : H - G the homomorphisms given by the definition of split extension. Then N= Ker(7r) a G and H = Im(a) is a subgroup of G. Suppose that a E N fl H. Then 7r(a) = e and a = a(h) for some h E H so that h = ir(a(h)) = 7r(a) = e. Therefore, a = a(e) = e, and we conclude that N fl H = (e}. Now let a E G and write

a = (a a(ir(a))-1) a(ir(a)) = nh. Clearly, h E H and ir(n) = ir(a - a(ir(a))-1) = x(a)lr(a(x(a))-1) = a(a)7r(a)-1 = e,

so n E N. Therefore, G is a semidirect product of N and k '_ H.

38

Chapter 1. Groups

(6.8) Remark. Comparing the definitions of semidirect product and direct

product, we see that if G is the semidirect product of N and H with H normal (in addition to N), then G is in fact the (internal) direct product of these subgroups. Of course, in an abelian group every subgroup is normal, so for abelian groups the notion of semidirect product reduces to that of direct product. In particular, we see from Proposition 6.7 that given a split exact sequence of abelian groups

1-+N e*G_1_4 H--+1 we have that G=NxH. We now consider a way to construct split extensions of N by H, which according to Proposition 6.7 is equivalent to constructing semidirect prod-

ucts. Let N and H be groups and let 0 : H - Aut(N) be a group homomorphism. We will write Oh E Aut(N) instead of 0(h). Then define

G = N x0 H = N )a H to be the set N x H with the multiplication defined by

(ni, hi) (n2, h2) = (nlOh, (n2), hi h2) We identify N and H with the subsets N x {e} and {e} x H, respectively. -

(6.9) Theorem. With the above notation,

(1) G = N x 0 H is a group, (2) H is a subgroup of G and N a G, (3) G is a split extension of N by H, and

(4) hnh-'=Oh(n) for all h E H C G and n E N C G. Proof. (1) (e, e) is easily seen to be the identity of C. For inverses, note that 1(n-'),h_')(n,h)

(Oh

= (Oh-1(n-')'Oh-l(n),h-'h) = (Oh-1(e),e) = (e,e)

and

(n,h)(Oh-1(n-'),h-') = (nOh(0h-1(n-'),hh-') = (nee (n') e) = (nn-', e) = (e, e) Thus, (n, h)-' = (Oh-, (n-') h-' ). To check associativity, note that ((ni, hi)(n2, h2))(n3, h3) _ (nIOh1(n2), hih2)(n3, h3) = (niOh,(n2)0h,h,(n3),hih2h3) = (nIOh,(n2)Oh,(Oh2(n3)),hih2h3)

= (nIOh,(n20h2(n3)),hih2h3)

= (nl,hi)(n20h2(n3),h2h3) = (n1, h1)((n2, h2) (n3, h3))

1.7 Groups of Low Order

39

(2) It is clear from the definition that N and H are subgroups of G.

Let r : N i 0 H -. H be defined by (b(n, h) = h. Then r is a group

homomorphism since a((n1,h1)(n2ih2)) = 1r(nl h,(n2),hlh2) = h1h2 = 7r(nl, hl)r(n2i h2) and N = Ker(7r), so N a G. (3) Let a : H -' G be defined by a(h) = (e, h). Then a is a homomorphism and r o a = 1H(4)

(e,h)(n,e)(e,h)-1 = (e,h)(n,e)(e,h'') _ (mh(n), h) (e, h-1)

_ (Oh(n)k (e),hh-1) _ (Oh (n), e).

0 (6.10) Examples.

(1) Let 0 : H - Aut(N) be defined by 0(h) = 1N for all h E H. Then N la 0 H is just the direct product of N and H.

(2) If 95: Z2 -, Aut(Zn) is defined by 1 i-+ 01(a) = -a where Z2 = 10, 1),

then Z )4 ,0 Z2 ti Den

(3) The construction in Example (2) works for any abelian group A in place of Zn and gives a group A A 0 Z2. Note that A m Z2 9t A X Z2 unless a 2 = e for all a E A. (4) Zpz is a nonsplit extension of Zp by Zp. Indeed, define r : Zpz -+ Zp by

r(r) = r (mod p). Then Ker(ir) is the unique subgroup of Zp2 of order

p, i.e., Ker(ir) = (p) C Zp2. But then any nonzero homomorphism a : Z, -, Zp2 must have IIm(a)I = p and, since there is only one subgroup of Zpa of order p, it follows that lm(a) = Ker(r). Therefore, r o a = 0 0 1Z, so that the extension is nonsplit. (6.11) Remark. Note that all semidirect products arise via the construction of Theorem 6.9 as follows. Suppose G = NH is a semidirect product. Define 0: H -. Aut(N) by Oh(n) = hnh-1. Then the map 4i : G --+ N xm H, defined by 4i(nh) = (n, h), is easily seen to be an isomorphism. Note that 4i is well defined by Lemma 6.5 and is a homomorphism by Theorem 6.9 (4).

1.7 Groups of Low Order This section will illustrate the group theoretic techniques introduced in this chapter by producing a list (up to isomorphism) of all groups of order at most 15. The basic approach will be to consider the prime factorization of

Chapter 1. Groups

40

IGI and study groups with particularly simple prime factorizations for their order. First note that groups of prime order are cyclic (Corollary 2.20) so that every group of order 2, 3, 5, 7, 11, or 13 is cyclic. Next we consider groups of order p2 and pq where p and q are distinct primes. (7.1) Proposition. If p is a prime and G is a group of order p2, then G =' Zp2

orC

Z,XZr.

Proof. If G has an element of order p2, then G Zp2. Assume not. Let e96 a E G. Then o(a) = p. Set N = (a). Let b E G with b V N, and set H = (b). Then N Z, and H Zp, and by Corollary 4.6, N a G and H a G; so

G°-NxH?5ZpxZp 0

by Proposition 6.3.

(7.2) Proposition. Let p and q be primes such that p > q and let G be a group of order pq. (1) If q does not divide p - 1, then G Z. (2) If q I p - 1, then G Zpq or G Zp x m

Z. when

0: Zq -' Aut(Zp) = Z. is a nontrivial homomorphism. All nontrivial homomorphisms produce isomorphic groups.

Proof. By Cauchy's theorem (Theorem 4.7) G has a subgroup N of order p and a subgroup H of order q, both of which are necessarily cyclic. Then N aG since (G : N) = q and q is the smallest prime dividing IGI (Corollary 4.6). Since it is clear that N n H = (e) and NH = G, it follows that G is the semidirect product of N and H. The map 0 : H - Aut(N) given by Oh (n) = hnh is a group homomorphism, so if q does not divide I Aut(N)I = I Aut(Zp)l = 1Z;1 = p - 1,

then 0 is the trivial homomorphism. Hence Oh = IN for all h E H, i.e., nh = hn for all h E H, n E N. Hence H a G and G Zp x Z. ?! Zpq (see Exercise 11). If q I p - 1 then there are nontrivial homomorphisms

0: Zq - Aut(N)

Zp

and for some homomorphism 0, G

ZP 14m Z.

Therefore, if N = (a) and H = (b), then G = (a, b), subject to the relations ap = e,

bq = e,

b-lab = ar

where rq - 1 (mod p). If r = 1 then 0 is trivial, H is normal, and G ZP X Z. Otherwise, G is nonabelian. We leave it as an exercise to


41

that all choices of r 54 1 produce isomorphic groups. Thus, if q I p - 1, then there are exactly two nonisomorphic groups of order pq.

(7.3) Corollary. If IGI = 2p, where p is an odd prime, then G G

Z2p or

D2p.

Proof. The only nontrivial homomorphism 0: Z2 -i Aut(Zp) = Zp is the homomorphism 1 -* 01 with ¢1(a) = -a. Apply Example 6.10 (2). (7.4) Remark. The results obtained so far completely describe all groups of order < 15, except for groups of order 8 and 12. We shall analyze each of these two cases separately.

Groups of Order 8 We will consider first the case of abelian groups of order 8.

(7.5) Proposition. If G is an abelian group of order 8, then G is isomorphic to exactly one of the following groups: (1) Z8,

(2) Z4 X Z2, or (3) Z2 X Z2 X Z2.

Proof. Case 1: Suppose that G has an element of order 8. Then G is cyclic and, hence, isomorphic to Z8. Case 2: Suppose every element of G has order 2. Let {a, b, c} C G\ {e} with c j4 ab. Then H = (a, b) is a subgroup of G isomorphic to Z2 X Z2. Furthermore, H n (c) _ (e) and H(c) = G so that

GL, Hx (c)S: Z2xZ2xZ2. Case 3: If G does not come under Case 1 or Case 2, then G is not cyclic and not every element has order 2. Therefore, G has an element a of order 4. We claim that there is an element b V (a) such that b2 = e. To see this, let c be any element not in (a). If c2 = e, take b = c. Otherwise, we must have o(c) = 4. Since IG/(a)l = 2, it follows that c2 E (a). Since a2 is the only element of (a) of order 2, it follows that c2 = a2. Let b = ac. Then b2 = a2c2 = a4 = e.

Proposition 6.3 then shows that

C

Z4xZ2.

Since every abelian group of order 8 is covered by Case 1, Case 2, or Case 3, the proof is complete.

42

Chapter 1. Groups

Now consider the case of nonabelian groups of order 8.

(7.6) Proposition. If G is a nonabelian group of order 8, then G is isomorphic to exactly one of the following two groups:

(1) Q = the quaternion group, or (2) D8 = the dihedral group of order 8. Proof. Since G is not abelian, it is not cyclic so G does not have an element of order 8. Similarly, if a2 = e for all a E G, then G is abelian (Exercise 8); therefore, there is an element a E G of order 4. Let b be an element of G not in (a). Since (G : (a)] = 2, the subgroup (a) a G. But IG/(a)I = 2 so that b2 E (a). Since o(b) is 2 or 4, we must have b2 = e or b2 = a2. Since (a) a C, b-lab is in (a) and has order 4. Since G is not abelian, it follows that b-lab = a3. Therefore, G has two generators a and b subject to one of the following sets of relations: (1) a4 = e,

b2 = e,

(2) a4 = e,

b2 = a2,

b-lab = a3;

b-lab = a3.

In case (1), G is isomorphic to Ds, while in case (2) G is isomorphic to Q. We leave it as an exercise to check that Q and D8 are not isomorphic.

0 (7.7) Remarks. (1) Propositions 7.5 and 7.6 together show that there are precisely 5 distinct isomorphism classes of groups of order 8; 3 are abelian and 2 are nonabelian. (2) D8 is a semidirect product of Z4 and Z2 as was observed in Example 6.10 (2). However, Q is a nonsplit extension of Z4 by Z2, or of Z2 by Z2 X Z2.

In fact Q is not a semidirect product of proper subgroups.

Groups of Order 12 To classify groups of order 12, we start with the following result.

(7.8) Proposition. Let G be a group of order p2q where p and q are distinct primes. Then G is the semidirect product of a p-Sylow subgroup H and a q-Sylow subgroup K.

Proof. If p > q then H a G by Corollary 4.6.

If q > p then 1 + kq p2 for some k > 0. Since q > p, this can only occur if k = 0 or 1 + kq = p2. The latter case forces q to divide p2-1 = (p+l)(p-1). Since q > p, we must have q = p+l. This can happen I

only if p _ 2 and q = 3. Therefore, in the case q > p, the q-Sylow subgroup K is a normal subgroup of G, except possibly when JGI = 22 - 3 = 12.


43

To analyze this case, let K be a 3-Sylow subgroup of a group G of order 12. If K is not normal in G, then the number of 3-Sylow subgroups of G is 4. Let these 3-Sylow subgroups be K1, K2, K3, and K4. Then K1 U K2 U K3 U K4 s for 9 distinct elements of G.

The remaining elements, together with the identity e, must form the 2-Sylow subgroup H of G. Hence, we must have H< G. Therefore, we have shown that at least one of H (a p-Sylow subgroup of G) or K (a q-Sylow subgroup of G) is normal in G. Since it is clear that H fl K = (e) and HK = G, it follows that G is a semidirect product of H

and K. (7.9) Proposition. A nonabelian group G of order 12 is isomorphic to exactly one of the following groups: (1) A4,

(2) D12, or

(3) T = Z3 AS Z4 where

:

Z4

Aut(Z3)

Z2 is the nontrivial

homomorphism.

Proof. Let H be a 2-Sylow subgroup and K a 3-Sylow subgroup of G. By Proposition 7.8 and the fact that G is nonabelian, exactly one of H and K is normal in G. Case 1: Suppose H 4 G. Then K is not normal in G. Since [G : K] = 4, G - S4. By Proposition 4.3, there is a permutation representation Ker(IiK) is the largest normal subgroup of G contained in K. Since K has prime order and is not normal, it follows that G is injective so that IM(4)K) C S4.

G

It is an easy exercise to show that A4 is the only subgroup of S4 of order 12; therefore, G A4 if the 2-Sylow subgroup is normal in G. Case 2: Suppose K< G and H Z4. In this case G

Z3 is 0 Z4

where 0: Z4 Aut(K) is a nontrivial homomorphism, but the only nontrivial automorphism of Z3 is a ' a-1 where K = (a). In this case G T. Case 3: Suppose K,* G and H °_5 Z2 X Z2. Let K = (a) and let

0:H-.Aut(K)2- Z2 be the conjugation homomorphism. Then H = (Ker(O)) X Z2, so let

Ker(0) = (c) and let d E H with 0(d) 34 1K. Then c-lac = a and d-lad = a-1 = a2. Let b = ac. Then o(b) = 6, d V (b), and

d-'bd = d-lacd = d-lade = a2c = (ac)-1 = b-1. Thus, G

D12.

44

Chapter 1. Groups

It remains to consider the case of abelian groups of order 12.

(7.10) Proposition. If G is an abelian group of order 12, then G is isomorphic to exactly one of the following groups: (1) Z12, or

(2) Z2 X Z6. Proof. Exercise.

By combining the results of this section we arrive at the following table

of distinct groups of order at most 15. That is, every group of order < 15 is isomorphic to exactly one group in this table.

Table 7.1. Groups of order < 15

Order

Abelian Groups

Nonabelian Groups

Total Number

I

{e}

1

2

Z2 Z3 Z4

1

3 4

1

2

Z2xZ2 5

6 7

8

Z5 Z6 Z7 Z8

Z4 X Z2

1

S3

2

Q

5

1

D8

Z2
Z9

2

Z3 X Z3 10 11

Z1o Z11

D1o

2

12

Z12

A4

5

Z2 X Z6

D12 Z3 X O Z4

13

Z13

14

Z14

15

Z15

1

1

D14

2 1

1.8 Exercises

45

1.8 Exercises 1.

2.

3. 4. 5.

6.

Prove that Z;, is a group. (See Example 1.2 (5).) Prove that P(X) (Example 1.2 (8)) with the symmetric difference operation is a group. Write the Cayley diagram for the group S3. Write the Cayley diagram for the group Z12. Let G be a group, g E G, and define a new multiplication on G by the formula a b = agb for all a, b E G. Prove that G with the multiplication is a group. What is the identity of G under ? If a E G what is the inverse of

a under.? Suppose that G is a set and is an associative binary operation on G such that there is an element e E G with e a = a for all a E G and such that for each a E G there is an element b E G with b a = e. Prove that (C, ) is a group. The point of this exercise is that it is sufficient to assume associativity, a left identity, and left inverses in order to have a group. Similarly, left can be replaced with right in the hypotheses.

7.

Prove that R' x R is a group under the multiplication defined by (a, b)(c, d) = (ac, ad + b).

8. 9.

Is this group abelian? Prove that if a2 = e for all a in a group G, then G is abelian. Let V C GL(2, R) be the set

V= 1[0

01

[

[ 01

0 1

10

[0

OiJ' ] Prove that V is a subgroup of GL(2, R) that is isomorphic to the Klein 01

4-group. 10.

For fixed positive integers bo,mo, and no consider the subset S C GL(3, Z) defined by

S=

11 m n [0

1

b

0

0

1

:molm, noln, bolb

ll

When is S a subgroup? The notation a 16 for integers a and b means that a divides b. 11.

Let G be a group and let a, b E G be elements such that ab = ba.

(a) Prove that o(# J o(a)o(b). b If ab = be and a n (b) = (e), show that

o(ab) = lcm{o(a), o(b)).

(Icm{n, m} refers to the least common multiple of the integers n and

M.)

(c) If ab = ba and o(a) and o(b) are relatively prime, then o(ab) = o(a)o(b). (d) Give a counterexample to show that these results are false if we do not assume commutativity of a and b. 12. If v : G -+ H is a group homomorphism then o(a(a)) I o(a) for all a E G with o(a) < oo. If o is an isomorphism then o(o(a)) = o(a). 13. (a) A group G is abelian if and only if the function f : G - G defined by f (a) = a-1 is a group homomorphism.

46

14.

15. 16.

17.

Chapter 1. Groups

(b) A group G is abelian if and only if the function g : G G defined by g(a) = a" is a group homomorphism. Let G be the multiplicative group of positive real numbers and let H be the additive group of all reals. Prove that G H. (Hint: the properties of the logarithm function.) Write all the subgroups of S3. Let G be a group and let H1, Hz be subgroups of G. Prove that H1 U H2 is a subgroup of G if and only if H1 C H2 or H2 C H1. Is the analogous result true for three subgroups H1, H2, H3? If G is a finite group and H and K are subgroups, prove that IHIIKI = IH n KIIHKI.

Prove that the intersection of two subgroups of finite index is a subgroup of finite index. Prove that the intersection of finitely many subgroups of finite index is a subgroup of finite index. 19. Let X be a finite set and let Y C X. Let C be the symmetric group Sx and define H and K by 18.

H={f G: f(y)=y forallyEY} K={f EG: f(y)EYforall yEY}. 20.

If IXI = n and IYI = m compute [G: H(, [G: KI, and [K: HI. If G is a group let Z(C) = {a E G : ab = ba for all b E G}. Then prove that Z(G) is an abelian subgroup of C. Z(C) is called the center of G. If G = GL(n, R) show that

Z(G) = {aI : a E R'}. Let G be a group and let H C Z(G) be a subgroup of the center of G. Prove that H a G. 22. (a) If G is a group, prove that the commutator subgroup G' is a normal subgroup of G, and show that G/C is abelian. (b) If H is any normal subgroup of G such that G/H is abelian, show that G'C_H. 23. If G is a group of order 2n show that the number of elements of G of order 21.

24.

2 is odd. Let Q be the multiplicative subgroup of GL(2, C) generated by

A= [0

U]

and

B= [ 0l 0].

(a) Show that A and B satisfy the relations A4 = I,A2 = B2,B-'AB = A-1. (Thus, is a concrete representation of the quaternion group.) b) Prove that IQI = 8 and list all the elements of Q in of A and B. c) Compute Z(Q) and prove that Q/Z(Q) is abelian. d Prove that every subgroup of Q is normal. 25.

Let n be a fixed positive integer. Suppose a group G has exactly one subgroup

H of order n. Prove that H a C. Let HaG and assume that G/H is abelian. Show that every subgroup K C G containing H is normal. 27. Let G. be the multiplicative subgroup of GL(2, C) generated by 26.

A=[

'

1]

and B=[? ']

1.8 Exercises

47

where(

exp(21ri/n). that Gn is isomorphic to the dihedral group Dan. (See Example 2.8 (13).) 28. Let G be a group of order n. If G is generated by two elements of order 2,

Dn ifn>4.

showthatG -e Zs xZ2

29. Let G be a nonabelian group of order 6. Prove that G °_° S3. 30. (a) b J If H a C and (G : H) = n, then show that an E H for all a E G.

that the result in part (a) is false if H is not normal in G.

31. Show that the alternating group A4 of order 12 does not have a subgroup

of order 6. (Hint: Find at least 8 elements of A4 that are squares, and apply Exercise 30.)

Recall (Definition 4.13) that a group C is called a p-group if IGI = pn for some integer n > 1. (a) If G is a p-group, show that Z(G) 0 (e). (Hint: Use the class equation Corollary 2.28).) (b) if ICI = pn, show that G has a subgroup of order p'n for every 0 < m < n. 33. Let G = { [ o b ] E GL(2, R) }. Prove that G is a subgroup of GL(2, R) and 32.

that G is isomorphic to the group R' x R with the multiplication defined in Exercise 7. 34.

(a) Find all homomorphisms 0: Z -. Zn. b Find all homomorphisms 0: Z7 Z16c What is a condition on finite cyclic groups G and H that ensures there is a homomorphism 0: G H other than the zero homomorphism?

Let Hom(Zn, Zn,) be the set of all group homomorphisms from Zn to Z. Let d be the greatest common divisor of m and n. Show that I Hom(Zn, Z,)I = d. 36. If n is odd, show that Don 2-1 Den X Z237. Write the class equations (Corollary 2.28) for the quaternion group Q and the dihedral group Ds. 35.

38.

that the alternating group A5 has no nontrivial normal sub oups.

((Hint: The class equation.) (The trivial subgroups of a group G are {e} and

G.) A group with no nontrivial normal subgroups is called simple. It is

known that A. is simple for all n 34 4. Suppose that G is an abelian group of order n. If m I n show that G has a subgroup of order m. Compare this result with Exercise 31. 40. (a) Write each of the following permutations as a product of dist cycles: 39.

a= (16

2

3

4

5

61

5

4

1

2

3

(1

2

6

7

8)

1

4 6

5

8

3 3

5

7

4

2

1

2

3

5

3

4

6

6 7

7 8

8 9

9)

(2

4 5

(1

2

3

5

6

7

8

9

1

4

3

8 6

91

5

R= 7=

4 2

1) 7

(b) Let a E Slo be the permutation

a_(1 2

3

4

5

6

5

4

1

7

10

7 2

8 6

9 10)3

9

8

Compute o(a) and calculate aloo. 41. 42.

Let H C S. be defined by H = {f ESn:j1)=1}. Prove that His a subgroup of S. that is isomorphic to Sn_1. Is H a Sn? (a) Prove that an r-cycle is even (odd) if and only if r is odd (even).

48

Chapter 1. Groups

(b) Prove that a permutation a is even if and only if there are an even number of even order cycles in the cycle decomposition of or.

43. Show that if a subgroup G of Sn contains an odd permutation then G has a

normal subgroup H with [C : HI = 2.

44. For a E Sn, let

f(a)={(i,j):1
(2 5

1

4

a(j) < a(i)}I.

3)ESS,

then j(a) = 5.) Show that sgn(a) = 1 if j(a) is even and agn(a) = -1 if T (o,) is odd. Thus, f provides a method of determining if a permutation is even or odd without the factorization into dist cycles. 45.

a Prove that S. is generated by the transpositions 12), (13), ..., (1 n). b Prove that Sn is generated by (12) and (12 n).

In the group S4 compute the number of permutations conjugate to each of the following permutations: e = (1), a = (12), Q = (12 3), ry = (12 3 4), and 6 = (12)(3 4). 47. (a) Find all the subgroups of the dihedral group Da. b Show that Ds is not isomorphic to the quaternion group Q. Note, however, that both groups are nonabelian groups of order 8. (Hint: Count the number of elements of order 2 in each group.) 48. Construct two nonisomorphic nonabelian groups of order p3 where p is an odd prime. 49. Show that any group of order 312 has a nontrivial normal subgroup. 50. Show that any group of order 56 has a nontrivial normal subgroup. 51. Show Aut(Zs X Z3) °_r S3. 52. How many elements are there of order 7 in a simple group of order 168? (See Exercise 38 for the definition of simple.) 53. Classify all groups (up to isomorphism) of order 18. 54. Classify all groups (up to isomorphism) of order 20. 46.

Chapter 2

Rings

2.1 Definitions and Examples (1.1) Definition. A ring (R, +, ) is a set R together with two binary opR (addition) and : R x R R (multiplication) erations +: R x R satisfying the following properties.

(a) (R, +) is an abelian group. We write the identity element as 0. (b) a (b c) = (a b) c ( is associative). and right (c) distributive over +). As in the case of groups, it is conventional to write ab instead of a b. A ring will be denoted simply by writing the set R, with the multiplication and addition being implicit in most cases. If R # {0} and multiplication on R has an identity element, i.e., there is an element 1 E R with al = la = a for all a E R, then R is said to be a ring with identity. In this case 1 0

(see Proposition 1.2). We emphasize that the ring R = {0} is not a ring with identity.

Convention. In Sections 2.1 and 2.2, R will denote an arbitrary ring. In the rest of this book, the word "ring" will always mean "ring with identity. "

If the multiplication on R is commutative, i.e., ab = ba for all a, b E R, then R is called a commutative ring. The standard rules of sign manipulation, which are familiar from the real or complex numbers, are also true in a general ring R. The verification of the following rules are left as an exercise.

(1.2) Proposition. Let R be a ring. Then if a, b, c E R the following rules are valid.

(1) aO=0a=0.

50

Chapter 2. Rings

(2) (-a)b = a(-b) = -(ab). (3) ab = (-a)(-b). (4) a(b - c) = ab - ac and (a- b)c = ac - bc. (5) If R has an identity then (-1)a = -a. (6) If R has an identity then 1 # 0. If a # 0 and b # 0 are elements of R such that ab = 0 then a and b are called zero divisors of the ring R. Note that 0 is not a zero divisor. If R has an identity, an element a E R is a unit if a has a multiplicative inverse, that is, if there is a b c R with ab = 1 = ba. R' will denote the set of all units of R. R' is a group, called the group of units of R.

(1.3) Lemma. Let R be a ring and a # 0 an element of R that is not a zero divisor. If b, c E R satisfy ab = ac, then b = c. Similarly, if ba = ca, then b

c.

Proof. If ab = ac, then

0=ab-ac=a(b- c), and since a is not a zero divisor and a b = c. The other half is similar.

0, this implies that b - c = 0, i.e.,

(1.4) Definition. (1) A ring R is an integral domain if it is a commutative ring with identity such that R has no zero divisors. (2) A ring R with identity is a division ring if R' = R\ {0}, i.e., every nonzero element of R has a multiplicative inverse. (3) A field is a commutative division ring. Thus, a commutative ring R with identity is an integral domain if and only if the equation ab = 0 (for a, b E R) implies that a = 0 or b = 0. R is a division ring if and only if the equations ax = b and ya = b are solvable

in Rfor every bERanda54 0ER. (1.5) Proposition. A finite integral domain is a field. Proof. Let R be a finite integral domain and let a 54 0 E R. Define 0a

R -. R by ma(b) = ab. Suppose that Oa(b) = Oa(c). Then ab = ac, so b = c by Lemma 1.3. Therefore, fa is an injective function on the finite set R so that JRI = 10a(R)I, and hence, 4a(R) = R. In particular, the equation ax = 1 is solvable for every a 0 0 and R is a field. (1.6) Remark. The conclusion of Proposition 1.5 is valid under much weaker

hypotheses. In fact, there is a theorem of Wedderburn that states that the commutativity follows from the finiteness of the ring. Specifically, Wedderburn proved that any finite division ring is automatically a field. This result

2.1 Definitions and Examples

51

requires more background than the elementary Proposition 1.5 and will not be presented here.

(1.7) Definition. If R is a ring with identity, then the characteristic of R, denoted char(R), is the smallest natural number n such that n 1 = 0. If

0 forallnEN, then we set char(R) = 0. (1.8) Proposition. If R is an integral domain, then char(R) = 0 or char(R) is prime.

Proof. Suppose that char(R) = n j4 0. If n is composite, then we may write n = rs where 1 < r < n and 1 < s < n. Then, by the definition of characteristic, r 1 96 0 and s 1 0. But 0 = n 1 = (r 1)(s 1). Therefore, the ring R has zero divisors. This contradicts the fact that R is an integral 0 domain, so n must be prime.

(1.9) Definition. A subset S of a ring R is a subring if S, under the operations of multiplication and addition on R, is a ring. Thus S is a subring of R if and only if S is an additive subgroup of R that is closed under multiplication.

We will now present a number of examples of rings. Many of the math-

ematical systems with which you are already familiar are rings. Thus the integers Z are an integral domain, while the rational numbers Q, the real numbers R, and the complex numbers C are fields. (1.10) Examples.

(1) 2Z = {even integers} is a subring of the ring Z of integers. 2Z does not have an identity and thus it fails to be an integral domain, even though it has no zero divisors. (2) Z,,, the integers under addition and multiplication modulo n, is a ring with identity. Z has zero divisors if and only if n is composite. Indeed,

ifn=rs for 1
we arrive at an equation rs = n where 0 < r < n and 0 < s < n, i.e., n is composite. (3) Example (2) combined with Proposition 1.5 shows that Z is a field if and only if n is a prime number. In particular, we have identified some finite fields, namely, Z. for p a prime. (4) There are finite fields other than the fields ZP. We will show how to construct some of them after we develop the theory of polynomial rings.

52

Chapter 2. Rings

Table 1.1. Multiplication and addition for a field with four elements 0

1

a

b

0

1

a

b

0

0

1

a

b

0

0

0

0

0

1

1

0

b

a

1

1

a

b

a

a

b

0

1

a

0 0

a

b

1

b

b

a

1

0

b

0

b

1

a

For now we can present a specific example via explicit addition and multiplication tables. Let F = {0, 1, a, b} have addition and multiplication defined by Table 1.1. One can check directly that (F, +, ) is a field with 4 elements. Note that the additive group (F, +) °-° Z2 X Z2 and that the multiplicative group (F', ) Z3(5) Let Z[i] = {m + ni : m, n E Z}. Then Z[i( is a subring of the field of complex numbers called the ring of gaussian integers. As an exercise, check that the units of Z[ij are (±l, ±i).

(6) Let d # 0,1 E Z be square-free (i.e., n2 does not divide d for any

n> 1)andletQ(fd]={a+bf :a,bEQ}c C. Then Q[fd(isa subfield of C called a quadratic field.

(7) Let X be a set and P(X) the power set of X. Then (P(X),A,n) is a commutative ring with identity where addition in P(X) is the symmetric difference (see Example 1.2 (8)) and the product of A and B is A n B. For this ring, 0 = 0 E P(X) and 1 = X E P(X ).

(8) Let R be a ring with identity and let M,n,n(R) be the set of m x n matrices with entries in R. If m = n we will write M,,(R) in place of MM,n(R). If A = [aid( E M,n,n(R) we let enti3(A) = aid denote the entry of A in the ith row and j1h column for 1 < i < m, 1 < j < n. If A, B E M,n,n(R) then the sum is defined by the formula

enti3(A + B) = enti?(A) + ent,,(B), while if A E M,n,n(R) and B E M,,,p(R) the product AB E Mm,p(R) is defined by the formula n

enti,(AB) = E entik (A) entkj (B). k=1

In particular, note that addition and multiplication are always defined for two matrices in M,,(R), and with these definitions of addition and multiplication, M,,(R) is a ring with identity, called a matrix ring. The identity of M,, (R) is the matrix In defined by entij (I,) = bid where bid is the kronecker delta

b"

_

1

0

ifi=j, ifi0j.


53

There are mn matrices E1,, (1 < i < m, 1 < j < n) in Mm,n that are particularly useful in many calculations concerning matrices. E;, is defined by the formula entkl (E 3) = bk;61j i that is, E;; has a 1 in the ij position and 0 elsewhere. Therefore, any A = (a;,] E M,n,n(R) can be written as m [nom

A=EEa'JE,,. i=1

j=1

Note that the symbol E,,; does not contain notation indicating which Mm,n(R) the matrix belongs to. This is determined from the context. There is the following matrix product rule for the matrices E;j (when the matrix multiplications are defined): EE,Ekl =bjkE1.

(1.2)

In case m = n, note that Eili = E,,, and when n > 1, E11E12 = E12 while E12E11 = 0. Therefore, if n > 1 then the ring Mn(R) is not commutative and there are zero divisors in Mn(R). The matrices E1, E M,,(R) are called matrix units, but they are definitely not (except for

n = 1) units in the ring Mn(R). A unit in the ring Mn(R) is an

(9)

invertible matrix, so the group of units of M,(R) is called the general linear group GL(n, R) of degree n over the ring R. There are a number of important subrings of M,,(R). To mention a few, there is the ring of diagonal matrices

Dn(R)={AEMn(R):ent11(A)=0 if i54 j}, the ring of upper triangular matrices

Tn(R)={AEMn(R):enti,(A)=0if i > j}, and the ring of lower triangular matrices

Tn(R)={A EMn(R):ent1,(A)=0if i<j}. All three of these subrings of Mn(R) are rings with identity, namely the identity of Mn(R). The subrings of strictly upper triangular matrices STn and strictly lower triangular matrices STn do not have an identity.

A matrix is strictly upper triangular if all entries on and below the diagonal are 0 and strictly lower triangular means that all entries on and above the diagonal are 0. (10) Let F be a subfield of the real numbers R and let x, y E F with x > 0 and y > 0. Define a subring Q(-x, -y; F) of M2(C) by

Q(-x, -y; F) =

a+b// c/+d cv-d a-b U

a, b,c,d EF}.

54

Chapter 2. Rings

(In these formulas

and v/'--y denote the square roots with positive imaginary parts.) It is easy to check that Q(-x, -y; F) is closed under matrix addition and matrix multiplication so that it is a subring of M2(C). Let

[

1- [0

j=

i-

1]'

0

1,

[

0

-,/=x]'

and k= [- 0

0

Then Q(-x, -y; F) = {al + bi + cj + dk : a, b, c, d E F}. Note that

i2 = -XI, j2 = -yl, k2 = -xyl ij

- -ji=k

ik = -ki = xj

kj=-jk=yi. Ifh=al+bi+cj+dkEH,leth=al-bi-cj-dk.Then hh = (a2 + xb2 + yc2 + xyd2)1

so that if h 96 0 then h is invertible, and in particular, h-' = ah where a = 1/(a2 + xb2 + yc2 + xyd2). Therefore, Q(-x, -y; F) is a division ring, but it is not a field since it is not commutative. Q(-x, -y; F) is called a definite quaternion algebra. In case F = R, all these quaternion algebras are isomorphic (see Exercise 13 for some special cases of this fact) and H = Q(- 1, -1; R) is called the ring of quaternions. (The notation is chosen to honor their discoverer, Hamilton.) Note that the subset

{±1, ±1, ±j, ±k} of H is a multiplicative group isomorphic to the quaternion group Q of Example 1.2.8 (12). (Also see Exercise 24 of Chapter 1.) In case F is

a proper subfield of R, the quaternion algebras Q(-x, -y; F) are not all mutually isomorphic, i.e., the choice of x and y matters here. (11) Let A be an abelian group. An endomorphism of A is a group homoA. Let End(A) denote the set of all endomorphisms morphism f : A of A and define multiplication and addition on End(A) by

(f + g)(a) = f (a) + g(a) (fg) (a) = f(g(a))

(addition of functions), (functional composition).

With these operations, End(A) is a ring (exercise). In general, End(A) is not commutative. The group of units of End(A) is the automorphism group of A. (12) Let Z+ denote the set of all nonnegative integers. If R is a ring with identity, let R[X] denote the set of all functions f : Z+ -. R such that


55

f (n) = 0 for all but a finite number of natural numbers n. Define a ring structure on the set RIX] by the formulas (f + 9)(n) = f (n) + 9(n) n

f (m)g(n - m).

(fg)(n) = M=0

It is easy to check that RIX] is a ring with these operations (do it). RIX] is called the ring of polynomials in the indeterminate X with coefficients in R. Notice that the indeterminate X is nowhere mentioned in our definition of RIX]. To show that our description of RIX] agrees with that with which you are probably familiar, we define X as a function on Z+ as follows

X (n) _

1

ifn = 1,

0

ifn54l.

Then the function Xn satisfies

Xn(m) _

1

if m = n,

0 ifmon.

Therefore, any f E RIX] can be written uniquely as 00

f = 1: f(n) Xn n[=0

where the summation is actually finite since only finitely many f (n) #

0. Note that X° means the identity of RIX], which is the function 1 : Z+ --+ R defined by

1(n)=

J1 ifn=0, l0 ifn>O.

We do not need to assume that R is commutative in order to define the polynomial ring RIX], but many of the theorems concerning polynomial rings will require this hypothesis, or even that R be a field. However, for some applications to linear algebra it is convenient to have polynomials over noncommutative rings. (13) We have defined the polynomial ring RIX] very precisely as functions from Z+ to R that are 0 except for at most finitely many nonnegative integers. We can similarly define the polynomials in several variables. Let (Z+)n be the set of all n-tuples of nonnegative integers and, if R is a commutative ring with identity, define RIX,,... , Xn] to be the set

of all functions f : (Z+)n -+ R such that f (a) = 0 for all but at most finitely many a E (Z+)n. Define ring operations on RIX,,... , Xn] by

56

Chapter 2. Rings

(f +9)(a) = f(a) +9(a) (fg)(a) = f(Q)9(,)

for all a E (Z+)n

Define the indeterminate X; by

X, (a) = J

1

0 otherwise.

If a = (al,...,an) E (Z+)n we write X° =

and we leave it as an exercise to check the following formula, which corresponds to our intuitive understanding of what a polynomial in several variables

is. If f E R[X,,... , Xn] then we can write

f = [1 a°Xa oE(Z++)n

where as = 0 except for at most finitely many a E (Z+)n. In fact, a° = f (a). Note that R is a subring of R[X,,... , Xn) in a natural way, and, more generally, R[X1, ... , Xn_1] is a subring of R[X1, ... , Xn]. (14) If R is a ring with identity then the ring of formal power series R[[X]] is defined similarly to the ring of polynomials. Specifically, R[[X]] is the set of all functions f : Z+ R with the same formulas for addition and multiplication as in R[X]. The only difference is that we do not assume that f (n) = 0 for all but a finite number of n. We generally write a formal power series as an expression 00

f(X) = EanXn n=0

Since we cannot compute infinite sums (at least without a topology and a concept of limit), this expression is simply a convenient way to keep track of f (n) for all n. In fact, an = f (n) is the meaning of the above equation. With this convention, the multiplication and addition of formal power series proceeds by the rules you learned in calculus for manipulating power series. A useful exercise to become familiar with algebra in the ring of formal power series is to that f E R[[X]J

is a unit if and only if f (0) is a unit in R. (15) Let G be a group and let R be a ring with identity. Let R(G) be the set of all functions f : G - R such that f (a) 0 0 for at most a finite number of a E G. Define multiplication and addition on R(G) by the formulas (f + 9)(a) = f(a) + 9(a)

f(b)9(b-la).

(fg)(a) =

bEG

Note that the summation used in the definition of product in R(G) is a finite sum since f (b) 16 0 for at most finitely many b E G. The ring


57

R(G) is called the group ring of C with coefficients from R. This is a ring that is used in the representation theory of groups.: he product in R(G) is called the convolution product. If S is only a semigroup (a set with an associative binary operation) then one can form a similar ring called the semigroup ring R(S). If S is a monoid, R(S) is a ring with identity. We leave the details to the reader, but we point out that the semigroup ring R(Z+) is nothing more than the polynomial ring R[X].

This list of examples should be referred to whenever new concepts for rings are introduced to see what the new concepts mean for some specific rings.

We conclude this introductory section by commenting that the generalized associative laws proved for groups in Proposition 1.1.4 is also valid for multiplication in a ring since the proof of the group theoretic result used only the associative law for groups; inverses and the group identity were not used. In particular, if R is a ring and a1,.. . , a,, are elements of R, then the product 111 ai is well defined so that we can define a" (n > 1), and if R has an identity we can also define a° = 1. Since a ring has two operations related by the distributive laws, there should be some form of generalized distributive law valid for rings; this is the content of the following result:

(1.11) Proposition. Let R be a ring and let a1, ..., am, Then

m

b1i

..., bn E R.

n

+ b")

(a1 + ... + am)(bl +

i=1 j=1

a'bj.

Proof. For m = 1 the proof is by induction on n using the left distributive law. Then proceed by induction on m. 0 Recall that the binomial coefficients are given by (T) = n!/(r!(n - r)!). The binomial theorem is proved by induction on n, exactly the same as the proof for real numbers; we leave the proof as an exercise.

(1.12) Proposition. (Binomial theorem) Let R be a ring with identity and let a, b E R with ab = ba. Then for any n E N n (a + b)" = k= 0

Proof. Exercise.

()a1bT_1c. k

0

Chapter 2. Rings

58

2.2 Ideals, Quotient Rings, and Isomorphism Theorems A function f : R -' S, where R and S are rings, is a ring homomorphism if

f(a + b) = f(a) + f(b) and

f(ab) = f(a)f(b)

for all

a,bE R.

If f is invertible (i.e., there exists a ring homomorphism g : S R such that fog = 1S and g o f = 1R), then we say that f is a ring isomorphism. As with group homomorphisms, f is invertible as a ring homomorphism if it

is a bijective function. If f is a ring homomorphism, then we let Ker(f) =

{a E R : f (a) = 01 and Im(f) = {b E S : b = f (a) for some a E R}. Thus Ker(f) and Im(f) are the kernel and image, respectively, of f when viewed as a group homomorphism between the abelian group structures on R and S so that Ker(f) and Im(f) are abelian subgroups of R and S respectively. Moreover, since f also preserves multiplication, it follows that Ker(f) and Im(f) are subrings of R and S respectively. From our study of groups we know that not every subgroup of a group can be the kernel of

a group homomorphism--the subgroup must be normal. In the case of a ring homomorphism f : R - S, Ker(f) is automatically normal since R is an abelian group under addition, but the multiplicative structure imposes a restriction on the subring Ker(f). Specifically, note that if a E Ker(f)

and r E R, then j(ar) = f(a)f(r) = Of(r) = 0 and f(ra) = f(r)f(a) = f (r)0 = 0 so that ar E Ker(f) and ra E Ker(f) whenever a E Ker(f) and r E R. This is a stronger condition than being a subring, so we make the following definition.

(2.1) Definition. Let R be a ring and let I C R. We say that I is an ideal of R if and only if (1) 1 is an additive subgroup of R,

(2) rl C I for all r E R, and (3) Ir C I for all r E R. A subset I C R satisfying (1) and (2) is called a left ideal of R, while if I satisfies (1) and (3), then I is called a right ideal of R. Thus an ideal of R is both a left and a right ideal of R. Naturally, if R is commutative then the concepts of left ideal, right ideal, and ideal are identical, but for noncommutative rings they are generally distinct concepts. We have already observed that the following result is true.

(2.2) Lemma. If f : R -. S is a ring homomorphism, then Ker(f) is an ideal of R.

2.2 Ideals, Quotient Rings, and Isomorphism Theorems

59

Every ring R has at least two ideals, namely, {0} and R are both ideals of R. For division rings, these are the only ideals, as we ee from the following observation.

(2.3) Lemma. If R is a division ring, then the only ideals of R are R and {0}.

Proof. Let I C R be an ideal such that I# {0}. Let a 0 0 E I and let b E R. Then the equation ax = b is solvable in R, so b E I. Therefore, 0 I=R. (2.4) Corollary. If R is a division ring and f : R - S is a ring homomorphism then f is injective or f = 0. Proof. If Ker(f) = {0} then f is injective; if Ker(f) = R then f = 0.

0

(2.5) Remarks. (1) In fact, the converse of Lemma 2.2 is also true; that is, every ideal is the kernel of some ring homomorphism. The proof of this requires the construction of the quotient ring, which we will take up next. (2) The converse of Lemma 2.3 is false. See Remark 2.28.

If I C R is an ideal then the quotient group R/1 is well defined since I is a subgroup (and hence a normal subgroup) of the additive abelian

group R. Let a : R -+ R/I, defined by rr(r) = r + I, be the natural projection map. We will make the abelian group R/I into a ring by defining

a multiplication on R/I. First recall that coset addition in R/I is defined by (r + I) + (s + 1) = (r + s) + I. Now define coset multiplication by the formula (r + I)(s + I) = rs + I. All that needs to be checked is that this definition is independent of the choice of coset representatives. To see this,

supposer+I=r'+I and s+I = s'+I. Then r' =r+a ands'=s+b where a, b E I. Thus, r's' = (r + a)(s + b)

=rs+as+rb+ab = rs+c where c = as + rb + ab E I because I is an ideal. Therefore, rs + I = r's' + I and multiplication of cosets is well defined. By construction the natural map rr : R -+ R/I is a ring homomorphism so that I = Ker(rr) is the kernel of a ring homomorphism. Note that if R is commutative then R/I is commutative for any I, and if R has an identity and I is a proper ideal then 1 + I is the identity of the quotient ring R/I. If a, b E R we will use the notation a =_ b (mod I) to mean a - b E I, i.e., a + I = b + I. This is a generalization of the concept of congruence of integers modulo an integer n.

The noether isomorphism theorems for groups, Theorems 1.3.11 to 1.3.15, have direct analogues for rings.

60

Chapter 2. Rings

(2.6) Theorem. (First isomorphism theorem) Let f : R - S be a ring homomorphism. Then R/Ker(f)

Im(f). (°_5 means ring isomorphism.)

Proof. Let K = Ker(f). From Theorem 1.3.11 we know that 7 R/K -+ Im(f), defined by 7(a + K) = f (a), is a well-defined isomorphism of groups. It only remains to check that multiplication is preserved. But

f ((a + K)(b + K)) = f(ab+K) = f(ab) = f(a)f(b) = f (a + K)f (b + K), so 7 is a ring homomorphism and hence an isomorphism.

(2.7) Theorem. (Second isomorphism theorem) Let R be a ring, I C R an ideal, and S C R a subring. Then S + I is a subring of R, I is an ideal of S + I, S n I is an ideal of S, and there is an isomorphism of rings

(S + 1)/i

S/(S n i).

Proof. Suppose that s, s' E S, a, a' E I. Then (s + a)(s' + a') = ss' + (as' + sa' + aa') E S + I,

so S + I is closed under multiplication and hence is a subring of R. (It is already an additive subgroup from the theory of groups.) The fact that I is an ideal of S + I and S n I is an ideal of S is clear. Let it : R - R/I be the natural homomorphism and let 7ro be the restriction of 7r to S. Then iro is a ring homomorphism with Ker(rro) = S n I and the first isomorphism

theorem gives S/(S n I) = S/ Ker(rro) Im(rro). But Im(ao) is the set of all cosets of I with representatives in S. Therefore, Im(rro) = (S+I)/I. (2.8) Theorem. (Third isomorphism theorem) Let R be a ring and let I and

J be ideals of R with I C J. Then J/I is an ideal of R/1 and

R/J

(R/I)/(J/I )

Proof. Define a function f : R/I - R/J by f (a + I) = a + J. It is easy to check that this is a well-defined ring homomorphism. Then

Ker(f)={a+I:a+J=J}={a+I:aEJ}=J/I. The result then follows from the first isomorphism theorem.

(2.9) Theorem. (Correspondence theorem) Let R be a ring, I C R an ideal

of R, and it : R - R/I the natural map. Then the function S

S/I

defines a one-to-one correspondence between the set of all subrings of R containing I and the set of all subrings of R/I. Under this correspondence, ideals of R containing I correspond to ideals of R/1.

Proof. According to the correspondence theorem for groups (Theorem 1.3.15) there is a one-to-one correspondence between additive subgroups


61

of R/I and additive subgroups of R containing I. It is only necessary to check that under this correspondence (which is H s-+ H/I) suLrings correspond to subrings and ideals correspond to ideals. We leave this to check 0 as an exercise.

(2.10) Lemma. Let R be a ring and let {SQ}QEA be a family of subrings (reap., ideals) of R. Then s = fQEAS0 is a subring (reap., ideal) of R.

Proof. Suppose a, b E S. Then a, b E SQ for all a E A so that a - b and ab E SQ for all a E A. Thus, a - b and ab E S, so S is a subring of R. If each

S. is an ideal and r E R then ar and ra E S. for all aEAsoar,raES and S is an ideal.

O

(2.11) Definition. If X C R is a subset then the subring generated by X is the smallest subring of R containing X and the ideal generated by X is the smallest ideal of R containing X. By Lemma 2.10 this is just the intersection of all subrings (resp., ideals) containing X. We will use the notation (X) to denote the ideal generated by X. (2.12) Lemma. Let X C R be a nonempty subset of a ring R. (1)

(2)

The subring of R generated by X is the sum or difference of all finite products of elements of X. If R is a ring with identity, then the ideal of R generated by X is the set

RXR= (3)

rixisi:ri,siER,xiEX,n>1

.

If R is a commutative ring with identity, then the ideal of R generated by X is the set

RX

rixi:r1ER,xiEX,n>1

.

Proof. (2) Every ideal containing X certainly must contain RXR. It is only necessary to observe that RXR is indeed an ideal of R, and hence it is the smallest ideal of R containing X. Parts (1) and (3) are similar and are left 0 for the reader.

(2.13) Remarks. (1) The description given in Lemma 2.12 (2) of the ideal generated by X is valid for rings with identity. There is a similar, but more complicated description for rings without an identity. We do not present it since we shall not have occasion to use such a description.

(2) If X = {a} then the ideal generated by X in a commutative ring R with identity is the set Ra = {ra : r E R}. Such an ideal is said to be principal. An integral domain R in which every ideal of R is principal is

62

Chapter 2. Rings

called a principal ideal domain (PID). The integers Z are an example of a PID. Another major example of a PID is the polynomial ring F[XJ where F is a field. This will be verified in Section 2.4.

There is another useful construction concerning ideals. If R is a ring and I1, ... , In are ideals of R then we define the sum of the ideals by n

1
We also define the product of the ideals I1, .

1i={ali.ani:aiiEIi i=1

. .

, In by

(1 < i < n)

and m is arbitrary

.

1j=1

(2.14) Lemma. If I1i ... , In are ideals of R then E 1 Ii is an ideal of R.

In fact, E i= I i is the ideal generated by the union U J= 1Ii . 1

Proof. Exercise.

(2.15) Definition. An ideal M in a ring R is called maximal if M R and M is such that if 1 is an ideal with M C I C R then I = M or I = R. (2.18) Theorem. Let R be a ring with identity and let I Ik R be an ideal of R. Then there is a maximal ideal of R containing I.

Proof. Let S be the set of all ideals J of R that contain I and are not equal to R. Then S 96 0 since I E S. Partially order S by inclusion and let C = { Ja } 4E A be a chain in S. Let J = UQE A JQ. Then J is an ideal of R since

if a, b E J, r E R, it follows that a, b E JQ for some a because C is a chain (i.e., JQ C Jp or Jp C J. for every Q,,3 E A). Thus a-b, ab, ar, ra E J. C J, so J is an ideal. Furthermore, J 36 R since 1 J. for any a. Thus J is an upper bound for the chain C so Zorn's lemma implies that S has a maximal element M and a maximal element of S is clearly a maximal ideal of R. 0 (2.17) Corollary. In a ring with identity there are always maximal ideals. Proof.

(2.18) Theorem. Let R be a commutative ring with identity. Then an ideal M of R is maximal if and only if RIM is a field.

Proof. Suppose RIM is a field. Then 0 0 1 E RIM, so M 0 R. But the only ideals in a field are {0} and the whole field, so the correspondence theorem shows that there are no ideals of R properly between M and R. Thus, M is maximal.


63

Conversely, suppose that M is a maximal ideal. It is necessary to show

that every a = a + M E RIM has an inverse if as 54 0 E RIM, i.e., if a V M. Consider the ideal Ra + M of R. Since a E Ra + M and a V M, it follows that Ra + M = R since M is assumed to be maximal. In particular,

ra+m= 1 forsomerE R,mE M. Letf =r+M. Then

ra=(r+M)(a+M)=ra+M=(1-m)+M=1+M=1ERIM. Thus, a-1 = r and RIM is a field. (2.19) Definition. An ideal P in a commutative ring R is said to be prime

if P 0 R and P is such that if ab E P then a E P or b E P. An element p E R is prime if the ideal Rp = (p) is a prime ideal. (2.20) Theorem. Let R be a commutative ring with identity. Then an ideal P of R is prime if and only if RIP is an integral domain. Proof. Suppose P is a prime ideal. Then P 96 R, so R/P {0}. Hence RIP is a ring with identity 1 + P. Given a = a + P, b = b + P E RIP, suppose

ab=O.Then (a+P)(b+P)=P, so ab+P= Pand abE P. SinceP is a prime ideal, this implies that a E P or b E P, i.e., a = 0 orb = 0. Therefore, RIP is an integral domain. Conversely, suppose RIP is an integral domain and suppose that ab E

P. Then ab = 0 E RIP. Therefore, a = 0 orb = 0, i.e., a E P or b E P. Thus P is a prime ideal. (2.21) Theorem. Let R be a commutative ring with identity and let I be an ideal. If I is maximal, then 1 is prime.

Proof. If I is maximal then R/I is a field and hence an integral domain. Therefore, I is prime by Theorem 2.20. More directly, suppose that I C R is not prime and let a, b E R with a V I, b V I, but ab E I. Let

J = {xER:axEI}. Then I C J since I is an ideal, and J is clearly an ideal. Also, J 54 I since b E J but b I, and J 34 R since 1 V J. Therefore, I is not maximal. The theorem follows by contraposition.

(2.22) Corollary. Let f : R --' S be a surjective homomorphism of commutative rings with identity.

(1) If S is a field then Ker(f) is a maximal ideal of R. (2) If S is an integral domain then Ker(f) is a prime ideal of R. Proof. R/ Ker(f)

Im(f) = S. Now apply Theorems 2.20 and 2.18.

64

Chapter 2. Rings

(2.23) Examples.

(1) We compute all the ideals of the ring of integers Z. We already know

that all subgroups of Z are of the form nZ = {nr : r E Z}. But if s E Z and nr E nZ then s(nr) = nrs = (nr)s so that nZ is an ideal of Z. Therefore, the ideals of Z are the subsets nZ of multiples of a fixed integer n. The quotient ring Z/nZ is the ring Zn of integers modulo n. It was observed in the last section that Zn is a field if and only if n is a prime number. (2)

Define 0 : Z[XJ - Z by 0(ao + a1X +

+ anXn) = ao. This is a

surjective ring homomorphism, and hence, Ker(¢) is a prime ideal. In fact, Ker(O) = (X) = ideal generated by X. Now define 0: Z[XJ Z2[XJ by O(ao+a1X

+...+anXn) = do+a1X+...+(1nXn

where a; = lr(a;) if a : Z

Z2 is the natural projection map. Z2[XJ is an integral domain (see Section 2.4), so Ker(t/i) is a prime ideal. In fact, Ker(O) = {aa + a1X + - - - +anXn : a; is even for all i}

= (2) = ideal generated by 2. Next consider the map ip' : Z[XJ -+ Z2 defined by

V(ao + a1X + ... + anXn) = so = ir(ao) Then Ker(O') is a maximal ideal since 0' is a surjective ring homomorphism to the field Z2. In fact,

Ker(tp') = (2, X) = ideal generated by {2, X}.

Note that (X)(2, X) and (2) 5 (2, X). Thus, we have some exam(3)

ples of prime ideals that are not maximal. Let G be a group and R a ring. Then there is a map

aug : R(G) -+ R called the augmentation map defined by

aug E rg g = J rg gEC

gEG

where we have denoted elements of R(G) by the formal sum notation EgEG rg - g rather than the formally equivalent functional notation f : G R with f (g) = re. We leave it as an exercise to check that aug is a ring homomorphism. The ideal I = Ker(aug) 9 R(G) is called the


65

augmentation ideal. If R is an integral domain, then I is prime, and if R is a field, then I is maximal. (4) Let F be a field and let T n(F) be the ring of n x n upper triangular

matrices with entries in F, and let ST(F) be the subring of strictly upper triangular matrices. Then ST" (F) is an ideal in the ring 7"(F) and the quotient

T"(F)/ST"(F)

Dn(F)

where Dn(F) is the ring of n x n diagonal matrices with entries in F. To see this define a ring homomorphism m : T(F) -' Dn(F) by a12

0

a22

a22

0

0

ann

It is easy to check (do it) that is a ring homomorphism and that Ker(b) = ST"(F) and Im(p) = Dn(F), so the result follows from the first isomorphism theorem.

The next result is an extension to general commutative rings of the classical Chinese remainder theorem concerning simultaneous solution of congruences in integers. For example, the Chinese remainder theorem is concerned with solving congruences such as

x = -1 (mod 15)

x-3 x=6

(mod 11) (mod 8).

(2.24) Theorem. (Chinese remainder theorem) Let R be a commutative ring with identity and let It, ... , In be ideals of R such that I; + Ij = R for all i j. (A collection of such ideals is said to be coprime or relatively prime.) Given elements a1,... , an E R there exists a E R such that

a . a; (mod l;) for 1 1. F o r each j > 1, since 1 + 1j = R, we can find bj E 11 and cj E Ij with bj + Cj = 1. Then fl 2(b; + cj) = 1, and since each b j E It, it follows that 1 = [J .2(bi + cj) E 11 + [I 2 I. Therefore, there

Chapter 2. Rings

66

is a1 E I1 and /31 E 11 =2 Ii such that al + 01 = 1. Observe that j31 solves the required congruences in the special case under consideration. That is,

(31-1 (mod I,)

/31-0 (mod1j)

for

j

1

since /31 - 1 E I1 and /31 E Ijj=2 Ii C II for i 0 1. By a similar construction we are able to find /32,. .. , /3n such that ,Ui =-- 1

Qi - 0 Now let a = a1131 +

(mod Ii) (mod II) for j # i.

+ a-

ai (modli) (1

Now suppose also that b - ai (mod Ii) for all i. Then b - a - 0 (mod Ii) so that b - a E Ii for all i, i.e.,

b - aEnli. i=1

0

The converse is clear.

There is another version of the Chinese remainder theorem. In order to state it we will need to introduce the concept of cartesian, or direct, product of rings. We will only be concerned with finite cartesian products. Thus let x Rn denote R1,. .. , Rn be finitely many rings and let f 1 Ri = R1 x Ri we may define addition and the cartesian product set. On the set f 1 multiplication componentwise, i.e., (a1,...,a.) + (b1,...,bn) _ (a1 + b1,...,an + bn)

(al....,an)(bl,...,bn) _ (albl,...,anbn),

to make r[° Ri into a ring called the direct product of R11. .. , Rn. Given 1

I < i < n there is a natural projection homomorphism iri defined by iri(a1,... , an) = a1.

:

Rj

Ri

(2.25) Corollary. Let R be a commutative ring with identity and let I1, ... ,1

be ideals of R such that Ii + I = R if i 0 j. Define Ft

f:R-.fiR/li i=1

by f (a) = (a+I1 i ..., a+In ). Then f is surjective and Ker(f) = I1 n...flln. Thus

n

f R/Ii. i=1


67

Proof. Surjectivity follows from Theorem 2.24, and it is clear from the definition of f that Ker(f) = I1 fl . . . fl In.

Sutpose that we take R = Z in Corollary 2.25. Let m E Z and let m = fl, p;' be the factorization of m into distinct prime powers. Then if (> Ik I; = (p;') it follows that Ii + Ij = Z if i # j. Moreover, (m) = I1 fl so that Corollary 2.25 applies to give an isomorphism of rings k

Tim = fl i=l

A practical method of applying the Chinese remainder theorem in Euclidean domains will be presented in Section 2.5. (This class of rings, defined there, includes the integers Z.) We conclude this section by computing all the ideals of the full matrix ring Mn(R) where R is a commutative ring with identity. To do this we will

make use of the matrix units Eij introduced in Example 1.10 (8). Recall that Eij has a 1 in the ij position and 0 elsewhere.

(2.26) Theorem. Let R be a ring with identity and let I be an ideal of Mn(R). Then there is an ideal J C R such that I = Mn(J).

Proof. First note that if J is any ideal of R then M,,(J) is an ideal of Mn(R). This follows immediately from the definition of multiplication and addition in M,,(R). Now suppose that I is an ideal in Mn(R), and define

J = {a E R: there exists A E I with ent11(A) = a}.

Observe that J is an ideal. Indeed, if a, b E J, r E R then a = entll(A), b = entll(B) for some A, B E I. Then a - b = entll(A - B), ra = entll((rE11A)), and ar = entll(A(rEll)). But A - B, (rEll)A, and A(rE11) are all in I since I is an ideal. Therefore, J is an ideal of R. Now we show that I = Mn(J). First observe that n

Elk

aijEij Eel = akeEll E I.

Thus, whenever A E I every entry of A is in J since the above calculation shows that every entry of A can be moved to the 1, 1 position of some matrix in I. Hence, I C M,,(J). But Mn(J) is generated by elements aEij where a = ent l l (A) for some A E I. Since aEi j = Ei 1 AEl j E I because I is an ideal, it follows that each generator of M,,(J) is in I, so Mn(J) C_ I, and we conclude that Mn(J) = I.

(2.27) Corollary. If D is a division ring then M,,(D) has no nontrivial proper ideals, i.e., the only ideals are {0} and Mn(D).

68

Chapter 2. Rings

Proof. The only ideals of D are {0} and D so that the only ideals of M"(D) are {0} = and

(2.28) Remark. Note that Corollary 2.27 shows that commutativity is a crucial hypothesis in Theorem 2.18. One might conjecture that an ideal M of any ring R (with identity) is maximal if and only if RIM is a division ring. But this is false since (0) is a maximal ideal of (by Corollary 2.27), but clearly, is not a division ring.

2.3 Quotient Fields and Localization If F is a field and R is a subring of F then R is an integral domain and the smallest subfield of F containing R is the subset

Q(R)={a/bEF:a,bER, b00}. Of course, if c 54 0 E R then a/b = (ac)/(bc) E F since a/b just means ab-1. Thus Q(R) is obtained from R in the same manner in which the rational numbers Q are obtained from the integers Z. We will now consider the converse situation. Suppose we are given an integral domain R. Can we find a field F that contains an isomorphic copy of R as a subring? The answer is yes, and in fact we will work more generally by starting with a commutative ring R and a subset S of R of elements that we wish to be able

to invert. Thus we are looking for a ring R' that contains R as a subring and such that S C (R')', i.e., every element of S is a unit of R'. In the case of an integral domain R we may take S = R \ {0} to obtain Q(R). (3.1) Definition. If R is a commutative ring, a subset S of R is said to be multiplicatively closed if the product of any two elements of S is an element of S.

Note that if S' is any nonempty subset of R, then the set S consisting of all finite products of elements of S' is multiplicatively closed. If S' contains no zero divisors, then neither does S.

(3.2) Definition. Let R be a commutative ring and let 0 96 S C R be a multiplicatively closed subset of R containing no zero divisors. The localization of R away from S is a commutative ring RS with identity, and an injective ring homomorphism 0 : R RS such that for all a E RS there are b E R

and c E S such that as(c) is a unit in RS and a = 0(b)O(c)-1. If R is an integral domain and S = R \ {0} then we call Rs the quotient field of R and we denote it Q(R).

2.3 Quotient Fields and Localization

69

(3.3) Remark. If S C R', then R together with the identity map 1R : R - R is the localization RS. In particular, if R is a field then any localization of R just reproduces R itself.

(3.4) Definition. Let S C R \ {0} be a nonempty multiplicatively closed subset of R containing no zero divisors, and let (RS, 0) and (R's, 0') be localizations of R away from S. RS and RS are said to be equivalent if there is an isomorphism Q : RS RS such that Q o 0 i.e., the following diagram of rings and ring homomorphisms commutes: R

f

m'

m

RS

IWS

(3.5) Theorem. Let R be a commutative ring. If S C R \ {0} is a multiplicatively closed subset containing no zero divisors, then then exists a localization RS away from S, and RS is unique up to equivalence.

Proof. Define a relation - on R x S by setting (a, b) - (c, d) if and only if ad = be. Observe that - is an equivalence relation. Symmetry and reflexivity are clear. To check transitivity suppose that (a, b) ' (c, d) and

(c, d) - (e, f). Then ad = be and cf = de. Therefore, adf = bcf and bcf = bde so that adf = bde. But d is not a zero divisor and d 0 0 since d E S so by Lemma 3.1, a f = be. Therefore, (a, b) - (e, f) and - is transitive and, hence, is an equivalence relation. Let RS = R X S/ - be the set of equivalence classes of the equivalence relation -. We will denote the equivalence class of (a, b) by the suggestive

symbol a/b. Note that (a, b) - (ac, bc) whenever c E S (S is multiplicatively closed, so be E S whenever b, c E S). Thus a/b = (ac)/(bc) for every c E S. Define ring operations on RS by the formulas

a cac _ b.d

a

c

b+d

bd

_ ad+bc bd

Note that bd E S since S is multiplicatively closed. Since the symbol a/b denotes an equivalence class, it is necessary to check that these oper& tions do not depend upon the choice of representative of the equivalence class used in their definition. To check this, suppose that a/b = a'/b' and

c/d = c'/d'. Then ab' = ba' and cd = dc' so that acb'd' = bda'c' and thus (ac)/(bd) = (a'c')/(b'd') and the multiplication formula is well defined. Similarly,

(ad + bc)b'd' =ab'dd'+bb'cd'

=ba'dd'+bb'dc' = (a'd' + b'c )bd,

so addition is also well defined.

70

Chapter 2. Rings

We leave to the reader the routine check that these operations make RS

into a commutative ring with identity. Observe that 0/s = 0/8', s/s = s'/s'

for any s, s' E S and 0/s is the additive identity of RS, while s/s is the multiplicative identity. If 8, t E S then (s/t)(t/s) = (st)/(at) = 1 E RS so

that (s/t)-1 = t/s.

Now we define : R - RS. Choose s E S and define R -a RS by 0,(a) = (as)/s. We claim that if a' E S then 0,. = 0,. Indeed, 0,(a) = (as)/s and -0,' (a) = (as')/s', but (as)/s = (as')/s' since ass' = as's. Thus

we may define 0 to be 0, for any s E S.

Claim. 0 is a ring homomorphism.

Indeed, if a, b E R and s E S then

O(a)4(b)

and (a + b)s

0(a + b)

=

as2 + bs2 s2

s

as

bs

s

s

_ 0(a) +.O(b).

Note that 0(1) = s/s = 1R,, and

Ker(O) = {a E R : Q = -11 s

s

={aER:as'=0 for some s'ES} = {0}

since S contains no zero divisors. Therefore, 0 is an injective ring homomorphism. Suppose that a/b E RS where a E R and b E S. Then a b

=

as

s

s

bs

= 0(a)(O(b))

Therefore, we have constructed a localization of R away from S. It remains only to check uniqueness up to equivalence.

Suppose that 0 : R -+ RS and 0' : R -' R'S are two localization of R away from S. Define 0 : RS -+ RS as follows. Let a E RS. Then

2.3 Quotient Fields and Localization

71

we may write a = 4(b)(O(c))-1 where b E R and c E S, and we define 13(a) _ O'(b)(O'(c))-'. Claim. ,3 is well defined.

then O(b)O(c) = O(Y)O(c) and =a= If hence ¢(bc') = 0(b'c). But 0 is injective so bc' = b'c. Now apply the homoso /3 is = (3(a) = morphism m' to conclude well defined. ¢(b)(O(c))-'

O(b')(O(c'))-'

4'(b')(O'(c))-1

¢'(b)(4'(c))-1

Claim. 3 is bijective.

RS by y(a') = q6(b')(O(c'))-1 if a' Define y : R'S for some b' E R, c' E S. Exactly as for /3, one checks that -y is well defined. O(b)(¢(c))-' and compute Then if a E RS, write a = y(Q(a)) = y(O'(b)(O'(c))-t)

= a. Similarly, one shows that /3(y(a')) = a' so that 33 is bijective. It remains to check that, 3 is a homomorphism. We will check that ,Q preserves multiplication and leave the similar calculation for addition as an exercise. Let a1 = a2 = -O(b2)(O(c2))-1. Since !b is a ring homomorphism, aja2 = so that 4(b1)(4(c1))-t and

O(blb2)(O(clc2))-1

f(ata2) _ O'(btb2)(O'(ctc2))-t

_ a(at)Q(a2)

This completes the proof of Theorem 3.5. (3.6) Examples.

(1) Q(Z) = Q.

(2) Let p E Z be a prime and let Sp = {1, p, p2, ...} C Z. Then a ZS' = b E Q: b is a power of p (3)

Let R be an integral domain, let P C R be a prime ideal and let S = R \ P. Then the definition of prime ideal shows that S is a multiplicatively closed subset of R and it certainly contains no zero divisors since

R is an integral domain. Then RS is isomorphic to a subring of the quotient field Q(R). In fact

Rs={b EQ(R):bP}.

72

Chapter 2. Rings

Rs is said to be R localized at the prime ideal P. It is an example of a local ring, that is, a ring with a unique maximal ideal. For RS the maximal ideal iscb(P) C RS. (4) Let d # 0,1 E Z be an integer that is square-free, i.e., whose prime factorization contains no squares, and then define Z[ ,/d-] = {a + bf a, b E Z}. It is easy to check that Z[ / is a subring of the complex numbers C and thus the quotient field Q(Z[fdJ) can be identified with a subfield of the complex numbers. In fact,

Q(Z[v']) = Q[vrd] = {a + bf : a, b E Q}. (5) Let F be a field and let F[X] be the ring of polynomials in the indeterminate X (see Example 1.10 (12)). Then F[X] is an integral domain (see Section 2.4 for a proof) and its quotient field is denoted F(X). It consists formally of quotients of polynomials in one variable, called rational functions over F.

2.4 Polynomial Rings Let R be a commutative ring with identity. The polynomial ring R[X] was defined in Example 1.10 (12). Recall that an element f E R[X] is a function

f : Z+ -a R such that f (n) # 0 for at most finitely many nonnegative integers n. If X E R[X] is defined by

X(n) _

1

0

if n = 1, if n 1,

then every f E R[X] can be written uniquely in the form 00

n=O

where the sum is finite since only finitely many an = f (n) E R are not 0. With this notation the multiplication formula becomes

1:(anvu') where

(fbx1)

_ n='o0

n

cn=Eanbn-m M=0

It is traditional to denote elements of R[X] by a symbol f (X) but it is important to recognize that f (X) does not mean a function f on the set

2.4 Polynomial Rings

73

R evaluated at X; in fact, X is not an element of R. However, there is a concept of polynomial function, which we now describe.

Let S be another commutative ring with identity, let : R - S be a ring homomorphism such that 0(1) = 1, and let u E S. Define a map Ou : R[X] --+ S by the formula 46u(ao + a1 X + ... + anX n) = O(ao) + O(a1)u + ... + O(an)un

We leave it as an exercise to check that mu is a ring homomorphism from

R[X] to S. It is called the substitution homorphism determined by m : R -+ S and u E S. If R is a subring of S and 0 : R -. S is the inclusion homomorphism (i.e., O(r) = r), then 46u just substitutes the element u E S

for the indeterminate X in each polynomial f(X) E R[X]. If Ker(¢u) = {0} then R[X] can be identified with a subring of S, and this is a way we frequently think of polynomials. However, it may not be the case that Ker(Ou) = {0}. For example, if R = R, S = C and u = i, then Ker(Ou) _ (X2 + 1). In general, we let R[u] = Im(ou). The above discussion is formalized in the following result, the details of which are left to the reader. Note that R can be viewed as a subring of R[X] via the identification a +-+ a 1. (4.1) Theorem. (Polynomial substitution theorem) Let R and S be commu-

tative rings with identity, let 0 : R -+ S be a ring homomorphism with O(1) = 1, and let u E S. Then there is a unique ring homomorphism Ou : R[X] -+ S such that ,u(X) = u and muIR = .

0

Proof. Exercise.

If f (X) _ deg(f (X)), by

0 an X- # 0 E R[X] we define the degree of f , denoted

deg(f (X)) = max{m : an 96 0}. Thus if n = deg(f (X)) then we may write f (X) = Eni_0 a,,Xm. We define deg(O) = -oo, and for convenience in manipulating degree formulas, we set

-oo < n and -oo + n = -oo for any n E Z+. The coefficient an is called the leading coefficient of f (X), while a0 is called the constant term. If the leading coefficient of f (X) is 1, then f (X) is a monic polynomial.

(4.2) Lemma. Let R be a commutative ring and let f (X), g(X) E R[X]. Then

(1) deg(f (X) + g(X)) < max{deg(f(X)),deg(g(X))}; (2) deg(f (X)g(X)) < deg(f (X)) + deg(g(X)); and (3) equality holds in (2) if the leading coefficient of f(X) or g(X) is not a zero divisor. In particular, equality holds in (2) if R is an integral domain.

Proof. (1) is clear from the addition formula for polynomials.

74

Chapter 2. Rings

For (2) and (3), suppose deg(f (X)) = n _> 0 and deg(g(X)) = m > 0. Then

with an#0 with bm

0.

Therefore,

f (X)g(X) = aobo + (aobl + a,bo)X + .

+ a"bmXm+n

so that deg(f (X)g(X )) < n + m with equality if and only if anbm -A 0. If an or bn is not a zero divisor, then it is certainly true that anbm # 0. The case for f (X) = 0 or g(X) = 0 is handled separately and is left to the reader. (4.3) Corollary. If R is an integral domain, then (1) R[X] is an integral domain, and (2) the units of R[X] are the units of R.

Proof (1) If f(X) A 0,g(X) 34 0, then deg(f (X)g(X )) = deg(f (X)) + deg(g(X)) > 0 > -co, and thus, f (X)g(X) A 0.

(2) If f (X)g(X) = 1 then deg(f (X)) + deg(g(X)) = deg(1) = 0. Thus, f (X) and g(X) are both polynomials of degree 0, i.e., elements of R. Therefore, they are units not only in R[X] but in R also.

We now consider the division algorithm for R[X] where R is a commutative ring. Let f (X) = ao + a1X + + anX" E R[X] and let g(X) = b0+b1X + +bm_1Xm_1 +Xm be a monic polynomial in R[X] of degree

m > 1. If n > m, let q1(X) = anX"-m. Then f1(X) = f(X) - g(X)gl(X) has degree < n - 1. If deg(fl (X)) > m then repeat the process with f (X) replaced by f 1(X ). After a finite number of steps we will arrive at a poly-

nomial f. (X) of degree < m. Letting q(X) = ql (X) + + q,(X) and r(X) = f (X) - g(X)q(X) we obtain an equation f (X) = g(X)q(X) + r(X) where deg(r(X)) < deg(g(X)). What we have described is the familiar long division process for polynomials.

(4.4) Theorem. (Division algorithm) Let R be a commutative ring, let f (X) E R[X ], and let g(X) E R[X J be a monic polynomial. Then there are

unique polynomials q(X) and r(X) in R[X] with deg(r(X)) < deg(g(X)) such that f (X) = g(X)q(X) + r(X). Proof. Existence follows from the algorithm described in the previous paragraph. Now consider uniqueness. Suppose there are two such decomposi-

tions f(X) = g(X)q(X) + r(X) and f(X) = g(X)gl(X) + rl(X) with deg(r(X)) < deg(g(X)) and deg(rl(X)) < deg(g(X)). Then


75

g(X)(gi(X) - q(X)) = r(X) - rl(X) Since g(X) is a monic polynomial, taking degrees of both sides gives

deg(g(X)) +deg(gl(X) - q(X)) = deg(r(X) - rl(X)) < deg(g(X)). This forces deg(gl (X) - q(X)) = -oo, i.e., ql (X) = q(X). It then follows that rl (X) = r(X), so uniqueness is established. (4.5) Corollary. (Remainder theorem) Let R be a commutative ring and let a E R. Then for any f (X) E R[X]

f (X) = (X - a)q(X) + f (a) for some q(X) E R[X]-

Proof. By the division algorithm we may write f (X) = (X - a)q(X) + r(X)

where deg(r(X)) < 0. Therefore, r(X) = r E R. Apply the substitution homomorphism X a to get f (a) = (a - a)q(a) + r so that r = f (a). (4.6) Corollary. Let R be a commutative ring, f (X) E R[X], and let a E R. Then f (a) = 0 if and only if X - a divides f (X). Proof.

(4.7) Corollary. Let R be an integral domain and let f (X) # 0 E R(X] be a polynomial of degree n. Then there are at most n roots of f (X) in R, i.e., there are at most n elements a E R with f (a) = 0.

Proof. If n = 0 the result is certainly true since f (X) = ao 54 0 and thus f (a) = ao 54 0 for every a E R, i.e., f (X) has no roots. Now let n > 0 and suppose, by induction, that the result is true for polynomials of degree < n. If there are no roots of f (X) in R, then there is nothing to prove. Thus suppose that there is at least 1 root a E R. Then by Corollary 4.6, X - a divides f (X), so we may write f (X) = (X - a)q(X) where deg(q(X)) = n - 1. By the induction hypothesis, there are at most n - 1 roots of q(X) in R. Now let b be any root of f (X) so that 0 = f (b) = (b - a)q(a). Hence b = a or b is a root of q(X). We conclude that f (X) has at most (n - 1) + 1 = n roots in R. (4.8) Remark. This result is false if R is not an integral domain. For example

if R = Z2 X Z2 then all four elements of R are roots of the quadratic polynomial X2 - X E R[X].

(4.9) Corollary. Let R be an integral domain and let f (X), g(X) E R[X] be polynomials of degree - < n. If f (a) = g(a) for n + 1 distinct a E R, then

f(X) =g(X)

Chapter 2. Rings

76

Proof. The polynomial h(X) = f (X) - g(X) is of degree < n and has 0

greater than n roots. Thus h(X) = 0 by Corollary 4.7.

In the case R is a field, there is the following complement to Corollary 4.9.

(4.10) Proposition. (Lagrange Interpolation) Let F be afield and let ao, a1, ..., an be n + 1 distinct elements of F. Let co, c1, ..., Cn be arbitrary (not necessarily distinct) elements of F. Then there exists a unique polynomial f (X) E F[X] of degree < n such that f (a,) = ci for 0 < i < n.

Proof. Uniqueness follows from Corollary 4.9, so it is only necessary to demonstrate existence. To see existence, for 0 c:Pi(X)

(4.2)

i=o

0

to conclude the proof.

(4.11) Remark. The polynomial given in Equation (4.2) is known as Lagrange's form of the interpolation polynomial. Of course, the interpolation polynomial is unique, but there is more than one way to express the interpolation polynomial; in the language of vector spaces (see Chapter 3), this is simply the observation that there is more than one choice of a basis of the vector space Pn(F) of polynomials in F[X] of degree at most n. The set {Pi(X) : 0 <_i < n} is one such basis. Another basis of Pn(F) is the set of polynomials Po(X) = 1 and

P,(X)=(X-ao)

(X-ai_1)

for 1 < i < n. Any polynomial f(X) E FIX] of degree at most n can be written uniquely as

= (4.3)

n

f(X) EoiPi(X) i=o

The coefficients cri can be computed from the values f (ai) for 0 < i < n. The details are left as an exercise. The expression in Equation (4.3) is known as Newton's form of interpolation; it is of particular importance in numerical computations.


77

(4.12) Theorem. Let F be a field. Then F[X] is a principal ideal domain (PID).

Proof. Let I be an ideal of F[X]. If I = {O} then I is a principal ideal, so suppose that 10 {0}. Choose g(X) E I such that g(X) 96 0 and such that deg(g(X)) < deg(f (X)) for all f (X) E I \ {0}. We claim that I = (g(X)). Since F is a field, we may multiply g(X) by an element of F to get a monic polynomial, which will also be in I. Thus we may suppose that g(X) is a monic polynomial. Let f (X) E I. Then by the division algorithm we may write f (X) = g(X)q(X)+r(X) where deg(r(X)) < deg(g(X)). But r(X) = f(X) - g(X)q(X) E I, so r(X) must be 0 since g(X) was chosen to have minimal degree among all nonzero elements of I. Thus f (X) = g(X)q(X) and we conclude that I = (g(X)). Since I was an arbitrary ideal of F[X], 0 it follows that F[X] is a PID. (4.13) Remarks. (1) If I is a nonzero ideal of F[X], then there is a unique monic polynomial of minimal degree in I. This is an immediate consequence of the proof of Theorem 4.12. (2) Both the statement and proof of Theorem 4.12 will be generalized to the case of Euclidean domains in Section 2.5 (Theorem 5.19).

If R is a PID that is not a field, then it is not true that the polynomial ring R[X] is a PID. In fact, if I = (p) is a nonzero proper ideal of R, then J = (p, X) is not a principal ideal (see Example 2.23 (2)). It is, however, true that every ideal in the ring R[X] has a finite number of generators. This is the content of the following result.

(4.14) Theorem. (Hilbert basis theorem) Let R be a commutative ring in which every ideal is finitely generated. Then every ideal of the polynomial ring R[X] is also finitely generated. Proof. The ideal (0) C R[X] is certainly finitely generated, so let I C R[X] be a nonzero ideal. If f (X) E R[X ] is not zero, we will let lc(f (X)) denote the leading coefficient of f (X). For n = 0, 1, 2, ..., let

In = {a E R : lc(f (X)) = a

for some f (X) E I of degree n} U {0}.

Note that In is an ideal of R for all n. Since lc(f (X)) = lc(X f (X)), it follows that In C for all n. Let J = U,°,° 0I,,. Then J is an ideal of R and hence is finitely generated, and since the sequence of ideals In is increasing, it is easy to see that J = I,, for some n. Also, In is finitely generated for any m. For 0 < m < n, let {aml, ... , amk- }

generate 1,,, and choose fmj (X) E I such that deg(fmi (X )) = m

and

lc(fmi (X)) = ami

78

Chapter 2. Rings

for all 0 < m < n and 1 <j < kin. Let

I =(fm,(X):0<m 0 and proceed by induction on r. Let a = lc(f (X)). If r < n, then a E Ir, so we may write a = Clad + ... + Ck_ark,..

Then lc(E ;'l Cifri X)) = a so that

deg(f(X)

k,

-

cifri(X)) < r i=1

and f (X) E If r > n then a E I, = In so that a = clam + -

E I in case r < n. '

-

Then, as above, deg

(1(x)

- C"1 CiXr-nfniX ) i=1

and hence f (X) E I by induction on r. Thus, I C I and the other inclusion is clear. Hence I is finitely generated. 0 (4.15) Corollary. Let R be a commutative ring in which every ideal is finitely generated. Then every ideal of R[X1, ... , Xn] is finitely generated.

Proof. This follows from Theorem 4.14 by induction on it since it is an easy exercise to that

R[X1, ... , Xn]

(R[X 1, ... , Xn-1]) [X.]-

0 (4.16) Corollary. Let F be a field. Then every ideal in the polynomial ring F[X1,

... , Xn]

is finitely generated. Proof

0

2.5 Principal Ideal Domains and Euclidean Domains

79

The theory of principal ideal domains will be studied in Section 2.5. At the present time we have two major examples of PIDs, namely, Z and F[X) for F a field, to which the theory will apply. Further examples will be given in Section 2.5. We will conclude this section with the following concept, which is defined by means of polynomials.

(4.17) Definition. Let F be a field. F is said to be algebraically closed if every nonconstant polynomial f (X) E FIX) has a root in F. (4.18) Remarks.

(1) According to Corollary 4.6, F is algebraically closed if and only if the nonconstant irreducible polynomials in FIX) are precisely the polynomials of degree 1. (2) The complex numbers C are algebraically closed. This fact, known

as the fundamental theorem of algebra, will be assumed known at a number of points in the text. We shall not, however, present a proof. (3) If F is an arbitrary field, then there is a field K such that F C K and

K is algebraically closed. One can even guarantee that every element

of K is algebraic over F, i.e., if a E K then there is a polynomial f (X) E FIX) such that f (a) = 0. Again, this is a fact that we shall not prove since it involves a few subtleties of set theory that we do not wish to address.

2.5 Principal Ideal Domains and Euclidean Domains The fundamental theorem of arithmetic concerns the factorization of an integer into a unique product of prime numbers. In this section we will show that the fundmental theorem of arithmetic is also valid in an arbitrary principal ideal domain. At present we have only two major examples of PIDs, namely, Z and F[X] for F a field. Some examples will be presented of other PIDs. We will start by defining precisely the concepts of factorization

needed to state and prove the extension of the fundamental theorem of arithmetic.

(5.1) Definition. Let R be an integral domain and let a, b E R \ {0).

(1) a and b are associates if a = ub for some unit u E R. We can define an equivalence relation on R by setting a - b if and only if a and b are associates in R. (2) a divides b (written a I b) if b = ac for some c E R.

(3) A nonunit a is irreducible if a = be implies that b or c is a unit. (4) A nonunit a is prime if a be implies that a I b or a c.

80

Chapter 2. Rings

(5.2) Remark. Let R be an integral domain and let a, b E R \ {0}. (1) a and b are associates if and only if a I b and b I a. (2) Recall that (a) denotes the ideal of R generated by a. Then a I b if and only if (b) C (a) and a and b are associates if and only if (a) = (b). (3) a is a prime element of R if and only if (a) is a prime ideal. (4) If a I b then au by for any units u, v. (5) If a l b and a is not a unit, then b is not a unit. Indeed, if b is a unit then be = 1 for some c E R and b = ad since a I b. Then (ad)c = 1 = a(dc) so, a is a unit also. an then p I ai for some i (exercise). (6) If p is a prime in R and p ( al

(5.3) Lemma. Let R be an integral domain. If a E R is prime, then a is irreducible.

Proof. Let a E R be prime and suppose that a = be. Then a I b or a I c. To be specific, suppose that a I b so that b = ad for some d E R. Then a = be = adc, and since R is an integral domain, Lemma 1.3 shows that 0 dc = 1 so that c is a unit. Thus a is irreducible. If R = Z then the concepts of prime and irreducible are the same, so that the converse of Lemma 5.3 is also valid. In fact, we shall show that the converse is valid in any PID, but it is not valid for every integral domain as the following examples show. (5.4) Examples. (1) Let F be a field and let

R = F[X2, X3J = {p(X) E F[XJ : p(X) = as + a2X 2 + a3 X3 +

+ anX"}.

Then X2 and X3 are irreducible in R, but they are not prime since X2 I (X3)2 = X6, but X2 does not divide X3, and X3 I X4X2 = X6, but X3 does not divide either X4 or X2. All of these statements are easily seen by comparing degrees. (2)

Let Z[/1 = {a + b/ :

a, b E Z}. In Z[f-] the element 2 is

irreducible but not prime.

Proof. Suppose that 2 = (a + bs)(c + dy) with a, b, c, d E Z. Then taking complex conjugates gives 2 = (a - bs)(c - d%), so multiplying these two equations gives 4 = (a2 + 3b2)(c2 + 3d2). Since the equation in integers a2 + 302 = 2 has no solutions, it follows that we must have a2 +3b2 = 1 or c2 +3d2 = 1 and this forces a = ±1, b = 0 or c = ±1, d = 0. Note that 2 is not a unit in Z[v/-3] Therefore, 2 is irreducible in Z[

since the equation 2(a + b/) = 1 has no solution with a, b E Z.


81

To see that 2 is not prime, note that 2 divides 4 = (1 + V" 3)(1- -../--3)

but 2 does not divide either of the factors 1 + vJ or 1 We conclude that 2 is not a prime in the ring Z[-3J.

in Z[/).

D

(5.5) Definition. Let R be an integral domain and let A be a subset of R containing at least one nonzero element. We say that d E R is a greatest common divisor (gcd) of A if (1) d l a for all a E A, and (2)

if eERand eI a forallaEA, then eI d.

If 1 is a gcd of A, then we say that the set A is relatively prime. We say that m 54 0 E R is a least common multiple (1cm) of A if 0 V A, if

(1) al m forallaEA, and (2)

if eE R and aIe for all aE R, then mIe.

Note that any two gcds of A are associates. Thus the gcd (if it exists) is well defined up to multiplication by a unit. The following result shows that in a PID there exists a gcd of any nonempty subset of nonzero elements.

(5.6) Theorem. Let R be a PID and let A be a subset of R containing at least one nonzero element.

(1) An element d E R is a gcd of A if and only if d is a generator for the ideal (A) generated by A. ... , an} is finite and a; 0 for 1 < i < n, then an element m E R is a lcm of A if and only if m is a generator of the ideal

(2) If A = {al,

(al) n ... (1 (an),

Proof. (1) Suppose that d is a generator of the ideal (A). Certainly d I a for each a E A since a E (A) = (d). Also, since d E (A), it follows that d = E , riai for rl,... , rn E R and al, ... , an E A. Therefore, if e I a for

allaE A, thene I dsothatdisagcdofA. Conversely, suppose that d is a gcd of the set A and let (A) = (c). Then d I a for all a E A so that a E (d). Hence, (c) = (A) C (d).

But, for each a E A, a E (c) so that c I a. Since d is a gcd of A, it follows that c I d, i.e., (d) C (c). Hence (c) = (d) and d is a generator of the ideal (A).

(2) Exercise.

0

(5.7) Corollary. Let R be a PID and a E R \ {0}. Then a is prime if and only if a is irreducible.

82

Chapter 2. Rings

Proof. Lemma 5.3 shows that if a is prime, then a is irreducible. Now assume that a is irreducible and suppose that a I be. Let d = gcd{a, b}. Thus a = de

and b = df . Since a = de and a is irreducible, either d is a unit or e is a unit. If e is a unit, then a I b because a is an associate of d and d I b. If d is a unit, then d = ar' + bs' for some r', s' E R (since d E (a, b)). Therefore,

I = ar + bs for some r, 8 E R, and hence, c = arc + bsc. But a I arc and a ( bsc (since a I be by assumption), so a I c as required.

(5.8) Corollary. Let R be a PID and let I C R be a nonzero ideal. Then I is a prime ideal if and only if I is a maximal ideal. Proof. By Theorem 2.21, if I is maximal then 1 is prime. Conversely, suppose that I is prime. Then since R is a PID we have that I = (p) where p is a prime element of R. If I C J = (a) then p = ra for some r E R. But p is prime and hence irreducible, so either r or a is a unit. If r is a unit then

(p) = (a), i.e., I = J. If a is a unit then J = R. Thus 1 is maximal. (5.9) Definition. Let R be a ring. We say that R satisfies the ascending chain condition (ACC) on ideals if for any chain

Il 9129139... of ideals of R there is an n such that Ik = In for all k > n, i.e., the chain is eventually constant. A ring that satisfies the ascending chain condition is said to be Noetherian. The following characterization of Noetherian rings uses the concept of maximal element in a partially ordered set. Recall what this means (see the appendix). If X is a partially ordered set (e.g., X C P(Y) where the partial order is set inclusion), then a maximal element of X is an element m E X such that if m < x for some x E X, then m = x. That is, a element m E X is maximal if there is no element strictly larger than m in the partial order of X. For example, if X = {(2), (3), (12) } is a set consisting of the given ideals of the ring Z with the partial order being inclusion of sets, then both (2) and (3) are maximal elements of X.

(5.10) Proposition. Let R be a ring. The following conditions on R are equivalent.

(1) R is Noetherian. (2) Every ideal of R is finitely generated. (3) Every nonempty subset of ideals of R has a maximal element. In particular, a PID is Noetherian. Proof. (1) . (3) Suppose that S = {IQ}QEA is a nonempty set of ideals of R that does not have a maximal element.


83

Then choose 11 E S. Since S does not have a maximal element, there is an element 12 E S such that 11 5 12. Similarly, 12 is not a maximal element

so there is 13 E S such that 12 5 13. In this manner we can construct a strictly increasing chain of ideal in R, which contradicts the assumption that R satisfies the ACC. Therefore, S must contain a maximal element if R satisfies the ACC.

(3) = (2) Let I be an ideal of R and consider the family S of all finitely generated ideals of R that are contained in I. By hypothesis, there is a maximal element J E S. Let a E I. Then the ideal J + (a) E S and it contains J. Since J is maximal in S, it follows that J = J + (a), i.e., a E J. Therefore, I = J and I is finitely generated. (2) * (1) Suppose that every ideal of R is finitely generated. Let

119129139 ... be a chain of ideals of R and let I = U°_1 I. Then I is an ideal of R so that I = (a1i ... , am) for some ai E R. But ai E I = U°O_1 In for 1 n there is an inclusion

(al, ... , am) C Ik C 1 = (al, ... , am) so that Ik = I = In for all k > n and R satisfies the ACC on ideals.

0

(5.11) Remark. In light of Definition 5.9 and Proposition 5.10, the Hilbert basis theorem (Theorem 4.14) and its corollary (Corollary 4.15) are often stated:

If R is a commutative Noetherian ring, then the polynomial ring RIX1i

... ,XnJ is also a commutative Noetherian ring.

(5.12) Theorem. (Fundamental theorem of arithmetic) Let R be a PID. Then any nonzero a E R can be written as a = up1 . . . pn where u is a unit and each pi is a prime. Moreover, this factorization is essentially unique. That is, if a = vQ1 . . . qm where v is a unit and each qi is a prime, then m = n and for some permutation a E Sn, qi is an associate of po(;) for

1

Proof. We first prove existence of the factorization. Let a # 0 E R. Then if a is a unit we are done and if a is a prime we are done. Otherwise write a = albs where neither a1 nor b1 is a unit. (Recall that in a PID, prime and irreducible elements are the same.) Thus (a) 5 (b1). If b1 is a prime, stop. Otherwise, write b1 = a2b2 with neither a2 nor b2 a unit. Continue in this way to get a chain (a) 5 (61) 5 (b2) 5 ...

84

Chapter 2. Rings

By the ascending chain condition, this must stop at some (bn). Therefore b is a prime and we conclude that every a 0 E R that is not a unit is divisible by some prime. Therefore, if a 0 is not a unit, write a = p1c1 where p1 is a prime. Thus (a) 5 (c1). If c1 is a unit, stop. Otherwise, write c1 = p2c2 with p2 a prime so that (c1) (c2). Continue in this fashion to obtain a chain of ideals ...

(a) 5 (cl) 5 (CO

By the ACC this must stop at some c, and by the construction it follows that this c = u is a unit. Therefore, a = P1c1 = P1P2c2 =

=

so that a is factored into a finite product of primes times a unit. Now consider uniqueness of the factorization. Suppose that a = pop, ...Pn

where Pi.... , pn, Q1i ... , qm are primes while p0 and qo are units of R. We will use induction on k = min{m, n}. If n = 0 then a is a unit, so a = qo and hence m = 0. Also m = 0 implies n = 0. Thus the result is true for k = 0.

Suppose that k > 0 and suppose that the result is true for all elements b E R that have a factorization with fewer than k prime elements. Then qm, so pn divides some qj since pn is a prime element. After reordering, if necessary, we can assume that pn I qm. But q,n is prime, so qm = Pnc implies that c is a unit. Thus, pn and qn are associates. Let Pn I go91

a'

Then a' has a factorization with fewer than k prime factors, so the induction

hypothesis implies that n - 1 = m - 1 and qi is an associate of p,(;) for some a E Sn _ 1 i and the argument is complete.

(5.13) Corollary. Let R be a PID and let a 0 0 E R. Then

a=upli ...pkk where P' .. , pA, are distinct primes and u is a unit. The factorization is essentially unique. Proof.

(5.14) Remark. Note that the proof of Theorem 5.12 actually shows that if R is any commutative Noetherian ring, then every nonzero element a E R has a factorization into irreducible elements, i.e., any a E R can be factored as a = up1 ... pn where u is a unit of R and p1, ... , pn are irreducible (not prime) elements, but this factorization is not necessarily unique; however, this is not a particularly useful result.


85

For non-Noetherian rings we do not even have the factorization into irreducible elements. Examples 5.15 (4) and (5) are examples. (5.15) Examples. (1)

(2)

In F[X2, X3] there are two different factorizations of X6 into irreducible elements, namely, (X2)3 = Xs = (X3)2.

In Z[v there is a factorization

4=2.2=(1+v)(1--v/'---3) into two essentially different products of irreducibles. (3)

If F is a field and p(X) E F[X) is an irreducible polynomial, then p(X) is prime since F[X] is a PID. Thus the ideal (p(X)) is maximal according to Corollary 5.8. Hence the quotient ring F[X]/(p(X)) is a field. If p E Z is a prime number let Fp denote the field Zp. Then F2[X)/(X2+X+1) is a field with 4 elements, F3[X]/(X2+1) is a field with 9 elements, while F2[X]/(X3 + X + 1) is a field with 8 elements. In fact, one can construct for any prime p E Z and n > 1 a field F. with q = p" elements by producing an irreducible polynomial of degree

n in the polynomial ring Fp[X]. (It turns out that F. is unique up to isomorphism, but the proof of this requires Galois theory, which we do

not treat here.) (4) Let H(C) be the ring of complex analytic functions on the entire complex plane. (Consult any textbook on complex analysis for verification of the basic properties of the ring H(C).) The units of H(C) are precisely the complex analytic functions f : C C such that f (z) 0 0

for all z E C. Furthermore, if a E C then the function z - a divides f (z) E H(C) if and only if f (a) = 0. From this it is easy to see (exercise) that the irreducible elements of H(C) are precisely the functions (z - a) f (z) where f (z) 34 0 for all z E C. Thus, a complex analytic function g(z) can be written as a finite product of irreducible elements if and only if g has only finitely many zeros. Therefore, the complex analytic function sin(z) cannot be written as a finite product of irreducible elements in the ring H(C). (Incidentally, according to Remark 5.14, this shows that the ring H(C) is not Noetherian.) (5) Let R be the subring of Q[X] consisting of all polynomials whose constant term is an integer, i.e.,

R are the constant polynomials ±1. Note that for any nonzero integer k, there is a factorization X = k(X/k) E R, and neither

factor is a unit of R. This readily implies that X does not have a factorization into irreducibles. (Again, Remark 5.14 implies that R is not Noetherian, but this is easy to see directly:

86

Chapter 2. Rings

...

(X) 5 (X/2) 5 (X/4) 5 (X/8) is an infinite ascending chain.)

Both of the examples of PIDs that we currently have available, namely,

Z and FIX] for F a field, actually have more structure than just that of a PID. Specifically, both Z and F(X] have a measure of the size of elements, Inl for n E Z and deg(p(X)) for p(X) E FIX], together with a division algorithm, which allows one to divide one element by another with a remainder "smaller" than the divisor. In fact the division algorithm was

precisely what was needed to prove that Z and F(X] are principal ideal domains. We formalize this property of Z and FIX] with the following definition.

(5.16) Definition. An integral domain R is a Euclidean domain if there is a Z+ = N U {0} such that function v : R \ {0} (1) for all a, b E R \ {0}, v(a) < v(ab); and (2) given a, b E R with a # 0, there are q, r E R with b = aq + r such that

r = 0 or v(r) < v(a). (5.17) Examples.

(1) Z together with v(n) = Ink is a Euclidean domain. (2)

If F is a field, then FIX] together with v(p(X)) = deg(p(X)) is a Euclidean domain (Theorem 4.4).

(5.18) Lemma. If R is a Euclidean domain and a E R\{0}, then v(1) < v(a). Furthermore, v(1) = v(a) if and only if a is a unit.

Proof. First note that v(1) < v(1-a) = v(a). If a is a unit, let ab = 1. Then v(a) < v(ab) = v(1), so v(1) = v(a). Conversely, suppose that v(1) = v(a) and divide 1 by a. Thus, 1 = aq+r where r = 0 or v(r) < v(a) = v(1). But v(1) < v(r) for any r 0 0, so the latter possibility cannot occur. Therefore,

0

r= 0 so l= aq and a is a unit. (5.19) Theorem. Let R be a Euclidean domain. Then R is a PID.

Proof. Let I C R be a nonzero ideal and let S={v(a):aEI\{0}}CZ+.

This set has a smallest element no. Choose a E I with v(a) = no. We claim

that I = Ra. Since a E I, it is certainly true that Ra C I. Now let b E I.

Then b=aq+rforq,rE R with r = 0 or v(r) < v(a). Butr=b - aqE I, so v(a) < v(r) if r 0. Therefore, we must have r = 0 so that b = aq E Ra. 0 Hence, I = Ra is principal.


87

In a Euclidean domain R the classical Euclidean algorithm for finding the gcd of two integers works for finding the gcd of two elemen a of R. This algorithm is the following. Given al, a2 E R \ {0} write al = a2gl + a3

with

a3 = 0 or v(a3) < v(a2)

a2 = a3q2 + a4

with

a4 = 0 or v(a4) < v(a3)

a3 = a4q3 + a5

with

a5 = 0 or v(a5) < v(a4)

Since v(a2) > v(a3) > v(a4) > , this process must terminate after a finite number of steps, i.e., an+1 = 0 for some n. For this n we have

an-1 = angn-1 + 0. Claim. an = gcd{al,a2}. Proof. If a, b E R then denote the gcd of {a, b} by the symbol (a, b). Theorem 5.5 shows that the gcd of {a, b} is a generator of the ideal generated by {a, b}. Now we claim that (a;, a;+1) = (at+1, ai+2) for 1 < i < n - 1. Since a; = a;+1qi + ai+2i it follows that xa, + ya;+l = x(ai+lq, + ai+2) + yai+1 = (xqi + y)a;+1 + xa;+2

Thus, (a;, ai+1) S (at+1, ai+2), and similarly

ra;+1 + sat+2 = rat+i + s(a; - ai+lqi) = sat + (r - gi)at+1 so that (a;+1, at+2) C (at, ai+1) Hence, (a;, a;+1) = (at+1, ai+2), and we conclude (al, a2) = (a2, a3) = ... = (an-1, an)-

But (an-,, an) = an since an (an-1, and the claim is proved. This result gives an algorithmic procedure for computing the gcd of two elements in a Euclidean domain. By reversing the sequence of steps used to compute d = (a, b) one can arrive at an explicit expression d = ra + sb for the gcd of a and b. We illustrate with some examples. (5.20) Examples.

(1) We use the Euclidean algorithm to compute (1254, 1110) and write it as r1254 + x1110. Using successive divisions we get

Chapter 2. Rings

88

1254=1110.1+144 1110=144.7+102 144=102.1+42

102=42.2+18

42=18.2+6 18=6.3+0. Thus, (1254, 1110) = 6. Working backward by substituting into successive remainders, we obtain

6=42-18.2 =42-(102-42.2).2=5.42-2.102

=54.1254-61.1110.

(2) Let f (X) = X2 -X+ l and g(X) = X3 + 2X2 + 2 E Z3[X]. Then g(X) = Xf (X) + 2X + 2 and

f (X) = (2X + 2)2.

(3)

Thus, (f(X), g(X)) = (2X + 2) = (X + 1) and X + 1 = 2g(X) 2X f (X). We now give an example of how to solve a system of congruences in a Euclidean domain by using the Euclidean algorithm. The reader should

refer back to the proof of the Chinese remainder theorem (Theorem 2.24) for the logic of our argument. Consider the following system of congruences (in Z): x

-1 (mod 15)

x=3 x

6

(mod 11) (mod 8).

We apply the Euclidean algorithm to the pair (88, 15):

88=15.5+13 15=13.1+2 13=2.6+1

2=1.2. Now we substitute backward to obtain


89

1=13-2.6 =-15.6+(88-15.5).7=88.7-15.41 =616-15.41, so 616 - 1 (mod 15) and 616 - 0 (mod 88). Similarly, by applying the

Euclidean algorithm to the pair (120, 11), we obtain that -120 - 1 (mod 11) and -120 - 0 (mod 120), and by applying it to the pair (165, 8), we obtain -495 - 1 (mod 8) and -495 = 0 (mod 165). Then our solution is

x=--I.(616)+3-(-120)+6-(-495)=-3946 (mod 1320) or, more simply,

x - 14 (mod 1320). We will now give some examples of Euclidean domains other than Z and FIX). If d 36 0,1 E Z is square-free, we let Z[f = {a+b/ : a, b E Z}. Then Z[f is a subring of the field of complex numbers C and the quotient field of Z[f is the quadratic field

Q[f] ={a+bfd:a,bEQ}. (5.21) Proposition. If d E {-2, -1, 2, 3 ) then the ring domain with v(a + bf) = Ia2 - db21.

is a Euclidean

Proof. Note that v(a + brd) = Ia2 - 621 > 1 unless a = b = 0. It is a straightforward calculation to check that v(a/3) = v(a)v(Q) for every a, E Z[-./a-l. Then

v(a/3) = v(a)v(0) > v(a), so part (1) of Definition 5.16 is satisfied. with /3 # 0. Then in the quotient field Now suppose that a, 0 E

of Z[f we may write a//3 = x + yf where x, y E Q. Since any rational number is between two consecutive integers and within 1/2 of the nearest integer, it follows that there are integers r, a E Z such that Ix - rl < 1/2

and ly - sI < 1/2. Let y = r + sv and b = /3((x - r) + (y - s)f). Then

a=/3(x+yv)=f3y+6. Since r, s E Z, it follows that y E Z[fd) and b = a - Qy E Z[ / also. Then

v(b) = v(p)v((x - r) + (y - s),Ad-) = v(Q)I (x - r)2 - d(y -

But

8)21.

Chapter 2. Rings

90

I(x-r)2-d(y-s)2I < Ix-rl2+Idlly-812 < (1/2)2 + 3(1/2)2 = 1.

The only possibility for equality is when Ix - rl = ly - sl = 1/2 and d = 3. But in this case I(x - r)2 - d(y - 8)2I = 11/4 - 3/41 = 1/2 < 1.

Therefore, we always have I(x - r)2 - d(y - 8)21 < 1 and we conclude that v(6) < v(/3). Hence Z[vQ] is a Euclidean domain. Complementary to Proposition 5.21, we have the following result:

(5.22) Proposition. If d < 0 then Zlvfd-l is a PID if and only if d = -1 or

d= -2. Proof. If d = -1 or d = -2 then by Proposition 5.21, Z1.1 rd-] is a Euclidean domain and hence a PID. For the converse we need the following lemma.

(5.23) Lemma. 2 is not a prime in Z[fd).

Proof. Either d or d - 1 is even so that 2 I d(d - 1). But

d(d-1)=d2-d=(d+f)(d-vd), so 21 (d+vd)(d- v'd) but neither (d+ f)/2 nor (d- v/ d-)/2 are in Z[fJ. Thus 2 divides the product of two numbers, but it divides neither of the numbers individually, so 2 is not a prime element in the ring Z[f . We now return to the proof of Proposition 5.22. We will show that if d < -3 then 2 is an irreducible element of Z[f1. Since in a PID, irreducible and prime are the same concept (Corollary 5.7), it will follow that Z/ cannot be a PID because Lemma 5.23 shows that 2 is not a prime in Z[v' .

Suppose 2 = a/3 for a, /3 E Z['/d] with a and 3 not units. Then we must have v(a) > 1 and v(0) > 1. Therefore, 4 = v(2) = v(a)v($),

and since v(a), v(/3) E N it follows that v(a) = vL/3) = 2. Thus if 2 is not there is a number a = a + Wd E Z['//J such that irreducible in

v(a) = a2 - db2 = ±2.

But ifd<-3 andbo0then a2-db2=a2+(-d)b2>0+3.1>±2, while if b = 0 then a2 - db2 = a2 96 ±2


91

when a E Z. Hence, if d < -3 there is no number a E Z[V' with v(c) = 2, Therefore, Proposition 5.22 and we conclude that 2 is irreducible in is proved.

(5.24) Remarks. (1)

A complex number is algebraic if it is a root of a polynomial with integer coefficients, and a subfield F C C is said to be algebraic if every element of F is algebraic. If F is algebraic, the integers of F are those elements of F that are roots of a monic polynomial with integer coefficients. In the quadratic field F = Q(v/--31, every element of the ring Z[v/-3) is an integer of F, but it is not true, as one might expect, that Z(v/-3) is all of the integers of F. In fact, the following can be shown: Let d # 0, 1 be a square-free integer. Then the ring of integers is of the field

Z[v' ] {Zr

if if

d

d

2, 3 (mod 4), 1 (mod 4).

l

In particular, the ring of integers of the field Q[vf---3]l is the ring

R={a+bl

1

2:a,bEZ}.

We leave it as an exercise for the reader to prove that R is in fact a

Euclidean domain. (Compare with Proposition 5.22.) (2) So far all the examples we have seen of principal ideal domains have been Euclidean domains. Let R={a+b(1+2-19)

:a,bEZ} -11119

be the ring of integers of the quadratic field Q[ . Then it can be shown that R is a principal ideal domain but R is not a Euclidean domain. The details of the verification are tedious but not particularly

difficult. The interested reader is referred to the article A principal ideal ring that is not a Euclidean ring by J.C. Wilson, Math. Magazine (1973), pp. 34-38. For more on factorization in the rings of integers of quadratic number fields, see chapter XV of Theory of Numbers by G. H. Hardy and E. M. Wright (Oxford University Press, 1960).

Chapter 2. Rings

92

2.6 Unique Factorization Domains We have seen in Section 2.5 that the fundamental theorem of arithmetic holds for any PID. There are, however, rings that are not PIDs for which the fundamental theorem of arithmetic holds. In this section we will give a result that allows for the construction of such rings.

(6.1) Definition. An integral domain R is a unique factorization domain (UFD) if every nonzero element a of R can be written essentially uniquely as a = upl pr where u is a unit and each pi is an irreducible element of R. Essentially uniquely means that if a = vql . q, where v is a unit and each q3 is irreducible, then r = 8 and, after reordering (if necessary), qi is an associate of pi. By collecting associate primes together, we may write (essentially uniquely) a=uipi'...pi+4

where pi is not an associate of p_, if i A j. This is called the prime factorisation of a and the primes pi are called the prime factors or prime divisors of a.

(6.2) Lemma. Let R be a UFD.

(1) An element a E R is irreducible if and only if it is prime. (2) Any nonempty set of nonzero elments of R has a greatest common divisor.

Proof. (1) Suppose a E R is irreducible and a I be. Thus ad = be for some d E R. Writing b, c, and d as a product of units and irreducible elements gives

auldl ... dr = u2b1 ... b,u3c1 ... ct where each bi, c,, and dk is irreducible and each ui is a unit. By uniqueness of factorization of be, it follows that the irreducible element a is an associate of some b, or c2 and, hence, a I b or a I c. (2) We prove this in the case that the set in question is {a, b} consisting

of two elements and leave the general case for the reader. Let pl, .... pr denote all the primes that are a prime factor of either a or b. Then we may write a=Pi'...Pr

and b=Pi'...pr

where 0 < mi and 0 < n, for each i. Let ki = min{mi,ni} for l < i < r and let

d=pii ...pr'

We claim that d is a gcd of a and b. It is clear that d I a and d I b, so let e be any other common divisor of a and b. Since e I a, we may write a = ec

2.6 Unique Factorization Domains

93

for some c E R. Taking prime factorizations of e, a, and c and applying the unique factorization assumption, we conclude that any pri.ae factor of e must also be a prime factor of a and the power of the prime that divides e can be no more than the power that divides a. Thus, since e also divides b, we may write

e = pi' ...

p

where P; < min{mi, ni} = ki. Therefore, every prime factor of e is also a prime factor of d and the power of the prime factor dividing e is at most 0 that which divides d. Thus e I d and d is a gcd of a and b. Our goal is to prove that if R is a UFD then the polynomial ring R(X] is also a UFD. This will require some preliminaries.

(6.3) Definition. Let R be a UFD and let f (X) 0 0 E R(X]. A gcd of the coefficients of f (X) is called the content of f (X) and is denoted cont(f (X)). The polynomial f (X) is said to be primitive if cont(f (X)) = 1.

Note that the content of f (X) is only uniquely defined up to multiplication by a unit of R. If f (X) is a nonzero polynomial then we can write f (X) = c f1(X) where f, (X) is primitive and c is the content of f (X).

(6.4) Lemma. (Gauss's lemma) Let R be a UFD and let f (X), g(X) be nonzero polynomials in R(X]. Then cont(f (X)g(X )) = cont(f (X)) cont(g(X)).

In particular, if f (X) and g(X) are primitive, then the product f (X)g(X ) is primitive.

Proof. Write f(X) = cont(f(X))fi(X) and g(X) = cont(g(X))91(X) where f1(X) and gi (X) are primitive polynomials. Then

f (X)g(X) = cont(f (X)) cont(g(X)) fl(X)gj(X), so it is sufficient to check that f1(X)g1(X) is primitive. Now let

fi(X) =ao+a1X +...+anX'n and

gl(X) and suppose that the coefficients of fl (X )gi (X) have a common divisor d other than a unit. Let p be a prime divisor of d. Then p must divide all of the coefficients of fi (X)g1(X ), but since f1(X) and gi (X) are primitive, p does not divide all the coefficients of f, (X) nor all of the coefficients of gi (X). Let a, be the first coefficient of fi (X) not divisible by p and let b, be the first coefficient of g1(X) not divisible by p. Consider the coefficient of X'+° in f i (X )gi (X ). This coefficient is of the form

Chapter 2. Rings

94

arbs + (ar+lba-1 + ar+2bs-2 +

) + (ar-lbs+1 + ar-2ba+2 + ' -)-

By hypothesis p divides this sum and all the in the first parenthesis are divisible by p (because p I bi for j < a) and all the in the second parenthesis are divisible by p (because p I aj for i < r). Hence p I arb, and since p is prime we must have p I ar or p I b contrary to our choice of ar and b,. This contradiction shows that no prime divides all the coefficients of f, (X)gl (X), and hence, f, (X)gl (X) is primitive. 0 (6.5) Lemma. Let R be a UFD with quotient field F. If f (X) 34 0 E F[X], then f (X) = of, (X) where a E F and f, (X) is a primitive polynomial in R[X]. This factorization is unique up to multiplication by a unit of R. Proof. By extracting a common denominator d from the nonzero coefficients

of f (X), we may write f (X) = (1 /d) f (X) where f (X) E R(X]. Then let a = cont(f(X))/d = c/d E F. It follows that f(X) = of, (X) where f1(X) is a primitive polynomial. Now consider uniqueness. Suppose that we can also write f (X) =,3f2(X) where f2 (X) is a primitive polynomial in R(X ] and Q = a/b E F. Then we conclude that (6.1)

adf2(X) =cbf1(X)

The content of the left side is ad and the content of the right side is cb, so ad = ucb where u E R is a unit. Substituting this in Equation (6.1) and dividing by cb gives

uf2(X) = f1(X) Thus the two polynomials differ by multiplication by the unit u E R and 0 the coefficients satisfy the same relationship 3 = a/b = u(c/d) = ua. (6.6) Lemma. Suppose R is a UFD with quotient field F. If f(X) E R[X] has positive degree and is irreducible in R[X], then f(X) is irreducible in FIX]. Proof, If f (X) E R[X] has positive degree and is irreducible in R(X ] then J (X) is primitive since cont(f (X)) I f (X) in R(X]. Suppose that f (X) is reducible in FIX]. Thus f (X) = g1(X)g2(X) where g;(X) E F[X] and

degg;(X) > 0 for i = 1 and 2. Then g;(X) = a;f;(X) where a, E F and f1(X) E R[X] is primitive. Thus,

P X) = a1a2f1(X)f2(X ),

and the product f1(X)f2(X) is primitive by Gauss's lemma. Thus, by Lemma 6.5, f(X) and f1(X)f2(X) differ by multiplication by a unit of R, which contradicts the irreducibility of f (X) in R[X). Thus, we conclude 0 that f (X) is irreducible in F(X].


95

(6.7) Corollary. Let R be a UFD with quotient field F. Then the irreducible elements of R[X] are the irreducible elements of R and the primitive polynomials f(X) E R[X] which are irreducible in FIX]. Proof.

(6.8) Theorem. If R is a UFD, then R[X] is also a UFD.

Proof. Let F be the quotient field of R and let f(X) 96 0 E R[X]. Since F[X] is a UFD, we can write

f(X) =P1(X) ... pr(X) where pi(X) E F[X] is an irreducible polynomial for 1 < i < r. By Lemma

6.5, pi(X) = aigi(X) where a; E F and qi(X) E R[X] is a primitive polynomial. Thus, we have

f(X)

=cg1(X)...gr(X)

where c = al . ar E F. Write c = a/b where a, b E R. Then taking contents we get cont(bf (X)) = cont(agi (X) . . . qr(X)) = a

by Gauss's lemma. Thus, b cont (f (X)) = a, so b ( a and cont (f (X)) = c E R. Each qi(X) is irreducible in F[X], and hence, it is irreducible in R[X].

Since R is a UFD, write c = udi ... d, where each di is prime in R and u E R is a unit. Thus we have a factorization

f(X) = udi...d9g1(X)...gr(X) of f(X) into a product of irreducible elements of R[X]. It remains to check uniqueness. Thus suppose we also have a factorization

f(X) = vbi ...btq'(X)...gk(X) where each q;(X) is a primitive polynomial in R[X] and each bi is an irreducible element of R. By Corollary 6.7, this is what any factorization into irreducible elements of R[X] must look like. Since this also gives a factorization in FIX] and factorization there is unique, it follows that r = k and qi(X) is an associate of q;(X) in FIX] (after reordering if necessary). But if primitive polynomials are associates in FIX], then they are associates in R[X] (Lemma 6.5). Furthermore,

cont(f(X)) =vbi ...bt =ud1 ... d so s = t and bi is an associate of d, (after reordering if necessary) since R is a UFD. This completes the proof.

96

Chapter 2. Rings

(6.9) Corollary. Let R be a UFD. Then R[X,, ... , X,,J is also a UFD. In particular, FIX,, ... , XnJ is a UFD for any field F.

0

Proof. Exercise.

(6.10) Example. We have seen some examples in Section 2.5 of rings that

are not UFDs, namely, F[X2, X3J and some of the quadratic rings (see Proposition 5.22 and Lemma 5.23). We wish to present one more example of a Noetherian function ring that is not a UFD. Let

S1={(x,y)ER2:z2+y2=1} be the unit circle in R2 and let I C R[X, YJ be the set of all polynomials such that f (x, y) = 0 for all (x, y) E S'. Then I is a prime ideal of R[X, Y) and R[X, YJ/I can be viewed as a ring of functions on S' by means of

f (X, Y) +I i f where 7(x, y) = f (x, y). We leave it for the reader to check that this is well defined. Let T be the set of all f (X, Y) + I E R[X, YJ/I such that 7(x, y) j4 0 for all (x, y) E S'. Then let the ring R be defined by localizing at the multiplicatively closed set T, i.e., R = (R[X, YJ/I )T .

Thus, R is a ring of functions on the unit circle, and a function in R is a unit if and only if the function never vanishes on S1. Claim. R is not a UFD. Proof. Let g(x, y) = x2 + (y - 1)2 and h(x, y) = x2 + (y + 1)2 E R. Then

gh(x,y) = (x2 + (y - 1)2)(x2 + (y + 1)2) x4 + x2((y - 1)2 + (y + 1)2) + (y2 = = x4 + x2((y - 1)2 + (y + 1)2) + x4

- 1)2

so that x divides gh in R, but clearly x does not divide either g or h (since neither g nor h is zero at both (0, 1) and (0, -1) E S', but x is). Therefore the ideal (x) C R is not prime. The proof is completed by showing

that x is an irreducible element of the ring R. To see this suppose that x = f 1 f 2 where neither is a unit. Then we must have V(fl) = 1(0, 1)) and V(f2) = { (0, -1) } (or vice versa), where V (g) means the set of zeros of the function g. Since fl and f2 are continuous functions, it follows that fl does not change sign on S1 \ {(O, 1) } and f2 does not change sign on S' \ ((0, -1)).

Therefore, x = fl f2 will not change sign on the set S1 \ 1(0, 1), (0, -1)}, and this contradiction shows that x must be an irreducible element of R. 0


97

In general, the problem of explicitly determining if a polynomial in R[X] is irreducible is difficult. For polynomials in Z[X] (and hence in Q[X])

there is a procedure due to Kronecker that in principle can determine the factors of an integral polynomial in a finite number of steps. For this method see, for example, Modern Algebra by B.L. van der Waerden, Vol. I, p. 77. We shall content ourselves with the following simple criterion for irreducibility, which is sometimes of use. An example of its use will be in Corollary 6.12.

(6.11) Theorem. (Eisenstein's criterion) Let R be a UFD with quotient field

F. Let f (X) = ao + a1X +

+ anX'i (an 0 0) be in R[X] and suppose

that p E R is a prime such that p does not divide an,

pIai for 0 Si
Then f (X) is irreducible in FIX]. Proof. If we write f (X) = cont(f (X )) f 1(X) then f, (X) is a primitive polynomial that also satisfies the hypotheses of the theorem. Thus without loss of generality we may suppose that f (X) is a primitive polynomial. If there

exists a factorization of f (X) into factors of degree > 1 in FIX] then by Lemma 6.6 there is also a factorization in R[X]. Thus, suppose that we can

write f (X) = g(X)h(X) where g(X), h(X) E R[X]. From Gauss's lemma we must have that g(X) and h(X) are primitive polynomials. Suppose

with f, m > 1, btc,,, 0, and e + m = n. Since p I ao = b0co but p2 does not divide a0, it follows that p I b0 or p I co but not both. To be specific, suppose that p I b0 but that p does not divide co. Not all the coefficients of g(X) are divisible by p since g(X) is primitive. Let bi be the first coefficient of g(X) that is not divisible by p so that 0 < i < t < n. Then we have a; = bico + bi_ic1 + ... + b0ci.

But p I a, and p I bj for j < i, so p I bic0. But p does not divide bi and p does not divide co, so we have a contradiction. Therefore, we may not

write f (X) = g(X)h(X) with deg g(X) > 1, deg h(X) > 1, i.e., f (X) is irreducible in FIX]. 0 The following is a useful consequence of Eisenstein's criterion:

(6.12) Corollary. Let p be a prime number and let f9(X) E Q[X] be the polynomial

98

Chapter 2. Rings

XP-1

fp(X) = X P-1 + XP-2 + ... + X + l = X-1 Then fp(X) is irreducible.

Proof. Since the map g(X) ' g(X + 1) is a ring isomorphism of Q[X], it is sufficient to that fp(X + 1) is irreducible. But p (by Exercise 1) divides every binomial coefficient (k) for 1 < k < p, and hence, P -

fP(X+1)= (X+l1)-1

XP-1+pXP-2+...+p.

Thus, fp(X + 1) satisfies Eisenstein's criterion, and the proof is complete. 0

2.7 Exercises 1.

2.

Prove the binomial theorem (Proposition 1.12). Give a counterexample if a and b do not commute. If p is a prime number, prove that the binomial coefficient (k) is divisible by p for 0 < k < p. Give an example to show that this result need not be true if p is not prime. Let R be a ring with identity and let a E R. The element a is said to be nilpotent if an = 0 for some n E N. It is said to be idempotent if a2 = a. Prove the following assertions.

3.

4.

5. 6. 7.

8. 9.

(a) If R has no zero divisors, then the only nilpotent element of R is 0 and the only idempotent elements of R are 0 and 1. b) No unit of R is nilpotent. The only idempotent unit of R is I. (c) If a is nilpotent, then 1 - a is a unit. (Hint: Geometric series.) If a is idempotent, then 1 - a is idempotent. (d) If R is commutative and N = {a E R : a is nilpotent}, show that N is an ideal of R. (e) Provide a counterexample to part (d) if R is not commutative. (Note that in parts (b)-(e), the ring R is allowed to have zero divisors.) For the ring P(X) of Example 1.10 (7), show that every A E P(X) satisfies the equation A2 = A. If P(X) is an integral domain, show that IXI = 1.

Continuing with the ring P(X), let a E X and define I. = A E P(X) a Al. Prove that la is a maximal ideal of P(X). What is P X)/I,? For a finite set X determine all of the ideals of P(X) and show that every maximal ideal is an ideal I. for some a E X. Prove Lemma 2.12 (1) and (3). Let R be a ring with identity. Show that R is a division ring if and only if it has no left or right ideals other than {0} and R. a Solve the equation 6x = 7 in the ring Z19, if possible. (b) Solve the equation 6x = 7 in the ring Z2o, if possible.

If R and S are rings with identities, prove that (R x S)' = R' x S. (Recall that R' denotes the group of units of the ring R.) Compute all the ideals, prime ideals, and maximal ideals of the ring Z60. What are the nilpotent elements of Z6e?

2.7 Exercises 10.

99

Let R be a ring and let R°P ("op" for opposite) be the abelian group R, together with a new multiplication a b defined b a b = ba, where ba

denotes the given multiplication on R. that R°" is a ring and that the identity function 1R : R R°P is a ring homomorphism if and only if R is a commutative ring. 11.

(a) Let A be an abelian group. Show that End(A) is a ring. (End(A) is defined in Example 1.10 (11).) (b) Let F be a field and V an F-vector space. Show that EndF(V) is a ring. Here, Endp(V) denotes the set of all F-linear endomorphisms of V, i.e.,

EndF(V)={hEEnd(V):h(av)=ah(v)forallvEV,aEF}. In this definition, End(V) means the abelian group endomorphisms of V,

12.

and the ring structure is the same as that of End(V) in part (a). (a) Let R be a ring without zero divisors. Show that if ab = 1 then ba = 1 as well. Thus, a and b are units of R. (b) Show that if a, b, c E R with ab = 1 and ca = 1, then b = c, and thus a (and b) are units. Conclude that if ab = 1 but ba $ 1, then neither a nor b are units. (c) Let F be a field and FIX] the polynomial ring with coefficients from F. F[X] is an F-vector space in a natural way, so by Exercise 11, R = EndF(FF[X]) is a ring. Give an example of a, b E R with ab = 1 but ba54 1.

13.

(a) Let x and y be arbitrary positive real numbers. Show that the quaternion

algebras Q(-x, -y; R) and Q(-1, -1; R) are isomorphic. (b) Show that the quaternion algebras Q(-1, -3;Q), Q(-1, -7;Q), and Q(-1, -11; Q) are all distinct.

(c) Analogously to Example 1.10 (10), we may define indefinite quaternion algebras by allowing x or y to be negative. Show that, for any nonzero

real number x and any subfield F of R, Q(1, x; F) and Q(l, -x; F) are isomorphic.

(d) Show that for any nonzero real number x, Q(1, x; R) and Q(1, 1; R) are isomorphic.

14. 15.

(e) Show that for any subfield F of R, Q(1, 1; F) is isomorphic to M3(F), the ring of 2 x 2 matrices with coefficients in F. (Thus, Q(1, 1; F) is not a division ring.) that Z(i)/(3 + i) Zio. Let F be a field and let R C F(XJ x F[Y] be the subring consisting of all pairs (f (X), g(Y)) such that f (0) = g(0 . that FIX, YJ/(XY) 5 R.

16.

Let R be a ring with identity and let I be an ideal of R. Prove that

17.

Let F be a field and let R = { [ o 0] E M2 (F) 1. that R is a ring. Does

5_

R have an identity? Prove that the set I = { o o] E R} is a maximal ideal of R. 18.

19.

(a) Given the complex number z = I + i, let o : R(X) -. C be the substitution homomorphism determined by z. Compute er(m). (b) Give an explicit isomorphism between the complex numbers C and the quotient ring R[XJ/(X' - 2X + 2). Let R = C11[0, 1]) be the ring of continuous real-valued functions on the interval (0,. Let T C [0, 11, and let

I(T)={f E R: f(x)=0forallxET}. a Prove that I(T) is an ideal of R. b If x E 10, 1] and M. = I({x}), prove that M. is a maximal ideal of R and R`M= ?f R.

100

Chapter 2. Rings (c) If S C R let

V(S)=(xEI: f(s)=0forall f ES). Prove that S is a closed subset of 10,1]. (You may quote appropriate

theorems on continuous functions.)

(d) If I R is any ideal, then prove that V(1) j4 0. (Hint: Suppose that V(I) = 0 and construct a function f E I such that f (x) 96 0 for all x E [0,1 . Conclude that I = R. At some point in your argument you

will need to use the compactness of 0,11.) (e) Prove that any maximal ideal M of is M. for some x E 10, 1]. 20. Let R C S be commutative rings with identity and let d E R be an element

such that the equation a2 = d is solvable in S but not in R, and let f denote a solution to the equation in S. Define a set R[ fd] by

R[vd]=(a+bf:a,bER) CS. (a) Prove that R[fd] is a commutative ring with identity. (b) Prove that Z[d] is an integral domain. (c) If F is a field, prove that F[fd] is a field. 21.

(a) If R = Z or R = Q and d is not a square in R, show that R[ fd] °_° R[X ]/ (X 2 - d) where (X2 - d) is the principal ideal of R(X] generated by X2 - d. (b) If R = Z or R = Q and di, d2, and di/d2 are not squares in R \ {0}, show that R[ f j ] and R[ v/ 2 are not isomorphic.

(The most desirable proof of these assertions is one which works for both R = Z and R = Q, but separate proofs for the two cases are acceptable.)

(c) Let R, = ZD[X]/(X2 - 2) and R2 = ZD[X]/(X2 - 3). Determine if

Rl°°R2in case p=2,p=5,or p= 11.

22.

Recall that R' denotes the group of units of the ring R. (a) Show that (Z(vr--1])' = {±1,

(b) If d < -1 show that (Z[f])' _ {t1). c) Show that

Z((1+" ] =ltl }1+ 2

2

l+ 2

(d) Let d > 0 E Z not be a perfect square. Show that if Z(d) has one unit other than ±1, it has infinitely many. (e) It is known that the hypothesis in part (d) is always satisfied. Find a unit in Z[%/-dl other than ±1 fcr 2 < d < 15, d A 4, 9. 23. Let F Q be a field. An element a E F is said to be an algebraic integer if for some monic polynomial p(X) E Z[X], we have p(a) = 0. Let d E Z be a nonsquare.

(a) Show that if a E F is an algebraic integer, then a is a root of an irreducible monic polynomial p(X) E Z[X]. (Hint: Gauss's lemma.)

(b) that a + bf E Q[fd] is a root of the quadratic polynomial p(X) = X2 - 2aX + (a2 - bed) E Q[X]. (c) Determine the set of algebraic integers in the fields Q[f] and Q(f]. 24.

(See Remark 5.24 (1).) If R is a ring with identity, then Aut(R), called the automorphism group of R, denotes the set of all ring isomorphisms m : R R.

2.7 Exercises

101

(a) Compute Aut(Z) and Aut(Q). b Compute Aut(Z[v' J) and Aut Q 'Q[X11) if d is not a square in Z or Q. -' Q[X] be the substitution homo(c) If a j4 0 E Q,b E Q let 0. s : morphism determined by V '--# aX + b. Prove that 4,,6 E Aut(Q(X]).

What is 0?

(d) If 0 E Aut(Q[XI) prove that there are a # 0, b E Q such that 0 = O,,s. 25. Let m = exp(2ai/n) E C and let R = Q[w). Show that Aut(R) °_e Z. 26.

Let R be a commutative ring with identity. Prove that Mn(R[X]) ?°

27.

Mn(R)[X] Let R be a commutative ring with identity and let f = ao + a1X +

+ a,X" E R[X] where an 96 0. We know that if R is an integral domain then the units of R[X] are the units of R (Corollary 4.3). This exercise will

investigate when f is a unit, a zero divisor, or is nilpotent when R is not assumed to be an integral domain.

(a) Prove that f is a unit in R[X) if and only if ao is a unit in R and

+ b,nX'n is the inverse a1, ... , an are nilpotent. (Hint: If bo + biX + of f, prove by induction on r that a'n+lb,_,. = 0. Conclude that an is nilpotent and apply Exercise 2 (c).) (b) Prove that f is nilpotent if and o if ao, al , ... , an are nilpotent. c Prove that f is a zero divisor in R[X] if and only if there is a nonzero

a E R such that of = 0. (Hint: Choose a nonzero polynomial g =

+ b,. X"' of least degree m such that f = 0. If m = 0 we are done. Otherwise, anb,n = 0 and hence ang = 0 (because (an9)f = an (g f) = 0 and deg(ang) < degg). Show by induction that an-,.g = 0 for (0 < r < n). Conclude that 6,,, f = 0.) 28. Factor the polynomial X2 + 3 into irreducible polynomials in each of the bo + bl X +

following rings.

a Z5 X.

(b) 29.

1X I.

Let F = Z5 and consider the following factorization in FIX]:

3X3+4X2+3=(X+2)2 (3X+2) _ (X + 2)(X + 4)(3X + 1).

Explain why (*) does not contradict the fact that F[X] is a UFD. 30. For what fields ZD is X 3 + 2X2 + 2X + 4 divisible by X 2 + X + I? 31. In what fields Zy is X2 + 1 a factor of X 3 + X 2 + 22X + 15? 32. Find the gcd of each pair of elements in the given Euclidean domain and mess the gcd in the form ra + aA .

a 189 and 301 in Z.

(b 1261 and 1649 in Z.

(c) X4-X3+4X2-X+3 and X3-2X2 +X-2 inQ[X]. d) X4 + 4 and 2X3 + X2 - 2X - 6 in Z3 [X ].

e 2 + IN and 1 + 3i in Z[i].

f) -4 + 7i and 1 + 7i in Z(i]. 33. Express X4 - X3 - 2 as a product of irreducible polynomials in each of the fields Q, 11, C, and Z5. 34.

Let F be a field and suppose that the polynomial X'n - 1 has m distinct roots in F. If k [ m, prove that X" - 1 has k distinct roots in F.

35.

If R is a commutative ring, let F(R, R) be the set of all functions f : R -+ R and make F(R, R) into a ring by means of addition and multi lication of functions, i.e.,

function 4ti : R[

g)(r) = f(r) g(r) and (f + 9)(r) = f (r) + g(r). Define a ] -- (R, ) by

102

Chapter 2. Rings

4,(f (X)) (r) = Or(f (X)) = f (r) for all r E R. a Show that 40 is injective if R is an infinite integral domain. b Show that 0 is not injective if R is a finite field. 36. 37

Let f (X) E Z [X]. Show that f (X') = (f (X ))P and that the map O(f M) = (f OF W is a ring homomorphism. Let F be a field and consider the substitution homomorphism

m: FIX,YI -.FIT] such that m(X) = T2 and O(Y) = T3. Show that Ker(O) is the principal ideal generated by y2 - X3. What is Im(m)? 38.

Prove Proposition 4.10 (Lagrange interpolation) as a corollary of the Chinese remainder theorem.

39.

Let F be a field and let ao, ai, ... , an be n + 1 distinct elements of F. If f : F - F is a function, define the successive divided differences of f, denoted f [ao, ... , a,], by means of the inductive formula: flao[ = f(ao)

flao, all =

f[ao,...,an]=

f[ai] - flan] al - ao

f[ao, ... ,an-2, an] - f[ao, ... ,an-2, an-i] an - an-i

Prove that the coefficient of Pi (X) in Equation (4.3) (Newton's interpolation formula) is the divided difference f [ao, ... , a,]. 40. (a) If f (X) E RIX] and z E C satisfies f (z) = 0, show that f ('z) = 0 (where a is the complex conjugate of z). (b) Let ro, ri, ..., rk be distinct real numbers, let zi, ... , z,,, be distinct complex numbers with z., All f o r 1 < j, f < m, and let 80, ... , sk E R,

Wi, ... , wn, E C. Prove that there is a polynomial f (X) E R[X] such

that f(r,) = s; for 0 1. Show that there are unique fo, f i ,

, fd E FIX I such that deg f, < deg g and

f

=fo+fig+...+fd9d.

Let K and L be fields with K C L. Suppose that f (X), g(X) E K[X). (a) If f (X) divides g(X) in L[X], prove that f (X) divides g(X) in XJ. b Prove that the greatest common divisor of f (X) and g(X) in KJX is the same as the greatest common divisor of f (X) and g(% in L X . (We will always choose the monk generator of the ideal (f (X), g(X) as the greatest common divisor in a polynomial ring over a field.) 43. (a) Suppose that R is a Noetherian ring and I C R is an ideal. Show that Rh is Noetherian. (b) If R is Noetherian and S is a subring of R, is S Noetherian? c) Suppose that R is a commutative Noetherian ring and S is a nonempty multiplicatively closed subset of R containing no zero divisors. Prove that Rs (the localization of R away from S) is also Noetherian. 42.

2.7 Exercises 44.

103

If R is a ring (not necessarily commutative) and f (X) = ao + aiX + + a,, X" E R[X], then we say that f (X) is regular o degree n if an is a unit of R. Note, in particular, that monic polynomials are regular and if R is a field then all nonzero polynomials in R[X) are regular. Prove the following version of the division algorithm:

Let f(X) E R(XJ and let g(X) E R[X] be a regular polynomial of degree n. Then there are unique polynomials ql (X ), ri (X), q2 (X ), and r2 (X) E R[X] such that deg rl (X) < n, deg r2 (X) < n,

f(X) = gi(X)g(X) +rl(X) and

f(X) =g(X)g2(X)+r2(X) The two equations represent the left and right divisions of f (X) by 9(X). In the special case that g(X) = X - a for a E R, prove the following version of these equations (noncommutative remainder theorem):

+ anX" E R[X] and let a E R. Then there are

Let f (X) = a1 + a1X +

unique qc(X) and qR (X) E R(X ] such that

f(X) = gR(X)(X - a) + fx(a) and

f(X) = (X -a)gc(X)+ft(a) where

n

fn (a) = 1:

n

akak

akak

and f c (a) _

k=0

k=0

are, respectively, the right and left evaluations of f (X) at a E R. (Hint: Use the formula Xk

- ak = (X k-1 + X k-2a + ... + Xak-2 + ak-1) (X - a) = (X - a)(Xk-1 + aXk-2 + ... + ak-2X + ak-1).

Then multiply on either the left or the right by at, and sum over k to get the division formulas and the remainders.) 45. 46.

47.

Let R be a UFD and let a and b be nonzero elements of R. Show that ab = (a, bJ(a, b) where [a, b) = lcm{a, b} and (a, b) = gcd{a, b}. Let R be a UFD. Show that d is a gcd of a and b (a, b E R \ {0}) if and only if d divides both a and b and there is no prime p dividing both a/d and b/d. In particular, a and b are relatively prime if and only if there is no prime p dividing both a and b.)

Let R be a UFD and let {r1}1 be a finite set of pairwise relatively prime nonzero elements of R (i.e., ri and r3 are relatively prime whenever i 36 j).

Let a = f 1 ri and let a, = a/ri. Show that the set {ai} 1 is relatively prime.

Let R be a UFD and let F be the quotient field of R. Show that d E R is a square in R if and only if it is a square in F (i.e., if the equation a2 = d has a solution with a E F then, in fact, a E R). Give a counterexample if R is not a UFD. 49. Let x, y, z be integers with gcd x, y, z = 1. Show that there is an integer a such that cd{x + ay, z} = 1. (Hint: The Chinese remainder theorem may be helpful.) 48.

104

50.

51.

Chapter 2. Rings

According to the Chinese remainder theorem there is an isomorphism of rings ¢ : Z60 - Z3 X Z4 X Z5. Compute 0(26), 4)(35), 4-1(2, 3, 4), and m-1(1, 2, 2). Solve the system of simultaneous congruences:

-3 (mod 13) x - 16 (mod 18) x -2 (mod 25) x

x=0 52.

(mod 29).

Solve the system of simultaneous congruences:

x-6 (mod21)

x9 (mod33) x

2

(mod 37).

53. For what values of a (mod 77) does the following system of simultaneous

congruences have a solution?

x-6 (mod21) x - 9 (mod 33) x =-a (mod 77). 54.

(a) Solve the following system of simultaneous congruences in Q(XI:

f(X) f (X) f (X)

-3 (mod X + 1) 12X

-4X

(mod X2 - 2) (mod X3).

b) Solve this system in Z5 X . c) Solve this system in Z3 X . d Solve this system in Z2 X . 55. Suppose that ml, m2 E Z are not relatively prime. Then prove that there are integers al, a2 for which there is no solution to the system of congruences:

x - a1 (mod ml) x - a2 (mod M2)56.

Let R be a UFD. Prove that

f(X,Y)=X4+2Y2X3+3Y3X2+4YX+5Y+6Y2 is irreducible in the polynomial ring R(X, Y].

Prove that if R is a UFD and if j(X) is a monic polynomial with a root in the quotient field of R, then that root is in R. (This result is usually called the rational root theorem.) 58. Suppose that R is a UFD and S C R\ {0} is a multiplicatively closed subset. Prove that Rs is a UFD. 59. Let F be a field and let F((X)j be the ring of formal power series with coefficients in F. If j = F,°° u a X" 0 0 E F, let o(f) = min{n # 0} and define o(0) = oo. o(f) is usually called the order of the power series j. 57.

Prove the following facts:

(a) o(f9) = o(f) + o(9).

2.7 Exercises

105

b) of +g) >_ min{o(f), o(g) }.

). cd ffis9ifandonlyifo(f)
(e) If f 96 0 then f is an associate of X0(J). Conclude that X is the only irreducible (up to multiplication by a unit) element of F((X]]. F[[X]] is a PID. In fact, every ideal is generated by Xk for some k. (g) Is [[ ]] a Euclidean domain? f

60.

If F is a field, let F((X)) denote the set of all formal Laurent series with coeffients in F, i.e.,

0 F((X)) =

{>anX'

F,m E Z is arbitrary

:a

n=m

where the ring operations are defined as for F[[X]]. Prove that F((X)) is isomorphic to the quotient field of the integral domain F[IX]]. 61.

Let R be a commutative ring and let S = R[T1 , ... , Tn]. Define f (X) E S[X] by n

f(X) = ll(X -Ti) i=1 n

= (-1)ror(Ti, ... Tn)Xf

r.

r=0

(a) Show that

Tit ...Tir.

ar(T1, ... Tn) = Thus, a, = Ti +

+Tn and on = Ti .. Tn. or is called the r`h elemen-

tary symmetric function. Therefore, the coefficients of a polynomial are obtained by evaluating the elementary symmetric functions on the roots of the polynomial.

(b) If g(X) = [I-,=, (1 -T;X), show that the coefficient of Xr is i.e., the same as the coefficient of X"-r in f(X). (c) Define sr(T1, ... Tn) = T; + + Tn for r > 1, and let so = n. the following identities (known as Newton's identities) relating the power sums sr and the elementary symmetric functions ok. r-1

(-1)kaksr-k + (-1)rror

=0

(1 < r < n)

k=0 n

J(- I)kakSr-k = 0

(r > n).

k=0

(Hint: Do the following calculation in S[[X]], where' means derivative with respect to X: g' (X)

g(X)

n n

i=s

1-TX

00

i=1 i=o

00

-ELT'+1Xj -Esj+1Xj. i=o

Now multiply by g(X) and compare coefficients of X `1 . )

Chapter 2. Rings

106

62.

Let P E Z be a prime and define

R = {a = (al, as, a3,

.

.) :

ak E (Z/pkZ), ak+1 = ak

(mod pk )

for all

k>1).

(a) Show that R is a ring under the operations of componentwise addition and multiplication. (b) Show that R is an integral domain. (Note that R contains Z as a subring so that char(R) = 0.) (c) Let

P={aER:a1=0EZ/pZ}.

Show that every element of R \ P is invertible. Show that P is a proper ideal of R. (Thus, P is the unique maximal ideal in R and so R is a local ring.)

(d) For a E R \ {0} define v(a) to be p^' 1 if n is the smallest value of k such that ak j4 0 E Z/pkZ. Show that v : R \ {0} -+ Z+ makes R into a Euclidean domain.

Remark. The ring R plays an important role in mathematics; it is known as the ring of p-adic integers and its quotient field is known as the field of p-adic numbers.

Chapter 3

Modules and Vector Spaces

3.1 Definitions and Examples Modules are a generalization of the vector spaces of linear algebra in which the "scalars" are allowed to be from an arbitrary ring, rather than a field. This rather modest weakening of the axioms is quite far reaching, including, for example, the theory of rings and ideals and the theory of abelian groups as special cases.

(1.1) Definition. Let R be an arbitrary ring with identity (not necessarily commutative). (1)

A left R-module (or left module over R) is an abelian group M together with a scalar multiplication map

:RxM-M that satisfy the following axioms (as is customary we will write am in place of (a, m) for the scalar multiplication of m E 1LI by a E R). In these axioms, a, b are arbitrary elements of R and m, n are arbitrary elements of M.

(ai)a(m + n) = am + an. (bi)(a + b)m = am + bm. (ci) (ab)m = a(bm).

(di)lm = m.

(2) A right R-module (or right module over R) is an abelian group M together with a scalar multiplication map

:MxR-+M that satisfy the following axioms (again a, b are arbitrary elements of R and m, n are arbitrary elements of M).

(&)(m + n)a = ma + na. (b,)m(a + b) = ma + mb.

108

Chapter3. Modules and Vector Spaces

(cr)m(ab) = (ma)b. (dr)m1 = m.

(1.2) Remarks. (1) If R is a commutative ring then any left R-module also has the struc-

ture of a right R-module by defining mr = rm. The only axiom that requires a check is axiom (cr). But m(ab) = (ab)m = (ba)m = b(am) = b(ma) = (ma)b.

(2) More generally, if the ring R has an antiautomorphism (that is, an additive homomorphism 0: R -' R such that 0(ab) = m(b)4(a)) then any left R-module has the structure of a right R-module by defining ma = ¢(a)m. Again, the only axiom that needs checking is axiom (cr):

(ma)b = .(b)(ma) = O(b)(O(a)m) = (m(b)O(a))m

= 0(ab)m

= m(ab).

An example of this situation occurs for the group ring R(G) where R is a ring with identity and G is a group (see Example 2.1.10 (15)). In this case the antiautomorphism is given by

-O(Ea99) _ 9EG

a99-' 9EG

We leave it as an exercise to check that : R(G) - R(G) is an antiautomorphism. Thus any left R(G)-module M is automatically a right R(G)-module.

(3) Let R be an arbitrary ring and let R°P ("op" for opposite) be the ring whose elements are the elements of R, whose addition agrees with

that of R, but whose multiplication - is given by a b = ba (where the multiplication on the right-hand side of this equation is that of R). Then any left R-module is naturally a right R°P-module (and viceversa). In fact, if M is a left R-module, define a right multiplication

of elements of R°P (which are the same as elements of R) on M by am. As in Remark 1.2 (1), the only axiom that requires checking is axiom (Cr). But

The theories of left R-modules and right R-modules are entirely parallel, and so, to avoid doing everything twice, we must choose to work on


109

one side or the other. Thus, we shall work primarily with left R-modules unless explicitly indicated otherwise and we will define an R-module (or module over R) to be a left R -module. (Of course, if R is commutative, Remark 1.2 (1) shows there is no difference between left and right R-modules.) Applications of module theory to the theory of group representations will, however, necessitate the use of both left and right modules over noncommutative rings. Before presenting a collection of examples some more notation will be introduced.

(1.3) Definition. Let R be a ring and let M, N be R-module8. A function f : M -+ N is an R-module homomorpbism if

(1) A MI + m2) = A MO + P M2) for all ml, m2 E M, and (2) f (am) = a f (m) for all a E R and m E M.

The set of all R-module homomorphisms from M to N will be denoted HomR(M, N). In case M = N we will usually write EndR(M) rather than HomR(M, M); elements of EndR(M) are called endomorphisms. If f E EndR(M) is invertible, then it is called an automorphism of M. The group of all R-module automorphisms of M is denoted AutR(M) (Aut(M)

if R is implicit). If f E HomR(M, N) then we define Ker(f) C M and Im(f) C N to be the kernel and image of f considered as an abelian group homomorphism. (1.4) Definition. (1) Let F be a field. Then an F-module V is called a vector space over F. (2) If V and W are vector spaces over the field F then a linear transfor-

mation from V to W is an F-module homomorphism from V to W. (1.5) Examples. (1)

Let G be any abelian group and let g E G. If n E Z then define the scalar multiplication ng by

(n ) if n > 0,

ng=

0

(-g) +

ifn=0, + (-g) (-n ) ifn < 0.

Using this scalar multiplication G is a Z-module. Furthermore, if G and H are abelian groups and f : G -a H is a group homomorphism, then f is also a Z-module homomorphism since (if n > 0) f (ng) = f (9 + ... + 9) = f (9) + ... + f(9) = nf(9) and

f(-g) = -f(g).

Chapter 3. Modules and Vector Spaces

110

(2) Let R be an arbitrary ring. Then R" is both a left and a right R-module via the scalar multiplications

a(bl, ... b") = (abi, ... ,ab") and

(bi, ...

, b" )a

= (bia, ... , b"a).

(3) Let R be an arbitrary ring. Then the set of matrices M,,,,"(R) is both a left and a right R-module via left and right scalar multiplication of matrices, i.e., ent;j (aA) = a ent,J(A) and

ent,3(Aa) = (ent,j(A))a.

(4) As a generalization of the above example, the matrix multiplication maps

Mm(R) x M,,,,"(R)

Mm."(R)

(A, B) .-- AB and

Mm,"(R) x M"(R)

Mm,"(R)

(A, B) - AB make M,,,,"(R) into a left Mm(R)-module and a right M"(R)-module. (5) If R is a ring then a left ideal I C R is a left R-module, while a right

ideal J C R is a right R-module. In both cases the scalar multiplication is just the multiplication of the ring R. (6) If R is a ring and I C R is an ideal then the quotient ring R/I is both a left R-module and a right R-module via the multiplication maps

(a, b + I)

ab + I

and

R/IxR-+R/I (a+I,b).--sab+l. (7)

M is defined to be an R-algebra if M is both an R-module and a ring, with the ring addition being the same as the module addition, and the multiplication on M and the scalar multiplication by R satisfying the following identity: For every r E R, m1, m2 E M,

(1.1)

r(mlm2) = (rmi)m2 = ml(rm2).


111

For example, every ring is a Z-algebra, and if R is a commutative ring,

then R is an R-algebra. Let R and S be rings and let 4 : R -+ S be a ring homomorphism with Im(m) C C(S) = (a E S : ab = ba for all b E S), the center of S. If M is an S-module, then M is also an R-module using the scalar multiplication am = (4(a))m for all a E R and m E M. Since S itself is an S-module, it follows that S is an R-module, and moreover, since Im(¢) C C(S), we conclude that S is an R-algebra. As particular cases of this construction, if R is a commutative ring, then the polynomial ring R[X] and the matrix ring

M are R-modules then HomR(M, N) is an abelian group via the operation (f + g)(m) = f (m) + g(m). However, if we try to make HomR(M, N) into an R-module in the natural way by defining of by the formula (a f)(m) = a(f (m)) we find that the function a f need not be an R-module homomorphism unless R is a commutative ring. To see this, note that (af)(rm) = a(f (rm)) = a(r(f (m))) = ar f (m).

This last expression is equal to r(a f) (m) = ra f (m) if R is a commutative ring, but not necessarily otherwise. Thus, if R is a commutative ring, then we may consider HomR(M, N) as an R-module for all M, N, while if R is not commutative then HomR(M, N) is only an abelian group. Since EndR(M) is also a ring using composition of R-module homomorphisms as the multiplication, and since there is a ring homomorphism 0 : R - EndR(M) defined by O(a) = a 1M where IM denotes the identity homomorphism of M, it follows from Example 1.5 (7) that EndR(M) is an R-algebra if R is a commutative ring. (9) If G is an abelian group, then Homz(Z, G) G. To see this, define

$ : Homz(Z, G) - G by 4i(f) = f (1). We leave it as an exercise to check that 4) is an isomorphism of Z-modules. (10) Generalizing Example 1.5 (9), if M is an R-module then

HomR(R,M)^='M as Z-modules via the map 4) : HomR(R, M) -+ M where 0(f) = f (1).

(11) Let R be a commutative ring, let M be an R-module, and let S C EndR(M) be a subring. (Recall from Example 1.5 (8) that EndR(M) is a ring, in fact, an R algebra.) Then M is an S-module by means of the scalar multiplication map S x M M defined by (f, m) '--' f (m). (12) As an important special case of Example 1.5 (11), let T E EndR(M) and define a ring homomorphism 0 : R[X] EndR(M) by sending X to T and a E R to alM. (See the polynomial substitution theorem (Theorem 2.4.1).) Thus, if

f (X) =ao+a1X

+...+aX"


112 then

O(f(X)) We will denote O(f (X)) by the symbol f (T) and we let Im(O) = R[T]. That is, R[T] is the subring of EndR(M) consisting of "polynomials" in T. Then M is an R[TJ module by means of the multiplication

f (T)m = f (T)(m).

Using the homomorphism 0: R(X) -, R[T] we see that M is an R[X]module using the scalar multiplication

f (X)m = f (T)(m). This example is an extremely important one. It provides the basis for applying the theory of modules over principal ideal domains to the study of linear transformations; it will be developed fully in Section 4.4.

(13) We will present a concrete example of the situation presented in Ex-

ample 1.5 (12). Let F be a field and define a linear transformation

T : F2 -+ F2 by T(ul, u2) = (u2, 0). Then T2 = 0, so if f (X) = + amXm E F[XJ, it follows that f (T) = ao1Fa + a1T. ao + a1X + Therefore the scalar multiplication f (X )u for u E F2 is given by f(X) . (U1, U2) = f(T)(u1,u2) = (ao1F2 +a,T)(u1,u2) = (aoui +a1u2,aou2)

3.2 Submodules and Quotient Modules Let R be a ring and M an R-module. A subset N C M is said to be a submodule (or R-submodule) of M if N is a subgroup of the additive group of M that is also an R-module using the scalar multiplication on M. What this means, of course, is that N is a submodule of M if it is a subgroup of M that is closed under scalar multiplication. These conditions can be expressed as follows.

(2.1) Lemma. If M is an R-module and N is a nonempty subset of M, then N is an R-submodule of M if and only if amt + bm2 E N for all

m1, m2 N and a, bE R. Proof. Exercise.

3.2 Submodules and Quotient Modules

113

If F is a field and V is a vector space over F, then an F-submodule of V is called a linear subspace of V. (2.2) Examples. If R is any ring then the R-submodules of the R-module R are precisely the left ideals of the ring R. If G is any abelian group then G is a Z-module and the Z-submodules of G are just the subgroups of G.

Let f : M - N be an R-module homomorphism. Then Ker(f) C M and Im(f) C N are R-submodules (exercise).

Continuing with Example 1.5 (12), let V be a vector space over a field F and let T E EndF(V) be a fixed linear transformation. Let VT denote V with the FIX)-module structure determined by the linear transformation T. Then a subset W C V is an F[X]-submodule of the module VT if and only if W is a linear subspace of V and T (W) C W, i.e., W must be a T-invariant subspace of V. To see this, note that

X w = T(w), and if a E F, then a w = aw-that is to say, the action of the constant polynomial a E F[X] on V is just ordinary scalar multiplication, while the action of the polynomial X on V is the action of T on V. Thus, an F[X]-submodule of VT must be a Tinvariant subspace of V. Conversely, if W is a linear subspace of V

such that T(W) C W then Tm(W) C W for all m > 1. Hence, if f (X) E F[X] and W E W then f (X) w = f (T)(w) E W so that W is closed under scalar multiplication and thus W is an F[X]-submodule of V.

(2.3) Lemma. Let M be an R-module and let (NQ}QEA be a family of submodules of M. Then N Z n0EA N. is a submodule of M. Proof. Exercise.

0

We now consider quotient modules and the noether isomorphism theorems. Let M be an R-module and let N C_ M be a submodule. Then N is a subgroup of the abelian group M, so we can form the quotient group M/N. Define a scalar multiplication map on the abelian group M/N by

a(m + N) = am + N for all a E R, m + N E M/N. Since N is an Rsubmodule of M, this map is well defined. Indeed, if m + N = m' + N then

m-m' E N so that am-am'= a(m-m') E N so that am+N = am'+N. The resulting R-module M/N is called the quotient module of M with respect to the submodule N. The noether isomorphism theorems, which we have seen previously for groups and rings, then have direct analogues for R-modules.

(2.4) Theorem. (First isomorphism theorem) Let M and N be modules over the ring R and let f : M N be an R-module homomorphism. Then Im(f).

114


Proof. Let K = Ker(f). From Theorem 1.3.10 we know that f : M/K Im(f) defined by f (m+K) = f (m) is a well-defined isomorphism of abelian groups. It only remains to check that f is an R-module homomorphism. But

f(a(m+K)) = f(am+K) = f(am) = af(m) = af(m+K) for all m E M and a E R, so we are done. 0 (2.5) Theorem. (Second isomorphism theorem) Let M be an R-module and let N and P be submodules. Then there is an isomorphism of R-modules

(N+P)/PAN/(NnP). Proof. Let rr : M , M/P be the natural projection map and let rro be the restriction of rr to N. Then ao is an R-module homomorphism with Ker(rro) = N n P and Im(iro) = (N + P)/P. The result then follows from the first isomorphism theorem. 0

(2.6) Theorem. (Third isomorphism theorem) Let M be an R-module and let N and P be submodules of M with P C N. Then

111/N = (M/P)/(N/P). Proof. Define f : Al/P -. MIN by f (m+ P) = m+N. This is a well-defined R-module homomorphism and

Ker(f)={m +P:m+N=N}={m +P:mEN}=NIP. The result then follows from the first isomorphism theorem (Theorem 2.4).

0 (2.7) Theorem. (Correspondence theorem) Let Al be an R-module, N a submodule, and A : M - MIN the natural projection map. Then the function P -- P/N defines a one-to-one correspondence between the set of all submodules of M that contain N and the set of all submodules of M/N. Proof. Exercise.

0

(2.8) Definition. If S is a subset of an R-module M then (S) will denote the intersection of all the submodules of M that contain S. This is called the submodule of M generated by S, while the elements of S are called generators of (S). Thus, (S) is a submodule of M that contains S and it is contained in every submodule of M that contains S, i.e., (S) is the smallest submodule of M containing S. If S = {x1, ... ,xn} we will usually write (xi, ... ,xn)


115

rather than ({x1, ... , for the submodule generated by S. There is the following simple description of (S).

(2.9) Lemma. Let M be an R-module and let S C M. If S = 0 then (S) =

{0},while (S)={E 1a;si:nEN,a;ER, s,ES,1

Proof. Exercise.

(2.10) Definition. We say that the R-module M is finitely generated if

M = (S) for some finite subset S of M. M is cyclic if M = (m) for some element m E M. If M is finitely generated then let µ(M) denote the minimal number of generators of M. If M is not finitely generated, then let µ(M) = oo. We will call u(M) the rank of M. (2.11) Remarks. (1)

(2)

(3) (4)

We have uc({O}) = 0 by Lemma 2.9 (1), and M # {0} is cyclic if and only if µ(M) = 1. The concept of cyclic R-module generalizes the concept of cyclic group. Thus an abelian group G is cyclic (as an abelian group) if and only if it is a cyclic Z-module. If R is a PID, then any R-submodule M of R is an ideal, so µ(M) = 1. For a general ring R, it is not necessarily the case that if N is a submodule of the R-module M, then µ(N) < µ(M). For example, if R is

a polynomial ring over a field F in k variables, M= R, and N C M is the submodule consisting of polynomials whose constant term is 0, then u(M) = 1 but µ(N) = k. Note that this holds even if k = oo. We shall prove in Corollary 6.4 that this phenomenon cannot occur if R is a PID. Also see Remark 6.5.

If M is a finitely generated R-module and N is any submodule, then

M/N is clearly finitely generated, and in fact, µ(M/N) < µ(M) since the image in M/N of any generating set of M is a generating set of M/N. There is also the following result, which is frequently useful for constructing arguments using induction on µ(M).

(2.12) Proposition. Suppose M is an R-module and N is a submodule. If N and M/N are finitely generated, then so is M and

µ(M) < µ(N) + µ(M/N). Proof. Let S = {x1, ... , xk} C N be a minimal generating set for N and if rr : M -+ M/N is the natural projection map, choose T = {yl, ... , yt} C M so that {rr(yl), ... ,7r(ye)} is a minimal generating set for M/N. We claim

that S U T generates M so that u(M) < k + I = µ(N) + µ(M/N). To see this suppose that x E M. Then rr(x) = al7r(y1) + + atrr(yt). Let y =

116


alyl+ +atyt E (T). Then ir(x-y) =O so that x-y E Ker(ir) = N = (S). It follows that x = (x - y) + y E (S U T), and the proof is complete. 0 (2.13) Definition. If {NQ}QEA is a family of R-submodules of M, then the submodule generated by {NQ}QEA is (UQEA NQ). This is just the set of all sums nQ, + + n(kk where nQ, E NQ, . Instead of (UQEA N") , we will use the notation EQEA NQ; if the index set A is finite, e.g., A = {1, ... , m}, we will write N1 + + Nn for the submodule generated by N1, ... , Nn.

(2.14) Definition. If R is a ring, M is an R-module, and X is a subset of M, then the annihilator of X, denoted Ann(X), is defined by

Ann(X)={aE R:ax=O forallxEX}. It is easy to check that Ann(X) is a left ideal of R, and furthermore,

if X = N is a submodule of M, then Ann(N) is an ideal of R. If R is commutative and N = (x) is a cyclic submodule of M with generator x, then

Ann(N)={aER:ax=O}.

This fact is not true if the ring R is not commutative. As an example, let R = M,, (R) = M and let x = Ell be the matrix with a 1 in the 11 position and 0 elsewhere. It is a simple exercise to check that Ann(Ell) consists of all matrices with first column 0, while Ann((E11)) = (0). If R is commutative and N is cyclic with generator x then we will

usually write Ann(x) rather than Ann((x)). In this situation, the ideal Ann(x) is frequently called the order ideal of x. To see why, consider the example of an abelian group G and an element g E G. Then G is a Z-module and

Ann(g)={nEZ:ng=0}

_ (p) where p = o(g) if o(g) < oo and p = 0 if (g) is infinite cyclic. Example. Let F be a field, V a vector space, T E EndF(V) a linear transformation, and let VT be the F[X] module determined by T (Example 1.5 (12)). If v E V then

Ann(v) = if (X) E F[X] : f (T)(v) = 0}. Note that this is a principal ideal (g(X)) since F[X] is a PID. (2.15) Proposition. Let R be a ring and let M = (m) be a cyclic R-module.

Then M R/ Ann(m). Proof. The function f : R -+ M defined by f (a) = am is a surjective Rmodule homomorphism with Ker(f) = Ann(m). The result follows by the 0 first isomorphism theorem.


117

(2.16) Corollary. If F is a field and M is a nonzero cyclic F-module then

Mn-iF. Proof. A field has only the ideals {0} and F, and 1 m = m t 0 if m 36 0 is a generator for M. Thus, Ann(m) 36 F, so it must be {0}.

If M is an R-module and I C R is an ideal then we can define the product of I and M by

IM={>aimi:nEN,aiEI,miEM}. ,_ 1

JJJ

The set IM is easily checked to be a submodule of M. The product IM is a generalization of the concept of product of ideals. If R is commutative and I C Ann(M) then there is a map

R/IxM -+M defined by (a + I )m = am. To see that this map is well defined, suppose

that a + I = b + I. Then a - b E I C Ann(M) so that (a - b)m = 0, i.e., am = bm. Therefore, whenever an ideal I C Ann(M), M is also an R/I module. A particular case where this occurs is if N = M/IM where I is any ideal of R. Then certainly I C Ann(N) so that M/IM is an R/I-module. (2.17) Definition. Let R be an integral domain and let M be an R-module. We say that an element x E M is a torsion element if Ann(x) ,-f {0}. Thus

an element x E M is torsion if and only if there is an a # 0 E R such that ax = 0. Let M, be the set of torsion elements of M. M is said to be torsion-free if M, = {0}, and M is a torsion module if M = M,. (2.18) Proposition. Let R be an integral domain and let M be an R-module.

(1) M, is a submodule of M, called the torsion submodule. (2) M/M, is torsion-free. Proof. (1) Let x, y E M, and let c, d E R. There are a 34 0, b 36 0 E R such that ax = 0 and by = 0. Since R is an integral domain, ab # 0. Therefore, ab(cx + dy) = bc(ax) + ad(by) = 0 so that cx + dy E M.

(2) Suppose that a 0 0 E R and a(x + M,) = 0 E (M/M,),. Then there is ab00ERwith (ba)x=b(ax)=0.Since ba54 0, it follows that x E M i.e., x + M, = 0 E M/M,. ax

(2.19) Examples. (1) If G is an abelian group then the torsion Z-submodule of G is the

set of all elements of G of finite order. Thus, G = G, means that every element of 0 is of finite order. In particular, any finite abelian

118


group is torsion. The converse is not true. For a concrete example, take

G = Q/Z. Then IGI = oo, but every element of Q/Z has finite order since q(p/q + Z) = p + Z = 0 E Q/Z. Thus (Q/Z),. = Q/Z. (2) An abelian group is torsion-free if it has no elements of finite order other than 0. As an example, take G = Z" for any natural number n. Another useful example to keep in mind is the additive group Q.

(3) Let V = F2 and consider the linear transformation T : F2 -+ F2 defined by T(ul, u2) = (u2i 0). See Example 1.5 (13). Then the F[X) module VT determined by T is a torsion module. In fact Ann(VT) = (X2). To see this, note that T2 = 0, so X2 u = 0 for all u E V. Thus, (X2) C Ann(VT). The only ideals of F[XI properly containing (X2) are (X) and the whole ring F[X], but X 0 Ann(VT) since X (0, 1) _ (1, 0) 54 (0, 0). Therefore, Ann(VT) _ (X2). The following two observations are frequently useful; the proofs are left as exercises:

(2.20) Proposition. Let R be an integral domain and let M be a finitely generated torsion R-module. Then Ann(M) 34 (0). In fact, if M = (x1i ... , xn) then Ann(M) = Ann(x1) n . . . n Ann(xn) 34 (0). Proof. Exercise.

(2.21) Proposition. Let F be afield and let V be a vector space over F, i.e., an F-module. Then V is torsion-free. Proof. Exercise.

3.3 Direct Sums, Exact Sequences, and Hom Let M1, ... , AIn be a finite collection of R-modules. Then the cartesian product set M1 x

.

x Mn can be made into an R-module by the operations

(x1, ... , xn) + (y1, ... , yn) _ (xl +Y1, ... , xn +Y.) a(x1i ... , xn) = (axi, ... , axn)

where the 0 element is, of course, (0, ... , 0). The R-module thus constructed is called the direct sum of M1, ... , Mn and is denoted

M1 ®...®Mn

n (or

®M;).

=1

3.3 Direct Sums, Exact Sequences, and Horn

119

The direct sum has an important homomorphism property, which, in fact, can be used to characterize direct sums. To describe this, suppose that fi : Mi -+ N are R-module homomorphisms. Then there is a map

f:Ml®...ED Mn-+N defined by

n

f(xi, ... ,x,) _

fi(xi)i=1

We leave it as an exercise to check that f is an R-module homomorphism. Now consider the question of when a module M is isomorphic to the direct sum of finitely many submodules. This result should be compared with Proposition 1.6.3 concerning internal direct products of groups.

(3.1) Theorem. Let M be an R-module and let Ml, ... , M,, be submodules of M such that

(1) M=M1+

+Mn, and

(2) for 1 < i < n,

Then

M1®... ®Mn. Proof. Let fi : Mi -+ M be the inclusion map, that is, fi(x) = x for all x E Mi and define

f:Ml®...(D MMn-+M

by f(xl,...,xn) = xl + ... + xn.

f is an R-module homomorphism and it follows from condition (1) that f is surjective. Now suppose that (x1, ... , xn) E Ker(f ). Then x1 +- +xn = 0 so that for 1 < i < n we have

xi = -(XI + ... + xi_1 + xi+l + ... + xn)Therefore,

xi E

so that (x1, ... ,xn) = 0 and f is an isomorphism.

=0 O

Our primary emphasis will be on the finite direct sums of modules just

constructed, but for the purpose of allowing for potentially infinite rank free modules, it is convenient to have available the concept of an arbitrary direct sum of R-modules. This is described as follows. Let {Mj)jEj be


120

a family of R-modules indexed by the (possibly infinite) set J. Then the cartesian product set fl J Mj is the set of all the indexed sets of elements (xj)jEJ where xj is chosen from Mi. This set is made into an R-module by the coordinate-wise addition and scalar multiplication of elements. More precisely, we define (xj)jEJ'+' (yj)JEJ = (x, +Yj)jE

a(xj)jEJ = (axj)jEJ For each k E J there is an R-rnodule homomorphism ak : IIjEJ 'V'i - Mk defined by 7rk((xj)jEJ) = xj, that is, irk picks out the element of the indexed set (xj)JEJ that is indexed by k. We define the direct sum of the indexed family { Mj }jEJ of R-modules to be the following submodule ®jEJ Mj of jEJ Mj:

®Mj = {(xj)jEJ : xj = 0 except for finitely many indices j E J}. jEJ

It is easy to check that ®jEJMj is a submodule of IIjEJ Mj. To get a feeling for the difference between direct sums and direct products when the index set is infinite, note that the polynomial ring R[XJ, as an R-module (ignoring the multiplicative structure), is just a countable direct sum of copies of R, in fact, the nth copy of R is indexed by the monomial X". However, the formal power series ring R[[X]], as an R-module, is just a countable direct product of copies of R. Again, the nth copy of R is indexed by the monomial X". Each element of the polynomial ring has only finitely many monomials with nonzero coefficients, while an element of the formal power series ring may have all coefficients nonzero.

The homomorphism property of the finite direct sum of R-modules extends in a natural way to arbitrary direct sums. That is, suppose that N is an arbitrary R-module and that for each j E J there is an R-module homomorphism fj : Mj N. Then there is a map f : ®jEJMj - N defined by f ((xj) jE J) = >jE J f j (x j ). Note that this sum can be considered as a well-defined finite sum since xj = 0 except for finitely many indices j E J. (Note that this construction does not work for infinite direct products.) We leave it as an exercise to check that f is an R-module homomorphism.

The characterization of when an R-module M is isomorphic to the direct sum of submodules is essentially the same as the characterization provided in Theorem 3.1. We state the result, but the verification is left as an exercise.

(3.2) Theorem. Let M be an R-module and let {Mj}jEJ be a family of submodules such that -,E.,Mj = (UjEJ Mj), and (1) M = >F,EJ Mj = {0} for every k E J. (2) Mk n

3.3 Direct Sums, Exact Sequences, and Hom

Then

121

MMi. jEJ

0

Proof. Exercise.

(3.3) Definition. If M is an R-module and MI C M is a submodule, we say that MI is a direct summand of M, or is complemented in M, if there is a submodule M 2 C M such that M ? 5 MI ®M2.

(3.4) Example. Let R = Z and M = Z e. If MI = (p) then MI is not complemented since MI is the only subgroup of M of order p, so condition (2) of Theorem 3.1 is impossible to satisfy.

The concept of exact sequences of R-modules and R-module homomorphisms and their relation to direct summands is a useful tool to have available in the study of modules. We start by defining exact sequences of R-modules.

(3.5) Definition. Let R be a ring. A sequence of R-modules and R-module homomorphisms

is said to be exact at Mi if Im(fi) = Ker(fi+I ). The sequence is said to be exact if it is exact at each Mi. As particular cases of this definition note that

(1) 0 -+ MI --f-+ M is exact if and only if f is injective,

(2) M -+ M2 - 0 is exact if and only if g is surjective, and (3) the sequence 0 - . MI

(3.1)

_ M 9-+ M2'0

is exact if and only if f is injective, g is surjective, and Im(f) = Ker(g). Note that the first isomorphism theorem (Theorem 2.4) then shows

that M2 ?t M/ Im(f). M/ Im(f) is called the cokernel of f and it is denoted Coker(f). (3.6) Definition. (1) The sequence (3.1), if exact, is said to be a short exact sequence. (2) The sequence (3.1) is said to be a split exact sequence (or just split)

if it is exact and if Im(f) = Ker(g) is a direct summand of M.


122

In the language of exact sequences, Proposition 2.12 can be stated as follows:

(3.7) Proposition. Let 0 -. M1 -+ M -i M2 - 0 be a short exact sequence of R-modules. If Ml and M2 are finitely generated, then so is M, and moreover, µ(M) <_ µ(M1) + µ(M2)

0

Proof

(3.8) Example. Let p and q be distinct primes. Then we have short exact sequences

0 - ZP

(3.2)

Zpq

Zq

0

and

0,Z,- Zp2-ZP- 0

(3.3)

where 0(m) = qm E Zpg, f (m) = pm E Zp2, and 0 and g are the canonical projection maps. Exact sequence (3.2) is split exact while exact sequence (3.3) is not split exact. Both of these observations are easy consequences of the material on cyclic groups from Chapter 1; details are left as an exercise. There is the following useful criterion for a short exact sequence to be split exact.

(3.9) Theorem. If

O- M1 -I.M 9M2--_0

(3.4)

is a short exact sequence of R-modules, then the following are equivalent: (1) (2) (3)

Them exists a homomorphism a : M M, such that a o f = 1 s,, . There exists a homomorphism 3: M2 -. M such that g o R = 1nt, The sequence (3.4) is split exact. If these equivalent conditions hold then M = Im(f) Eli Ker(a) Ker(g) EH Im(p)

M1 e M2.

The homomorphisms a and ,3 are said to split the exact sequence (3.4) or be a splitting.

Proof. Suppose that (1) is satisfied and let x E M. Then

a(x - f(a(x))) = a(x) - (a o f)((k(x)) = 0 since a o f

Therefore, x - f(a(x)) E Ker(a) so that Al - Ker(a) + Im(f ).


123

Now suppose that f (y) = x E Ker(a) fl Im(f). Then

0 = a(x) = a(f(y)) = y, and therefore, x = f (y) = 0. Theorem 3.1 then shows that

M = Im(f) ® Ker(a). Define,Q : M2 -- M by /3(u) = v - f(a(v))

(3.5)

where g(v) = u. Since g is surjective, there is such a v E M, but it may be possible to write u = g(v) for more than one choice of v. Therefore, we

must that /3 is well defined. Suppose that g(v) = u = g(v'). Then V - v' E Ker(g) = Im(f) so that

(v - f(a(v))) - (v' - f(a(v'))) = (v - v') + (f(a(v') - f(a(v))) E lm(f) n Ker(a) = {0}. We conclude that /3 is well defined. Since it is clear from the construction

of /3 that g o /3 = ln! we have verified that (1) implies (2) and that M Im(f) ® Ker(a), i.e., that (3) holds. The proof that (2) implies (1) and (3) is similar and is left as an exercise.

Suppose that (3) holds, that is, M =' M' (D M" where M' = Ker(g) _ Im(f). Let 7r1 : M - M' and 7r2 : M M" be the projections, and c : M" M be the inclusion. Note that 7r1 o f : M1 M' and g o t : M" -y M2 are isomorphisms. Define a : M All by a = (7r1 o f)-1 o 7r1

and/3:M2-+Mby/3 to(got)-l. Then aof=1M, and go/=1412, 0

so (1) and (2) hold.

If M and N are R-modules, then the set HomR(M, N) of all R-module homomorphisms f : M -' N is an abelian group under function addition. According to Example 1.5 (8), HomR(M, N) is also an R-module provided that R is a commutative ring. Recall that EndR(M) = HomR(M) denotes the endomorphism ring of the R-module M, and the ring multiplication is composition of homomorphisms. Example 1.5 (8) shows that EndR(M) is an R-algebra if the ring R is commutative. Example 1.5 (10) shows that HomR(R, M) ^_' M for any R-module M. Now consider R-modules M, M1, N, and N1, and let : N , N1, i(i : M -. M1 be R-module homomorphisms. Then there are functions

0.: HomR(M, N)

HomR(M, N1)

and

0*: HomR(Ml, N) -' HomR(M, N)


124

defined by

0.(f) = 0 of

for all

f E HomR(M, N)

and

tP* (g) = go 10

for all g E HomR(Ml, N).

It is straightforward to check that 0.(f +g) = 0. (f) +0. (g) and ?P* (f +g) _ Ii' (f) + t/5' (g) for appropriate f and g. That is, 0. and +f are homomorphisms of abelian groups, and if R is commutative, then they are also R-module homomorphisms. Given a sequence of R-modules and R-module homomorphisms lVij-1 .m`.

(3.6)

A-+: M,+i

...

and an R-module N, then HomR( , N) and HomR(N, ) produce two sequences of abelian groups (R-modules if R is commutative): HomR(N, Mi-1)

(3.7)

(m.).

HomR(N, M:)

HomR(N, M1+1) - .. . and

Hom M

(3.8)

(00,

HomR(Mi, N)

i01+i.

HomR(Ms+1, N)

A natural question is to what extent does exactness of sequence (3.6) imply exactness of sequences (3.7) and (3.8). One result along these lines is the following. (3.10) Theorem. Let (3.9)

be a sequence of R-modules and R-module homomorphisms. Then the sequence (3.9) is exact if and only if the sequence (3.10)

0 -. HomR(N, M1) m'+ HomR(N, M) 1P' HomR(N, M2)

is an exact sequence of Z-modules for all R-modules N.

if (3.11)

M1

0

is a sequence of R-modules and R-module homomorphisms, then the sequence (3.11) is exact if and only if the sequence (3.12)

0

HomR(M2, N) --+ HomR(M, N) m . HomR(Ml, N)

is an exact sequence of Z-modules for all R-modules N.


125

Proof. Assume that sequence (3.9) is exact and let N be an arbitrary Rmodule. Suppose that f E HomR(N, M) and -0.(f) = 0. Then 0 = 0 o f (x) = g5(f (x))

for all x E N. But 0 is injective, so f (x) = 0 for all x E N. That is, f = 0, and hence, 0. is injective. Since o 0 = 0 (because sequence (3.9) is exact at M), it follows that

IMO-W) =+G-0.(f) =0000f =0 for all f E HomR(N, M). Thus Im(q.) C Ker(t/'.). It remains to check the other inclusion. Suppose that g E HomR(N, M) with t1.(g) = 0, i.e., 0(g(x)) = 0 for all x E N. Since Ker(tP) = Im(¢), for each x E N, we may write g(x) = 4(y) with y E Ml. Since 0 is injective, y is uniquely determined by the equation g(x) = 0(y). Thus it is possible to define a function f : N - Ml by f (x) = y whenever g(x) = ¢(y). We leave it as an exercise to check that f is an R-module homomorphism. Since 0.(f) = g, we conclude that Ker(t/'.) = Im(O.) so that sequence (3.10) is exact. Exactness of sequence (3.12) is a similar argument, which is left as an exercise.

Conversely, assume that sequence (3.10) is exact for all R-modules N. Then 0. is injective for all R-modules N. Then letting N = Ker(O) and t : N - Ml be the inclusion, we see that 0.(t) = 0 o t = 0. Since 0.: HomR(N, Ml) --- HomR(N, M) is injective, we see that t = 0, i.e., N = (0). Thus, 0 is injective. Now letting N = MI we see that

0= (0.00.)(1m.) =V,00. Thus Im(O) C Ker(O). Now let N = Ker(t(i) and let t : N -+ M be the inclusion. Since tp. (t) = 7P o t = 0, exactness of Equation (3.10) implies that

t = ¢.(a) for some a E HomR(N, MI). Thus, Im(O)

Im(t) = N = Ker(tP),

and we conclude that sequence (3.9) is exact. Again, exactness of sequence (3.11) is left as an exercise.

Note that, even if

0--.MI-.M 1P+M2-i0 is a short exact sequence, the sequences (3.10) and (3.12) need not be short exact, i.e., neither tG. or 0' need be surjective. Following are some examples to illustrate this.


126

(3.11) Example. Consider the following short exact sequence of Z-modules: (3.13)

0- Z--0+ Z_4Z,n--+ 0

where 4(i) = mi and tai is the canonical projection map. If N = Z,, then sequence (3.12) becomes 0 ---+ Homz(Z,n, Z,,) ----+ Homz(Z, Zn) 0' Homz(Z, Zn),

which, by Example 1.5 (10), becomes 0 --+ Homz(Zm, Zn) ----* Z" ----+ Z"

so that Homz(Z,n, Zn) = Ker(O*).

Let d = gcd(m, n), and write m = m'd, n = n'd. Let f E Homz(Z, Zn). Then, clearly, 0* (f) = 0 if and only if 0* (f) (1) = 0. But

0*(f)(1) = f(m 1) = mf(1) = m'df(1). Since m' is relatively prime to n, we have m'df (1) = 0 if and only if df(1) _ 0, and this is true if and only if f (1) E n'Zn. Hence, Ker(O') = n'Zn Zd, i.e.,

(3.14)

Homz(Zm, Zn) = Zd.

This example also shows that even if

0,Af,-+M- M2- 0 is exact, the sequences (3.10) and (3.12) are not, in general, part of short exact sequences. For simplicity, take m = n. Then sequence (3.12) becomes (3.15)

0-4 Zn-+ Zn -4Zn

with 0' = 0 so that 0' is not surjective, while sequence (3.10) becomes !3.16)

0 -' Homz(Z,,, Z) -+ Homz(Z,,, Z)

Since Homz(Zn, Z) = 0 and Homz(Zn, Zn)

V-4

Homz(Zn, Zn).

Zn, sequence (3.16) becomes

0-+0--a0 ±- +Zn and ii. is certainly not surjective.

These examples show that Theorem 3.10 is the best statement that can be made in complete generality concerning preservation of exactness under application of HomR. There is, however, the following criterion for the preservation of short exact sequences under Hom:

3.3 Direct Sums, Exact Sequences, and Horn

127

(3.12) Theorem. Let N be an arbitrary R-module. If (3.17)

0-4 M1-0- M 0-+ M2

is a split short exact sequence of R-modules, then

(3.18) 0 -. HomR(N, M1) 0.i HomR(N, M) - HomR(N, M2)

0

and

(3.19) 0

HomR(M2, N)

HomR(M, N)

HorR(M1, N)

0

are split short exact sequences of abelian groups (R-modules if R is commutative). Proof. We will prove the split exactness of sequence (3.18); (3.19) is similar

and it is left as an exercise. Given Theorem 3.10, it is only necessary to show that t&, is surjective and that there is a splitting for sequence (3.18). Let /3 : M2 -+ M split the exact sequence (3.17) and let f E HomR(N, M2). Then

00f) = (WOQ)0f 0.aa.(f)=IM,,,

= (1M2) ° f = (1HomR(N,Ma)) (f)

Thus, r/). o)3. = 1HomR(N,M2) so that r/i. is surjective and /i. is a splitting of exact sequence (3.18).

(3.13) Corollary. Let M1, M2, and N be R-modules. Then (3.20)

HomR(N, M1 ® M2) 25 HomR(N, M1) ® HomR(N, M2)

and (3.21)

HomR(Ml (B M2, N) 25 HomR(M1, N) ED HorR(M2, N).

The isomorphisms are Z-module isomorphisms (R-module isomorphisms if R is commutative). Proof. Both isomorphisms follow by applying Theorems 3.12 and 3.9 to the split exact sequence

0-+ M1 `+

0

where c(m) = (m, 0) is the canonical injection and rr(ml, m2) = m2 is the canonical projection.

128


(3.14) Remarks. (1)

Notice that isomorphism (3.20) is given explicitly by

't(f)=(r1of,7rzof) where f E HomR(N, M1 ® M2) and Jri(ml, m2) = mi (for i = 1, 2); while isomorphism (3.21) is given explicitly by

4'(f)=(fo4,foL2) where f E HomR(Ml ® M2, N), Li : M1 - M1 ® M2 is given by

G1(m)=(m,0)andi2:M2- M1®M2is given by t2(m)=(0,m). (2) Corollary 3.13 actually has a natural extension to arbitrary (not nec-

essarily finite) direct sums. We conclude this section by stating this extension. The proof is left as an exercise for the reader.

(3.15) Proposition. Let {Mi},EI and {Nj}jEJ be indexed families (not necessarily finite) of R-modules, and let M = ®,EIMi, N = ®jEJNj. Then

HomR(M, N) = [I(®HomR(Mi, Nj)). iEI jEJ Proof. Exercise.

0

3.4 Free Modules (4.1) Definition. Let R be a ring and let M be an R-module. A subset S C M is said to be R-linearly dependent if there exist distinct x1, ... , xn in S and elements a1, ... , an of R, not all of which are 0, such that

A set that is not R-linearly dependent is said to be R-linearly independent. When the ring R is implicit from the context, we will sometimes write

linearly dependent (or just dependent) and linearly independent (or just independent) in place of the more cumbersome R-linearly dependent or R-linearly independent. In case S contains only finitely many elements x1, x2i ... , xn, we will sometimes say that x1, x2, ... , x, are R-linearly dependent or R-linearly independent instead of saying that S = {x1, ... , xn} is R-linearly dependent or R-linearly independent.

3.4 Flee Modules

129

(4.2) Remarks.

(1) To say that S C M is R-linearly independent means that whenever there is an equation

=0 where x1 i ... , xn are distinct elements of S and a1, ... , an are in R,

then a1 =

=an =0.

(2) Any set S that contains a linearly dependent set is linearly dependent. (3) Any subset of a linearly independent set S is linearly independent. (4) Any set that contains 0 is linearly dependent since 1 -0 = 0. (5) A set S C M is linearly independent if and only if every finite subset of S is linearly independent.

(4.3) Definition. Let M be an R-module. A subset S of M is a basis of M if S generates M as an R-module and if S is R-linearly independent. That is, S C M is a basis if and only if M = {0}, in which case S = 0 is a basis, or M 0 {0} and (1) every x E M can be written as

x = a1x1 + ... + anxn for some x1i ... , xn E S and a1, ... , an E R, and (2) whenever there is an equation a1x1 + ... + anxn = 0

where x1i ... , xn are distinct elements of S and a1, ... , an are in R, then a1 = ... = an = 0.

It is clear that conditions (1) and (2) in the definition of basis can be replaced by the single condition: (1') S C M is a basis of M 36 {0} if and only if every x E M can be written uniquely as

x=alxl+...+anxn foral,...,anERandx1,...,xnES.

(4.4) Definition. An R-module M is a free R-module if it has a basis.

(4.5) Remark. According to Theorem 3.2, to say that S = {xj }jE J is a basis of M is equivalent to M being the direct sum of the family {Rx,)jEJ

130


of submodules of M, where Ann(xj) = {0} for all j E J. Moreover, if J is any index set, then N = (DjE JRj, where Rj = R for all j E J, is a free Rmodule with basis S = {ej}jEJ, where ej E N is defined by ej = (bjk)kEJ. Here, bjk is the kronecker delta function, i.e., bjk = 1 E R whenever j = k and bjk = 0 E R otherwise. N is said to be fine on the index set J. (4.6) Examples. (1) If R is a field then R-linear independence and R-linear dependence in

a vector space V over R are the same concepts used in linear algebra. (2) R' is a free module with basis S = {el, ... , en} where e; = (0,...,0,1,0,...,0)

with a 1 in the ith position. (3) Mm,n(R) is a free R-module with basis

S={E,j:1
subset of G is Z-linearly independent. Indeed, if g E G then IGI g = 0 but Cl I34 0. Therefore, finite abelian groups can never be free Zmodules, except in the trivial case G = {0} when 0 is a basis. (6)

If R is a commutative ring and I C R is an ideal, then I is an R-

module. However, if I is not a principal ideal, then I is not free as an R-module. Indeed, no generating set of I can be linearly independent since the equation (-a2)al +ala2 = 0 is valid for any al, a2 E R. (7) If M1 and M2 are free R-modules with bases Sl and S2 respectively, then Ml ® M2 is a free R-module with basis S; U S2, where

Sj ={(x,0):iES1}

and

S2={(0,y):yeS2}.

(8) More generally, if {Mj}jEJ is a family of free R-modules and Sj C ltlj is a basis of Mj for each j E J, then M = ®jEJMj is a free R-module and S = UjEJS is a basis of M, where S' C M is defined by Sj = {Sjo = (bjksjt)kEJ Sj, E Sj}. Informally, Si consists of all elements of M that contain an element of Sj in the jth component and 0 in all other components. This example incorporates both Example 4.6 (7) and Example 4.6 (2). Example 4.6 (5) can be generalized to the following fact.

3.4 Free Modules

131

(4.7) Lemma. Let M be an R-module where R is a commutative ring. Then an element x E M is R-independent if and only if Ann(x) = {0}. In particular, an element a E R is an R-independent subset of the R-module R if and only if a is not a zero divisor. Proof. Exercise.

(4.8) Proposition. Let R be an integral domain and let M be a free Rmodule. Then M is torsion-free.

Proof. Let M have a basis S = {x) }jEJ and let x E M,. Then ax = 0 for some a 96 0 E R. Write x = /jEJ ajxj. Then 0 = ax = 1:(aaj)xj. jEJ

Since S is a basis of M, it follows that aaj = 0 for all j E J, and since a 34 0 and R is an integral domain, we conclude that aj = 0 for all j E J. Therefore, x = 0, and hence, M, = (0) so that M is torsion-free.

The existence of a basis for an R-module M greatly facilitates the construction of R-module homomorphisms from M to another R-module N. In fact, there is the following important observation. (4.9) Proposition. Let M be a free R-module with basis S, let N be any

R-module, and let h : S - N be any function. Then there is a unique f E HomR(M, N) such that f Is = h. Proof. Let S = {xj}jEJ. Then any x E M can be written uniquely as

x=

jEJ ajxj where at most finitely many aj are not 0. Define f : M - N

by

f (x) = > ajh(xj) jEJ

It is straightforward to check that f E HomR(M, N) and that f is = h. Remark. The content of Proposition 4.9 is usually expressed as saying that the value of a homomorphism can be arbitrarily assigned on a basis. (4.10) Corollary. Suppose that M is a free R-module with basis S = {xj}jEJ. Then HomR(M, N) II Nj jEJ

where N. = N for all j E J. Proof. Define 4i : HomR(M, N) rjjEJ Nj by 4i(f) = (f(xj)),EJ. Then 4? is an isomorphism of abelian groups (R-modules if R is commutative).


132

(4.11) Theorem. Let R be a commutative ring and let M and N be finitely generated free R-modules. Then HomR(M, N) is a finitely generated free R-module.

Proof. Let B = {v1, ... , v,, } be a basis of M and C = {w1,

of N. Define f1 E HomR(M, N) for is 15 m and 1< j

... , wn} a basis n by

_ fw, if k = i, f',(vk)

-

if k 76 i.

0

fij is a uniquely defined element of HomR(M, N) by Proposition 4.9. We claim that { fij : 1 < i < m; 1 < j < n} is a basis of HomR(M, N). To see this suppose that f E HomR(M, N) and for 1 < i < m write

f(vi) = ai1w1 + ... + ainwn. Let

rm

g = L,

n

E aij fij.

i=1 j=1

Then

g(vk) = ak1wl + ... +aknwn = f(vk) for 1 < k < m, so g = f since the two homomorphisms agree on a basis of M. Thus, {fij 1 < i < m; I < j < n} generates HomR(M, N), and :

we leave it as an exercise to check that this set is linearly independent and, hence, a basis. (4.12) Remarks. (1)

A second (essentially equivalent) way to see the same thing is to write M ®;" 1R and N L" ®j=1R. Then, Corollary 3.13 shows that m

n

HomR(M, N) = ®®HomR(R, R). i=1 j=1

But any f E HomR(R, R) can be written as f = f(1) . 1R. Thus HomR(R, R)

R so that m

n

HomR(M, N) L

R. i=1 j=1

(2) The hypothesis of finite generation of M and N is crucial for the validity of Theorem 4.11. For example, if R = Z and M = ®i°Z is the free Z-module on the index set N, then Corollary 4.10 shows that 00

HomR(M, Z) L fl Z.

3.4 Free Modules

133

But the Z-module r IT Z is not a free Z-module. (For a proof of this fact (which uses cardinality arguments), see I. Kaplansky, Infinite Abelian Groups, University of Michigan Press, (1968) p. 48.)

(4.13) Proposition. Let M be a free R-module with basis S = {xj}jEJ. If I is an ideal of R, then IM is a submodule of M and the quotient module M/IM is an R/I-module. Let it : M --+ M/IM be the projection map. Then M/IM is a free R/I-module with basis tr(S) = {Tr(xj)}jEJ.

0

Proof. Exercise.

(4.14) Proposition. Every R-module M is the quotient of a free module and if M is finitely generated, then M is the quotient of a finitely generated free R-module. In fact, we may take µ(F) = u(M).

Proof. Let S = {xj}jEJ be a generating set for the R-module M and let F = ®jEJRj where Rj = R be the free R-module on the index set J. Define the homomorphism TG : F -+ M by

,P((aj)jEJ) = Eajx3. jEJ

Since S is a generating set for M, 0 is surjective and hence M LY F/ Ker(tp). Note that if ISI < oo then F is finitely generated. (Note that every module has a generating set S since we may take S = M.) Since M is a quotient of

F, we have µ(M) < µ(F). But F is free on the index set J (Remark 4.5), so µ(F) < IJI, and since J indexes a generating set of M, it follows that µ(F) < µ(M) if S is a minimal generating set of M. Hence we may take F 0 with µ(F) = µ(M). (4.15) Definition. If M is an R-module then a short exact sequence

0---+

O

where F is a free R-module is called a free presentation of M.

Thus, Proposition 4.14 states that every module has a free presentation.

(4.16) Proposition. If F is a free R-module then every short exact sequence

0-+Ml -+M-f+F--i0 of R-modules is split exact. Proof. Let S = {xj }jEJ be a basis of the free module F. Since f is surjective,

for each j E J there is an element yj E M such that f (yj) = x j . Define M by h(xj) = yj. By Proposition 4.9, there is a unique 6 E h:S

134


HomR(F, M) such that f3ls = h. Since f o Q(xl) = xj = 1F(x,) for all j E J, it follows that f o O = IF, and the result follows from Theorem 3.9. 0 (4.17) Corollary. (1) Let M be an R-module and N C M a submodule with M/N free. Then

MAN®(M/N).

(2) If M is an R-module and F is a free R-module, then ML, Ker(f) ® F

for every surjective homomorphism f : M - F. Proof. (1) Since M/N is free, the short exact sequence

0 --y N -- ' M

M/N ---. 0

is split exact by Proposition 4.16 Therefore, M N ® (M/N) by Theorem 3.9.

(2) Take N = Ker(f) in part (1). (4.18) Corollary. Let N be an arbitrary R-module and F a free R-module. If (4.1)

is a short exact sequence of R-modules, then

0 -. HomR(N, Ml) - HomR(N, M) -V1. HomR(N, F) -- 0 is a (split) short exact sequence of abelian groups (R-modules if R is commutative). Proof. By Proposition 4.16, the sequence (4.1) is split exact, so the corollary follows immediately from Theorem 3.12.

(4.19) Remark. It is a theorem that any two bases of a free module over a commutative ring R have the same cardinality. This result is proved for finite-dimensional vector spaces by showing that any set of vectors of cardinality larger than that of a basis must be linearly dependent. The same procedure works for free modules over any commutative ring R, but it does require the theory of solvability of homogeneous linear equations over a commutative ring. However, the result can be proved for R a PID without the theory of solvability of homogeneous linear equations over R; we prove this result in Section 3.6. The result for general commutative rings then follows by an application of Proposition 4.13.

The question of existence of a basis of a module, that is, to ask if a given module is free, is a delicate question for a general commutative ring R. We have seen examples of Z-modules, namely, finite abelian groups, which

3.4 Free Modules

135

are not free. We will conclude this section with the fact that all modules over division rings, in particular, vector spaces, are free modules. In Section 3.6 we will study in detail the theory of free modules over a PID.

(4.20) Theorem. Let D be a division ring and let V be a D-module. Then V is a free D-module. In particular, every vector space V has a basis. Proof. The proof is an application of Zorn's lemma.

Let S be a generating set for V and let Bo C S be any linearly independent subset of S (we allow Bo = 0). Let T be the set of all linearly independent subsets of S containing Bo and partially order T by inclusion. If {Bi} is a chain in T, then UBi is a linearly independent subset of S that contains Bo; thus, every chain in T has an upper bound. By Zorn's lemma, there is a maximal element in T, so let B be a maximal linearly independent subset of S containing Bo. We claim that S C (B) so that V = (S) C (B). Let v E S. Then the maximality of B implies that V U {v} is linearly dependent so that there is an equation m

Eaivi+bv=0 i=i

where v1, ... , v,,, are distinct elements of B and al, ... , am, b E D are not all 0. If b = 0 it would follow that Ei"_1 aivi = 0 with not all the scalars ai = 0. But this contradicts the linear independence of B. Therefore, b 0 0 and we conclude m

v = b-'(bv) =

E(-b-lai)vi E (B).

i=i Therefore, S C (B), and as observed above, this implies that B is a basis of V.

The proof of Theorem 4.20 actually proved more than the existence of a basis of V. Specifically, the following more precise result was proved.

(4.21) Theorem. Let D be a division ring and let V be a D-module. If S spans V and Bo C S is a linearly independent subset, then there is a basis

BofVsuch that BOCBCS. Proof.

(4.22) Corollary. Let D be a division ring, and let V be a D-module. (1) Any linearly independent subset of V can be extended to a basis of V. (2) A maximal linearly independent subset of V is a basis. (3) A minimal generating set of V is a basis. Proof. Exercise.


136

Notice that the above proof used the existence of inverses in the division ring D in a crucial way. We will return in Section 3.6 to study criteria that ensure that a module is free if the ring R is assumed to be a PID. Even when R is a PID, e.g., R = Z, we have seen examples of R modules that are not free, so we will still be required to put restrictions on the module M to ensure that it is free.

3.5 Projective Modules The property of free modules given in Proposition 4.16 is a very useful one,

and it is worth investigating the class of those modules that satisfy this condition. Such modules are characterized in the following theorem. (5.1) Theorem. The following conditions on an R-module P are equivalent. (1) Every short exact sequence of R-modules

0- M, -+M-'P-.0 splits. (2) There is an R-module P' such that P ® P is a free R-module.

(3) For any R-module N and any surjective R-module homomorphism ?p :

M - P, the homomorphism tb.: HomR(N, M) -' HomR(N, P) is surjective.

(4) For any surjective R-module homomorphism ¢ : M

N, the homo-

morphism

0.: HomR(P, M)

HomR(P, N)

is surjective.

Proof. (1) = (2). Let 0 -i K ---* F ---I P - 0 be a free presentation of P. Then this short exact sequence splits so that F

P ® K by Theorem

3.9.

(2)

.

(3). Suppose that F = P ® P' is free. Given a surjective R-

module homomorphism t : M P, let ?i' = V) ®1 P' : M ® P P ® P' _ F; this is also a surjective homomorphism, so there is an exact sequence

0--+Ker(V"') -i M®P' - F-+0. Since F is free, Proposition 4.16 implies that this sequence is split exact; Theorem 3.12 then shows that ?i; : HomR(N, M ® P)

HomR(N, P 9 P')

3.5 Projective Modules

137

is a surjective homomorphism. Now let f E HomR(N, P) be arbitrary and

let f = i o f , where c : P -+ P ® P is the inclusion map. Then there is an f E HomR(N, M ® P') with 10' (f) = f'. Let 7r : M ® P M and 7r' : P ® P' P be the projection maps. Note that 7r' o c = lp and 0 o 7r = r' o V. Then (7ro f) =V1o(?ro f)

=7r'o,'o f =7r'0 f' = (7rI oi.) o f

= f. Therefore, j'. is surjective. (3)

(4). Let 0 --- K -. F "' P -+ 0 be a free presentation of

P. By property (3), there is a g E HomR(P, F) such that t',(Q) = 1p, i.e., loof3 = lp. Let 46: M -+ N be any surjective R-module homomorphism and

let f E HomR(P, N). Then there is a commutative diagram of R-module homomorphisms

F - P -+ 0

If

M --' N -+ 0 with exact rows. Let S = {x,}jEJ be a basis of F. Since 0 is surjective,

we may choose yj E M such that 0(yj) = f o v/i(xj) for all j E J. By Proposition 4.9, there is an R-module homomorphism g : F -+ M such that g(xi) = y, for all j E J. Since 0 o g(x,) = 0(yy) = f o v/1(x, ), it follows that D o g f o 0. Define f E HomR(P, M) by f= g -,3 and observe that

0.(f)=0o(go0) =fo'YOQ

=folF = f. Hence, O.: HomR(P, M) - HomR(P, N) is surjective. (4) => (1). A short exact sequence

0 - MI

M -'p s P -+ 0,

in particular, includes a surjection V : M -i P. Now take N = P in part (4). Thus,

ii.: HomR(P, M) -+ HomR(P, P) is surjective. Choose /3 : P -+ M with t/i.(,0) = lp. Then /3 splits the short exact sequence and the result is proved. 0


138

(5.2) Definition. An R-module P satisfying any of the equivalent conditions of Theorem 5.1 is called projective. As noted before Theorem 5.1, projective modules are introduced as the class of modules possessing the property that free modules were shown to possess in Proposition 4.16. Therefore, we have the following fact:

(5.3) Proposition. Free R-modules are projective. Proof.

(5.4) Corollary. Let R be an integral domain. If P is a projective R-module, then P is torsion-free.

Proof. By Theorem 5.1 (2), P is a submodule of a free module F over R. According to Proposition 4.8, every free module over an integral domain is torsion-free, and every submodule of a torsion-free module is torsion-free. (5.5) Corollary. An R-module P is a finitely generated projective R-module if and only if P is a direct summand of a finitely generated free R-module.

Proof. Suppose that P is finitely generated and projective. By Proposition 4.14, there is a free presentation

0- K - F , P - 0 such that F is free and p(F) = µ(P) < oo. By Theorem 5.1, P is a direct summand of F. Conversely, assume that P is a direct summand of a finitely generated free R-module F. Then P is projective, and moreover, if P ® P - F then

F/P - P so that P is finitely generated. (5.6) Examples. (1) (2)

(3)

Every free module is projective.

Suppose that m and n are relatively prime natural numbers. Then as abelian groups Z,,,,, ?5 Z,,, ® Z,,. It is easy to check that this isomorphism is also an isomorphism of Z,,,,,-modules. Therefore, Z,,, is a direct summand of a free Z,,,,,-module, and hence it is a projective Z,,,,,-module. However, Z,,, is not a free Z,,,,, module since it has fewer than mn elements. Example 5.6 (2) shows that projective modules need not be free. We will present another example of this phenomenon in which the ring R is an integral domain so that simple cardinality arguments do not suffice.

Let R = Z(v5] and let I be the ideal I = (2, 1 +

= (al, a2). It

is easily shown that I is not a principal ideal, and hence by Example 4.6 (6), we see that I cannot be free as an R-module. We claim that I


139

E R, let F be a is a projective R-module. To see this, let b = 1 free R-module with basis Is,, 82}, and let 0: F - I be the R-module homomorphism defined by c(rjs1 + r2s2) = r1al + r2a2.

Now define an R-module homomorphism a : 1 -+ F by

a(a) = -as, + ((ab)/2)s2. Note that this makes sense because 2 divides ab for every a E I. Now for a E I, 0 o a(a) = b(-asl + ((ab)/2)s2)

= -aal + ((ab)/2)a2

= -aal + aa2b/2 = -2a + 3a =a so that a is a splitting of the surjective map 0. Hence, F and by Theorem 5.1, I is a projective R-module.

Ker(O) ® I

Concerning the construction of new projective modules from old ones, there are the following two simple facts:

(5.7) Proposition. Let { Pj }jEJ be a family of R-modules, and let P = ®jEJ Pi. Then P is projective if and only if Pj is projective for each j E J.

Proof. Suppose that P is projective. Then by Theorem 5.1, there is an R-module P' such that P ® P = F is a free R-module. Then

F=P®P'=(®Pj)ED P', jEJ

and hence, each Pj is also a direct summand of the free R-module F. Thus, Pj is projective. Conversely, suppose that Pj is projective for every j E J and let P,' be an R-module such that Pj ® Pj' = F, is free. Then

P®((DPj')(P,®P;) jEJ

jEJ

® Fj. jEJ

Since the direct sum of free modules is free (Example 4.6 (8)), it follows that P is a direct summand of a free module, and hence P is projective.

140


(5.8) Proposition. Let R be a commutative ring and let P and Q be finitely generated projective R-modules. Then HomR(P, Q) is a finitely generated projective R-module.

Proof. Since P and Q are finitely generated projective R-modules, there are R-modules P and Q' such that P®P' and Q®Q' are finitely generated free modules. Therefore, by Theorem 4.11, HomR(P ® P, Q (D Q') is a finitely generated free R-module. But HomR(P (D P', Q ® Q') ?' HomR(P, Q) ® HomR(P, Q') ® HomR(P', Q) ® HomR(P, Q')

so that HomR(P, Q) is a direct summand of a finitely generated free Rmodule, and therefore, it is projective and finitely generated by Corollary

0

5.5.

Example 5.6 (3) was an example of an ideal in a ring R that was projective as an R-module, but not free. According to Example 4.6 (6), an ideal I in a ring R is free as an R-module if and only if the ideal is principal.

It is a natural question to ask which ideals in a ring R are projective as R-modules. Since this turns out to be an important question in number theory, we will conclude our brief introduction to the theory of projective modules by answering this question for integral domains R.

(5.9) Definition. Let R be an integral domain and let K be the quotient field of R. An ideal I C R is said to be invertible if there are elements a1, ... , an E I and bl, ... , bn E K such that (5.1) (5.2)

b=I C_ R for 1 < i < n, and albs + + anbn = 1.


If I C R is the principal ideal I = (a) where a 0 0, then I is an invertible ideal. Indeed, let b = 1/a E K. Then any x E I is divisible by a in R so that bx = (1/a)x E R, while a(l/a) = 1.

(2) Let R = Z[vr---5] and let I = (2, 1 + ). Then it is easily checked that I is not principal, but I is an invertible ideal. To see this, let a1 =2,a2=1+-,,/--5,b1 =-1, and b2 = (1 - -,/--5)/2. Then albs + a2b2 = -2 + 3 = 1.

Furthermore, a1b2 and a2b2 are in R, so it follows that b2I C R, and we conclude that I is an invertible ideal. The following result characterizes which ideals in an integral domain

R are projective modules. Note that the theorem is a generalization of Example 5.6 (3):


141

(5.11) Theorem. Let R be an integral domain and let I C R be an ideal. Then I is a projective R-module if and only if I is an invertible ideal. Proof. Suppose that 1 is invertible and choose al, ... , an E I and b1, ... , bn in the quotient field K of R so that Equations (5.1) and (5.2) are satisfied.

Let 0: Rn -' I be defined by

O(xi, ... , xn) = alxl + ... + anxn, and define 6 : I -+ Rn by /3(a) = (abl, ... , abn).

Note that abi E R for all i by Equation (5.1). Equation (5.2) shows that n

O o /3(a) = E ai(abi) = a t aibi =a i_1

i=1

for every a E I. Therefore 0 o /j = 1 p and Theorem 3.9 implies that I is a direct summand of the free R-module Rn, so I is a projective R-module. Conversely, assume that the ideal 19 R is projective as an R-module. Then I is a direct summand of a free R-module F, so there are R-module

homomorphisms 0 : F -+ I and 3 : I -' F such that 0 o /3 = 11. Let S = {xj}jEJ be a basis of F. Given x E 1, /3(x) E F can be written uniquely as

cjxj.

/3(x) _

(5.3)

jEJ

For each j E J, let tlj(x) = cj. This gives a function Oj : I R, which is easily checked to be an R-module homomorphism. If aj = O(xj) E I, note that for each x E I, Oj(x) = 0 except for at most finitely many j E J; for each x E I, Equation (5.3) shows that

(5.4) (5.5)

x = -0(0(x)) = 1: 'pj(x)aj jEJ

Given x j4 0 E I and j E J, define b, E K (K is the quotient field of R) by (5.6)

bj = Oj(x)

X

The element bj E K depends on j E J but not on the element x 54 0 E I. To see this, suppose that x' 54 0 E I is another element of I. Then X'pj(x) = Oj(x'x) =''j(xx) = x1pj(x )

142


so that >' 1(x)/x = 1(ii(x')/x'. Therefore, for each j E J we get a uniquely defined bi E K. By property (5.4), at most finitely many of the bi are not 0. Label the nonzero bi by b1, ... , bn. By property (5.5), if x $ 0 E I then n

n

x= E V)i(x)a.i =>(bix)ai =x .i=1

.i=1

n

(bia)). i=1

Cancelling x $ 0 from this equation gives a1b1 + ... + anbn = 1

where a1, ... , an E I and b1 , bn E K. It remains to check that b,I C R for 1 < j < n. But if x $ 0 E I then bi = tki(x)/x so that bix = V)j(x) E R. Therefore, I is an invertible ideal and the theorem is proved. 0 (5.12) Remark. Integral domains in which every ideal is invertible are known

as Dedekind domains, and they are important in number theory. For example, the ring of integers in any algebraic number field is a Dedekind domain.

3.6 Free Modules over a PID In this section we will continue the study of free modules started in Section 3.4, with special emphasis upon theorems relating to conditions which ensure that a module over a PID R is free. As examples of the types of theorems to be considered, we will prove that all submodules of a free Rmodule are free and all finitely generated torsion-free R-modules are free, provided that the ring R is a PID. Both of these results are false without the assumption that R is a PID, as one can see very easily by considering an integral domain R that is not a PID, e.g., R = Z[X], and an ideal I C R that is not principal, e.g., (2, X) C Z[XJ. Then I is a torsion-free submodule of R that is not free (see Example 4.6 (6)). Our analysis of free modules over PIDs will also include an analysis of which elements in a free module M can be included in a basis and a criterion

for when a linearly independent subset can be included in a basis. Again, these are basic results in the theory of finite-dimensional vector spaces, but the case of free modules over a PID provides extra subtleties that must be carefully analyzed. We will conclude our treatment of free modules over PIDs with a fundamental result known as the invariant factor theorem for finite rank submodules of free modules over a PID R. This result is a far-reaching generalization of the freeness of submodules of free modules, and it is the basis

3.6 Free Modules over a PID

143

for the fundamental structure theorem for finitely generated modules over PIDs which will be developed in Section 3.7. We start with the following definition:

(6.1) Definition. Let M be a free R-module. Then the free rank of M, denoted free-rankR(M), is the minimal cardinality of a basis of M.

Since we will not be concerned with the fine points of cardinal arithmetic, we shall not distinguish among infinite cardinals so that free-rankR(M) E Z+ U {oo}.

Since a basis is a generating set of M, we have the inequality u(M) < free-rankR(M). We will see in Corollary 6.18 that for an arbitrary commutative ring R and for every free R-module, free-rankR(M) = p(M) and all bases of M have this cardinality.

(6.2) Theorem. Let R be a PID, and let M be a free R-module. If N C M is a submodule, then N is a free R-module, and free-rankR(N) < free-rankR(M). Proof. We will first present a proof for the case where free-rankR(M) < 00. This case will then be used in the proof of the general case. For those who are only interested in the case of finitely generated modules, the proof of the second case can be safely omitted.

Case 1. free-rankR(M) < 00.

We will argue by induction on k = free-rankR(M). If k = 0 then M = (0) so N = (0) is free of free-rank 0. If k = 1, then M is cyclic so M = (x) for some nonzero x E M. If N = (0) we are done. Otherwise, let

I = {a ER: ax EN). Since! is an ideal ofRandRisaPID,I=(d); since N j4 (0), d 54 0. If Y E N then y = ax = rdx E (dx) so that N = (dx) is a free cyclic R-module. Thus free-rankR(N) = 1 and the result is true for k = 1. Assume by induction that the result is true for all M with free-rank k, and let M be a module with free-rankR(M) = k+l. Let S = {x1, ... xk+l } be a basis of M and let Mk = (xl, ... ,xk). If N C Mk we are done by induction. Otherwise N n Mk is a submodule of Mk which, by induction, is free of free-rank e < k. Let {yl, ... , ye} be a basis of N n Mk. By Theorem 2.5

N/(N n Mk) 25 (N + Mk)/Mk C M/Mk = (xk+I + Mk)By the k = 1 case of the theorem, (N + Mk)/Mk is a free cyclic submodule of M/Mk with basis dxk+1 + Mk where d 54 0. Choose ye+l E N so that ye+i = dxk+i + x' for some x' E Mk. Then (N + Mk) lMk = (ye+i + Mk).

144


We claim that S' = {yl, ... , ye, yt+1 } is a basis of N. To see this, let y E N. Then y + Mk = ae+i(ye+i + Mk) so that y - at+lyt+1 E N n Mk, which implies that y - at+Iyt+1 = a1y1 + aeye. Thus S' generates N. Suppose

that aly, + +ae+lyt+1 = 0. Then at+1(dxk+1 +x')+aIy1 + +aeye = 0 so that ae+ldxk+l E Mk. But S is a basis of M so we must have ae+ld = 0; since d 34 0 this forces at+1 = 0. Thus a1y1 + + aeye = 0 which implies that a1 = = at = 0 since {y1, . . , yt} is linearly independent. Therefore S' is linearly independent and hence a basis of N, so that N is free with free-rankR(N) < e + 1 < k + I. This proves the theorem in Case 1. Case 2. free-rankn(M) = oo.

Since (0) is free with basis 0, we may assume that N # (0). Let S =

{xj}jEJ be a basis of M. For any subset K C J let MK = ({xk}kEK) and let NK = N n MK. Let T be the set of all triples (K, K', f) where K' C K C J and f : K' -+ NK is a function such that (f (k)}kEK' is a basis of NK We claim that T # 0. .

Since N $ (0) there is an x j4 0 E N, so we may write x = aixj, + + }. But MK is a free R-module with free-rankR(MK) < k < oo and NK is a nonzero submodule. By Case 1, NK is free with free-rankR(NK) = e < k. Let {y1, ... yt} be a basis of NK, and let K' = (j 1, ... , jt}, and define f : K' NK by f (j;) = y, for 1 < i < e. Then (K, K', f) E T so that T A 0, as claimed. Now define a partial order on T by setting (K, K', f) < (L, L', g) if K C L, K' C L', and 9IK' = f. If {(KQ, KQ, fQ)}QEA C T is a chain, then (UQEA KQ, UQEA KQ, F) where FIK' = fQ is an upper bound in T for the chain. Therefore, Zorn's lemma applies and there is a maximal element akxjk. Hence X E NK where K = {{ j 1 ,---,

(K, K', f) of T. Claim. K = J. Assuming the claim is true, it follows that MK = M, NK = N n MK = N, and {f(k)}kEK' is a basis of N. Thus, N is a free module (since it has a basis), and since S was an arbitrary basis of M, we conclude that N has a basis of cardinality < free-rankR(M), which is what we wished to prove. It remains to the claim. Suppose that K j4 J and choose j E

J \ K. Let L = K U {j}. If NK = NL then (K, K', f)

(L, K', f),

contradicting the maximality of (K, K', f) in T. If NK 36 NL, then NL/(NL n MK)

(NL + MK)/MK C ML/MK = (x3 + MK)-

By Case 1, (NL + 1MIK)/MK is a free cyclic submodule with basis dxj + MK

where d # 0. Choose z E NL so that z = dxj + w for some w E MK. Then (NL + MK)/MK = (z + MK). Now let L' = K' U {j} and define L' -+NL by

f,(k) = J f (k) if k E K', Z

ifk=j.


145

We need to show that { f'(k)}kEL' is a basis of NL. But if x E NL then x + MK = cz + MK for some c E R. Thus x - cz E MK n N = NK so that

x - cz =

bkf(k) kE K'

where bk E R. Therefore, { f (k)}kEL' generates NL. Now suppose EkEL' bkf'(k) = 0. Then

b,z+1: bkf(k)=0 kE K'

so that

db,x,+b,w+ E bkf(k) =0. kEK'

That is, db,x, E MK n (x,) = (0), and since S = {xt}tEJ is a basis of M, we must have db, = 0. But d j4 0, so b, = 0. This implies that EkE K, bk f (k) = 0. But { f (k) }kE K' is a basis of NK, so we must have bk = 0 for all k E K'. Thus { f'(k)}kEL' is a basis of NL. We conclude that (K, K', f) (L, L', f'), which contradicts the maximality of (K, K', f). Therefore, the claim is verified, and the proof of the theorem is complete. (6.3) Corollary. Let R be a PID and let P be a projective R-module. Then P is free. Proof. By Proposition 4.14, P has a free presentation

0--+K-+F--+P-+0. P ® K. Since P is projective, this exact sequence splits and hence F Therefore, P is isomorphic to a submodule of F, and Theorem 6.2 then shows that P is free. (6.4) Corollary. Let M be a finitely generated module over the PID R and let N C M be a submodule. Then N is finitely generated and

µ(N) < µ(M). Proof. Let

0- K- F 0+M-y0

be a free presentation of M such that free-rank(F) = p(M) < oo, and let NI = 0-1(N). By Theorem 6.2, N1 is free with µ(N1) < free-rank(NI) < free-rank(F) = µ(M). Since N = O(NI ), we have µ(N) < µ(N1), and the result is proved.

146


(6.5) Remark. The hypothesis that R be a PID in Theorem 6.2 and Corollaries 6.3 and 6.4 is crucial. For example, consider the ring R = Z[X] and

let M = R and N = (2, X). Then M is a free R-module and N is a submodule of M that is not free (Example 4.6 (6)). Moreover, R = Z[V/-5],

P = (2, 1 + /) gives an example of a projective R-module P that is not free (Example 5.6 (3)). Also note that 2 = µ(N) > µ(M) = 1 and 2 = µ(P) > I = u(R). Recall that if M is a free module over an integral domain R, then M is torsion-free (Proposition 4.8). The converse of this statement is false even

under the restriction that R be a PID. As an example, consider the Zmodule Q. It is clear that Q is a torsion-free Z-module, and it is a simple exercise to show that it is not free. There is, however, a converse if the module is assumed to be finitely generated (and the ring R is a PID).

(6.6) Theorem. If R is a PID and M is a finitely generated torsion-free R-module, then M is free and free-rankR(M) = µ(M).

Proof. The proof is by induction on µ(M). If µ(M) = 1 then M is cyclic with generator {x}. Since M is torsion-free, Ann(x) = {0}, so the set {x} is linearly independent and, hence, is a basis of M.

Now suppose that µ(M) = k > 0 and assume that the result is true for all finitely generated torsion-free R-modules M' with µ(M') < k. Let {x1, ... , xk } be a minimal generating set for M, and let

M1={xERI:axE(x1)

for some a0OER}.

Then M/M1 is generated by {x2 + M1, ... , xk + M1 } so that µ(M/Ml) _ j < k - 1. If ax E M1 for some a 54 0 E R, then from the definition of M1, b(ax) E (xl) for some b 0 0. Hence x E M1 and we conclude that M/Ml is torsion-free. By the induction hypothesis, M/M1 is free of free-rank j. Then Corollary 4.17 shows that ML, M1® (M/Ml ). We will show that M1 is free of free-rank 1. It will then follow that

k=µ(l41)

It remains to show that M1 is free of rank 1. Note that if R is a field then M1 = R xl and we are done. In the general case, M1 is a submodule of M, so it is finitely generated by t < k elements. Let {y', ... , yt} be

a generating set for M1 and suppose that a;y; = bix1 with a, # 0 for 1 < i < t. Let qo = a1 ... at. Claim. If ax = bxl with a 34 0 then a I bqo


147

To see this note that x = Ei=1 ciyi so that cjgoy;

qox = ;=1

_

ci(go/ai)aiyi i=1

_

G(qo/a;)b;xi ;=1

I

Ec;(go/a;)b; x1 i=1 Therefore.

I

X1. bgox, = agox = (c(o/ai)bi) aE i=1

Since M1 is torsion-free, it follows that t bqo = a

(ci(o/a)bi) i=1

and the claim is proved. Using this claim we can define a function : M1 --- R by t(x) _ (bqo)/a whenever ax = bx1 for a 36 0. We must show that 0 is well defined.

That is, if ax = bx1 and a'x = b'x, then (bqo)/a = (b'qo)/a'. But ax = bx1 and a'x = b'x1 implies that a'bx1 = a'ax = ab'x1 so that alb = all because M is torsion-free. Thus a'bqo = ab'qo so that (bqo)/a = (b'qo)/a' and 0 is well defined. Furthermore, it is easy to see that 0 is an R-module homomorphism so that Im(O) is an R-submodule of R, i.e., an ideal. Suppose that O(x) = 0. Then ax = bx1 with a 0 0 and O(x) = (bqo)/a = 0 E R. Since R is an integral domain, it follows that b = 0 and hence ax = 0. Since M is torsion-free we conclude that x = 0. Therefore, Ker(O) = {0} and M1 = Im(O) = Rc. Hence, M1 is free of rank 1, and the proof is complete.

(6.7) Corollary. If M is a finitely generated module over a field F, then M is free.

Proof. Every module over a field is torsion-free (Proposition 2.20). (6.8) Remark. We have already given an independent proof (based on Zorn's lemma) for Corollary 6.7, even without the finitely generated assumption (Theorem 4.20). We have included Corollary 6.7 here as an observation that

148


it follows as a special case of the general theory developed for torsion-free finitely generated modules over a PID.

(6.9) Corollary. If M is a finitely generated module over a PID R, then (M/M,). Proof. There is an exact sequence of R-modules

0-+M,--iM-'M/M,

0.

Hence, M/M, is finitely generated and by Proposition 2.18, it is torsionfree, so Theorem 6.6 shows that M/M, is free. Then Corollary 4.17 shows

that M M, ® (M/M,). The main point of Corollary 6.9 is that any finitely generated module over a PID can be written as a direct sum of its torsion submodule and a free submodule. Thus an analysis of these modules is reduced to studying the torsion submodule, once we have completed our analysis of free modules. We will now continue the analysis of free modules over a PID R by studying when an element in a free module can be included in a basis. As a corollary of this result we will be able to show that any two bases of a finitely generated free R-module (R a PID) have the same number of elements.

(6.10) Example. Let R be a PID and view R as an R-module. Then an element a E R forms a basis of R if and only if a is a unit. Thus if R is a field, then every nonzero element is a basis of the R-module R, while if R = Z then the only elements of Z that form a basis of Z are 1 and -1. As a somewhat more substantial example, consider the Z-module Z2. Then the element u = (2, 0) E Z2 cannot be extended to a basis of Z2 since if v is any element of Z2 with {u, v} linearly independent, the equation

au+f3v=(1,0) is easily seen to have no solution a, 0 E Z. Therefore, some restriction on elements of an R-module that can be included in a basis is necessary. The above examples suggest the following definition. (6.11) Definition. Let M be an R-module. A torsion-free element x # 0 E M

is said to be primitive if x = ay for some y E M and a E R implies that a is a unit of R. (6.12) Remarks. (1) If R is a field, then every nonzero x E M is primitive.

(2) The element x E R is a primitive element of the R-module R if and only if x is a unit.


149

(3) The element (2, 0) E Z2 is not primitive since (2, 0) = 2 (1, 0). (4) If R = Z and M = Q, then no element of M is primitive. (6.15) Lemma. Let R be a PID and let M be a free R-module with basis

S = {xj}jEJ. If X = E,EJajxj E M, then x is primitive if and only if gcd(jai }jEJ) = 1.

Proof. Let d = gcd({aj}jEJ). Then x = d(>2jEJ(aj/d)xj), so if d is not a unit then x is not primitive. Conversely, if d = 1 and x = ay then

E ajxj = x jEJ = ay(

=

l

a(Ebjxj)

jEJ

_ E abjxj. jEJ

Since S = {xj}jEJ is a basis, it follows that aj = abj for all j E J. That is, a is a common divisor of the set {aj}jEJ so that a I d = 1. Hence a is a unit and x is primitive. (6.14) Lemma. Let R be a PID and let M be a finitely generated R-module.

If x E M has Ann(x) = (0), then we may write x = ax' where a E R and x' is primitive. (In particular, if M is not a torsion module, then M has a primitive element.) Proof. Let xo = x. If xo is primitive we are done. Otherwise, write xo = alxl

where al E R is not a unit. Then (xo) 5 (x1). To see this, it is certainly true that (xo) C (xi). If the two submodules are equal then we may write

xl = bxo so that xo = aixl = albxo, i.e., (1 - alb) E Ann(xo) = (0). Therefore, 1 = alb and al is a unit, which contradicts the choice of al.

Now consider xl. If xl is primitive, we are done. Otherwise xl = a2x2 where a2 is not a unit, and as above we conclude that (xl) (x2). Continuing in this way we obtain a chain of submodules (6.1)

(xo) 5 (xi) 5 (x2) 5 ...

Either this chain stops at some i, which means that xi is primitive, or (6.1) is an infinite properly ascending chain of submodules of M. We claim that the latter possibility cannot occur. To see this, let N = Ui_° I (xi). Then N is a submodule of the finitely generated module M over the PID R so that N is also finitely generated by {yl,... , yk } (Corollary 6.4). Since (xo) C (xl) C , there is an i such that {yi,...,yk} C (xi). Thus N = (xi) and hence (x,) = (xi+i) = , which contradicts having an infinite properly

150


ascending chain. Therefore, xi is primitive for some i, and if we let x' = xi we conclude that x = ax' where a = ala2 . ai. 0 (6.15) Remark. Suppose that M is a free R-module, where R is a PID, and x E M. Then Ann(x) = (0), so x = ax' where x' is a primitive element of

M. If S = {xj }jEJ is a basis of M, then we may write x' = E,E J bjxj so that x = ax=Eabjxj= cjxj. jEJ

jEJ

Since gcd({bj}jEJ) = 1 (by Lemma 6.13) we see that a = gcd({cj}jEJ). The element a E R, which is uniquely determined by x up to multiplication by a unit of R, is called the content of x E M and is denoted c(x). (Compare with the concept of content of polynomials (Definition 2.6.3).) Thus, any x E M can be written

x = c(x) i

(6.2)

where x' is primitive.

(6.16) Theorem. Let R be a PID and let M be a fee R-module with

rank(M) = k = µ(M) = free-rank(M). If x E M is primitive, then M has a basis of k elements containing x.

Proof. Assume first that k < oo and proceed by induction on k. Suppose k = 1 and let M have a basis {x1 }. Then x = ax1 for some a E R. Since x is primitive, it follows that a is a unit so that (x) = (x1) = M, hence {x} is a basis of M. The case k = 2 will be needed in the general induction step, so we present it separately. Thus suppose that M has a basis {x1, x2} and let x = rx1 + sx2 where r, s E R. Since x is primitive, gcd{r, s} = 1, so we may write ru + sv = 1. Let x2 = -vxl + ux2. Then

x1 =ux-sx2 and

x2 = vx + rx2.

Hence, (x, x2) = M. It remains to show that {x, x2} is linearly independent. Suppose that ax + bx2 = 0. Then a(rx1 + sx2) + b(-vx1 + UX2) = 0. Since {x1i x2} is a basis of M, it follows that

ar-bv=0 and


151

as+bu=0. Multiplying the first equation by u, multiplying the second by v, and adding shows that a = 0, while multiplying the first by -s, multiplying the second by r, and adding shows that b = 0. Hence, {x, x2} is linearly independent and, therefore, a basis of M. Now suppose that u(M) = k > 2 and that the result is true for all free R-modules of rank < k. By Theorem 6.6 there is a basis {x1, ... , xk } of M.

Letx=Ekla;x;.Ifak=0then xEM1=(x1,...,xk_1),sobyinduction there is a basis {x, x2, ... ,xk_1} of M1. Then {x,x2, ... ,xk_1, xk} is a basis of M containing x. Now suppose that ak # 0 and let y = Ei=1 a;x;. If y = 0 then x = akxk, and since x is primitive, it follows that ak is a unit

of R and {x1, ... ,xk_1, x} is a basis of M containing x in this case. If 0 then there is a primitive y' such that y = by' for some b E R. In y particular, y' E M1 so that M1 has a basis {y', x2, ... , x'ti-11 and hence M has a basis {y', x2, ... , xk_1, xk}. But x =akxk + y = akxk + by' and gcd(ak, b) = 1 since x is primitive. By the previous case (k = 2) we conclude

that the submodule (xk, y') has a basis {x, y"}. Therefore, M has a basis {x, x2, ... , xk_1, y"} and the argument is complete when k = µ(M) < oo. If k = oo let {x, }jE J be a basis of M and let x = 1 aixj, for

some finite subset I = 01, ... , j } c J. If N = (xj...... xi,,) then x is a primitive element in the finitely generated module N, so the previous argument applies to show that there is a basis {x, x2, ... , x' } of N. Then {x, x2, ... ,xn} U {x,}iEJ\I is a basis of M containing x. 0 (6.17) Corollary. If M is a free module over a PID R, then every basis of M contains µ(M) elements. Proof. In case µ(M) < oo, the proof is by induction on µ(M). If µ(M) = 1 then M = (x). If {x1, x2} C M then x1 = a1x and and x2 = a2x so that a2x1 -a, X2 = 0, and we conclude that no subset of M with more than one element is linearly independent. Now suppose that p(M) = k > 1 and assume the result is true for all free R-modules N with µ(N) < k. Let S = {x) }JE J c M be any basis of M and choose x E S. Since x is primitive (being an element of a basis), Theorem 6.16 applies to give a basis {x, y2i ... , yk } of M with precisely µ(M) = k elements. Let N = M/(x) and let it : M - N be the projection map. It is clear that N is a free R-module with basis 7r(S) \ {ar(x)}. By Proposition 2.12 it follows that µ(N) > k -1, and since {7r(y2), ... , ir(yk)} generates N, we conclude that µ(N) = k - 1. By induction, it follows that ISI - 1 < oo and ISO - 1 = k - 1, i.e., ISI = k, and the proof is complete in case µ(M) < oo. In case µ(M) = oo, we are claiming that no basis of M can contain a finite number k E Z+ of elements. This is proved by induction on k, the proof being similar to the case µ(M) finite, which we have just done. We leave the details to the reader. 0

152


(6.18) Corollary. Let R be any commutative ring with identity and let M be a free R-module. Then every basis of M contains µ(M) elements.

Proof. Let I be any maximal ideal of R (recall that maximal ideals exist by Theorem 2.2.16). Since R is commutative, the quotient ring R/I = K is a field (Theorem 2.2.18), and hence it is a P1D. By Proposition 4.13, the quotient module M/IM is a finitely generated free K-module so that Corollary 6.17 applies to show that every basis of M/IM has µ(M/IM) elements. Let S = {xj}JEJ be an arbitrary basis of the free R-module M and let it : M - M/IM be the projection map. According to Proposition 4.13, the set 7r(S) = {7r(xJ)}3EJ is a basis of M/IM over K, and therefore,

14(m)< V1 = µ(M/IM) 5 AM. Thus, µ(M) = IJI, and the corollary is proved.

0

(6.19) Remarks. (1) If M is a free R-module over a commutative ring R, then we have

proved that free-rank(M) = µ(M) = the number of elements in any basis of M. This common number we shall refer to simply as the rank of M, denoted rankR(M) or rank(M) if the ring R is implicit. If R is a field we shall sometimes write dimR(M) (the dimension of M over

R) in place of rankR(M). Thus, a vector space M (over R) is finite dimensional if and only if dimR(M) = rankR(M) < oo. (2) Corollary 6.18 is the invariance of rank theorem for finitely generated free modules over an arbitrary commutative ring R. The invariance of rank theorem is not valid for an arbitrary (possibly noncommutative) ring R. As an example, consider the Z-module M = ®nENZ, which is the direct sum of countably many copies of Z. It is simple to check

that M M ® M. Thus, if we define R = Endz(M), then R is a noncommutative ring, and Corollary 3.13 shows that

R = Endz(M) = Homz(M, M) Homz(M, M (D M) Homz(M, M) ® Homz(M, M)

R®R. The isomorphisms are isomorphisms of Z-modules. We leave it as an exercise to check that the isomorphisms are also isomorphisms of R-

modules, so that R a, R2, and hence, the invariance of rank does not hold for the ring R. There is, however, one important class of noncommutative rings for which the invariance of rank theorem holds, namely, division rings. This will be proved in Proposition 7.1.14.


153

(6.20) Corollary. If M and N are free modules over a PID R, at least one of which is finitely generated, then M N if and only if rank(M) -= rank(N).

Proof. If M and N are isomorphic, then p(M) = µ(N) so that rank(M) = rank(N). Conversely, if rank(M) = rank(N), then Proposition 4.9 gives a homomorphism f : M - N, which takes a basis of M to a basis of N. It is easy to see that f must be an isomorphism.

(6.21) Remark. One of the standard results concerning bases of finitedimensional vector spaces is the statement that a subset S = {x1, ... , x") of a vector space V of dimension n is a basis provided that S is either a spanning set or linearly independent. Half of this result is valid in the current context of finitely generated free modules over a PID. The set (2) C Z is linearly independent, but it is not a basis of the rank 1 Z-module Z. There is, however, the following result.

(6.22) Proposition. Let M be a finitely generated free R-module of rank = k

where R is a PID. If S = {x1, ... ,xk} generates M, then S is a basis. Proof. Let T = {e1}?-1 be the standard basis of Rk. Then there is a homo-

morphism 0: Rk - M determined by 4,(e,) = xj. Since (S) = M, there is a short exact sequence

RkM ---+ O where K = Ker(¢). Since M is free, Corollary 4.16 gives Rk -- M ®K, and according to Theorem 6.2, K is also free of finite rank. Therefore,

k = rank(M) + rank(K) = k + rank(K) and we conclude that rank(K) = 0. Hence 0 is an isomorphism and S is a basis.

We will conclude this section with a substantial generalization of Theorem 6.2. This result is the crucial result needed for the structure theorem for finitely generated modules over a PID.

(6.23) Theorem. (Invariant factor theorem for submodules) Let R be a PID, let M be a free R-module, and let N C M be a submodule (which is automatically free by Theorem 6.2) of rank n < oo. Then there is a basis S of M, a subset {x1, ... , xn} C S, and nonzero elements ,, ..., sn E R such that (6.3)

{s1x1,

...

,snx,a}

is a basis of N

and (6.4)

si I si+1

for

1 < i < n - 1.

154


Proof. If N = (0), there is nothing to prove, so we may assume that N 96 (0)

and proceed by induction on n = rank(N). If n = 1, then N = (y) and {y} is a basis of N. By Lemma 6.14, we may write y = c(y)x where x E M is a primitive element and c(y) E R is the content of y. By Theorem 6.16, there is a basis S of M containing the primitive element x. If we let x1 = x and Si = c(y), then slxl = y is a basis of N, so condition (6.3) is satisfied; (6.4) is vacuous for n = 1. Therefore, the theorem is proved for n = 1. Now assume that n > 1. By Lemma 6.14, each y E N can be written as y = c(y) y' where c(y) E R is the content of y (Remark 6.15) and y' E M is primitive. Let S = {(c(y)) : y E N}. This is a nonempty collection of ideals of R. Since R is Noetherian, Proposition 2.5.10 implies that there is a maximal element of S. Let (c(y)) be such a maximal element. Thus, y E N and y = c(y) x, where x E M is primitive. Let s1 = c(y). Choose any basis T of M that contains x. This is possible by Theorem 6.16 since x E M is primitive. Let xl = x and write T = {x1}UT' = {xl}U{x'j}jEJ'. Let M1 = ({x'j}jEJ') and let N1 = M1f1N.

Claim. N = (slxl) ®N1. To see this, note that (six,) fl Nl C (x1) fl M1 = (0) because T is a basis of M. Let z E N. Then, with respect to the basis T, we may write z = alx1 +

b

jEJ'

Let d = (Si, al) = gcd{s1, al}. Then we may write d = us1 +val where u, V E R. If w = uy + vz, then Equation (6.5) shows that

w=uy+vz = (us1 + val)xl + E vbjx'j

jEJ'

= dx1 + = vbjxj. jEJ'

Writing w = c(w) w' where c(w) is the content of w and w' E M is primitive, it follows from Lemma 6.13 that c(w) I d (because c(w) is the greatest common divisor of all coefficients of w when expressed as a linear combination of any basis of M). Thus we have a chain of ideals (Si) C (d) C (c(w)),

and the maximality of (sl) in S shows that (Si) = (c(w)) = (d). In particular, (Si) = (d) so that s1 I a1, and we conclude that

z = b1(slxl) + E bjij jEJ'


155

That is, z E (s1x1) + N1. Theorem 3.1 then shows that

N = (six,) ® Ni, and the claim is proved. By Theorem 6.2, N1 is a free R-module since it is a submodule of the free R-module M. Furthermore, by the claim we see that

rank(N1) = rank(N) - 1 = n - 1. Applying the induction hypothesis to the pair N1 C M1, we conclude that there is a basis S' of M1 and a subset {x2, ... xn} of S', together with nonzero elements 82, ... , 8n of R, such that (6.6)

{82x2,

... , snxn }

is a basis of N1

and (6.7)

si I s;+1

for

2 < i < n - 1.

Let S = S' U {x,}. Then the theorem is proved once we have shown tha Si 132.

To that s1 $2, consider the element 82x2 E Ni C N and let z = 81x1 + 32x2 E N. When we write z = c(z) z' where z' E M is primitive and c(z) E R is the content of z, Remark 6.15 shows that 1

c(z) = (s1, 82). Thus, (Si) C (c(z)) and the maximality of (s1) in S shows that (c(z)) = (s1), i.e., s1 I s2i and the proof of the theorem is complete. 0 (6.24) Example. Let N C Z2 be the submodule generated by y1 = (2, 4), y2 = (2, -2), and y3 = (2, 10). Then c(y1) = c(y2) = c(y3) = 2. Furthermore, 2 divides every component of any linear combination of y1, y2, and

y3, so s1 = 2 in the notation of Theorem 6.23. Let v1 = (1, 2). Then y1 = 2v1. Extend v1 to a basis of Z2 by taking v2 = (0,1). Then (6.8)

Ni = N n ((0, 1)) = ((0, 6)).

To see this note that every z E Ni can be written as z = a1y1 + a2y2 + a3y3

where a1, a3, a3 E Z satisfy the equation 2a1 + 2a2 + 2a3 = 0.

Thus, 4a1 = -4a2 - 4a3, and considering the second coordinate of z, we see that z = (z1, z2) where z2 = 4a1 - 2a2 + 10a3 = -6a2 4. 6a3 = 6(a3 - a2).

Therefore, {v1i V2} is a basis of Z2, while {2v1, 6v2} is a basis of N. To check, note that y1 = 2v1, y2 = 2v1 - 6v2, and y3 = 2v1 + 6v2.

156


(6.25) Remark. In Section 3.7, we will prove that the elements {s1, ... , sn}

are determined just by the rank n submodule N and not by the particular choice of a basis S of M. These elements are called the invariant factors of the submodule N in the free module M.

3.7 Finitely Generated Modules over PIDs The invariant factor theorem for submodules (Theorem 6.23) gives a complete description of a submodule N of a finitely generated free R-module M over a PID R. Specifically, it states that a basis of M can be chosen so

that the first n = rank(N) elements of the basis, multiplied by elements of R, provide a basis of N. Note that this result is a substantial generalization of the result from vector space theory, which states that any basis of a subspace of a vector space can be extended to a basis of the ambient space. We will now complete the analysis of finitely generated R-modules (R a PID) by considering modules that need not be free. If the module M is not free, then, of course, it is not possible to find a basis, but we will still be able to express M as a finite direct sum of cyclic submodules; the cyclic submodules may, however, have nontrivial annihilator. The following result constitutes the fundamental structure theorem for finitely generated modules over principal ideal domains.

(7.1) Theorem. Let M # 0 be a finitely generated module over the PID R. If u(M) = n, then M is isomorphic to a direct sum of cyclic submodules

M-Rw1ED .®Rwn such that (7.1)

R # Ann(wl) J Ann(w2) D . . . D Ann(wn) = Ann(M).

Moreover, for 1 < i < n X7.2)

Ann(wi) = Ann (M/(Rwi}1 +

+ Rwn)) .

Proof. Since µ(M) = n, let {vi, ... vn} be a generating set of M and ,

define an R-module homomorphism 0: Rn -+ M by n O(a1,

... , an) _ E aivi. i=1

Let K = Ker(0). Since K is a submodule of Rn, it follows from Theorem 6.2 that K is a free R-module of rank in < n. By Theorem 6.23, there is a basis {yl, ... , yn} of Rn and nonzero elements sl, ... , s,n E R such that

3.7 Finitely Generated Modules over PIDs

{81y,,

(7.3)

157

is a basis for K

... , smym}

and si I si+l

(7.4)

for

1 < i < m - 1.

Let w; = ¢(yi) E M for 1 < i < n. Then {w1, ... ,wn} generates M since 0 is surjective and {yl, ... , yn } is a basis of

Rn. We claim that

By the characterization of direct sum modules (Theorem 3.1), it is sufficient

to check that if aiwi +.... + anwn = 0

(7.5)

where ai E R, then aiwi = 0 for all i. Thus suppose that Equation (7.5) is satisfied. Then

ai (yl) + ... + anO(yn) = -O(alyl + ... + anyn)

so that a1y1 + ... +anyn E Ker(O) = K = (sly,, ... , smym) Therefore,

aly, b1sly1 for some bl, ... , b,n E R. But {yl, ... , yn} is a basis of Rn, so we conclude that a; = bisi for 1 < i < m while ai =0 form + 1 < i < n. Thus, aiwi = bisiO(yi) = biO(siyi) = 0

for 1 < i < m because siyi E K = Ker(O), while aiwi = 0 for m+1 m, since (yi) fl Ker(O) = (0), it follows that Ann(wi) = (0). Since si I sn for all i and since Ann(wi) = (si), we conclude that s,,M = 0. Hence, Ann(wn) = (sn) = Ann(M) and Equation (7.1) is satisfied. Since

Equation (7.2) follows from Equation (7.1). The proof is now completed

by observing that Ann(wi) 54 R for any i since, if Ann(wi) = R, then

158


Rwi = (0), and hence, M could be generated by fewer than n elements. But n = µ(M), so this is impossible because µ(M) is the minimal number of generators of M. A natural question to ask is to what extent is the cyclic decomposition provided by Theorem 7.1 unique. Certainly, the factors themselves are not unique as one can see from the example Z2

Z.(1,0)ED More generally, if M is a free R-module of rank n, then any choice of basis {v1, ... ,v,a} provides a cyclic decomposition

Rv with Ann(vi) = 0 for all i. Therefore, there is no hope that the cyclic factors

themselves are uniquely determined. What does turn out to be unique, however, is the chain of annihilator ideals

Ann(wl) 2 ... where we require that Ann(w;) # R, which simply means that we do not allow copies of (0) in our direct sums of cyclic submodules. We reduce the uniqueness of the annihilator ideals to the case of finitely generated torsion R-modules by means of the following result. If M is an R-module, recall that the torsion submodule M, of M is defined by M, = {x E M : Ann(x) 34 (0)}-

(7.2) Proposition. If M and N are finitely generated modules over a PID R, then Ma, N if and only if M, N, and rank M/M, = rank N/N,.

Proof. Let 0 : M -. N be an isomorphism. Then if x E M there is an a 94 0 E R with ax = 0. Then aO(x) = O(ax) = 0(0) = 0 so that O(x) E N,. Therefore, 0(M,) C N,. Applying the same observation to 0' chows that O(M,) = N,. Thus, OIM, : M, -' N, is an isomorphism; if n : N -. N/N, is the natural projection, it follows that Ker(ao0) = M,. The first isomorphism theorem then gives an isomorphism M/M, = N/N,. Since M/M, and N/N, are free R-modules of finite rank, they are isomorphic if and only if they have the same rank. The converse follows from Corollary 6.20.

Therefore, our analysis of finitely generated R-modules over a PID R is reduced to studying finitely generated torsion modules M; the uniqueness of the cyclic submodule decomposition of finitely generated torsion modules is the following result.


159

(7.3) Theorem. Let M be a finitely generated torsion module over a PID R, and suppose that there are cyclic submodule decompositions

MRwi

(7.6)

and (7.7) where

(7.8)

Ann(wl) D

Ann(wk)

(0)

with

Ann(wl) 0 R

an d (7.9)

Ann(zl)

D Ann(zr.) # (0)

with

Ann(zi) # R.

Then k = r and Ann(wi) = Ann(zi) for 1 < i < k. Proof. Note that Ann(M) = Ann(wk) = Ann(zr). Indeed,

Ann(M) = Ann(Rwi + = Ann(wi) fl

+ Rwk) fl Ann(wk)

= Ann(wk)

since Ann(wi) D Ann(wk). The equality Ann(M) = Ann(zr) is the same argument. We will first show that k = r. Suppose without loss of generality that k > r. Choose a prime p E R such that (p) Ann(wi), i.e., p divides the generator of Ann(wi). Then (p) D Ann(wi) for all i. Since p E Ann(M/pM), it follows that M/pM is an R/pR-module and Equations (7.6) and (7.7) imply (7.10)

MIpM

Rwi/(pRwi) ®...

Rwk/(pRwk)

and (7.11)

M/pM = Rzi/(pRzi) ® ... ® Rz,./(pRz,.).

Suppose that pRwi = Rwi. Then we can write apwi = wi for some a E R.

Hence, ap - 1 E Ann(wi) C (p) by our choice of p, so 1 E (p), which contradicts the fact that p is a prime. Therefore, pRwi # Rwi for all i and Equation (7.10) expresses the R/pR-module M/pM as a direct sum of cyclic R/pR-modules, none of which is (0). Since R/pR is a field (in a PID prime ideals are maximal), all R/pR-modules are free, so we conclude that M/pM is free of rank k. Moreover, Equation (7.11) expresses M/pM as a direct sum of r cyclic submodules, so it follows that k = µ(M/pM) < r.

Thus, r = k, and in particular, Rzi/(pRzi) 0 0 since, otherwise, M/pM could be generated by fewer than k elements. Thus, (p) 2 Ann(i) for all i; if not, then (p) + Ann(zi) = R, so there are a E R and c E Ann(zi) such that ap + c = 1. Then zi = apzi + czi = apzi E pRzi, so Rzi/(pRzi) = 0, and we just observed that Rzi/(pRzi) 54 0.


160

We are now ready to complete the proof. We will work by induction

on t(Ann(M)) where, if I = (a) is an ideal of R, then t(I) is the number of elements (counted with multiplicity) in a prime factorization of a. This number is well defined by the fundamental theorem of arithmetic for PIDs. Suppose that t(Ann(M)) = 1. Then Ann(M) = (p) where p E R is prime. Since Ann(M) = Ann(wk) = Ann(zk) = (p) and since (p) is a maximal ideal, Equations (7.8) and (7.9) imply that Ann(wi) = (p) = Ann(zi) for all i, and the theorem is proved in the case t(Ann(M)) = 1. Now suppose the theorem is true for all finitely generated torsion Rmodules N with t(Ann(N)) < t(Ann(M)), and consider the isomorphisms (7.12)

pM'5 pRwl ® ... ® pRwk

pRw,+i ® ... ® pRwk

and (7.13)

pM L' pRzl ® ... ® pRzk =' pRzt+i ® ... ® pRzk

where Ann(wj) = ... = Ann(w,) = Ann(zi) = ... = Ann(zi) = (p) and Ann(w,+i) g6 (p), Ann(zt+i) 0 (p) (s and t may be 0). Then Ann(pM) = (a/p) where Ann(M) = (a), so t(Ann(pM)) = t(Ann(M))-1. By induction we conclude that k - s = k - t, i.e., s = t, and Ann(pwi) = Ann(pzi) for s < i < k. But Ann(pwi) = (ai/p) where Ann(wi) = (ai). Thus Ann(wi) _ 0 Ann(zi) for all i and we are done.

R/ Ann(wi) and since R/I and R/J are isomorphic RSince Rwi modules if and only if I = J (Exercise 10), we may rephrase our results as follows.

(7.4) Corollary. Finitely generated modules over a PID R are in one-to-one correspondence with finite nonincreasing chains of ideals

R#I,212_J...2In. Such a chain of ideals corresponds to the module

M=R/Il®...®R/In. Note that µ(M) = n and if Ik+i

In = (0) but Ik 96 (0), then

M^_'R/II®...®R/Ik®R"-k We will use the convention that the empty sequence of ideals (n = 0) corresponds to M = (0). Proof

0

(7.5) Definition. If M is a finitely generated torsion module over a PID R and M Rwi ® ®Rwn with Ann(wi) Ann(wi+i) (1 < i < n - 1) and Ann(wi) 0 R, then the chain of ideals Ii = Ann(wi) is called the chain of invariant ideals of M.

3.7 Finitely Generated Modules over PII)s

161

Using this language, we can express our results as follows:

(7.6) Corollary. Two finitely generated torsion modules over a PID are isomorphic if and only if they have the same chain of invariant ideals.

0

Proof.

(7.7) Remark. In some cases the principal ideals Ann(w3) have a preferred

generator aj. In this case the generators {ajIj=1 are called the invariant factors of M. The common examples are R = Z, in which case we choose aj > 0 so that aj = IZ/Ann(w,)I, and R = F[X], where we take monic polynomials as the preferred generators of ideals. (7.8) Definition. Let R be a PID, and let M be a finitely generated torsion R-module with chain of invariant ideals (31) ? (92) ? ... D (3n)-

We define me(M) = sn and co(M) = 81

sn.

Note that me(M) and co(M) are only defined up to multiplication by a unit, but in some cases (R = Z or R = F[X]) we have a preferred choice of generators of ideals. In these cases me(M) and co(M) are uniquely defined. Concerning the invariants me(M) and co(M), there is the following trivial but useful corollary of our structure theorems.

(7.9) Corollary. Let M be a finitely generated torsion module over a PID R. (1) If a E R with aM = 0, then me(M) I a.

(2) me(M) divides co(M). (3) If p E R is a prime dividing co(M), then p divides me(M).

Proof. (1) Since Ann(M) = (sn) = (me(M)) by Theorem 7.1 and the defintion of me(M), it follows that if aM = 0, i.e., a E Ann(M), then me(M) I a. (2) Clearly s,, divides s1 ... sn. (3) Suppose that p I 31 sn = co(M). Then p divides some si, but (si) 2 (sn), so si I sn. Hence, p I sn = me(M). 0

(7.10) Remark. There are, unfortunately, no standard names for these invariants. The notation we have chosen reflects the common terminology in

the two cases R = Z and R = F[X]. In the case R = Z, me(M) is the exponent and co(M) is the order of the finitely generated torsion Z-module

162


(= finite abelian group) M. In the case R = FIX] of applications to linear algebra to be considered in Chapter 4, me(VT) will be the minimal polynomial and co(VT) will be the characteristic polynomial of the linear transformation T E HomF (V) where V is a finite-dimensional vector space over the field F and VT is the FIX]-module determined by T (see Example 1.5 (12)).

There is another decomposition of a torsion R-module M into a direct sum of cyclic submodules which takes advantage of the prime factorization of any generator of Ann(M). To describe this decomposition we need the following definition.

(7.11) Definition. Let M be a module over the PID R and let p E R be a prime. Define the p-component Mp of M by M p = {x E M : Ann(x) = (p")

for some

n E Z+}.

If M = Mp, then M is said to be p-primary, and M is primary if it is p-primary for some prime p E R. It is a simple exercise to check that submodules, quotient modules, and direct sums of p-primary modules are p-primary (Exercise 54).

(7.12) Theorem. If M is a finitely generated torsion module over a PID R, then M is a direct sum of primary submodules.

Proof. Since M is a direct sum of cyclic submodules by Theorem 7.1, it is sufficient to assume that M is cyclic. Thus suppose that M = (x) and suppose that Ann(x) = (a) = (pr' ... pr,,) where pl, ... , pn are the distinct prime divisors of a. Let qi = a/p;'. Then 1 = (ql, ... , qn) = gcd{ql, ... , q,,}, so there are bl, ... , bn E R such that (7.14)

1 = bigi + ... + bngn

Let xi = bigix. Then Equation (7.14) implies that

x=xl +...+ xn so that Al = (x) Suppose that y E (x1) fl ((x2) +

(XI) + ... + (xn) + (x,)). Then

y=Clx1

C2x2+...+Cnxn

and hence, p 'y = clb1Pi'gix = c1blax = 0 and qly = C2g2P23x2 +

+CngnPn"xn = 0,


163

Therefore, {pl' , ql } C Ann(y), but (pi' , ql) = 1 so that where q = ql Ann(y) = R. Therefore, y = 0. A similar calculation shows that

(xi) n ((xl) + ... + (xi) + ... + (xn)) = (0), so by Theorem

C

Combining Theorems 7.1 and 7.12, we obtain the following result:

(7.13) Theorem. Any finitely generated torsion module M over a PID R is a direct sum of primary cyclic submodules.

Proof. Suppose M Rwl® . ®Rwn as in Theorem 7.1. Then if Ann(wi) _ (si), we have si I si+1 for 1 < i < n - 1 with sl 54 1 and sn # 0 (since M is torsion). Let p', ... pk be the set of distinct nonassociate primes that occur as a prime divisor of some invariant factor of M. Then Si =

ulpil,

... peklk

sn = unplnl ...

pknk

where the divisibility conditions imply that

0 < elf < e2j < ... < eni

for

1<j

Then the proof of Theorem 7.12 shows that M is the direct sum of cyclic eij > 0}, and the theorem is proved. submodules with annihilators 0

eij > 0, 1 < j < k} are called

(7.14) Definition. The prime powers the elementary divisors of M.

(7.15) Theorem. If M and N are finitely generated torsion modules over a

PID R, then M N if and only if M and N have the same elementary divisors.

Proof. Since M is uniquely determined up to isomorphism from the invariant factors, it is sufficient to show that the invariant factors of M can be recovered from a knowledge of the elementary divisors. Thus suppose that (sl)

(32)

...

(sn)

is the chain of invariant ideals of the finitely generated torsion module M. This means that si ( si+1 for 1 < i < n. Let pl, ... ,pk be the set of distinct nonassociate primes that occur as a prime divisor of some invariant factor of M. Then

164


$1 = U1P11" ... Pkik (7.15) P`"k

sn = unP1"' ...

k

where the divisibility conditions imply that (7.16)

< enj

0 < e1j < e2j <

for

1 < j < k.

Thus, the elementary divisors of M are (7.17)

We show that the set of invariant factors (Equation (7.15)) can be reconstructed from the set of prime powers in Equation (7.17). Indeed, if

ej= maxe;j, 1<.

1<j

then the inequalities (7.16) imply that sn is an associate of pi'

(Pi' , ...

, Pkk

pkk. Delete

}

from the set of prime powers in set (7.17), and repeat the process with the set of remaining elementary divisors to obtain sn_1. Continue until all prime powers have been used. At this point, all invariant factors have been recovered. Notice that the number n of invariant factors is easily recovered from the set of elementary divisors of M. Since s1 divides every si, it follows

that every prime dividing 81 must also be a prime divisor of every s,. Therefore, in the set of elementary divisors, n is the maximum number of 0 occurrences of p"i for a single prime p. (7.16) Example. Suppose that M is the Z-module M = Z22 X Z22 X Z3 X Z32 X Z5 X Z7 X Z7a .

Then the elementary divisors of M are 22, 22, 3,3 2 , 5, 7,7 2. Using the algorithm from Theorem 7.15, we can recover the invariant factor description of M as follows. The largest invariant factor is the product of the highest power of each prime occurring in the set of elementary divisors, i.e., the least common multiple of the set of elementary divisors. That is, 32 = 72 5 32 22 = 8820. Note that the number of invariant factors of M is 2 since powers of the primes 2, 3, and 7 occur twice in the set of elementary divisors, while no prime has three powers among this set. Deleting 72, 5,3 2 , 22 from the set of elementary divisors, we obtain s1 = 7.3.22 = 84. This uses all the elementary divisors, so we obtain Al

Z&4 X Zs82o


165

We now present some useful observations concerning the invariants me(M) and co(M) where M is a torsion R-module (R a PID). See Definition 7.8 for the definition of these invariants. The verification of the results that we wish to prove require some preliminary results on torsion R-modules, which are of interest in their own right. We start with the following lemmas. (7.17) Lemma. Let M be a module over a PID R and suppose that x E MT.

If Ann(x) = (r) and a E R with (a, r) = d (recall that (a, r) = gcd{a, r}), then Ann(ax) = (r/d).

Proof. Since (r/d)(ax) = (a/d)(rx) = 0, it follows that (r/d) C Ann(ax). If b(ax) = 0, then r I (ba), so ba = rc for some c E R. But (a, r) = d, so there are s, t E R with rs + at = d. Then rct = bat = b(d - rs) and we see that bd = r(ct + bs). Therefore, b E (r/d) and hence Ann(ax) = (r/d). 0 (7.18) Lemma Let M be a module over a PID R, and let x1, ... , xn E Mr with Ann(xi) = (ri) for 1 < i < n. If {rl, ... , rn} is a pairwise relatively + xn, then Ann(x) _ (a) _ (f 1 ri). prime subset of R and x = x1 + Conversely, if y E M, is an element such that Ann(y) _ (b) si) where {s1, . . . , s, } is a pairwise relatively prime subset of R, then we may + yn where Ann(yi) _ (si) for all i. write y = yl +

Let x = x1 + - + xn. Then a = rj 1 ri E Ann(x) so that (a) C Ann(x). It remains to check that Ann(x) C (a). Thus, suppose that bx = 0. By the Chinese remainder theorem (Theorem 2.2.24), there are c1, ... , cn E R such that 1 (mod (ri)), ci = 0 (mod (r3)), if j 0 i. P r o o f.

Then, since (rj) = Ann(xj), we conclude that cixj = 0 if i 96 j, so for each

iwith 1

Therefore, bci E Ann(xi) = (ri), and since ci - 1 (mod (ri)), it follows that ri I b for 1 < i < n. But {ri, ... , rn } is pairwise relatively prime and thus a is the least common multiple of the set {r1, ... ,rn}. We conclude that a I b, and hence, Ann(x) = (a).

f)

Conversely, suppose that y E M satisfies Ann(y) = (b) = (i=1

where the set {s1, ... , sn } is pairwise relatively prime. As in the above paragraph, apply the Chinese remainder theorem to get c1, ... , en E R such that

(I

(mod (si)),

c' = 0 (mod (sj)), if j , i. Since b is the least common multiple of {s1i ... , sn}, it follows that

1 - c1

(mod (b)),

166


and hence, if we set yi = cjy, we have

y1+...+yn=(c1+...+cn)y=y Since (b, c1) (se).

sj), Lemma 7.17 shows that Ann(yi) = Ann(ciy) _

0

(7.19) Proposition. Let R be a PID and suppose that M is a torsion Rmodule such that

M Rw1 ®... Rwn with Ann(wi) = (ti). Then the prime power factors of the t; (1 < i < n) are the elementary divisors of M.

Proof. Let pi, ... pk be the set of distinct nonassociate primes that occur as a prime divisor of some ti. Then we may write

t1 = ulpi" ...

pklk

(7.18)

to = unpl. ... where u1i ... , un are units in R and some of the exponents eij may be 0. The proof of Theorem 7.12 shows that Rwi

Rzi1 ® ... Rzik

where Ann(zij) For notational convenience we are allowing zij = 0 for those (i, j) with eij = 0. Therefore, (7.19)

Al f___ (3) Rzjj

is Let S = {pe'' } where we allow multiple occurrences of a prime power pe, and let

where Ann(zij) =

S = {Zij}. Let m be the maximum number of occurrences of positive powers of a single

prime in S. If (7.20)

fnj = max eij 1

for

1 < j < k,

we define (7.21)

sm = Pi m' ... Pkmk

Note that fmj > 0 for 1 < j < k. Delete {pf-1, ... , pf-* } from the set S and repeat the above process with the remaining prime powers until no further positive prime powers are


167

available. Since a prime power for a particular prime p is used only once at each step, this will produce elements al, ... , sm E R. From the inductive description of the construction of si, it is clear that every prime dividing s; also divides s;+1 to at least as high a power (because of Equation (7.21)). Thus, s; s;+1 for 1 0}={pWs:eo,6 > 0}

(7.23)

where repetitions of prime powers are allowed and where (7.24)

for

0:5

1<j

by Equation (7.20). For each (1 < i < m), choose w;j E S with Ann(w;j) = and let xi = wit + + wik. Lemma 7.18 shows that Ann(x;) = (si) for

1 < i < m, and thus, k

k

Rx;

Rl (si) °-` ®Rl (Pf ") j=1

®Rw:j j=1

Equation (7.19) then shows that

M ®RzaQ 0.0

m

k

i=1

j=1

?®

Rw;j

! Rxl®...0Rzm where Ann(x;) = (s;). Since si s;+1 for 1 < i < m, it follows that {Si, ... , am} are the invariant factors of M, and since the set of prime I

power factors of {Si, ... , sm} (counting multiplicities) is the same as the set of prime power factors of {t1i ... , t,,} (see Equation (7.23)), the proof is complete.

(7.20) Corollary. Let R be a PID, let M1, ... , Mk be finitely generated torsion R-modules, and let M = ®k ,M;- If {d;1, ... , die;} } is the set of elementary divisors of Mi, then

168


S={d,,:1
Mi ?' Rwi1 ®... ® Rwi,.,

where Ann(wi1) = (sib) and si, si,j+l for 1 < j < ri. The elementary divisors of Mi are the prime power factors of {sit, ... , si,., }. Then I

k

M=®MiL, ®Rwij i=1

i,j

where Ann(wij) = (sjj). The result now follows from Proposition 7.19.

O

(7.21) Proposition. Let R be a PID, let M1, ... , Mk be finitely generated torsion R-modules, and let M = ®;` 1Mi. Then (7.25)

me(M) = lcm{me(M1), ... ,me(Mk)) k

(7.26)

co(M) _ [Ico(Ms). i=1

Since Ann(M) = Ik 1 Ann(MM), Equation (7.25) follows since (me(Mi)) = Ann(M5). Since co(M) is the product of all invariant factors of M, which is also the product of all the elementary divisors of M, Equation (7.26) follows from Corollary 7.20. 0 Proof.

I

The special case R = Z is important enough to emphasize what the results mean in this case. Suppose that M is an abelian group, i.e., a Z-

module. Then an element x E M is torsion if and only if nx = 0 for some n > 0. That is to say, x E MT if and only if o(x) < oo. Moreover, Ann(x) = (n) means that o(x) = n. Thus the torsion submodule of M consists of the set of elements of finite order. Furthermore, M is finitely generated and torsion if and only if M is a finite abelian group. Indeed, if M = (x1, ... , xk) then any x E M can be written x = nix1 + + nkxk where 0 < ni < o(xi) < oo f o r 1 < i < k. Therefore, IMI < n 1 o(xi). Hence, the fundamental structure theorem for finitely generated abelian groups takes the following form.

(7.22) Theorem. Any finitely generated abelian group M is isomorphic to Zr ®M1 where IMi I < oo. The integer r is an invariant of M. Any finite abelian group is a direct sum of cyclic groups of prime power order and these prime power orders, counted with multiplicity, completely characterize


169

the finite abelian group up to isomorpism. Also any finite abelian group is uniquely isomorphic to a group Z,, X . . X Z,k where 8i 1 si+i for all i.

Proof.

Given a natural number n it is possible to give a complete list of all abelian groups of order n, up to isomorphism, by writing n = pi' Pkk where pl, ... ,pk are the distinct prime divisors of n. Let M be an abelian group of order n. Then we may write M as a direct sum of its primary components M^='MM ED ...ED Mpk

where IMp, I

Then each primary component Mp, can be written as a

direct sum

Mp,?'Zp'-,,® ®Zp:u where

1<ej,:5

e;t

and

Furthermore, the main structure theorems state that M is determined up to isomorphism by the primes pl, ... , pk and the partitions ei1, ... , eit of the exponents r;. This is simply the statement that M is determined up to isomorphism by its elementary divisors. Therefore, to identify all abelian groups of order n, it is sufficient to identify all partitions of ri, i.e., all ways to write r; = e;1 + + e;t as a sum of natural numbers.

(7.23) Example. We will carry out the above procedure for n = 600 = 23 3.52. There are three primes, namely, 2, 3, and 5. The exponent of 2

is3andwecanwrite 3=1+1+1,3=1+2,and3=3.Thus there are

three partitions of 3. The exponent of 3 is 1, so there is only one partition, while the exponent of 5 is 2, which has two partitions, namely, 2 = 1 + 1 and 2 = 2. Thus there are 3.1.2 = 6 distinct, abelian groups of order 600. They are Z2 X Z2 X Z2 X Z3 X Z5 X Z5 L Z2 X Z10 X Z30 Z2 X Z2 X Z2 X Z3 X Z25 = Z2 X Z2 X Z150

Z2XZ4XZ3XZ5XZ5~Z10XZ60


170

Z2 X Z4 X Z3 X Z25 = Z2 X Z300

z8xZ3xZ5xZ5Z5XZ120 z8xZ3XZ25Z600 where the groups on the right are expressed in invariant factor form and those on the left are decomposed following the elementary divisors. We will conclude this section with the following result concerning the

structure of finite subgroups of the multiplicative group of a field. This is an important result, which combines the structure theorem for finite abelian groups with a bound on the number of roots of a polynomial with coefficients in a field.

(7.24) Theorem. Let F be a field and let G C F' = F \ {0} be a finite subgroup of the multiplicative group F. Then G is a cyclic group. Proof. According to Theorem 7.1, G is isomorphic to a direct sum

G(z1)ED ...ED (zn) where, if we let ki = o(zi) = order of z,, then ki I ki+1 for 1 < i 5 n - 1 and

Ann(G) = Ann(zn) _ (kn)Z. In the language of Definition 7.8, me(G) = kn. This means that for all z E G. Now consider the polynomial

zk..

=I

P(X) = Xk^ - 1.

(7.27)

Since F is a field, the polynomial P(X) has at most kn roots, because degree P(X) = kn (Corollary 2.4.7). But, as we have observed, every element of

G is a root of P(X), and JGJ = k1k2 ... kn.

Thus, we must have n = 1 and G = (z1) is cyclic.

(7.25) Corollary. Suppose that F is a finite field with q elements. Then F' is a cyclic group with q - 1 elements, and every element of F is a root of the polynomial X9 - X. Proof. Exercise.

(7.28) Corollary. Let

Gn = {e2xi(k/n) : 0 < k < n- 1} C C. Then Gn is the only subgroup of C' of order n.

3.8 Complemented Submodules

171

Proof. Let H be a finite subgroup of C' with IHJ = n. Then every element

z of H has the property that z" = 1. In other words, z is a root of the equation X" = 1. Since this equation has at most n roots in C and since every element of G,, is a root of this equation, we have z E G". Thus, we conclude that H C G" and hence H = G" because n = IHI = IG"I. 0

3.8 Complemented Submodules We will now consider the problem of extending a linearly independent subset of a free R-module to a basis. The example {2} C Z shows that some restrictions on the subset are needed, while Theorem 6.16 shows that any primitive element of a finitely generated free R-module (R a PID) can be extended to a basis.

(8.1) Definition. Let M be an R-module and S C M a submodule. Then S is said to be complemented if there exists a submodule T C M with M be a finitely generated free R-module with basis {v1, ... , v,,) and let S = (v1, ... , v,). Then S is complemented by T = (v,+1, ... , v"). This example shows that if W = {w1, ... wk} is a linearly independent subset of M, then a necessary condition for W to extend to a basis of M is

that the submodule (W) be complemented. If R is a PID, then the converse is also true. Indeed, let T be a complement of (W) in M. Since R is a PID, T is free, so let {x1, ... , x,.} be a basis of T. Then it is easy to check that

{w1,...,wk,x1,...,x,.}is a basis of M. (8.2) Proposition. Let R be a PID, let M be a free R-module, and let S be a submodule. Consider the following conditions on S. (1) S is complemented.

(2) MIS is free.

(3) IfxESandx=ayforsomeyEM,aO0ER,thenyES. Then (1) s (2) and (2)

(3), while if M is finitely generated, then

(3) = (1).

Proof. (1) (2). If S is complemented, then there exists T C M such that S E T Q5 M. Thus, M/S se T. But T is a submodule of a free module over a PID R, so T is free (Theorem 6.2).

(2) =: (3). Suppose MIS is free. If x E S satisfies x = ay for some y E M, a 0 0 E R, then a(y + S) = S in MIS. Since free modules are torsion-free, it follows that y + S = S, i.e., y E S.

172


(3) = (1). Let M be a finite rank free R-module and let S C M be a submodule satisfying condition (3). Then there is a short exact sequence

O-S- M

(8.1)

M/S- O

where n is the projection map. Condition (3) is equivalent to the statement that MIS is torsion-free, so M/S is free by Theorem 6.6. But free modules are projective, so sequence (8.1) has a splitting or : MIS -. M and Theorem 3.9 shows that M - S ® Im(a), i.e., S is complemented. (8.3) Remarks.

(1) A submodule S of M that satisfies condition (3) of Proposition 8.2 is called a pure submodule of M. Thus, a submodule of a finitely generated module over a PID is pure if and only if it is complemented. (2) If R is a field, then every subspace S C M satisfies condition (3) so that

every subspace of a finite-dimensional vector space is complemented. Actually, this is true without the finite dimensionality assumption, but our argument has only been presented in the more restricted case. The fact that arbitrary subspaces of vector spaces are complemented follows from Corollary 4.21. (1) is false without the hypothesis that M be (3) The implication (3) finitely generated. As an example, consider a free presentation of the Z-module Q:

O-*S-M-+Q-0.

Since MIS - Q and Q is torsion-free, it follows that S satisfies condition (3) of Proposition 8.1. However, if S is complemented, then a complement T - Q; so Q is a submodule of a free Z-module M, and hence Q would be free, but Q is not a free Z-module. (8.4) Corollary. If S is a complemented submodule of a finitely generated R-module (R a PID), then any basis for S extends to a basis for M. Proof. This was observed prior to Proposition 8.2.

(8.5) Corollary. If S is a complemented submodule of M, then rank S = rank M if and only if S = M. Proof. A basis {vl, ... , v,n} of S extends to a basis {vl, ... , vn} of M. But

n=m,soS=(vl,...,vn)=M.

If AI = Z and S = (2), then rank S = rank M but M 0 S. Of course, S is not complemented. (8.6) Corollary. If S is a complemented submodule of M, then

rank M = rank S + rank(M/S).

3.8 Complemented Submodules

173

Proof. Let S = (v1, ... , v,,,) where m = rank S. Extend this to a basis {v1, ... , of M. Then T = (v,,,+i, ... , v,,) is a complement of S in M and T '_5 M/S. Thus,

rank M=n=m+(n-m)=rank S+rank(M/S). (8.7) Proposition. Let R be a PID and let f : M -. N be an R-module homomorphism of finite-rank free R-modules. Then

(1) Ker(f) is a pure submodule, but (2) Im(f) need not be pure. Proof. (1) Suppose X E Ker(f), a # 0 E R, and y E M with x = ay. Then 0 = f (x) = f (ay) = a f (y). But N is free and, hence, torsion-free so that f (y) = 0. Hence, condition (3) of Proposition 8.2 is satisfied, so Ker(f) is complemented.

(2) If f : Z -. Z is defined by f (x) = 2x, then Im(f) = 2Z is not a complemented submodule of Z.

(8.8) Proposition. Let R be a PID and let f : M - N be an R-module homomorphism of finite-rank free R-modules. Then

rank M = rank(Ker(f )) + rank(Im(f )). Proof. By the first isomorphism theorem, Im(f) M/ Ker(f ). But Ker(f) is a complemented submodule of M, so the result follows from Corollary 8.6.

(8.9) Corollary. Let R be a PID and let M and N be finite-rank free R-modules with rank(M) = rank(N). Let f E HomR(M, N). (1) If f is a surjection, then f is an isomorphism. (2) If f is an injection and Im(f) is complemented, then f is an isomor-

phism.

Proof. (1) By Proposition 8.8, rank(Ker(f)) = 0, i.e., Ker(f) = (0), so f is an injection. (2) By Proposition 8.8, rank N = rank M = rank(Im(f )). Since Im(f) is complemented by hypothesis, f is a surjection by Corollary 8.5. (8.10) Proposition. Let R be a field and let M and N be R-modules with rank(M) = rank(N) finite. Let f E HomR(M, N). Then the following are equivalent.

(1) f is an isomorphism. (8.11) Proposition. Let M be a finite-rank free R-module (R a PID). If S and T are pure submodules, then

174


(2) f is an injection. (3) f is a surjection. Proof. Since R is a field, Im(f) is complemented (by Remark 8.3 (2)), so this is an immediate consequence of Corollary 8.9. 0

(8.11) Proposition. Let M be a finite-rank free R-module (R a PID). If S and T are pure submodules, then rank(S + T) + rank(S n T) = rank S + rank T.

Proof Note that if S and T are pure submodules of M, then S n T is also pure. Indeed, if ay E S n T with a 0 0 E R, then y E S and y E T since these submodules are pure. Thus, y E S n T, so S n T is complemented by Proposition 8.2. Then

(S + T)/T

S/(S n T).

By Corollary 8.6, we conclude

rank(S + T) - rank(T) = rank(S) - rank(S n T). 0

(8.12) Remark. It need not be true that S + T is pure, even if S and T are both pure. For example, let S = ((1, 1)) C Z2 and let T = ((1, -1)) C Z2. Then S and T are both pure, but S+T 76 Z2, so it cannot be pure. In fact,

2.(1,0)=(2,0)=(1, 1)+(1, -1) ES+T,

3.9 Exercises 1.

If M is an abelian group, then Endz(M), the set of abelian group endomorphisms of M, is a ring under addition and composition of functions.

(a) If M is a left R-module, show that the function 0 : R -+ Endz(M)

2. 3. 4.

defined by 0(r)(m) = rm is a ring homomorphism. Conversely, show that any ring homomorphism 0 : R -. Endz(M) determines a left Rmodule structure on M. (b) Show that giving a right R-module structure on M is the same as giving a ring homomorphism 0: R°' -+ Endz(M). if and Show that an abelian group G its the structure of a only if nG = (0). Show that the subring Z(4 ] of Q is not finitely generated as a Z-module if 9 ¢ Z. Let M be an S module and suppose that R C S is a subring. Then 1Li is also an R-module by Example 1.5 10). Suppose Chat N C M is an R-submodule.

LetSN=(sn:sES, nEN . (a) If S =

and R = Z, show that SN is the S-submodule of M generated by N. (b) Show that the conclusion of part (a) need not hold if S = R and R = Q.

3.9 Exercises 5.

175

Let M be an R-module and let A, B, and C be submodules. If C C A, prove

that

An(B+C)=(AnB)+C.

This equality is known as the modular law. Show, by example, that this formula need not hold if C is not contained in A. 6.

Let R be a commutative ring and let S C R\ {0} be a multiplicatively closed subset of R containing no zero divisors. Let M be an R-module. Mimicking the construction of RS (Theorem 2.3.5), we define MS as follows. Define a

relation - on M x S by setting (x, s) - (y, t) if and only if utx = usy

for some u E S. that this is an equivalence relation (see the proof of Theorem 2.3.5). We will denote the equivalence class of (x, s) by the suggestive symbol x/s.

a Prove that MS is an Rs-module via the operation (a/s)(x/t) = (ax)/(st).

(b) If f : M -* N is an R-module homomorhism, show that Is : Ms

NS

defined by fs(x/s) = f(x)/s is an RS-module homomorphism. (c) If x E M, show that x/I = 0 in MS if and only if Ann(x) n S 76 0.

7.

Let R C FIX] be the subring

R = {f(X) E FIX] : f(X) =ao+a2X2+...+anX"}. Thus, f (X) E R if and only if the coefficient of X is 0. Show that FIX] is a finitely generated R-module that is torsion-free but not free. 8. Show that Q is a torsion-free Z-module that is not free. 9. (a) Let R be an integral domain, let M be a torsion R-module, and let N be a torsion-free R-module. Show that HomR M, N) = (0).

(b) According to part (a), Homz(Z,n, Z) = (0). If n = km, then Z,n is a Zn-module. Show that

Zn) °-` Zm10.

Let R be a commutative ring with 1 and let I and J be ideals of R. Prove that R/I °_° R/J as R-modules if and only if I = J. Suppose ask that R/I and /J be isomorphic rings. Is the same conclusion valid? id? (Hint: Consider F[X]/(X - a) where a E F and show that F[X]/(X - a) °_° F as

X

11. 12. 13.

rings.) Prove Theorem 2.7. Prove Lemma 2.9.

Let M be an R-module and let f E EndR(M) be an idempotent endomorphism of M, i.e., f o f = f. (That is, f is an idempotent element of the ring EndR(M).) Show that M °- (Ker(f)) ® (Im(f)).

14. 15.

16. 17.

Prove the remaining cases in Theorem 3.10.

Let R be a PID and let a and b E R be nonzero elements. Then show that HomR(R/Ra, R/Rb) ?f R/Rd where d = (a, b) is the greatest common divisor of a and b. Compute Homz(Q, Z). Give examples of short exact sequences of R-modules


176 and

O-.N1 -.N such that

a) Mi'`N1, M2tN, M2 N2; b) Mi° NI,M N,M22.N2;

18.

c) M1 N1, M °__ N, M2 °5 N2 Show that there is a split exact sequence

0 - . mZmn -

Zmn -+ nZmn - 0

of Zmn-modules if and only if (m, n) = 1. 19.

Let N1 and N2 be submodules of an R-module M. Show that there is an exact sequence

O-.N1nN2!N1®N2-N1+N2-+0 20.

21.

where p(x) = (x, x) and m(x, y) = x - y. Let R be an integral domain and let a and b be nonzero elements of R. Let M = R/R(ab) and let N = Ra/R(ab). Then M is an R-module and N is a subm ule. Show that N is a complemented submodule in M if and only if there are u, v E R such that ua + vb = I.

Let R be a ring, M a finitely generated R-module, and 4 : M -+ R" a

surjective R-module homomorphism. Show that Ker ¢) is finitely generated. (Note that this is valid even when M has submodules that are not finitely generated.) (Hint: Consider the short exact sequence:

0 -+ K -. M m-+ R" -+ 0. ) 22.

Suppose that

0 , M1 m M - M2 -+ 0 19 1h 0 -. Ni1' N N2 -0 is a commutative diagram of R-modules and R-module homomorphisms. Assume that the rows are exact and that f and h are isomorphisms. Then prove that g is an isomorphism. 23. Let R be a commutative ring and S a multiplicatively closed subset of R containing no zero divisors. If M is an R-module, then Ms was defined in Exercise 6. Prove that the operation of forming quotients with elements of S is exact. Precisely: (a) Suppose that M' f M 9+ M" is a sequence of R-modules and homomorphisms which is exact at M. Show that the sequence Ms Is Ms gs MS is an exact sequence of Rs-modules and homomorphisms. (b) As a consequence of part (a), show that if M' is a submodule of M, then Ms can be identified with an Rs-submodule of Ms.

(c) If N and P are R-submodules of M, show (under the identification

of part (b)) that (N + P)s = Ns + Ps and (N n P)s = Ns n Ps.

(That is, formation of fractions commutes with finite sums and finite intersections.)

(d) If N is a submodule of M show that

3.9 Exercises

177

(M/N)s'-` (Ms)/(Ns)-(That

is, formation of fractions commutes with quotients.) 24.

Let F be a field and let {f;(X)},:o be any subset of F[X] such that degf;(X) = i for each i. Show that {f;(X)} o is a basis of FIX) as an F-module.

25. Let R be a commutative ring and consider M = R[X] as an R-module. Then

N = R[X21 is an R-submodule. Show that M/N is isomorphic to R[X] as an R-module. 26.

Let G be a group and H a subgroup. If F is a field, then we may form the group ring F(G) (Example 2.1.9 (15)). Since F(G) is a ring and F(H) is a subring, we may consider F(G) as either a left F(H)-module or a right

F(H)-module. As either a left or right F( H)-module, show that F(G) is free of rank [G : H]. (Use a complete set {g; of coset representatives of H as a basis.) 27.

Let R and S be integral domains and let 01, ... , ¢, be n distinct ring are S-linearly indepenhomomorphisms from R to S. Show that dent in the S-module F(R, S) of all functions from R to S. (Hint: Argue by

induction on n, using the property O;(ax) = O;(a)O;(x), to reduce from a dependence relation with n entries to one with n - 1 entries.) 28. Let G be a group, let F be a field, and let O; : G -. F' for 1 < i < n be n distinct group homomorphisms from G into the multiplicative group F' of F. Show that 01, ..., 0n are linearly independent over F (viewed as elements of the F-module of all functions from G to F). (Hint: Argue by induction on n, as in Exercise 27.) 29. Let R = Z3o and let A E M2,3(R) be the matrix

A=

1

1

0

2

-11

.

3

Show that the two rows of A are linearly independent over R, but that any two of the three columns are linearly dependent over R. 30. Let V be a finite-dimensional complex vector space. Then V is also a vector space over R. Show that dime, V = 2dimc V. (Hint: If

B={vi,...,vn} is a basis of V over C, show that i3' = {vl, 31.

... vn, ivl, ... ivn)

is a basis of V over R.) Extend Exercise 30 as follows. Let L be a field and let K be a subfield of L. If V is a vector space over L, then it is also a vector space over K. Prove

that

dimK V = [L: K] dim,. V

where [L : K] = dimK L is the dimension of L as a vector space over K. (Note that we are not assuming that dimK L < oo.) 32.

Let K C L be fields and let V be a vector space over L. Suppose that

B = {ua },Er is a basis of V as an L-module, and let W be the K-submodule of V generated by B. Let U C W be any K-submodule, and let U,. be the L-submodule of V generated by U. Prove that

U,.nW=U. That is, taking L-linear combinations of elements of U does not produce any new elements of W.


178

That is, taking L-linear combinations of elements of U does not produce any new elements of W. 33. Let K C L be fields and let A E A, (K), b E M.,1(K). Show that the matrix equation AX = b has a solution X E M,,,1(K) if and only if it has a solution X E M,,,I(L). 34. Prove that the Lagrange interpolation polynomials (Proposition 2.4.10) and the Newton interpolation polynomials (Remark 2.4.11) each form a basis of the vector space of polynomials of degree < n with coefficients from F. 35.

Let F denote the set of all functions from Z+ to Z+, and let M be the

free Q-module with basis F. Define a multiplication on M by the formula (f g)(n) = f (n) + g(n) for all f, g E F and extend this multiplication by linearity to all of M. Let fm be the function f,,, n) = 6.,,,,, for all m, n > 0. Show that each fm is irreducible in fact, prime) as an element of the ring Al. Now consider' onsider the function f (n) = 1 for all it > 0. Show that f does not have a factorization into irreducible elements in Al. (Hint: It may help to think of f as the "infinite monomial" Xo ro)Xi (1)

36.

... Xm(-) ....

(Compare this exercise with Example 2.5.15.) Let F be a field, and let

2 = {p,(X) : pa(X) is an irreducible monic polynomial in FIXJ}.

We will say that a rational function h(X) = f (X)/g(X) E F(X) is proper if deg(f (X)) < deg(g(X)). Let F(X)pr denote the set of all proper rational functions in F[XJ. (a) Prove that F(X) (b) Prove that B=

FIX] tD F(X)pr as F-modules.

X'

E Z; 0 < j < deg(pa(X)), k > 1 }

(p. (X))'

is a basis of F(X)pr as an F-module. The expansion of a proper rational

function with respect to the basis B is known as the partial fraction

37. 38.

expansion; it should be familiar front elementary calculus. Prove that Q is not a projective Z-module. Let

R = { f : [0, 1] - R :

f is continuous and f (0) = f (1) }

and let M = {f : 10, 11

R:

f is continuous and f (0) = - f (1)}.

Then R is a ring under addition and multiplication of functions, and M is an R-module. Show that M is a projective R-module that is not free. (Hint: 39.

Show that M ®M ? -, R (D R.) Show that submodules of projective modules need not be projective. (Hint: Consider pZP2 C Z,2 as ZD2-modules.) Over a PID, show that submodules of projective modules are projective.

40.

(a) If R is a Dedekind domain, prove that R is Noetherian.

b If R is an integral domain that is a local ring (i.e., R has a unique

maximal ideal), show that any invertible ideal I of R is principal. (c) Let R be an integral domain and S C R \ {0} a multiplicatively closed subset. If I is an invertible ideal of R, show that Is is an invertible ideal of Rs.

3.9 Exercises

179

(d) Show that in a Dedekind domain R, every nonzero prime ideal is maximal. (Hint: Let M be a maximal ideal of R containing a prime ideal P, 41. 42. 43.

44.

and let S = R \ M. Apply parts (b) and (c).) Show that Z[f-3] is not a Dedekind domain. Show that Z[XJ is not a Dedekind domain. More generally, let R be any integral domain that is not a field. Show that RIX] is not a Dedekind domain. Suppose R is a PID and M = R(x) is a cyclic R-module with Ann M = (a) 9 (0). Show that if N is a submodule of M, then N is cyclic with Ann N = (b) where b is a divisor of a. Conversely, show that M has a unique submodule N with annihilator (b) for each divisor b of a.

Let R be a PID, M an R-module, x E M with Ann(x) = (a) 96 (0). Factor

a = upi' ... pk" with u a unit and pi, ..., pi, distinct primes. Let Y E M with Ann(y) = (b) 0- (0), where b = ti p1`' . . .pk ' with 0 < rn; < n; for 1 < i < k. Show that Ann(x + y) = (a). Let R be a PID, let M be a free R-module of finite rank, and let N C M be a submodule. If M/N is a torsion R-module, prove that rank(M) = rank(N). 46. Let R be a PID and let M and N be free R-modules of the same finite rank. Then an R-module homomorphism f : M - N is an injection if and only if N/ Im(f) is a torsion R-module. 45.

47.

Let u = (a, b) E Z2.

(a) Show that there is a basis of Z2 containing u if and only if a and b are relatively prime. (b) Suppose that u = (5,12). Find a v E Z2 such that {u, v} is a basis of

V.

48. Let M be a torsion module over a PID R and assume Ann(M) = (a) 0 (0).

If a = pi' pk'' where pi, ... , pk are the distinct prime factors of a, then show that MD, = qiM where q; = a/p;'. Recall that if p E R is a prime,

then M9 denotes the p-primary component of M. Let M be a torsion-free R-module over a PID R, and assume that x E M is a primitive element. If px = qx' show that q I p. 50. Find a basis and the invariant factors for the submodule of Z3 generated by 49.

x1 _ (1,0,-1), x2 = (4,3, -1), x3 = (0,9,3), and x4 = (3, 12,3).

51.

Find a basis for the submodule of Q(XJ3 generated by

f, = (2X-1,X,X2+3), 52.

f2 = (X,X,X2),

f3 = (X + 1, 2X, 2X2 -3).

Determine the structure of Z3/K where K is generated by xl = (2,1, -3) and x2 = (1,-1,2).

Let R = RIXJ and suppose that M is a direct sum of cyclic R-modules with annihilators (X - 1)3, (X2 + 1)2, (X - 1)(X2 + 1)4, and (X + 2)(X2 + 1)2. Determine the elementary divisors and invariant factors of M. 54. Let R be a PID and let p E R be a prime. Show that submodules, quotient modules, and direct sums of p-primary modules are p-primary. 53.

55. An R-module M is said to be irreducible if (0) and M are the only submodules of M. Show that a torsion module M over a PID R is irreducible if and only if M = R(x) where Ann(x) = (p) where p is prime. Show that

if M is finitely generated, then M is indecomposable in the sense that M is not a direct sum of two nonzero submodules if and only if M = R(x) where Ann(x) = (0) or Ann(z) = (p°) where p is a prime. 56. Let M be an R-module where R is a PID. We say that M is divisible if for each nonzero a E R, aM = M. (a) Show that Q is a divisible Z-module.


180

57. 58. 59. 60.

(b) Show that any quotient of a divisible R-module is divisible. It follows for example that Q /Z is a divisible Z-module. (c) If R is not a field, show that no finitely generated R-module is divisible. Determine all nonisomorphic abelian groups of order 360. Use elementary divisors to describe all abelian groups of order 144 and 168. Use invariant factors to describe all abelian groups of orders 144 and 168. If p and q are distinct primes, use invariant factors to describe all abelian groups of order

(a) p2g2, (b) p" q, (c) P5. 61. If p and q are distinct primes, use elementary divisors to describe all abelian groups of order p3q2.

62.

63. 64.

65. 66.

67.

Let G, H, and K be finitely generated abelian groups. If G x K °_° H x K, show that G I H. Show by example that this need not be true if we do not assume that the groups are finitely generated. Determine all integers for which there exists a unique abelian group of order n. Show that two finite abelian groups are isomorphic if and only if they have the same number of elements of each order. Let p be a prime and assume that a finite abelian group G has exactly k elements of order p. Find all possible values of k Find a generator for the cyclic group F' where F is each of the following fields (see Example 2.5.15 (3)): (a) F2(XJ/(X2 + X + 1). (b) F3[X]/(X2 + 1). Let 0

M1

-

M2

/s+

... "- Mn+l -+ 0

be an exact sequence of finite rank free modules and homomorphisms over a

PID R. That is, fl is injective, fn is surjective, and Im(f1) = Ker(f;+l) for

1
n+l

(-1)'}1 rank(M.) = 0. 68.

If f(XI, ... Xn) E R[X1, ... X 1 , , the degree off is the highest degree of a monomial in f with nonzero coe cient, where

deg(X" ...X'") = it +...+in. Let F be a field. Given any five points {v1, ... ,v5} C F2, show that there is a quadratic polynomial f (Xl, X2) E F[X1, X21 such that f (v;) = 0 for

1<5. 69.

Let M and N be finite-rank free R-modules over a PID R and let f E HomR(M, N). If S C N is a complemented submodule of N, show that

70.

f '(S) is a complemented submodule of M. Let R be a PID, and let f : M -. N be an R-module homomorphism of finite rank free R-modules. If S C N is a submodule, prove that rank(f -1(S)) = rank(S fl Im(f )) + rank(Ker(f)).

3.9 Exercises 71.

181

Let M1 f M L M2 be a sequence of finite-rank R-modules and R-

module homomorphisms, where R is a PID, and assume that Im(f) is a complemented submodule of M.

(a) Show that

rank(Im(g o f)) = rank(lm(f)) - rank(Im(f) fl Ker(g)). (b) Show that

rank(lm(g o f)) = rank(Im(f) + Ker(g)) - rank(Ker(g)). If R is a field, then all submodules of R-modules are complemented, so these formulas are always valid in the case of vector spaces and linear transformations. Show, by example, that they need not be valid if Im(f) is not complemented. 72. Let R be a PID, and let M, N, and P be finite rank free R -modules. Let

M - N and g : M - P be homomorphisms. Suppose that Ker(f) C Ker(g) and Im(f) is a complemented submodule of N. Then show that there is a homomorphism h: N -+ P such that g = h o f. 73. Let F be a field and let V be a vector space over F. Suppose that f, g E V' = HomF(V, F) such that Ker(f) C Ker(g). Show that there is a E F such that g = a f . this same result true if F is replaced by a PID? 74.

Let R be a PID and let M be a finite rank free R-module. Let Ck(M) denote the set of complemented submodules of M of rank k. Let G be the group of units of the ri EndR(M). (a) Show that-710, N) -. b(N) determines an action of the group C on the set Ck(M).

(b) Show that the action defined in part (a) is transitive, i.e., given N1, N2 E Ck(M) there is 0 E G that sends N1 to N2.

Chapter 4

Linear Algebra

The theme of the present chapter will be the application of the structure theorem for finitely generated modules over a PID (Theorem 3.7.1) to canonical form theory for a linear transformation from a vector space to itself. The fundamental results will be presented in Section 4.4. We will start with a rather detailed introduction to the elementary aspects of matrix algebra, including the theory of determinants and matrix representation of linear transformations. Most of this general theory will be developed over an arbitrary (in most instances, commutative) ring, and we will only specialize to the case of fields when we arrive at the detailed applications in Section 4.4.

4.1 Matrix Algebra We have frequently used matrix rings as a source of examples, essentially using an assumed knowledge of matrix algebra (particularly matrix multiplication). The present section will give the primary formulas of the more elementary aspects of matrix algebra; many of the proofs will be left as exercises. In this section, we do not assume that a ring R is commutative, except when explicitly stated. Let R be a ring. By an m x n matrix over R we mean a rectangular array

A=

all

a12

aln

a21

a22

a2n

aml

am2

. . .

=[si;[

amm

where a, E R for 1 < i < m, 1 < j < n. One can treat a matrix more

formally as a function f: I x J R where I = {1, 2, ..., m}, J= (1, 2, ..., n}, and f(i, j) = ai;, but we shall be content to think of a matrix in the traditional manner described above as a rectangular array consisting of m rows and n columns. Let Mm,n (R) denote the set of all m x n

matrices over R. When m = n we will write M,,(R) instead of Mn,n(R), which is consistent with the notation introduced in Example 2.1.10 (8). If A = [aij) E Mm,n(R), then ai, is called the ij1h entry of A; there are times

4.1 Matrix Algebra

183

when it will be convenient to denote this by entij(A) = aij. The index i is referred to as the row index, and by the ith row of A we mesa the 1 x n matrix rowi(A) = Iail ... ain], while the index j is the column index and the jth column of A is the m x 1 matrix

alj colj(A) _ a,n j

If A, B E M,n,n(R) then we say that A = B if and only if entij(A) _ entij(B) f o r alli, j with 1
entij (A + B) = entij (A) + entij (B)

whenever A, B E M,n,n(R). It is easy to see that this operation makes M,n,n(R) into an abelian group with identity O,n,n (the m x n matrix with all entries equal to 0 E R). We shall generally write 0 rather than O,n,n for the zero matrix in Mm,n(R). The additive inverse of A E Mm,n(R)

is the matrix -A defined by entij(-A) = -entij(A). Now let a E R and A E M,n,n(R), and define the scalar product of a and A, denoted aA E M,n,n(R), by entij(aA) = a entij(A). Similarly, Aa E M,n,n(R) is defined by entij(Aa) = entij(A)a. In the language to be introduced in Chapter 7, these scalar multiplications make M,n,n(R) into an (R, R)bimodule (Definition 7.2.1), i.e., M,n,n(R) is both a left R-module and a right R-module under these scalar multiplications, and a(Ab) = (aA)b for

alla,bERandAEM,n,n(R). The addition and scalar multiplication of matrices arise naturally from thinking of matrices as functions f : I x J - R, namely, they correspond to addition and scalar multiplication of functions. However, multiplication of matrices is motivated by the relationship of matrices to linear transformations, which will be considered in Section 4.3. For now we simply present matrix multiplication via an explicit formula. We can multiply a matrix A E M,n,n(R) and a matrix B E Mn,p(R) and obtain a matrix AB (note the order) in M,n,p(R) where AB is defined by the formula n

entij(AB) = > entik(A) entkj(B). k=1

Thus to multiply a matrix A by B (in the order AB) it is necessary that the number of columns of A is the same as the number of rows of B. Furthermore, the formula for entij (AB) involves only the ith row of A and the jth column of B. The multiplication formula can also be expressed as

184

Chapter 4. Linear Algebra

enti3(AB) = [rowi(A)][col3(B)]

where the formula for multiplying a row matrix by a column matrix is b1

[ a1

...

an ] bn

n

_ > akbkJ k=1

Let I E M. (R) be defined by ent,3 (In) = bid where bij is the kronecker 6 function, i.e., bii = 1 and bid = 0 if i 0 j. The following lemma contains some basic properties of matrix multiplication. In part (c), the concept of

center of a ring is needed. If R is a ring, then the center of R, denoted C(R), is defined by

C(R)={aER:ab=ba forallbER}. Note that C(R) is a subring of R and R is commutative if and only if R = C(R). (1.1) Lemma. The product map M,n,n(R) x MM,p(R) -+ M,,,,p(R) satisfies the following properties (where A, B, and C are matrices of appropriate sizes and a E R):

(1) A(B+C)=AB+AC. (2) (A + B)C = AC + BC. (3) a(AB) = (aA)B and both equal A(aB) when a E C(R). (4) I,nA = A and AIn = A. (5) rowi(AB) _ [rowi(A)]B. (6) col3(AB) = A[co1J(B)]. (7) The product map M,n,n(R) x Mn,p(R) x Mp,q(R) _+ M,,,,q(R) satisfies the associative law A(BC) = (AB)C. Proof. Exercise.

Remark. The content of the algebraic properties in Lemma 1.1 is that M,n,n(R) is a left Mm(R)-module and a right Mn(R)-module. Note that Mn(R) is a ring by (1), (2), and (7). The verification of (7) is an unenlightening computation, but we shall see in Remark 3.7 that this associativity is a consequence of the fact that composition of functions is associative. (Indeed, the basic reason for defining matrix multiplication as we have done is to make Proposition 3.6 true.) Also note that M,,(R) is an algebra over R by Lemma 1.1 (3) if R is commutative, but not otherwise.

Recall (from Example 2.1 (8)) that the matrix E3 E M,,(R) is defined to be the matrix with 1 in the ijtn position and 0 elsewhere. E1, is called a matrix unit, but it should not be confused with a unit in the matrix ring

4.1 Matrix Algebra

185

Mn(R); unless n = 1 the matrix unit Eii is never a unit in Mn(R). We recall the basic properties of the matrices Eii in the following :emma. (1.2) Lemma. Let {Eii : 1 < i, j < n} be the set of matrix units in MM(R) and let A = [aii] E Mn(R). Then

(1) EiiEki = bjkEsl.

(2) En Eii = I.. Ei.i n (3) A =

n

aiiEii = i.i=1 Eiiaij.

(4) EiiAEk1 = aikE 1. Proof. Exercise.

Remark. When speaking of matrix units, we will generally mean the mar trices Eii E Mn(R). However, for every m, n there is a set {&i}; ` 1 1 C Mm,n(R) where Eii has 1 in the (i, j) position and 0 elsewhere. Then, with appropriately adjusted indices, items (3) and (4) in Lemma 1.2 are valid. Moreover, {Eii}in 1J 1 is a basis of M,n,n(R) as both a left R-module and a right R-module. Hence, if R is commutative so that rank makes sense (Remark 3.6.19), then it follows that rankR(Mm,n(R)) = mn.

If a E R the matrix aIn E Mn(R) is called a scalar matrix. The set of scalar matrices {aIn : a E R} C Mn(R) is a subring of Mn(R) isomorphic

to R, via the isomorphism a -+ aIn. Note that (aIn)A = A(aIn) for all a E C(R), A E Mn(R), where C(R) is the center of R. Let v : R Mn(R) be the ring homomorphism v(a) = aIn. Then there is the following result. (1.3) Lemma. If R is a ring, then

C(Mn(R)) = v(C(R)). That is, the center of Mn(R) is the set of scalar matrices where the scalar is chosen from the center of R.

Proof. Clearly v(C(R)) C C(Mn(R)). We show the converse. Let A E C(Mn(R)) and let 1 < j < n. Then AE1i = E13A and Lemma 1.1 (5) and (6) show that colk(AE1i) = A[colk(E1i)J = bik coll(A) and

rowk(Eli A) = [rowk(E1i)]A = blk rowi(A).

Comparing entries in these n pairs of matrices shows that ai, = 0 if s # j and all = aii. Since j is arbitrary, this shows that A = a11In is a scalar matrix. Since A must commute with all scalar matrices, it follows that all E C(R).


186

A unit in the ring M,, (R) is a matrix A such that there is some matrix B with AB = BA = In. Such matrices are said to be invertible or unimodular. The set of invertible matrices in M,,(R) is denoted GL(n, R) and is called the general linear group over R of dimension n. Note that GL(n, R) is, in fact, a group since it is the group of units of a ring.

If A = (a,j] E M,n,n(R), then the transpose of A, denoted At E !VIn,,n(R), is defined by

entij(At) = aj, = ent,,(A). The following formulas for transpose are straightforward and are left as an exercise.

(1.4) Lemma. Let R be a ring. (1) If A, B E Mm,n(R), then (A + B)t = At + Be. (2) If R is commutative, if A E Mm,n (R) and B E Mn,p(R), then (AB)t =

BtAt. (3) If R is commutative and A E GL(n, R), then

(At)-' = (A-

Proof.

There is a particularly important R-module homomorphism from Mn(R) to R defined as follows:

(1.5) Definition. Define the R-module homomorphism Tr : Mn(R) -' R by n

n

a::

Tr(A) =

The element Tr(A) E R is called the trace of the matrix A E M (R). The following is a simple but important result: (1.6) Lemma. Let R be a commutative ring. (1) If A E lbl,n,n(R) and B E Aln,,n(R), then

Tr(AB) = Tr(BA). (2) If A E M,,(R) and S E GL(n, R), then

Tr(S-'AS) = Tr(A).

4.1 Matrix Algebra

187

Proof (1) entii(AB)

R (AB) i=1

M n

= E E entik(A) entki(B) i=1 k=1 n

m

= E E entki (B) entik (A) k=1 i=1 n

_ 1: entkk(B) k=1

= T-(BA).

(2) Tr(S-1AS) = Tr(SS-1A) = Tr(A).

0

(1.7) Definition. If R is a ring and A, B E Mn(R), then we say that A and B are similar if there is a matrix S E GL(n, R) such that S-1 AS = B.

Similarity determines an equivalence relation on Mn(R), the significance of which will become clear in Section 4.3. For now, we simply note that Lemma 1.6 (2) states that, if R is commutative, then similar matrices have the same trace.

(1.8) Definition. Let R be a ring and let {Eij};j=1 C Mn(R) be the set of matrix units. Then we define a number of particularly useful matrices in Mn(R). (1)

For a E R and i 36 j define

Tij(a) = In +aEj. The matrix Tij (a) is called an elementary transvection. Tij (a) differs from In only in the ijth position, where Tij(a) has an a.

(2) IfaER' is a unit ofRand1
(3)

is called an elementary dilation. Di(a) agrees with the identity matrix In except that it has an a (rather than a 1) in the ith diagonal position. The matrix

Pij =In -Eii-Ejj+Eij+Eji

is called an elementary permutation matrix. Note that Pij is obtained from the identity In by interchanging rows i and j (or columns i and

j)

188


(1.9) Definition. If R is a ring, then matrices of the form Di(,0), Tij(a), and Pig are called elementary matrices over R. The integer n is not included

in the notation for the elementary matrices, but is determined from the context.

Examples. Suppose n = 3 and a E R. Then

l 0a T13(a) =

0 0

1

0

0

1

1

D2(a) =

0

0

0a0 0

0

P13 =

1

0 0

0

1

1

0

1

0

0

Another useful class of matrices for which it is convenient to have an explicit notation is the set of diagonal matrices. (1.10) Definition. A matrix A E Mn(R) is diagonal if entij (A) = 0 whenever

i # j. Thus, a diagonal matrix A has the form A = E 1 ajEii in of the matrix units Ei3. We shall use the notation A = diag(al, ..., an) to denote the diagonal matrix A = E 1 aiEii. In particular, the elementary matrix Di(,3) = diag(1, ..., 1, 0, 1, ..., 1) where the )3 occurs in the ieh diagonal position.

Note the following formula for multiplication of diagonal matrices:

diag(al,..., an) diag(bl,..., bn) = diag(albl,..., anbn).

(1.1)

This observation will simplify the calculation in part (3) of the following collection of basic properties of elementary matrices. (1.11) Lemma. Let R be a ring. (1) If a, 3 E R and i # j, then Ti3(a)Tij (3) = Ti,(a + (3). (2) If a E R and i 36 j, then Ti3(a) is invertible and (3) If 3 E R' and 1 < i < n, then Di(i3) is invertible and [T1j(a)J-1

= Ti3(-a). Di(,3)-l =

Di(13-1). (4) Pi2j = In, so Pig is invertible and is its own inverse.

Proof. (1)

Ti3(a)Tij(/3) = (I +aEi3)(I +QEi3)

=I+(a+,3)Ei3 =T1 (a+03)

since E =0ifi96j. (2) Tij (a)Tij (-a) =Tij (0) = In.

(3) Dt(,3)Di(13-1) = diag(1, ..., 3, ..., 1) diag(1, ..., 0-1, ... ,1) _ In by Equation (1.1). (4) Exercise.

0

4.1 Matrix Algebra

189

The matrices Ti3(a), Di(a), and Pig are generalizations to arbitrary rings of the elementary matrices used in linear algebra. This fa.:t is formalized in the next result. To simplify the statement we introduce the following notation. Suppose R is a ring and A E Mm(R). The left multiplication by A is a function LA : Mm,n(R) . Mm.n(R) defined by GA(B) = AB. Similarly, if C E MM(R), the right multiplication by C is the function

Rc : Mm,n(R) - Mm,n(R) defined by RC(B) = BC. Note that LA is a right M,,(R)-module homomorphism, while Rc is a left Mm(R)-module homomorphism.

(1.12) Proposition. Let R be a ring, let a E R, 0 E R', and m, n E N be given, and let A E Mm,n(R). (1) LT,,(0)(A) = T;,(a)A is obtained from A by replacing rowi(A) by

a[row,(A)] + rowi(A) and leaving the other rows intact. (2) LD,(p)(A) = Di((3)A is obtained from A by multiplying rowi(A) by 'O on the left and leaving the other rows intact.

(3) Lp,,(A) = PijA is obtained from A by interchanging rowi(A) and rows (A) and leaving the other rows intact.

Proof. (1) Ti3(a)A = (I + aE1j)A = A + aE1jA. By Lemma 1.1 (5), rowk(Ei,A) = 0 if k 36 i and rowi(Ei3A) = [rowi(Ei,)]A = row,(A) since

row;(Eij) = [0

0

1

0

0]

where the 1 is in the jig` position. Thus, rowk(TTj(cr)A) = rowk(A) if k 76 i while rowi(Ti3(a)A) = rowi(A) +a[row,(A)].

(2) Di((3)A = (I-Eii+(3Eii)A = A+(Q-1)EiiA. The same calculation as in part (1) shows that rowk(Di((3)A) = rowk(A) if k # i while rowi(D;((3)A) = rowi(A) + ((3 - 1)[rowi(A)] = I3[rowi(A)].

(3) This is a similar calculation, which is left as an exercise.

0

There is a corresponding result that relates right multiplication by elementary matrices to operations on the columns of a matrix. We state the result and leave the verification (using Lemma 1.1 (6)) to the reader.

(1.13) Proposition. Let R be a ring, let a E R, (3 E R', and m, n E N be given, and let A E Mm,n(R).


190

(1) RT,,( )(A) = AT,j(a) is obtained from A by replacing colj(A) by [col,(A))a + colj(A) and leaving the other columns intact. (2) RD,(p)(A) = AD,(Q) is obtained from A by multiplying col,(A) by $ on the right and leaving the other columns intact.

(3) Rp,(A) = AP,j is obtained from A by interchanging col,(A) and colj(A) and leaving the other columns intact.

0

Proof.

We conclude this introductory section by introducing the notation of partitioned matrices. Partitioning matrices into smaller submatrices is a technique that is frequently useful for ing properties that may not be as readily apparent if the entire matrix is viewed as a whole. Thus, suppose then we that may think of A as an r x s block matrix

All

...

Ara

where each block A,j is a matrix of size mi x nj with entries in R. Two particularly important partitions of A are the partition by rows Al

where

A=

A; = rowi(A) E M1,n(R)

Am

and the partition by columns A = (B1

. .

.

Bn J

Bj = colj(A) E M,,,,1(R).

where

Consider the problem of multiplying two partitioned matrices. Thus, suppose that

A= and

j1l

... Bit

B= Bat

...

Bat

are partitioned matrices. Can the product C = AB be computed as a partitioned matrix C = [C,,( where C,, = Fk=1 AtkBkj? The answer is yes provided all of the required multiplications make sense. In fact, parts (5) and (6) of Lemma 1.1 are special cases of this type of multiplication for the partitions that come from rows and columns. Specifically, the equation

4.1 Matrix Algebra

191

rowi(AB) = [rowi(A)]B of Lemma 1.1 (5) translates into a product of partitioned matrices A1B

Al

B=

AB=

A,,,B

Am

where A, = rowi(A), while the equation colj(AB) = A[colj(B)] of Lemma 1.1 (6) translates into another product of partitioned matrices

AB=A[B1

By]=[AB1

ABp]

where Bj = colj (B). For the product of general partitioned matrices, there is the following result.

(1.14) Proposition. Let R be a ring, and let A E Mm,n(R), B E Mn,p(R). Suppose

that and that Aij E Mm;,n, (R) while Bij E Mn,,pl (R). Then the matrix C = AB has a partition C = [Cij] where Cij E Mm;,p, (R) and t

AikBkj

Cij = k=1

Proof. Suppose 15 a < m and 1 < Q < p. Then n

(1.2)

ent0R(C) _ 1: a,-y6-,O. ry=1

In the partition of Cgiven bym=

we

have that entQO(C) is in a submatrix Cij E (R) so that entaO(C) = entwr(Cij) where 1 < w < mi and 1 < r < pj. Thus we have a partition of rowa(A) and col0(B) as

row.(A) = [row. (Ai1)

...

and

col,. (B13 )

col,(B) col, ( Bij ) From Equation (1.2) we conclude that

row,,,(Ait)]

192


ent,r(Cij) = enta,s(C) n

=

aa7b70 y=1 n1

_

ni

n2

a,,,b0.,

+ E aatibck, + ... +

1:

aa7ba'Y

ry=ne-,+1

-f=n1+1

7=1

= entr(Ai1Bij) + ent,,,.(Ai2B2j) + ... + ent,.r(AitBtj) and the result is proved.

A particularly useful collection of partitioned matrices is the set of block diagonal matrices. A partitioned matrix

All

A1,

Arl

Ar,

A= is said to be a block diagonal matrix if r = s and if Aij = 0 whenever i 0 j. The submatrices Aii are the diagonal blocks, but note that the blocks Aii can be of any size. Generally, we will denote the diagonal blocks with the single subscript Ai. If Al 0

A=

0 AZ

..

0 0

.

0

0

...

Ar

is a block diagonal matrix, then we say that A is the direct sum of the matrices A1i ... , Ar and we denote this direct sum by A=A1ED ...ED A,.

®Ar E M,n,n(R) where m = E;=1 mi and n = F-'=1 ni. The following result contains some straightforward results concerning the algebra of direct sums of matrices: Thus, if Ai E Mm,,n; (R), then A

(1.15) Lemma. Let R be a ring, and let A1, ... , Ar and B1, ... , Br be matrices over R of appropriate sizes. (The determination of the needed size is left to the reader.) Then (1) (®i .1Ai) + (E Bi) (2) (®i IAi) (®i=1Bi) (3) (E D7 = ®i=1Ai

®i=1 (Ai + Bi),

(AiBi), 1

if Ai E GL(ni, R), and

(4) Tr(®i lAi) = i=11 (Ai) if Ai E M., (R). Proof. Exercise.

4.1 Matrix Algebra

193

The concept of partitioned matrix is particularly convenient for describing and ing various properties of the tensor product (also called the kronecker product) of two matrices. (1.16) Definition. Let R be a commutative ring, let A E Mm1 rn, (R), and let

B E Mm2,n,(R). Then the tensor product or kronecker product of A and B, denoted A® B E Mmon,,nin,(R), is the partitioned matrix

A®B=

(1.3)

where each block C,, E Mm2na is defined by

C;, = (ent:i(A)) B = a;3 B.

There is a second possibility for the tensor product. A ® B could be defined as the partitioned matrix D11

...

D1n2

D,n21

...

Dm2n 2

where each block D1 E Mm,n, (R) is defined by D13 = A(ent13 (B)) = Ab11.

We shall see in Section 7.2 that the two versions of the tensor product arise because of different possibilities of ordering standard bases on the tensor product of free modules of finite rank. Since there is no intrinsic difference between the two possibilities, we shall use the definition in Equation (1.3) as the definition of the tensor product of matrices. (1.17) Examples.

(1) Im ®In = Imn (2) I m ®B = ®;"_1B. (3)

ra c

bl d

_

®I2-

a 0 c

0

b

a

0

0 b

0

d 0'

0

c

0

d

The following result is an easy consequence of the partitioned multiplication formula (Proposition 1.14):

(1.18) Lemma. Let R be a commutative ring and let Al E Mm,,n, (R), A2 E Mn,,r1(R), B1 E Mms.n2(R), and B2 E Mn2,r,(R). Then


194

(A1 (9 B1) (A2 0 B2) = (A1A2) ® (B1B2)

Proof. By Proposition 1.14, Cij, the (i, j) block of (A1 ® B1)(A2 ® B2), is given by nl

(entik(A1)B1) (entki(A2)B2)

C1, _ k=1

= CY:(entik(A1)entki(A2)))

B1 B2,

k=1

which one recognizes as the (i, j) block of (A1A2) ® (B1B2).

0

(1.19) Corollary. Let R be a commutative ring, let A E M,n(R), and let B E Mn(R). Then A ®B = (A (9 In)(Im (9 B).

0

Proof.

(1.20) Lemma. If A E M,n(R) and B E Mn(R), then Tr(A ® B) = Tr'(A)Tt(B). 0

Proof. Exercise.

4.2 Determinants and Linear Equations Throughout this section, we will assume that R is a commutative ring.

(2.1) Definition. Let R be a commutative ring and let D : Mn(R) - R be a function. We say that D is n-linear (on rows) if the following two conditions are satisfied. (1) If B is obtained from A by multiplying a single row of A by a E R then

D(B) = aD(A). (2)

If A, B, C E Mn(R) are identical in all rows except for the ith row and

rowi(C) = rowi(A) + rowi(B), then

D(C) = D(A) + D(B).

4.2 Determinants and Linear Equations

195

Furthermore, we say that D is alternating if D(A) = 0 for any matrix A that has two rows equal. D : Mn(A) --+ R is said to be a determinant function if D is n-linear and alternating.

Note that property (2) does not say that D(A + B) = D(A) + D(B). This is definitely not true if n > 1. One may also speak of n-linearity on columns, but Proposition 2.9 will show that there is no generality gained in considering both types of n-linearity. Therefore, we shall concentrate on rows. (2.2) Examples.

(1) Let Dl and D2 be n-linear functions. Then for any choice of a and b in R, the function D : Mn(R) R defined by D(A) = aDi(A)+bD2(A) is also an n-linear function. That is, the set of n-linear functions on Mn(R) is closed under addition and scalar multiplication of functions, i.e., it is an R-module.

(2) Let o E S. be a permutation and define D, : Mn(R)

R by the

formula

D0(A) = al(l)a2(2) ... ano(n)

(3)

where A = [a,3). It is easy to check that D, is an n-linear function, but it is not a determinant function since it is not alternating. Let f : Sn -# R be any function and define Df : Mn(R) -+ R by the formula D f = EaES. f (a)D,. Applying this to a specific A = [a13] E Mn(R) gives

Df(A) =

aES

f (a)al o(,)a2a(2) .

ano(n).

By examples (1) and (2), Df is an n-linear function. (4) If n = 2 and c E R, then Dc(A) = c(a11a22 - a12a21) defines a deter-

minant function on M2(A). The first order of business is to prove that there is a determinant function for every n and for every commutative ring R and that this determinant function is essentially unique. More precisely, any determinant function is completely determined by its value on the identity matrix In. Note that for n = 1 this is clear since D([a]) = D(a[1]) = aD([1]) for every 1 x 1 matrix [a] by property (1) of n-linearity. The strategy for ing existence and essential uniqueness for determinant functions is to first a number

of basic properties that any determinant function must satisfy and then from these properties to derive a formula that must be used to define any determinant function. It will then only remain to check that this formula, in fact, defines a determinant function.

196


(2.3) Lemma. Let D : M,,(R) - R be n-linear. If rowi(A) = 0 for some i, then D(A) = 0. Proof. Since rowi(A) = 0 rowi(A), property (1) of n-linearity applies to show that D(A) = 0 D(A) = 0. 17 (2.4) Lemma. Let D : M,, (R) -a R be a determinant function. If i 96 j and Pij is the elementary permutation matrix determined by i and j, then

D(PijA) = -D(A) for all A E

(That is, interchanging two rows of a matrix multiplies

D(A) by -1.) Proof. Let Ak = rowk(A) for 1 < k < n, and let B be the matrix with rowk(B) = rowk(A) whenever k 0 i, j while rowi(B) = rowj(B) = Ai+Aj. Then since D is n-linear and alternating,

Ai+Aj

Aj

=D

0=D(B)=D

+D Ai+Aj

Aj + Aj

=D

+D

+D Aj

Ai

=D

Aj

Ai

Ai

Aj

+D

= D(A) + D(PijA) by Proposition 1.12. Thus, D(PijA) = -D(A), and the lemma is proved. 0

(2.5) Remark. Lemma 2.4 is the reason for giving the name "alternating"

to the property that D(A) = 0 for a matrix A that has two equal rows. Indeed, suppose D has the property given by Lemma 2.4, and let A be a matrix with rows i and j equal. Then PijA = A, so from the property of


197

Lemma 2.4 we conclude D(A) = -D(A), i.e., 2D(A) = 0. Thus, if R is a ring in which 2 is not a zero divisor, the two properties are equivalent, but in general the property of being alternating (as given in Definition 2.1) is stronger (Exercise 16).

(2.6) Lemma. Let D : Mn(R) - R be a determinant function. If i 34 j, a E R, and Tii (a) is an elementary transvection, then

D(Tii(a)A) = D(A). (That is, adding a multiple of one row of A to another row does not change the value of D(A).)

Proof. Let B be the matrix that agrees with A in all rows except row i, and assume that row; (B) = or rowi (A). Let A' be the matrix that agrees with A in all rows except row i and assume that rowi(A') = rowi(A). Then D is alternating so D(A') = 0 since rowi(A') = rowi(A) = rowi(A'), and thus, D(B) = aD(A') = 0 since D is n-linear. Since Tij(a)A agrees with A except in row i and rowi(Tii(a)A) = rowi(A) + a rowi(A) = rowi(A) + rowi(B) (see Proposition 1.12), it follows from property (2) of n-linearity that

D(Tii (a)A) = D(A) + D(B) = D(A). 0

Let E1 = row; (In) for 1 < i < n, and consider all n x n matrices formed by using the matrices E; as rows. To develop a convenient notation, let Stn denote the set of all functions w : {1, 2, ..., n} {1, 2, ..., n},

and if w E Stn, let P,,, denote the n x n matrix with rowi(P,) = E,(;). For example, if n = 3 and w(1) = 2, w(2) = 1, and w(3) = 2, then P,,, =

0

1

0

1

0

0

0

1

0

If W E fln is bijective so that w E Sn, then P,, is called a permutation matrix. If w = (i j) is a transposition, then P,, = Pii is an elementary permutation matrix as defined in Section 4.1. In general, observe that the product ,,,A is the matrix defined by rowi(P,A) = row,,(,)(A).

According to Proposition 1.12 (3), if w E Sn is written as a product of transpositions, say w = (i1 jl)(i2 j2) ... (it it), then PW

= Pilii Pisi2 ... Pi9i, In = InPiiit Pi2iz ... Piti,

198


Since w-1 = (it jt)(it-i jt-i)

(ii j1), the second equality, together with

Proposition 1.13 (3), shows that

coli(P.) = E,.,-,(i) Therefore, again using Proposition 1.13 (3), we see that AP,, is the matrix defined by coli(AP,,) = col,,-,(i)(A). To summarize, left multiplication by P,,, permutes the rows of A, following the permutation w, while right multiplication by P,,, permutes the columns of A, following the permutation w-1.

Recall that the sign of a permutation a, denoted sgn(o), is +1 if a is a product of an even number of transpositions and -1 if a is a product of an odd number of transpositions. (2.7) Lemma. Let R be a commutative ring and let D : MM(R) --. R be a determinant function. If w E Stn, then (1) D(P,,,) = 0 if w V Sn, and (2)

D(P,,,) = sgn(w)D(In) if w E Sn.

Proof. (1) If w V Sn then w(i) = w(j) for some i $ j so that P, has two rows that are equal. Thus, D(P,,) = 0. (2) If w E Sn, write w = (i1 ii) to get (by Proposition 1.12)

(it it) as a product of transpositions

Pw=Pi,i,...piti, By Lemma 2.4, we conclude that

(-1)tD(In) = sgn(w)D(In), 0

and the lemma is proved.

Now let A = [aii] E Aln(R) and partition A by its rows, i.e., Al A

An

where Ai = rowi(A) for 1 aijEj, j=1


199

and thus, Fn En,=1 aij1Ej1 E uj2=1 a2j2Ej3

A =

`n `', 1 anjn Ejn If D is a determinant function, we may compute D(A) using (2.1), the n-linearity of D, and Lemma 2.7 as follows:

En En =1 a1j1 Ej1 E a2j3Ejs D(A) = D uj2 [n

Lj

anjn Ejn

Ej2=1Ej1

n

a2j2Ejs

alj, D j1=1

ujn n

anjnEjn

Ej,

n j2

a1j1a2j3D 11=1 j,=1

En

nc 1 anjnEjn

n

n

= F ... E a1j1 a2jz ... anjn D ji=1

Ej1 Ej2

jn=1 Ejn

1: a1 (1)a2,a(2)...anw(n)D(PW) LaE12n

(2.2)

=

L:

sgn (w)a1,,(1) ...

l an,,()] D(In)

Thus, we have arrived at the uniqueness part of the following result since formula (2.2) is the formula which must be used to define any determinant function.

(2.8) Theorem. Let R be a commutative ring and let a E R. Then there is exactly one determinant function D. : Mn(R) --- R such that Da(In) = a. Thus D. = aD1, and we let det = D1.

Proof. If there is a determinant function D then according to Equation (2.2), Da must be given by


200

(2.3)

D0(A) = a E sgn(o )a1 a(1) ...ano(n)

I

J

oES

It remains to check that Equation (2.3) in fact defines a determinant func-

tion on Mn(R). It is sufficient to check this with a = 1 since a scalar multiple of a determinant function is also a determinant function. If we let

f (a) = sgn(a) 1, then Dl = Df as defined in Example 2.2 (3), and as observed in that example, Df is an n-linear function for each f : Sn -+ R. In particular, Dl is n-linear and it remains to check that it is alternating. To this, suppose that A E Mn(R) has row, (A) = rowj(A) with i 96 j.

Then for 1 < k < n we have aik = ajk. If a E Sn, let a' = a o (i j). We claim that (2.4)

al (1) "' ano(n) = a, ,(1) "' ano'(n)-

This is because a(k) = a'(k) if k 54 i, j so that ako(k) = ako'(k), while a'(i) = a(j) and a'(j) = a(i) so that a,,,(,) = a,,(j) = aj,(j) and aj,'(j) = aj(,) = a,,(;). Hence Equation (2.4) is valid. But sgn(a') = - sgn(a), so sgn(a)al ,(,) ... ano(n) + sgn(a')al,,(,) ... an al(n) = 0.

But a

a' gives a pairing of the even and odd permutations in Sn, and

hence, we conclude

sgn(a)al 0(1) ... an v(n)

D1(A) = oES

=

(sgn(a)al0(1)... ano(n) + sgn(a')al,,(1) ... ano'(n))

egn(o)=1

=0. Therefore, Dl is a determinant function on Mn(R) and Theorem 2.8 is proved.

(2.9) Corollary. Let A = [aij] E Mn(R) be an upper (resp., lower) triangular

matrix, i.e., a,j = 0 for i > j (reap., i < j). Then n

det(A) = all ... ann = flail. i=1

Proof. If a E Sn is not the identity permutation, then for some i and j, a(i) > i and a(j) <j, so in either case, a,,(1)...an0(n) = 0,

and the result follows by Equation (2.3).


201

Our determinant functions have been biased towards functions on the rows of matrices. But, in fact, we can equally well consider functions on the columns of matrices. Perhaps the simplest way to see this is via the following result:

(2.10) Proposition. If A E Mn(R) then det(A) = det(At). Proof. Let A = [aid). Then

det(A) _ > sgn(a)a1 ,(1) ... ano(n) oES., sgn(a)ao-1(,(1))c(1)...ao-1(C(R))C\f1)

oES,

_

1:

sgn(a)ao_,(1) 1 ...ao-1(n)n

oES.,

E Sgn(a-1)ao-1(1) I ... so-1(n) n cES sgn(r)a7(1) 1

... a1.(n) n

*ES.,

= det(Al).

Here we have used the fact that sgn(a-1) = sgn(a) and that av-1(o(1)) 0(1) "' ao-1(o(n)) o(n) = ao-,(1) 1 ... ao_,(n) n,

This last equation is valid because R is commutative and {a(1), ..., a(n)} 0 is just a reordering of {1, ..., n} for any a E Sn. (2.11) Theorem. If A, B E Mn(R) then

det(AB) = det(A) det(B). Proof. Define DB : M,, (R) -. R by DB(A) = det(AB). Since rowi(AB) = [rowi(A)]B, it is easy to check (do it) that DB is n-linear and alternating. Thus DB is a determinant function, so by Theorem 2.8, DB(A) = adet(A) where a = DB(In) = det(B). 0

This result is an example of the payoff from the abstract approach to determinants. To prove Theorem 2.11 directly from the definition of determinant as given by Equation (2.3) is a somewhat messy calculation. However, the direct calculation from Equation (2.3) is still beneficial in that a more general product formula is valid. This approach will be pursued in Theorem 2.34.

(2.12) Corollary. If R is a commutative ring, A E Mn(R), and S E GL(n, R), then

202


det(S-'AS) = det(A). That is, similar matrices have the same determinant. Proof. Exercise.

0

Similarly to the proof of Theorem 2.11, one can obtain the formula for the determinant of a direct sum of two matrices.

(2.13) Theorem. If R is a commutative ring, A; E Mn; (R) for 1 < i < r, and A = ® 1A;, then r

det(A) = rl det(Ai). i=1

Proof. It is clearly sufficient to prove the result for r = 2. Thus suppose that A = A1® A2. We may write A = (A1®I,,,) (In, ® A2). Then Theorem 2.11 gives

det(A) = (det(A1 ® I,,,)) (det(In, ® A2))

.

Therefore, if det(A1 ® In,) = det(A1) and det(In, ® A2) = det(A2), then we are finished. We shall show the first equality; the second is identical. Define D1 : Mn1(R) -+ R by D1(B) = det(B (D In3). Since det is (n1 + n2)-linear and alternating on Mn1+n2(R), it follows that D1 is n1linear and alternating on M,,,(R). Hence, (by Theorem 2.8), D1(A1) = adet(A1) for all Al E Mn1(R), where a = D1(In,) = det(In, (D In3) = 1, and the theorem is proved. 0

Remark. Theorem 2.13 and Corollary 2.9 can both be generalized to a formula for block triangular matrices. (See Exercise 13.) There is also a simple formula for the determinant of the tensor product of two square matrices.

(2.14) Proposition. Let R be a commutative ring, let A E Mm(R), and let B E Mn(R). Then det(A (9 B) = (det(A))n (det(B))m.

Proof. By Corollary 1.19, A ® B = (A (9 In)(I, ® B). By Example 1.17 (2), Im 0 B = ®i'_' 1B, so det(I,n ® B) = det(B)m by Theorem 2.13. We leave it as an exercise to that the rows and columns of the matrix A ® In can be permuted to obtain that A ® In is (permutation) similar to ® 1A (Exercise 50). The proof is then completed by another application of Theorem 2.13.

0


203

We will now consider cofactor expansions of determinants. If A E A,(R) and 1 < i, j < n, let Aid be the (n - 1) x (n - 1) matrix obtained by deleting the i1h row and jth column of A. (2.15) Theorem. (Laplace expansion) Let R be a commutative ring and let A E Mn(R). Then n

(1)

1:(-1)k+'aki det(Ak,) = bid det(A) k=1

and n

(2)

>(_1)k+iaik det(A,k) = bij det(A). k=1

Proof. If A E Mn(R) and 1 < i, j < n, let n

Dii(A) = 1: (-1)k+'akidet(Aki) k=1

That is, Di,(A) E R is defined by the left-hand side of equation (1). We claim that the function Did : Mn(R) -. R is a determinant function for all i, j. Note that the function A '-. akidet(Ak,) is n-linear on Mn(R). Since a linear combination of n-linear functions is n-linear, it follows that Did is n-linear. It remains to check that it is alternating. To see this, suppose that A E Mn(R) is a matrix with rows p and q equal. If k 96 p, q, then Ak; has two rows equal, so det(Ak,) = 0 in this case. Thus, (2.5)

Di,(A) = (-1)p+"a, det(Apj) + (-1)q+3agidet(Agi)

Note that the assumption that rowp(A) = rowq(A) means that api = aqi. To be explicit in our calculation, suppose that p < q. Then the matrix Aqj is obtained from App by moving row q - 1 of Aye to row p and row t of App

to row t + l for t = p, ... , q-2. In other words, Aqi

= PAp;

where w E Sn_1 is defined by w(t) = t fort q, while w(q-1) = p

and w(t) = t + 1 for t = p, ... , q - 2. That is, w is the (q - p)-cycle (p,p+1,...,q-1).Then byLemma 2.7 det(Aq,) = det(PA) = sgn(w) det(Apj)

(-1)9_p-1

det(Ap,).

Equation (2.5) then shows that Dij (A) = 0, and hence Dij is alternating. Therefore, Did is a determinant function, so by Theorem 2.8 (2.6)

Dii(A) = Dii(I.)det(A)

204


for all A E M,a(R). A direct calculation shows that Di1(In) = bid, so that Equation (2.6) yields (1) of the theorem. Formula (2) (cofactor expansion along row j) is obtained by applying formula (1) to the matrix At and using the fact (Proposition 2.10) that det(A) = det(A'). 0 (2.16) Definition. If A E MM(R), then we define the cofactor matrix of A, denoted Cofac(A), by the formula

enti,(Cofac(A)) _

(-1)i+.i det(Aij),

and we define the adt of A, denoted Adj(A), by Adj(A) = (Cofac(A))t.

The following result should be familiar for matrices with entries in a field, and the proof is the same: (2.17) Theorem. Let A E Mn(R).

(1) A(Adj(A)) = (Adj(A))A = det(A)In. (2) A is invertible if and only if det(A) is a unit in R, and in this case,

A-' = (det(A))-' Adj(A). Proof. Formula (1) follows immediately from Theorem 2.15. If A E Mn(R)

is invertible, then there is a matrix B such that AB = BA = In. Thus (det(A))(det(B)) = 1 so that det(A) E R'. Conversely, if det(A) is a unit, then B = (det(A))-' Adj(A) satisfies AB = BA = In by formula (1). 0 (2.18) Remark. The definition of inverse of a matrix requires that AB = BA = I,,, but as a consequence of Theorem 2.17, we can conclude that a matrix A E Mn(R) is invertible provided that there is a matrix B E MM(R)

such that AB = In. Indeed, if AB = In then (det(A))(det(B)) = 1, so det(A) is a unit in R, and hence, A is invertible. (2.19) Examples.

(1) A matrix A E Mn(Z) is invertible if and only if det(A) = ±1.

If F is a field, a matrix A E Mn(F[X]) is invertible if and only if det(A) E F' = F \ {0}. (3) If F is a field and A E Mn(F[X]), then for each a E F, evaluation of each entry of A at a E F gives a matrix A(a) E M,,(F). If det(A) = f (X) 34 0 E FIX], then A(a) is invertible whenever f (a) 0 0. Thus, (2)

A(a) is invertible for all but finitely many a E F.


205

As an application of our results on determinants, we shall present a determinantal criterion for the solvability of homogeneous linear equations

when the coefficients can be from an arbitrary commutative ring. This criterion will involve some ideals of R generated by various determinants of submatrices of matrices over R. We will start by defining these ideals.

To state the results in a reasonably compact form it is convenient to introduce some appropriate notation. If p, m E N with p _< m, let Qp,,,m denote the set of all sequences a = (i1 i ..., ip) of p integers with < ip < m. Note that the cardinality of the set Qp,m 1 < i1 < i2 < is IQp,ml = (p). Suppose A E M,,,,,,(R). If a E Qp,m and ,0 E Qj,n, let A[a I 0) denote the submatrix of A consisting of the elements whose row

index is in a and whose column index is in Q. If a E Qp,m then there is a complementary sequence a E Qp_m,m consisting of the integers in {1, 2,. .. , m} not included in a and listed in increasing order. To give some examples of these notations, let A =

all

a12

a13

a14

all

a22

a23

a24

a31

a32

a33

a34

If a = (1, 3) E Q2,3 and 0 = (2, 3) E Q2,4, then A[a 18]

a12 a32

a13

a33]

while a = 2 E Q1,3 and (3 = (1, 4) E Q2,4 so that A[& I)3j = [ a21

a24 } .

(2.20) Definition. If R is a commutative ring and A E Mm,n(R), then a t x t minor of A is the determinant of any submatrix A[a 1,0] where a E Qt,m, Q E Qt,,. The determinantal rank of A, denoted D-rank(A), is the largest t such that there is a nonzero t x t minor of A.

(2.21) Definition. If R is any commutative ring, A E Mm,n(R), and 1 < t < min{m, n}, let Ft (A) = ({det A[a I

/31

: a E Qt,m, Q E Qt,n}) C R.

That is, Ft (A) is the ideal of R generated by all the t x t minors of A. We set Fo(A) = R and Ft (A) = 0 if t > min {m, n}. Ft (A) is called the tth_

Fitting ideal of A. The Laplace expansion of determinants (Theorem 2.15) shows that Ft+l(A) C Ft (A). Thus there is a decreasing chain of ideals

R=Fo(A)2F1(A)_JF2(A)D .... If R is a PID, then Ft(A) is a principal ideal, say Ft (A) = (dt(A)) where dt(A) is the greatest common divisor of all the t x t minors of A. In this case, a generator of Ft (A) is called the tth-determinantal divisor of A.

206


(2.22) Definition. If A E M,n,n(R), then the M-rank(A) is defined to be the largest t such that Ann(Ft(A)) = 0. (2.23) Remarks.

(1) M-rank(A) = 0 means that Ann(F1(A)) 96 0. That is, there is a nonzero a E R with aa,j = 0 for all aid. Note that this is stronger than saying that every element of A is a zero divisor. For example, if A = [ 2 3 ] E M1,2(Z6), then every element of A is a zero divisor in Z6 but there is

no single nonzero element of Z6 that annihilates both entries in the matrix. (2) If A E then M-rank(A) = n means that det(A) is not a zero divisor of R.

(3) To say that M-rank(A) = t means that there is an a # 0 E R with a D = 0 for all (t + 1) x (t + 1) minors D of A, but there is no nonzero

b E R which annihilates all t x t minors of A by multiplication. In particular, if det Ala 101 is not a zero divisor of R for some a E Q,,,n, (3 E Q,,,,, then M-rank(A) > s. (2.24) Lemma. If A E Mm,n(R), then

0 < M-rank(A) < D-rank(A) < min{m, n}.

0

Proof. Exercise.

We can now give a criterion for solvability of the homogeneous linear equation AX = 0, where A E M,,,,n(R). This equation always has the trivial solution X = 0, so we are looking for solutions X # 0 E Mn,1(R).

(2.25) Theorem. Let R be a commutative ring and let A E Mm,n(R). The matrix equation AX = 0 has a nontrivial solution X 0 0 E Mn,1(R) if and only if M-rank(A) < n.

0, so choose b 0 0 E R with b Ft+1(A) = 0. Without loss of generality, we may assume that t < m since, if necessary, we may replace the system AX = 0 by an equivalent one (i.e., one with the same solutions) by adding some rows of zeroes to the bottom of A. If t = 0 then ba,, = 0 for all aid and we

Proof. Suppose that M-rank(A) = t < n. Then Ann(Ft+1(A))

b

. Then X -A 0 E M,1(R) and AX = 0.

may take X = b

Thus, suppose t > 0. Then b Ann(Ft(A)) = 0, so bdet Ala 1010 0 for some a E Qt.+,,,13 E Qt,,, By permuting rows and columns, which does not affect whether AX = 0 has a nontrivial solution, we can assume a =


207

F o r 1 < i < t + 1 let E Qt,t+l where i indicates that i is deleted. Let di = (-1)t+1+' det A[a 1,8j). Thus dl, ... dt+l are the cofactors of the matrix

Al = A[(1,...,t + 1) [ (1, ... ,t +

1)]

obtained by deleting row t + 1 and column i. Hence the Laplace expansion theorem (Theorem 2.15) gives 1t+i j=l aijdj = 0

1 < i < t,

(2.7)

(1,...,t,t+1)J t
Let X =

where xn

5xi=bdi if1
bEt+1 aljdj

AX= e+i bamidi

0 0

bdetA[(1,...,t,t+ 1) 1 (1,...,t,t+1)] L

bdetA((1,...,t,m) (1,...,t,t+1)]

0.

Thus, X is a nontrivial solution to the equation AX = 0. Conversely, assume that X 96 0 E Mn,1(R) is a nontrivial solution to AX = 0, and choose k with xk 96 0. We claim that Ann(Fn(A)) 0 0. If

n > m, then Fn(A) = 0, and hence, Ann(Fn(A)) = R 96 (0). Thus, we may assume that n < m. Let a = (1, ... , n) and for each 3 E Qn,m, let B0 = A[a 1 ,0). Then, since AX = 0 and since each row of B0 is a full row

of A, we conclude that B0X = 0. The adt matrix formula (Theorem 2.17) then shows that (det Bp)X = (Adj B0)BOX = 0,

208


from which we conclude that xk det Bp = 0. Since 6 E Q, ,n, is arbitrary, we conclude that xk - F,,(A) = 0, i.e., xk E But xk 96 0 so Ann(F,,(A)) 0 0, and we conclude that M-rank(A) < n and the proof is

0

complete.

In case R is an integral domain we may replace the M-rank by the ordinary determinantal rank to conclude the following: (2.26) Corollary. If R is an integral domain and A E Mm,n(R), then AX = 0 has a nontrivial solution if and only if D-rank(A) < n.

Proof. If I C R, then Ann(I) 0 0 if and only if I = 0 since an integral domain has no nonzero zero divisors. Therefore, in an integral domain, D-rank(A) = M-rank(A). 0 The results for n equations in n unknowns are even simpler: (2.27) Corollary. Let R be a commutative ring. (1) If A E M,,(R), then AX = 0 has a nontrivial solution if and only if

det A is a zero divisor of R. (2)

If R is an integral domain and A E M,, (R), then AX = 0 has a nontrivial solution if and only if det A = 0.

Proof. If A E Mn(R) then Fn(A) = (det A), so M-rank(A) < n if and only if det A is a zero divisor. In particular, if R is an integral domain then 0 M-rank(A) < n if and only if det A = 0.

There are still two other concepts of rank which can be defined for matrices with entries in a commutative ring. (2.28) Definition. Let R be a commutative ring and let A E Mm,n(R). Then we will define the row rank of A, denoted row-rank(A), to be the maximum number of linearly independent rows in A, while the column rank of A, denoted col-rank(A) is the maximum number of linearly independent columns. (2.29) Corollary. (1) If R is a commutative ring and A E Mm,,n(R), then

max{row-rank(A), col-rank(A)} < M-rank(A) < D-rank(A). (2) If R is an integral domain, then

row-rank(A) = col-rank(A) = M-rank(A) = D-rank(A). Proof. Exercise.

0


209

(2.30) Remarks. (1) If R is an integral domain, then all four ranks of a matrix A E M,,,,. (R)

are equal, and we may speak unambiguously of the rank of A. This will be denoted by rank(A).

(2) The condition that R be an integral domain in Corollary 2.29 (2) is necessary. As an example of a matrix that has all four ranks different, consider A E M4(Z210) defined by 0 2

A=

3 6 3 0

2

0 0 0

3

0

5 0 0 7

We leave it as an exercise to check that row-rank(A) = 1 col-rank(A) = 2 M-rank(A) = 3 D-rank(A) = 4.

As a simple application of solvability of homogeneous equations, we note the following result: (2.31) Proposition. Let R be a commutative ring, let M be a finitely generated R-module, and let S C M be a subset. If ISI > ji(M) = rank(M), then S is not R-linearly independent.

Proof. Let µ(M) = m and let T = {wl, ... , w,n } be a generating set for M consisting of m elements. Choose n distinct elements {vj, ... , v,} of S for some n > m, which is possible by our hypothesis ISI > µ(M) = m. Since

M=(w1,...,wm),wemaywrite vj = al jwl + ... + amjwm

with aij E R. Let A = faij] E Mm,n(R). Since n > m, it follows that M-rank(A) < m < n, so Theorem 2.25 shows that there is an X 36 0 E Mn,1(R) such that AX = 0. Then (2.8)

Exivo = Exo j=1

j=1

m

_ i=1

(atiwi) i=1

r n

(a*xi)

Wi

j=1=1

=0 since AX = 0. Therefore, S is R-linearly dependent.

0

210


(2.32) Corollary. Let R be a commutative ring, let M be an R-module, and let N C M be a free submodule. Then rank(N) < rank(M). Proof. If rank(M) = oo, there is nothing to prove, so assume that rank(M) _ m < oo. If rank(N) > m, then there is a linearly independent subset of M, namely, a basis of N, with more than m elements, which contradicts Propo-

0

sition 2.31.

(2.33) Remark. Corollary 2.32 should be compared with Theorem 3.6.2, concerning submodules of free modules over a PID. Also, note that the condition that N he free is necessary (see Remark 3.6.5). The following result, known as the Cauchy-Binet theorem, generalizes the formula for the determinant of a product (Theorem 2.11) to allow for products of possibly nonsquare matrices. One use of this formula is to investigate the behavior of the rank of matrices under products. (2.34) Theorem. (Cauchy-Binet formula) Let R be a commutative ring and

let A E Mm,n(R), B E Mn,p(R). Assume that 1 < t < min{m, n, p} and let a E Qt,m, Q E Qt,p. Then

det(AB[a) /3)) _ E det(A[a [ d)det(B[7 [ p1). 'YEQt.n

Proof. Suppose that a = (a1, ..., at), 13 = (Qt, ..., 3t) and let C = AB[a 10). Thus, entij (C) = row., (A) colp, (B) n

at:, kbko, k=1

so that Ek=1 a.,kbk3,

... >k.1 aQIkbkI,

C=

.

Ek=1

a.,k40,

aa,kbkO,

Using n-linearity of the determinant function, we conclude that n

(2.9)

detC= E k,=1

bk, Q,

n

Ea.,k, ke_1

bk, Ot

a., k, det bkt A,

...

bkt At

If k; = kj for i 34 j, then the ith and jth rows of the matrix on the right are equal so the determinant is 0. Thus the only possible nonzero determinants on the right occur if the sequence (k1, ... , kt) is a permutation of a sequence


211

-It) E Qt,,,. Let or E St be the permutation of {1, ..., t} such y = (yi, that y, = k,(,) for 1 < i < t. Then bkl 01

bkl 01

det

(2.10)

= sgn(a) det B[y 1,31. bk, a,

bk, Q,

Given y E Qt,,,, all possible permutations of -y are included in the summation in Equation (2.9). Therefore, Equation (2.9) may be rewritten, using Equation (2.10), as

)detBIV

detC = 'YEQe ,..

I p]

eES

J

E det A[a I y] det B[y 10], 7EQ*,+.

which is the desired formula.

11

(2.35) Examples.

(1) The Cauchy-Binet formula gives another verification of the fact that det(AB) = det A det B for square matrices A and B. In fact, the only element of Q,,,,, is the sequence -y = (1, 2, ... ,n) and A[y I y] = A, B[y yJ = B, and AB[y I y] = AB, so the product formula for determinants follows immediately from the Cauchy-Binet formula. (2) As a consequence of the Cauchy-Binet theorem, if A E Mm,n(R), and B E Mn,p(R) then I

(2.11)

D-rank(AB) < min{D-rank(A), D-rank(B)}.

To see this, let t > min{D-rank(A), D-rank(B)} and suppose that a E Qt,,n, Q E Qt,p. Then by the Cauchy-Binet formula

det(AB[a 1,31) = E det(A[a I y]) det(B[y 10]). -fEQt...

Since t > min{D-rank(A), D-rank(B)}, at least one of the determinants det A[a I -y] or det B[y I a] must be 0 for each y E Qt,,,. Thus, det(AB[a 10]) = 0, and since a and 3 are arbitrary, it follows that D-rank(AB) < t, as required. Equation (2.11) easily gives the following result, which shows that determinantal rank is not changed by multiplication by nonsingular matrices.

(2.36) Proposition. Let A E M..n(R) and let U E GL(m, R), V E GL(n, R). Then

212


D-rank(UAV) = D-rank(A).

Proof Any matrix B E Mm,n(R) satisfies D-rank(B) < min{m, n} and since D-rank(U) = n and D-rank(V) = m, it follows from Equation (2.11) that D-rank(UAV) < min{D-rank(A), n, m} = D-rank(A) and

D-rank(A) = D-rank(U-1(UAV)V-1) < D-rank(UAV).

0

This proves the result.

We will conclude this section by giving a version of the Laplace expansion theorem that allows for expansion of det A along a given set of rows rather than a single row. The choice of rows along which expansion takes place is given by an element a E Qt,,. Recall that if y E Qt,n then 7 E Q,,-t,,, denotes the complementary sequence. With these preliminaries out of the way, the general Laplace expansion theorem can be stated as follows:

(2.37) Theorem. (Laplace expansion) Let A E Mn(R) and let a E Qt,n (1 t < n) be given. Then (2.12)

det A =

(-1)e(o)+a(ti> det(A[a I -y]) det(A(a

where

ry])

t

s(-Y) = E yj j=1

for any -y = (-Y1, ... ,'Yt) E

Proof. The proof is essentially similar to the proof of Theorem 2.15 (which is a special case of the current theorem). If A E M,,(R), define (2.13)

D,,(A) = E (_i)88 (,) det(A(a I y])det(A[a 17]) 7EQt..

Then D,, : Mn(R) -- R is easily shown to be n-linear as a function on the columns of A E Mn(R). To complete the proof, it is only necessary to show that D,, is alternating and that D ,(In) = 1. Thus, suppose that colp(A) = colq(A), and to be specific, assume that p < q. If p and q are both in y E Qt,n, then A(a I yJ will have two columns equal so that det A [a I y] = 0,

while, if both p and q are in 3 E Qn_t,n, then det A[& 17J = 0. Thus, in the evaluation of D,,(A) it is only necessary to consider those y E Qt,n such that p E y and q E ^y, or vice-versa. Thus, suppose that p E y, q E ry and


213

define a new sequence 7' E Qt,,, by replacing p E 7 by q. Then ry' agrees with ry except that q has been replaced by p. Thus (2.14)

s(7') - s(7) = q - p

Now consider the sum

(-1)°(7) det(A[a 17]) det(A[a 17]) + (-1)"('Y') det(A[a 17']) det(A[a

which we denote by S(A). We claim that this sum is 0. Assuming this, since -y and 7' appear in pairs in Qt,,,, it follows that DQ(A) = 0 whenever two columns of A agree; thus DQ is alternating. It remains to check that S(A) = 0.

Suppose that p = 7k and q = it. Then 7 and 7' agree except in the range from p to q, as dory and ry'. This includes a total of q - p + 1 entries. If r of these entries are included in 7, then 71 < ... < 7k = p < 7k+1 < ... < 7k+,-1 < q < 7k+r < ... < 7t

and

A[aI7')=A[aI7]P,., where w is the r-cycle (k + r - 1, k + r - 2, ... , k). Similarly,

A[&I7J=A[aI7JPW' where w' is a (q - p + 1 - r)-cycle. Thus, det(A[a 17']) det(A[a 17 J)

_

(-1)°(1')+(r-1)+(q-D)-r

det(A[a 171) det(A[a I

Yl)

Since s(7') + (q - p) - 1 - s(7) = 2(q - p) - 1 is odd, we conclude that S(A) = 0. Thus DQ is alternating, and since it is straightforward to check that 1, the result is proved. 0 Applying formula (2.12) to At in place of A, gives the Laplace expansion in columns:

(2.15)

det A

=E

det(A[7 I aJ) det(A[ry I &]).

7EQt.

Note that if t = 1 then Ql,,, = {1, 2, ..., n} so that A[i I j] = a+i while Al? I j = A13, so Theorem 2.37 includes Theorem 2.15 as a special case.

214


4.3 Matrix Representation of Homomorphisms Before beginning with the procedure of associating a matrix with a homomorphism between free modules, we would like to make some remarks about the ticklish situation that arises for a noncommutative ring. We will only need this once, at the very end of Section 7.1, so the reader who is only interested in the commutative case may (and is well advised to) skip the more general situation. Difficulties already arise in the simplest case. Let R be a ring and let us consider a free R-module M of rank I with basis 5 = {v}. Then M = {rv : r E R}. We wish to give "coordinates" to the elements of M, i.e., identify the elements of M with the elements of R, and clearly, there is only one reasonable choice here, namely, that, in the basis 8, rv should have coordinate 1rv[8 = [r[. Now consider f E EndR(M). We wish to represent

f by a matrix with respect to the basis 5, and again there is only one reasonable choice: if f (v) = sv, then f should have coordinate matrix in the basis 5 given by [f]13 = [s[. (Note that f is not "left-multiplication by s" unless s E C(R), the center of R. Indeed, g : M M defined by g(m) = sm is not an R-endomorphism unless s E C(R), as then g(rm) = srm # rsmn = rg(m) in general. Of course, there is no problem if R is commutative.) Now, the theory we are about to develop will tell us that for any m E M, we may calculate f (m) by [f(m)1S = If1B[rn]a.

However, when we try to apply this to ni = rv, we get f (m) = f (rv) _ rf(v) = r(sv) = (rs)v, so [f(m)15 = [rs) while [f]a[m]a = [s][r] = [sr]. If R is commutative, these are equal, but in general they are not. On the other hand, this formulation of the problem points the way to its solution. Namely, recall that we have the ring R°P (Remark 3.1.2 (3)) whose elements are the elements of R, whose addition is the same as that of R, but whose multiplication is given by r s = sr, where on the right-hand side we have the multiplication of R. Then, indeed, the equation [rs1 = Is] [r]

is valid, and we may hope that this modification solves our problem. This hope is satisfied, and this is indeed the way to approach coordinatization of R-module homomorphisms when R is not commutative. Now we come to a slight notational point. We could maintain the above notation for multiplication in RP throughout. This has two disadvantages: the practical-that we would often be inserting the symbol which is

easy to overlook, and the theoretical-that it makes the ring RIP look special (i.e., that for any "ordinary" ring we write multiplcation simply by juxtaposing elements, whereas in RIP we do not), whereas R°P is a perfectly

4.3 Matrix Representation of Homomorphisms

215

good ring, neither better nor worse than R itself. Thus, we adopt a second solution. Let op : R -, RP be the function that is the identity ,n elements, i.e., op(t) = t for every t E R. Then we have op(sr) = op(r) op(s), where the multiplication on the right-hand side, written as usual as juxtaposition, is the multiplication in R°P. This notation also has the advantage of reminding us that t E R, but op(t) E R°P. Note that if R is commutative then RIP = R and op is the identity, which in this case is a ring homomorphism. In fact, op : R -+ RP is a ring homomorphism if and only if R is commutative. In most applications of matrices, it is the case of commutative rings that is of primary importance. If you are just interested in the commutative case (as you may well be), we advise you to simply mentally (not physically) erase "op" whenever it appears, and you will have formulas that are perfectly legitimate for a commutative ring R. After this rather long introduction, we will now proceed to the formal mathematics of associating matrices with homomorphisms between free modules.

If R is a ring, M is a free R-module ofrank n,andB={vl,...,vn} is a basis of M, then we may write v E M as v = al vi +

+ anvn for unique a1, ... , an E R. This leads us to the definition of coordinates. Define

io:M -+ Mn,1(R°P)by op(al) op(an)

The n x 1 matrix [v]8 is called the coordinate matrix of v with respect to the basis B.

Suppose that B' = {vi, ... ,v;,} is another basis of M and define the matrix Ps,, E Mn(R°P) by the formula (3.1)

pt = [[v1]8'

[v2]8,

...

[vn]8']-

That is, cold (Pg,) = [v3]13-. The matrix P8 is called the change of basis

matrix from the basis B to the basis B'. Since vj = E 1 b,jv;, it follows that if B = 8' then PB = In. (3.1) Proposition. Let M be a free R-module of rank n, and let 8, 8', and B" be bases of M. Then

(1) for any v E M, [via' = Pt IV] B; (2) PB , = PB, PB; and (3) PB is invertible and (PB) -' = PB '. Proof. (1) Note that Mn,l(R°P) is an R-module where the operation of R on Mn,l (R°P) is given by r A = A op(r) for r E R, A E Mn,l (R°P). Then

216


defined by ,b'(v) = [v]B,, is an R-module homomorphism, as is

0": M - MM,1(R°P),

defined by /"(v) = PB[v]B. To show that u" = 1/i' we need only show that they agree on elements of the basis B. If B = {vj, ... , then (vi]B = ei = Ei1 E M.,1(R) since vi = F,' 1 b;jv,. Thus, ti"(vi) = PB,ei = coli(PB,) =

='+G'(vi)

as required.

(2) For any v E M we have

(PB,PB,) [v]B = PX (PB'[v]B)

= PB %[v]B' = [v]B = PB [v]B. Therefore, PB;, PB, = P88,,.

(3) Take B" = B in part (2). Then

PB PB = PB = I,,. Thus, PB, is invertible and (PE) -1 =

PB,.

(3.2) Lemma. Let R be a ring and let M be a free R-module. If B' is any basis of M with n elements and P E GL(n, R°P) is any invertible matrix, then there is a basis B of M such that

P=Pg. Proof. Let B' = {vi, ... , vn} and suppose that P = (op(pii )]. Let v? _ 1 pijv;. Then B = {v1, ... , is easily checked to be a basis of M, and by construction, Pt = P.

-,

Remark. Note that the choice of notation for the change of basis matrix P8 has been chosen so that the formula in Proposition 3.1 (2) is easy to . That is, a superscript and an adjacent (on the right) subscript that are equal cancel, as in

pt pt = PB,, . The same mnemonic device will be found to be useful in keeping track of superscripts and subscripts in Propositions 3.5 and 3.6.


217

(3.3) Definition. Let M and N be finite rank free R-modules with bases

B = {vi, ... , v,n } and C = {w1, ... , w, ,J respectively. Fo, each f E HomR(M, N) define the matrix of f with respect to B, C, denoted [f]C, by

(3.2)

colj [ f ]C = [f (vj )]c

Thus, if f (vj)

for 1 < j:5 m.

aijwi for 1 < j < m, then [f]B = [op(aij)] E Mn,m(R°P)

If f E EndR(M) and B is a basis of M, then we will write [f]B in place of [f]8. (3.4) Remarks.

(1) Note that every matrix A E Mn,,n (RP) is the matrix [f) B for a unique f E HomR(M, N). Indeed, if B = {vi, ... , v,n} and C = {w1, ... , wn} are bases of M and N respectively, and if A = [op(aij )], then define f E

HomR(M, N) by f (vj) = E 1 aijwi. Such an f exists and is unique by Proposition 3.4.9; it is clear from the construction that A = [f]. Thus the mapping f '-+ [f] B gives a bijection between HomR(M, N) and Mn,,n(R°P). (2) Suppose that R is a commutative ring. Then we already know (see Theorem 3.4.11) that HomR(M, N) is a free R-module of rank mn, as is the R-module Mn,,n(R); hence they are isomorphic as R-modules. A choice of basis B for M and C for N provides an explicit isomorphism

HomR(M, N) - Mn,m(R),

defined by 4 (f) = (f). We leave it as an exercise for the reader to check that 4 ( f , ) = Eji where {E1}11 is the standard basis of Mn,m(R), while {fij}in1 is the basis of HomR(M, N) constructed 1

in the proof of Theorem 3.4.11.

Note that if 1M : M -+ M denotes the identity transformation and B, B' are two bases of M, then B

B

[lor]e- = PB,

so that the change of basis matrix from the basis B to the basis B' is just the matrix of the identity homomorphism with respect to the matrix B on the domain and the basis B' on the range. (3.5) Proposition. With the above notation, if v E M then [.f(v)IC = [f]C[v]B-

218

Chapter 4. Linear Algebra op(bi )

Proof. If [v)8 =

then op(bm)

f(v)=f E bjvj (j=1

M

_ >bjf(vj) j=1

- j=1 b' (atiwt) i=1 J

n

_ i=1

Therefore, If

m

(jaij

wi.

j=1

= [ E 1 op(biai,) ... F_j op(bjanj) I' = [flc [v] 8. 1

0

(3.6) Proposition. If M, N, and P are free R-modules with bases B, C, and P are R-module homomorphisms, then

D, respectively, and f : M -. N and g : N

[9°f11D1 = [91v[f1C-

Proof. By Proposition 3.5, if v E M then [91v ([f]B[v1B) _ [91v[f(v))c

[9(f(v))ly _ [(9 ° f)(v)1v _ [9°f11[v1B.

Choosing v = vj = P element of the basis B so that [v] 8 = Ej1 E Mn,1(R°P),

we obtain [91v (colj((f]c)) = [91v ([f1c[vj]13)

= [9°f1[vj1B

=colj([9°fIS) for 1 < j < n. Applying Lemma 1.1 (6), we conclude that [9 ° f1v = [91D[f]C

as claimed.


219

(3.7) Remark. From this proposition we can see that matrix multiplication is associative. Let M, N, P, and Q be free R-modules with ba.,es B, C, D,

and 6 respectively, and let f R-module homomorphisms. Then, by Proposition 3.6, L"ie (WJV[f]C) _ [hi9 ([9 ° f1D)

_ [h o (g o f )]B

_ [(hog)of]' [hog] [f 1 B

_ ([hlE [9]ai

[f J'3

By Remark 3.4 (1), every matrix is the matrix of a homomorphism, so associativity of matrix multiplication follows from the associativity of functional composition. (Actually, this proves associativity for matrices with entries

in RP, but then associativity for matrices with entries in R follows from the observation that R = (R°P)°P. Also observe that we used associativity in the proof of Proposition 3.1, but we did not use this proposition in the proof of Proposition 3.6, so our derivation here is legitimate.) (3.8) Corollary. Let M and N be free R-modules of rank n. Let B be a basis of M and let C be a basis of N. Then a homomorphism f E HomR(M, N) is an isomorphism if and only if the matrix [fl' E Mn(R°P) is invertible. C

Proof. Suppose g = f -1 E HomR(N, M). Then, by Proposition 3.6, In = [1M]8 = [9 o f]BB = [918[f1C,

and similarly, In = [f1C[91B

Thus, (f] C

is an invertible matrix. The converse is left as an exercise.

0

(3.9) Corollary. Let R be a ring and let M be a free R-module of rank n. Then EndR(M) is isomorphic (as a ring) to Mn(R°Q). If R is a commutative ring, then this isomorphism is an isomorphism of R-algebras. Proof. If B is a basis of M, let EndR(M)

Mn(R°P)

be defined by 4' (f) = [f]B. According to Proposition 3.6, 4iB is a ring homomorphism, while it is a bijective map by Remark 3.4. If R is commu0 tative, it is an R-algebra isomorphism by Lemma 1.1 (3). (3.10) Remark. From Lemma 1.3 and Corollary 3.9 we immediately see that

if R is a commutative ring and M is a free R-module of finte rank, then the center of its endomorphism ring is


220

C(EndR(M)) = R 1m. That is, a homomorphism f : M - M commutes with every other homomorphism g : M -+ M if and only if f =r-1m for some r E R. For the remainder of this section we shall assume that the ring R is commutative.

(3.11) Proposition. Let R be a commutative ring, let M and N be free Rmodules, and let f E HomR(M, N). (1) If rank(M) < rank(N), then f is not surjective. (2) If rank(M) > rank(N), then f is not injective. (3) If rank(M) = rank(N) is finite and f is injective,

then N/ Im(f) is a

torsion R-module. (4) If rank(M) = rank(N) is finite and R is an integral domain, then f is

injective if and only if N/ Im(f) is a torsion R-module. Proof. (1) By the definition of rank (Definition 3.2.9), if f were surjective, then we would have rank(N) < rank(M). (2) If f were injective, N would contain a free submodule Im(f) of

rank(Im(f)) = rank(M) > rank(N), contradicting Corollary 2.32. (3) Let f be an injection, and let 7r : N -p N/ Im(f) be the projection.

Suppose that N/ Im(f) is not torsion, and let n E N/ Im(f) be an element with Ann(n) _ (0). Let n E N with 7r (n) = n. Then Im(f) fl Rn = (0), and hence,

NDIm(f)®Rn, which is a free module of rank(M) + 1 = rank(N) + 1, contradicting Corollary 2.32. (4) Let R be an integral domain and assume that N/ Im(f) is a torsion module. Pick a basis {w1, ... , wn} of N. Since N/ Im(f) is a torsion mod-

ule, there exists vi E M and ci # 0 E R with f (vi) = ciwi for 1 < i < n. Suppose V E M and f (v) = 0. Then the set {V, V1, ... , Vn}

is linearly dependent by Proposition 2.31, so let

an equation of linear dependence with not all of {a, a1, ... , an } equal to zero. Then


221

0=af(v) = f(av)

i=1

Since {w1, ... , wn } is a basis of N and R is an integral domain, it follows

that ai = 0 for all i. Hence av = 0 and thus v = 0 (similarly), and we conclude that f is an injection. (3.12) Remark. The assumption that R is an integral domain in Proposition

3.11 (4) is necessary. Let R = Zmn and set M = N = R. Let f : M -. N be defined by f (v) = mv. Then Zmn/ Im(f) = Zn is a torsion Z,nn-module, but f is not injective. The relationship between invertibility of homomorphisms and invertibility of matrices allows one to conclude that a homomorphimm between free R-modules of the same finite rank is invertible if it has either a left or a right inverse.

(3.13) Proposition. Let M and N be free R-modules of finite rank n, let B be a basis of M, and let C be a basis of N. If f E HomR(M, N), then the following are equivalent.

(1) f is an isomorphism. (2) f has a right inverse, i.e., there is a homomorphism g E HomR(N, M) such that f g = 1 N .

(3) f has a left inverse, i.e., there is a homomorphism h E HomR(N, M) such that hf = 1M. (4) f is a surjection.

s is an invertible matrix.

(5) (6)

[f]

(7)

(f )C has a left inverse.

(f ]

has a right inverse.

Proof. The equivalence of (5), (6), and (7) follows from Remark 2.18, while

the equivalence of (1), (2), and (3) to (5), (6), and (7), respectively, is a consequence of Corollary 3.8.

Now clearly (1) implies (4). On the other hand, assume that f is a surjection. Then there is a short exact sequence

0-Ker(f),M

0.


222

This sequence splits since N is free so that there is an R-module homomorphism g : N -. M such that f9 = IN, i.e., g is a right inverse for f. Thus (4) implies (2), and the proof is complete. 0 (3.14) Remark. In Proposition 3.13 (4) it is not possible to replace surjective

by injective. It is true that if f has a left inverse, then f is injective, but the converse need not be true. For example, f : Z -' Z by f (x) = 2x is injective, but it is not left invertible. However, in case the ring R is a field, the converse is valid. This is the content of the following result.

(3.15) Proposition. Let F be a field and let M and N be vector spaces over F of dimension n. Then the following are equivalent.

(1) f is an isomorphism. (2) f is injective. (3) f is surjective. Proof. This is simply a restatement of Corollary 3.8.10.

0

(3.16) Proposition. Let M and N be free R-modules with bases 8, 8' and C, C' respectively. If f : M N is an R-module homomorphism, then [f ]c and [ f )c, are related by the formula

[f15 = Pc-[f1c (P8-)-1.

Proof. Since f = IN o f o 1M, Proposition 3.6 shows that (3.4)

(f1c = [lN1c [f1c(1M1g

But [1NJg, = PC, and [1M]B' = Pg' = (PS)-', so Equation (3.3) follows from Equation (3.4).

0

We now give a determinantal criterion for the various properties of a homomorphism.

(3.17) Proposition. Let M and N be free R-modules with rank(M) _ rank(N) finite. (1) f is surjective (and hence an isomorphism) if and only if in some (and, hence, in any) pair of bases B of M and C of N, det([ f ]B) is a unit of R.

(2) f is injective if and only if in some (and, hence, in any) pair of bases B of M and C of N, det([ f ]B) is not a zero divisor in R. Proof. (1) is immediate from Proposition 3.13 and Theorem 2.17 (2), while 0 part (2) follows from Corollary 2.27.


223

(3.18) Definition. (1)

Let R be a commutative ring. Matrices A, B E Mn,,n(R) are said to be equivalent if and only if there are invertible matrices P E GL(n, R) and Q E GL(m, R) such that

B = PAQ. Equivalence of matrices is an equivalence relation on Mn,,n(R). (2) If M and N are finite rank free R-modules, then we will say that R-

module homomorphisms f and g in HomR(M, N) are equivalent if there are invertible endomorphisms h1 E EndR(M) and h2 E EndR(N)

such that h2 f hi 1 = g. That is, f and g are equivalent if and only if there is a commutative diagram

M IM hll

+

N hs

9+ N

where the vertical maps are isomorphisms. Again, equivalence of homomorphisms is an equivalence relation on HomR(M, N). (3.19) Proposition.

(1) Two matrices A, B E Mn,m(R) are equivalent if and only if there are bases B, B' of a free module M of rank m and bases C, C' of a free module N of rank n such that A = [f] and B = [f]. That is, two c matrices are equivalent if and only if they represent the same R-module homomorphism with respect to different bases. (2) If M and N are free R-modules of rank m and n respectively, then homomorphisms f and g E HomR(M, N) are equivalent if and only if there are bases B, B' of M and C, C' of N such that [f ]c = (gilc

l'.

That is, f is equivalent to g if and only if the two homomorphisms are represented by the same matrix with respect to appropriate bases. Proof. (1) Since every invertible matrix is a change of basis matrix (Lemma 3.2), the result is immediate from Proposition 3.16. (2) Suppose that [f]c = [g]c,. Then Equation (3.3) gives (3.5)

[f ]c = [91 c' = l c, [9]c (PB) -'

The matrices Pg and PC are invertible so that we may write (by Corollary 3.9) Pg = [h1]5 and [h2]c where h1 E EndR(M) and h2 E EndR(N) are invertible. Thus Equation (3.5) gives

If)" = [h2]c[9]c ([h1]g)-1 = [h29hi 1]c c

224


Hence, f = hsghi 1 and f and g are equivalent. The converse statement is left as an exercise.

0

Using the invariant factor theorem for submodules (Theorem 3.6.23), it is possible to explicitly describe the equivalence classes under the equivalence relations of equivalence of homomorphisms and equivalence of ma-

trices if one restricts the ring R to be a PID. This is the content of the following result.

(3.20) Proposition. Let R be a PID and let f : M N be an R-module homomorphism between a free R-module M of rank m and a free R-module

N of rank n. Then there is a basis B = {vl, ... , v,,,} of M, a basis C = {wl, ... , of N, and nonzero elements sl, ... , ar of R, where r = rank Im(f ), such that s; I si+1 for 1 r.

That is, the matrix of f with respect to the bases B and C is [fJc=[Or

01

where Dr = diag(81, ... , sr). Proof. By the invariant factor theorem for submodules (Theorem 3.6.23), of N and elements al, ... ,sr E R such there is a basis C = {wl, ...

that si 1 8i+1 for 1 < i < r - 1 and {81w1i ... , srwr} is a basis for the submodule Im(f) C N. Now choose any subset {v1, ... , v,-) C M such that f (vi) = aiwi for 1:5 i < r. By Proposition 3.8.8, Ker(f) is a submodule of M of rank m - r. Thus, we may choose a basis {vr+1, ... , v,,,) of Ker(f).

Claim. B = {vi, ... , vr, Vr+1,

,

v,n} C M is a basis of M.

To the claim, suppose that v E M. Then f (v) E Im(f ), so we may write f(v) _ >2 aif(vi) i=1

f(aivi)

_ i=1

Therefore, f (v we may write

aivi) = 0 so that v r

aivi E Ker(f ), and hence,

m

V - E aivi = E aivi. i=1

i=r+1

It follows that 8 generates M as an R-module.


To check linear independence, suppose that m

225

Ein 1 aivi = 0. Then

r

0 = >aif(vi) _ >ai(siwi) :=1

{=1

Since {s1w1i ... , srwr} is linearly independent, this implies that ai = 0 for

1 < i < r. But then

m

aivi = 0, i-r+1

is a basis of Ker(f ), we conclude that ai = 0 for and since {vr+1, ... , all i, and the claim is verified.

It is clear from the construction that fflcs

=

Dr

0

0

o1

where Dr = diag(s1i ... , sr). This completes the proof.

0

(3.21) Remark. In the case where the ring R is a field, the invariant factors are all 1. Therefore, if f : M - N is a linear transformation between finitedimensional vector spaces, then there is a basis B of M and a basis C of N

such that the matrix of f is 1f1B

=

10r

0,

The number r is the dimension (= rank) of the subspace Im(f).

(3.22) Corollary. Let R be a PID and let M and N be free R-modules of rank m and n respectively. Then homomorphisms f and g E HomR(M, N) are equivalent if and only if the subspaces Im(f) and Im(g) have the same invariant factors as submodules of N. In particular, if R is a field, then f and g are equivalent if and only if rank(Im(f )) = rank(Im(g)).

0

Proof. Exercise.

If M = N, then Proposition 3.16 becomes the following result:

(3.23) Proposition. Let f E EndR(M) and let B, B' be two bases for the free R-module M. Then [f] B' = PB [f]B (PB,)-1.


226

Proof.

Proposition 3.23 applies to give a result analogous to Proposition 3.19 for similarity of matrices in M,,(R). Recall (Definition 1.7) that two matrices A and B in M,,(R) are similar if there is a matrix S E GL(n, R) such that B = S-' AS. For homomorphisms, the definition is the following: (3.24) Definition. Let R be a commuative ring and let M be a free R-module of rank n. If j r, g E EndR(M), then we say that f and g are similar if there is an invertible homomorphism h E EndR(M) such that g = h-' f h.

In this situation, Proposition 3.19 becomes the following result: (3.25) Corollary. (1) Two matrices A, B E Mn (R) are similar if and only if there are bases

B and B' of a free R-module M of rank n and f E HomR(M) such that A = [f)13 and B = if ]s- . That is, two n x n matrices are similar if and only if they represent the same R-module homomorphism with respect to different bases. (2) Let M be a free R-module of rank n and let f, g E EndR(M) be endomorphisms. Then f and g are similar if and only if there are bases B and B' of M such that [f]L; = [g)s'

That is, f is similar to g if and only if the two homomorphisms are represented by the same matrix with respect to appropriate bases. Proof. Exercise.

(3.26) Remark. Let R be a commutative ring and T a set. A function 0 : Mn(R) - T will be called a class function if O(A) =,O(B) whenever A and B are similar matrices. If M is a free R-module of rank n, then the class function 0 naturally yields a function 4' : EndR(M) - T defined by 4'(f) = m([fIB) where B is a basis of M. According to Corollary 3.25, the definition of m is independent of the choice of basis of M because 0 is a class function. The

most important class functions that we have met so far are the trace and the determinant (Lemma 1.6 (2) and Corollary 2.12). Thus, the trace and the determinant can be defined for any endomorphism of a free R-module of finite rank. We formally record this observation. (3.27) Proposition. Let R be a commutative ring and let Al be a finite rank free R-module.

4.3 Matrix Representation of Homomorphisms (1)

227

There is an R-module homomorphism Tr : EndR(M) -+ R defined by

Tr(f) = T VI is) where B is any basis of M. Tr(f) will be called the trace of the homomorphism f; it is independent of the choice of basis B. (2) There is a multiplicative function det : EndR(M) - R defined by

det(f) = det([fJB) where B is any basis of M. det(f) will be called the determinant of the homomorphism f ; it is independent of the choice of basis B. Proof.

Note that multiplicativity of the determinant means det(fg) = det(f) det(g).

Since det(1) = 1, it follows that f is invertible if and only if det(f) is a unit in R. Since similar matrices represent the same endomorphism with respect to different bases, one goal of linear algebra is to find a matrix B similar to a given matrix A such that B is as simple as possible. This is equivalent (by Corollary 3.25) to finding a basis of a free R-module so that the matrix [ f ]B of a given homomorphism f is as simple as possible. When R is a field, this is the subject of canonical form theory that will be developed in detail in the next section. For now we will only indicate the relationship between direct sum decompositions of free R-modules and decompositions of matrices. (3.28) Proposition. Let R be a commutative ring, and let M1, M2, N1, and N2 be finite rank free R-modules. If Bi is a basis of Mi and Ci is a basis of Ni (i = 1, 2), then let B1 U B2 and C1 U C2 be the natural bases of Ml ® M2 and N1 ® N2 respectively (see Example 3.4.6 (7)). If fi E HomR(M1, Ni) for i = 1, 2, then fl ® f2 E HomR(M1 ® M2, N1 ® N2) and UB2 = [f1JB' ® [f1® f2 JcBlUc,

[f2JB2

2.

Proof. Exercise.

We now specialize to the case of endomorphisms.

(3.29) Definition. Let M be an R-module and let f E EndR(M). A submod-

ule N C M is said to be invariant under f (or an invariant submodule of f) if f (x) E N whenever x E N.

228


(3.30) Proposition. Let R be a commutative ring, let M be a free R module

of rank m, and let f E HomR(M). If B = {v1, ... , vm } is a basis of M then the matrix [f] B has the block form [fit;

_

AB

0D

where A E Mr(R) if and only if the submodule N = (v1, ... , v,.) is an invariant submodule of f. Proof. If if JB = [tip) then the block form means that tip = 0 for r + 1 t,.,vi =

t3vi E N.

0

Since N is generated by v1, ... , V,., the result follows.

(3.31) Remark. As a special case of this result, a matrix [1 [B is upper triangular if and only if the submodule (v1, ... vk) (where B = {vj, ... ,v,,,}) is invariant under f for every k (1 < k < m).

In Proposition 3.30, if the block B = 0, then not only is (vi, ...

, vk)

an invariant submodule, but the complementary submodule (vk+1, ... , v,,,) is also invariant under f. From this observation, extended to an arbitrary number of blocks, we conclude:

(3.32) Proposition. Let M be a free R-module of rank m, let f E EndR(M), and let A = [fie E Mm(R). Then A is similar to a block diagonal matrix

B = Al ® ® Ak where Ai E Al, (R) (rl + - + rk = n) if and only if there are free submodules All, ... , Mk of M such that

(1) M; is an invariant submodule off which is free of rank ri, and (2) M All ED ...ED Mk. Proof

0

The case of this result when r, = 1 for all i is of particular importance. In this case we are asking when A is similar to a diagonal matrix. To state the result in the manner we wish, it is convenient to make the following definition.

(3.33) Definition. If M is a free R-module of rank m and f E EndR(M), then a nonzero x E M is called an eigenvector off if the cyclic submodule (x) of M is invariant under f. That is, x 0 E M is an eigenvector of Al if and only if f (x) = ax for some a E R. The element a E R is called


229

an eigenvalue of f . The eigenniodule (or eigenspace) off corresponding to the eigenvalue a is the submodule Ker(f - a1M). If A E Mn(R) then by an eigenvalue or eigenvector of A, we mean an eigenvalue or eigenvector of the R-module homomorphism

TA : M.,i(R) - Mn.1(R) defined by TA(v) = Av. We shall usually identify Mn,1(R) with R' via the standard basis {Eti1 : 1 < i < n} and speak of TA as a map from R" to R". In practice, in studying endomorphisms of a free R-module, eigenvalues and eigenvectors play a key role (for the matrix of f depends on a choice of basis, while eigenvalues and eigenvectors are intrinsically defined). We shall consider them further in Section 4.4.

(3.34) Corollary. Let M be a free R-module of rank m, let f E EndR(M),

and let A = [f]g E Mn(R). Then A is similar to a diagonal matrix diag(ai, ... ,a.) if and only if there is a basis of M consisting of eigenvectors of A.

0

Proof.

(3.35) Definition. A matrix which is similar to a diagonal matrix is said to be diagonalisable. An endomorphism f E EndR(M) is diagonalizable if M has a basis B such that [f]13 is a diagonal matrix. A set S = { fi }iE j of endomorphisms of M is said to be simultaneously diagonalizable if M has a basis B such that [fi]s is diagonal for all i E I. The concept of diagonalizability of matrices with entries in a field will

be studied in some detail in Section 4.4. For now we will conclude this section with the following result, which we shall need later.

(3.36) Theorem. Let R be a PID, let M be a free R-module of finite rank

n, and let S = { fi : M - M}$EJ be a set of diagonalizable R-module endomorphisms. Then the set S is simultaneously diagonalizable if and only

if S consists of commuting endomorphisms, i.e., there is a basis B of M with [fi]n diagonal for all i E I if and only if fi fi = f, fi for all i, j.

Proof. For simplicity, we assume that we are dealing with a pair of Rmodule endomorphisms {f, g}; the general case is no more difficult. First suppose that B = {v1, ... , vn } is a basis of M in which both [ f ]8 and [g]j; are diagonal. Then f (vi) = \;v; and g(vi) = µivi for 1 < i < n. Then f(9(vi)) = f(µivi) = µi9(vi) = /1i'\ivi

= '\wivi = Ai9(vi) = 9(A1vi)

= 9(f (vi)), so f g and g f agree on a basis of M and hence are equal.

230


Conversely, suppose that f g = g f and let Al, ... , Ak be the distinct eigenvalues of f, and let A,, ... , ,ut be the distinct eigenvalues of g. Let

Mi = Ker(f - Ailm) Nj = Ker(g - µj I t j)

(3.8)

(3.9)

(1 < i < k) (1 < j < Q).

Then the hypothesis that f and g are diagonalizable implies that (3.10)

MI®...(D Mk

and (3.11)

M'NI(D ...ED Nt.

First, we observe that Mi is g-invariant and Nj is f-invariant. To see this, suppose that v E Mi. Then f(g(v)) = Of M) = g(Aiv) = Aig(v),

i.e., g(v) E Ker(f - Ai1M) = Mi. The argument that Nj is f-invariant is the same.

Claim. Mi=®j=1(MinN3) fort
where wj E Nj. The claim will be proved once we show that wj E Mi for all j. Since v E Mi, (3.13)

Aiv=f(v)=f(w1)+...+f(wt).

But Nj is f-invariant, so f(wj) E Nj for all j. But (3.14)

Aiv = Aiw1 +

+ Aiwt.

Comparing Equations (3.13) and (3.14) and using Equation (3.11), we see

that Aiwj = f(wj) for 1 < j < t, i.e., wj E Mi for all j, and the claim is proved.

To complete the proof of the theorem, note that since R is a PID, each of the submodules Mi n Nj is a free R-module (Theorem 3.6.2), so let Bij = {vij} be a basis of Mi n Nj. According to the claim, we have k

t

®®(MinN.) i=1 j=1

so that B = uj jBij is a basis of M consisting of common eigenvectors of f and g, i.e., [f] B and [g]B are both diagonal. 0

4.4 Canonical Form Theory

231

(3.37) Remark. The only place in the above proof where we used that R is a PM is to prove that the t eigenspaces Mi n Nj are free submodules of M. If R is an arbitrary commutative ring and f and g are commuting diagonalizable endomorphisms of a free R-module M, then the proof of Theorem 3.36 shows that the t eigenspaces Mi n Nj are projective R-modules. There is a basis of common eigenvectors if and only if these submodules are in fact free. We will show by example that this need not be the case.

Thus, let R be a commutative ring for which there exists a finitely generated projective R-module P which is not free (see Example 3.5.6 (3) or Theorem 3.5.11). Let Q be an R-module such that P ®Q = F is a free R-module of finite rank n, and let M=FED F=P1ED Q1®P2ED Q2 where P1 = P2 = P and Q1 = Q2 = Q. Then M is a free R-module of rank 2n. Let \1 96 a2 and Pl # P2 be elements of R and define f, g E EndR(M) by

f (x1, yl, x2, 92) = (Aixl, )'lyl, A2x2, A212) and

9(x1, bl, x2, y2) = (P1x1, l'2Y1, P2x2, P1Y2)

where xi E Pi and yi E Qi for i = 1, 2. Then f is diagonalizable with eigenspaces

F M2 = Ker(f -A21M)=P2®Q2F M1 = Ker(f -a11M)=Pi ®Q1 and g is diagonalizable with eigenspaces

N, =Ker(g-µ11M)=PI

F F.

Moreover, f g = g f . However, there is no basis of common eigenvectors of f and g since the t eigenspace M1 n N1 = Pl is not free.

4.4 Canonical Form Theory The goal of the current section is to apply the theory of finitely generated modules over a PID to the study of a linear transformation T from a finitedimensional vector space V to itself. The emphasis will be on showing how this theory allows one to find a basis of V with respect to which the matrix representation of T is as simple as possible. According to the properties of matrix representation of homomorphisms developed in Section 4.3, this is equivalent to the problem of finding a matrix B which is similar to a given

232


matrix A, such that B has a form as simple as possible. In other words, we are looking for simply described representatives of each equivalence class of matrices under the equivalence relation of similarity. We will start by carefully defining the module structures that are determined by linear transformations. This has already been mentioned in Example 3.1.5 (12), but we will repeat it here because of the fundamental importance of this construction.

Let F be a field, let V be a finite-dimensional vector space over F (i.e., V is a free F-module of finite rank), and let T : V V be a linear transformation (i.e., T is an F-module homomorphism). Let R = F[X] be the polynomial ring with coefficients from F. Recall (Theorem 2.4.12) that R is a principal ideal domain. Let

0: R - HomF(V) be the F-algebra homomorphism determined by O(X) = T (see Section 2.4). To be explicit, if f (X) = ao + a1 X + + a,,, X" then (4.1)

4(f(X))=aoly+a1T+

+a"T".

Then V becomes an R-module via the scalar multiplication (4.2)

(f (X)) v = O(f (X))(v)

for each f(X) E R = F[X] and v E V. Combining Equations (4.1) and (4.2) we see that the R-module structure on V determined by T is given by (4.3)

(f (X)) v = aov + a1T(v) + ... + a"T"(v).

Note that each T E EndF(V) will induce a different R-module structure on the same abelian group V. To distinguish these different module structures, we will write VT for the vector space V with the R-module structure described by Equation (4.3). When there is no chance of confusion, we will sometimes write V for the R-module VT. Again we note that the module VT has been previously introduced in Example 3.1.5 (12).

Note that scalar multiplication of a vector v E VT by the constant polynomial ao E F(XJ is the same as the scalar multiplication aov, where ao is considered as an element of the field F and v E V. This is an immediate observation based on Equation (4.3). It is also worth pointing out explicitly that an R-submodule N of the R-module VT is just a subspace of V that is T-invariant. Recall (Definition 3.29 and also Example 3.2.2 (4)) that this means that T(v) E N for all v E N. We will begin our study by computing the R-module homomorphisms between two R-modules VT and Ws and by relating this computation to the similarity of linear transformations.

(4.1) Proposition. Let V and W be vector spaces over the field F, and suppose that T E EndF(V), S E EndF(W). Then

4.4 Canonical Form Theory (4.4)

233

HomF(x((VT, Ws) = {U E HomF(V, W) : UT = SU).

Proof Suppose U E HomF(xi(VT, Ws). As we observed above, the F[X]module action on V and W reduces to the F-module action (i.e., scalar multiplication). Thus U E HomF(V, W). Let v E V and w E W. Then X v = T(v) and X w = S(w). Then, since U is an F[XJ- module homomorphism, we have

U(T(v)) = U(X v) = X U(v) = S(U(v)). Since v E V is arbitrary, we conclude that UT = SU. Conversely, suppose that U : V - W is a linear transformation such that UT = SU. We claim that

U(f (X) - v) = f (X) U(v)

(4.5)

for all v E V and f (X) E F[X]. But Equation (4.5) is satisfied for polynomials of degree 0 since U is a linear transformation, and it is satisfied for f (X) = X since

U(X v) = U(T(v)) = S(U(v)) = X U(v). Since F[X] is generated by the constant polynomials and X, it follows that Equation (4.5) is satisfied for all polynomials f (X), and hence, U E HomF(x((VT, Ws)

(4.2) Theorem. Let V be a vector space over the field F, and let T1, T2 E EndF(V). Then the R-modules VT, and VT, are isomorphic if and only if T1 and T2 are similar. Proof. By Proposition 4.1, an R-module isomorphism (recall R = F[X])

P:VT2 -'VT1 consists of an invertible linear transformation P : V V such that PT2 = T1P, i.e., T1 = PT2P-1. Thus VT, and VT2 are isomorphic (as R-modules) if and only if the linear transformations T1 and T2 are similar. Moreover, we have seen that the similarity transformation P produces the R-module isomorphism VT, to VT,.

This theorem, together with Corollary 3.25, gives the theoretical underpinning for our approach in this section. We will be studying linear transformations T by studying the R-modules VT, so Theorem 4.2 says that, on the one hand, similar transformations are indistinguishable from this point of view, and on the other hand, any result, property, or invariant we derive in this manner for a linear transformation T holds for any transformation similar to T. Let us fix T. Then by Corollary 3.25, as we vary the basis B of V, we obtain similar matrices [TIs. Our objective will be to

234


find bases in which the matrix of T is particularly simple, and hence the structure and properties of T are particularly easy to understand. (4.3) Proposition. Let V be a vector space over the field F and suppose that dimF(V) = n < oo. If T E EndF(V), then the R-module (R = F[X]) VT is a finitely generated torsion R-module.

Proof. Since the action of constant polynomials on elements of VT is just

the scalar multiplication on V determined by F, it follows that any Fgenerating set of V is a priori an R-generating set for VT. Thus U(VT) < n = dimF(V). (Recall that u(M) (Definition 3.2.9) denotes the minimum number of generators of the R-module M.) Let v E V. We need to show that Ann(v) $ (0). Consider the elements v, T(v), ..., T"(v) E V. These are n+1 elements in an n-dimensional vector space V, and hence they must be linearly dependent. Therefore, there are

scalars ao, al, ... a E F, not all zero, such that (4.6)

apv + a1T(v) +

+

0.

If we let f (X) = ao + al X +

then Equation (4.6) and the .+ definition of the R-module structure on VT (Equation (4.3)) shows that f (X )v = 0, i.e., f (X) E Ann(v). Since f (X) # 0, this shows that Ann(v) 0 (0).

This innocent looking proposition has far reaching consequences, for it means that we may apply our results on the structure of finitely generated torsion modules over a PID R to the study of VT. Henceforth, we will fix a finite-dimensional vector space V over F and T E EndF(V). We begin with the following observation.

(4.4) Corollary. There is a polynomial f (X) E F[X] of degree at most n with f (T) = 0. Proof. By Theorem 3.7.1, Ann(VT) = Ann(v) for some v E VT. But the proof of Proposition 4.3 shows that Ann(v) contains a polynomial f (X) of degree at most n. Thus f (X) E Ann(VT) so that f (X )w = 0 for all w E VT, i.e., f (T) (w) = 0 for all w E V. Hence f (T) = 0 as required. (4.5) Remark. It is worth pointing out that the proofs of Proposition 4.3 and Corollary 4.4 show that a polynomial g(X) is in Ann(VT) if and only if g(T) = 0 E EndF(V). In Section 3.7 we had two decompositions of finitely generated torsion R-modules, namely, the cyclic decomposition (Theorem 3.7.1) and the cyclic primary decomposition (Theorem 3.7.13). Each of these decompositions will produce a canonical form for T, namely, the first will produce the rational canonical form and the second will give the Jordan canonical


235

form. Moreover, Theorem 3.7.12 applied to the R-module VT produces an important direct sum decomposition, which we will refer to as the primary decomposition theorem for the linear transformation T. We will begin by studying the cyclic decomposition of VT. According to Theorem 3.7.1, the torsion R-module VT can be written as a direct sum of k =,u(VT) cyclic R-submodules

VT = Rv1 ® ... ®Rvk

(4.7)

such that Ann(vi) = (fi(X)) for 1 < i < k and

(fl (X)) 2 (f2(X)) J ... 2 (fk(X))

(4.8)

Equation (4.8) is equivalent to the condition

fi(X) I fi+1(X)

(4.9)

for

1 < i < k.

Theorem 3.7.3 shows that the ideals (fi(X)) for 1 < i < k are uniquely determined by the R-module VT (although the generators {v1, ... , vk } are not uniquely determined), and since R = F[X], we know that every ideal contains a unique monic generator. Thus we shall always suppose that f1(X) has been chosen to be monic. (4.6) Definition. The monic polynomials f1(X ), ... , fk(X) in Equation (4.8) are called the invariant factors of the linear transformation T.

The invariant factor fk(X) of T is called the minimal polynomial MT(X) of T. The characteristic polynomial cT(X) of T is the product of all the invariant factors of T, i.e., (4.10)

CT(X) = f1(X)f2(X) ... fk(X)-

(4.7) Remark. In the language of Definition 3.7.8, we have (4.11)

mT(X) = me(VT)

and

CT(X) = Co(VT).

(4.8) Lemma. Ann(VT) = (mT(X)). Proof. This is immediate from Equation (3.7.1).

O

(4.9) Corollary. mT(X) is the unique monic polynomial of lowest degree with

MT(T) = 0.


236

Proof. An ideal I of F[X] is generated by a polynomial of lowest degree in I. Apply this observation to Ann(VT), recalling the description of Ann(VT) in Remark 4.5. 0 (4.10) Corollary. (1) If q(X) E FIX] is any polynomial with q(T) = 0, then

mT(X) I q(X).

(2) MT(X) divides cT(X). (3) If p(X) is any irreducible polynomial dividing cT (X ), then p(X) divides

mT(X).

0

Proof. This is a special case of Corollary 3.7.9. (4.11) Lemma. (1) If VT is cyclic, say VT = R/(f (X)), then

dimp(V) = deg(f(X)). (2) I f V T ° _ Rvi ® E )

where Ann(vi) = (fi(X)) as in Equation (4.7),

then k

(4.12)

> deg(fi(X )) = dim(V) = deg(cT(X )). i=1

(3)

The following are equivalent: (a) VT is cyclic. (b) deg(mT(X)) = dim(V).

(c) MAX) = CT(X)Proof. (1) Suppose that VT = Rv. Then the map q : R

VT defined by

q(T)(v)

R/ Ker(i) as F-modules. But Ker(rl) _ (f (X)), and as F-modules, R/(f(X)) has a basis {1, X, ... ,Xn-1} where n = is surjective and vT

deg(f (X)). Thus, dimF(V) = n = deg(f (X)). (2) and (3) are immediate from (1) and the definitions.

0

(4.12) Definition. Let f (X) = Xn + + a1X + a0 E F[X] + be a monic polynomial. Then the companion matrix C(f(X)) E Mn(F) of f (X) is the n x n matrix an_1Xn-1


(4.13)

C(f M) =

0

0

0

0

-ao

1

0

0

-al

0

1

0

0 0

I0

0 0

1

0

0

1

0

.

237

-a2

-an-1 -°n-2 J

(4.13) Examples.

(1) For each AEF,C(X-A)=[a]EM1(F). (2) diag(al, ... an) = ®i 1C(X - a,). (3) C((2 + 1) = (4)

r0 Ol l

If A = C(X - a) ®C(X2 - 1), then a 0

A= 0

0

0

0

1

1

0

(4.14) Proposition. Let f (X) E F[X] be a monk polynomial of degree n,

and let T : Fn - F" be defined by multiplication by A = C(f(X)), i.e., T(v) = Av where we have identified F" with Mn,1(F). Then mT(X) _

f(X). Proof. Let ej = col,(In). Then from the definition of A = C(f (X)) (Equation (4.13)), we see that T(e1) = e2, T(e2) = e3, ..., T(en-1) = en. There-

fore, T1(e1) = e,.+1 for 0 < r < n - 1 so that {e1,T(e1), ... ,T"-'(el)} is a basis of Fn and hence (F")T is a cyclic R-module generated by e1. Thus, by Lemma 4.11, deg(mT(X)) = n and

(mT(X)) = Ann((F")T) = Ann(e1). But T"(e1) = -aoe1 - a1T(e1) -

an-1Tn-1(e1),

i.e., T"(el)+an-1T"-1(el)+...+a1T(e1)+aoe1 =0.

Therefore, f(T)(e1) = 0 so that f(X) E Ann(e1) = (mT(X)). But deg(f (X )) = deg(mT(X)) and both polynomials are monic, so f (X) = mT(X).

238


(4.15) Corollary. In the situation of Proposition 4.14, let

8 = {v,T(v), ... ,T' '(v)} where v = el = coll (I"). Then B is a basis of F" and [T]B = C(f (X )). Prof. This is clear from the calculations in the proof of Proposition 4.14. 0 (4.16) Corollary. Let V be any finite-dimensional vector space over the field

F, let T E EndF(V), and suppose that the R-module VT is cyclic with Ann(VT) = (f (X )) . Then there is a basis 8 of V such that [T]13 = C(f (X)).

Proof. If VT = Rv then we may take

8 = {v,T(v), ... ,T"-1(v)}

0

where n = dim(V).

(4.17) Theorem. (Rational canonical form) Let V be a vector space of dimension n over a field F and let T E EndF(V) be a linear transformation. If {f1(X), ... , fk(X)} is the set of invariant factors of the F[X]-module VT, then V has a basis B such that (4.14)

[T)B = C(f1(X )) ® C(f2(X )) ® ... ED C(fk (X)).

Proof. Let VT ^_' Rv1 ® ... ® Rvk where R = F[XI and Ann(vi) = (fi(X))

and where fi(X) I fi+1(X) for 1 < i < k. Let deg(f;) = ni. Then Bi = {vi, T(vi), ... ,T1,-1(vi)} is a basis of the cyclic submodule Rvi. Since submodules of VT are precisely the T-invariant subspaces of V, it follows that TAR,,; E EndF(Rvi) and Corollary 4.16 applies to give (4.15)

(TIRvj , =

C(fi(X))

By Equation (4.12), ni + + nk = n, and hence, B = U;= I is a basis of V and Proposition 3.32 and Equation (4.15) apply to give Equation (4.14).

0 (4.18) Corollary. Two linear transformations T1 and T2 on V have the same rational canonical form if and only if they are similar.

Proof. Two linear transformations T1 and T2 have the same rational canonical form if and only if they have the same invariant factors, which occurs if and only if the R-modules VT, and VT, are isomorphic. Now apply Theorem 4.2.

0


239

(4.19) Corollary. Every matrix A E Mn(F) is similar to a unique matrix in rational canonical form.

Proof. Regard A E Mn(F) as defining a linear transformation TA Fn -. Fn by TA(v) = Av. Then A is similar to B E Mn(F) as matrices if and only if TA is similar to TB as elements of EndF(Fn). Thus we may apply Theorem 4.17 and Corollary 4.18. We now pause in our general development in order to see how to compute cT(X) and to prove a famous result. We will need the following simple lemma:

(4.20) Lemma. Let F be a field and let f (X) E FIX) be a monic polynomial

of degree n. Then the matrix XIn - C(f (X)) E Mn(F[XJ) and (4.16)

det (XIn - C(f (X))) = f(X).

Proof. Let f (X) = Xn + + a1X + ao E FIX]. The proof + is by induction on n = deg(f). If n = 1, the result is clear. Now suppose that n > 1, and compute the determinant of Equation (4.16) by cofactor an_1Xn-1

expansion along the first row; applying the induction hypothesis to the first summand.

det(XIn - C(f (X))) = det 1

X

0

-1 0

X -1

0 0

0 0

X

. .. . .. . ..

. .

0 0 0

0 0 0

ao a1

a2

-1 X

.

...

0

-1 ...

0

an-2

-1 X+an_1

0 0 0

a1

a2

= X det 0

-1

0

0

X an-2 -1 X +an_1

-1 X

...

0

-1

..

0 0

0 0

...

+ a0(_1)n+1 det

0 0

0 0

-1 X

= X (Xn-1 + an_1Xn-2 +... + al) +ao(-1)n+l(_1)n-1

= Xn+an_1Xn-1 +...+a1X +ao

0

-1

240


= f(X), and the lemma is proved.

(4.21) Definition. If A E Mn(F) then we will denote the polynomial det(XIn - A) E F[X] by CA(X) and we will call cA(X) the characteristic polynomial of the matrix A.

We will prove that the characteristic polynomial of a linear transformation T (as defined in Definition 4.6 (3)) is the characteristic polynomial of any matrix representation of T.

(4.22) Lemma If A and B E Mn(F) are similar, then cA(X) = CB(X)-

Proof. Suppose that B = P-'AP for some P E GL(n, F). Then det P 0 0 and (det P)-1 = det P-1. Hence,

CB(X) = det(XI - B) = det(XIn - P-'AP) = det(P-1(XIn - A)P) = (detP-1)(det(XIn - A))(detP) = det(XIn - A) = CA(X) 0 (4.23) Proposition. Let V be a finite-dimensional vector space over a field F and let T E EndF(V). If B is any basis of V, then (4.17)

cT(X) = c(Tie(X) = det(XIn - (T]B).

Proof. By Lemma 4.22, if Equation (4.17) is true for one basis, it is true for any basis. Thus, we may choose the basis B so that (T] B is in rational canonical form, i.e., (4.18)

(T]B = ®C(fi(X)) i=1

where f1(X), ... , fk(X) are the invariant factors of T. If deg(fi(X)) = n then Equation (4.18) gives k

(4.19)

XIn - (T]B = ®(XIn, - C(fi(X ))) i=1

Equation (4.19) and Theorem 2.11 imply


241

clTl8(X) = det(XI - [T18) k

det(XI,,, -C(fi(X)))

= i-1 k

_ fj fi (X)

by Lemma 4.20

i=1

= CT(X),

0

which gives Equation (4.17).

(4.24) Corollary. (Cayley-Hamilton theorem) Let T E EndF(V) be any linear transformation on a finite-dimensional vector space V and let B be any basis of V. Let A = [TJ8 E F[XJ. Then CA(T) = 0.

Proof. By Proposition 4.23, cA(X) = cT(X) and mT(X) I cT(X). Since 0 mA(T) = 0, it follows that cT(T) = 0, i.e., cA(T) = 0. (4.25) Remark. The Cayley-Hamilton theorem is often phrased as A linear transformation satisfies its characteristic polynomial. From our perspective, the fact that cT(T) = 0 is a triviality, but it is a nontrivial result that CT(X) = cA(X) where A = [TJ8. However, there is an alternate approach in which the characteristic polynomial of a linear transformation is defined to be CA(X) where A is some matrix representation of T. From this perspective, the Cayley-Hamilton theorem becomes a nontrivial result. It is worth pointing out that the Cayley-Hamilton theorem is valid for matrices with entries in any commutative ring; of course, the invariant factor theory is not valid for general rings so a different proof is needed (see Exercise 56). In fact, we shall sketch a second proof of the Cayley-Hamilton theorem in the exercises which is valid for any commutative ring R. From the point of view of the current section, the utility of Proposition 4.23 and Corollary 4.24 is that we have an independent method of calculating the characteristic polynomial. Further techniques for computing the invariant factors of a given linear transformation will be presented in Chapter 5.

(4.26) Example. Let V be an n-dimensional vector space over the field F,

and if A E F, define a linear transformation T : V - V by T(v) = av for all v E V. Then, considering the F[XJ-module VT, we have Xv = Av,

i.e., (X - A)v = 0 for every v E V. Thus X - A = mT(X) since X - A is the monic polynomial of lowest possible degree (namely, degree 1) with

MT(T) = 0. Then cT(X) = (X - A)' since deg(or(X)) = n and the only prime factor of cT(X) is (X - A) by Corollary 4.10 (3). Then the rational canonical form of T is

242

Chapter 4. Linear Algebra n

®C(X-A)=AIn. i=1

Of course, this is the matrix of T in any basis of V. Also note that VT 25

Rv1 ® ® Rvn has rank n over R with each cyclic submodule Rv, 25 R/(X - A) having dimension 1 over the field F.

(4.27) Example. Let B = { v1i ... , vn } be a basis of the n-dimensional

vector space V, and let T : V -. V be defined by T(vi) = A,v, where A, E F. Assume that the A, are all distinct, i.e., a, # A3 for i 0 j. Then each subspace (v,) is a T-invariant subspace and hence a submodule of the F[X )-module VT. Therefore,

VT =Rv,e...®Rvn where Ann(vi) = (X - A,). Note that

me(Rv,) = Ann(vi) = (X - A,), so Proposition 3.7.21 implies that MT(X) = me(VT)

= lcm{me(Rvl), ... , me(Rvn)} n

=

fl(X

- Ai)

i=1

= f(X) Also by Proposition 3.7.21, we see that cT(X) = f (X). Therefore MT(X) _ cT(X) and Lemma 4.11 (3) shows that the R-module VT is cyclic with annihilator (f (X)). Thus the rational canonical form of T is 0 1

0 0

0

1

0 0

0 0

... ...

0

0

0 0

0 0

-a0 -a1 -a2

1

0

-an_2

0

1

-an_,

[T)eo = C(f(X)) =

where f (X) = X' + an- 1 X n -1 + appropriately.

...

+ a 1 X + ao and the basis Bo is chosen

This example actually illustrates a defect of the rational canonical form. Note that [T)5 = diag(A1, A2, ... , An), which is a diagonal matrix. By comparison, we see that [T)13 is much simpler than [T] and it reflects

the geometry of the linear transformation much more clearly. Our next goal is to find a canonical form that is as "simple" as possible, the Jordan


243

canonical form. When a transformation has a diagonal matrix in some basis, this will indeed be its Jordan canonical form. This special case is important enough to investigate first. (4.28) Definition.

(1) A linear transformation T : V -* V is diagonalizable if V has a basis such that 171E is a diagonal matrix.

(2) A matrix A E Mn(F) is diagonalizable if it is similar to a diagonal matrix.

(4.29) Remark. Recall that we have already introduced the concept of diagonalizability in Definition 3.35. Corollary 3.34 states that T is diagonalizable if and only if V posseses a basis of eigenvectors of T. Recall that v 0 E V is an eigenvector of T if the subspace (v) is T-invariant, i.e., T(v) = Av for

some A E F. The element A E F is an eigenvalue of T. We will consider criteria for diagonalizability of a linear transformation based on properties of the invariant factors.

(4.30) Theorem. Let T : V - V be a linear transformation. Then T is diagonalizable if and only if mT(X) is a product of distinct linear factors, i. e., t

mT(X) = [J(X - Ai) i=1

where A1i ... , At are distinct elements of the field F.

Proof. Suppose that T is diagonalizable. Then there is a basis B of V such that [T]B = diag(a1i ... , an). By reordering the basis B, if necessary, we can assume that the diagonal entries that are equal are grouped together. That is, [T]B = ® Ailn i=1

where n = n1 +

+ nt and the Ai are distinct. If

B = {v11, ... t V1n,, V21, ...

, V2n2, .

. .

, vtl, ... , vine },

then let Bi = {vi1, ... , vin, } and let V = (B,). Then T(v) = Aiv for all v E Vi, so Vi is a T-invariant subspace of V and hence an F[X]-submodule of VT. from Example 4.26, we see that me(V) = X - A,, and, as in Example 4.27,

mT(X) = me(VT) = H(X - A1) i=1

as claimed.

244


Conversely, suppose that mT(X) = rj;=1(X - Ai ), where the Ai are distinct. Since the X -,\i are distinct irreducible polynomials, Theorem 3.7.13 applied to the torsion F[X]-module provides a direct sum decomposition (4.20)

VT

ED

ED

t

where Ann(Vi) = (X - Ai). In other words, Vi is a T-invariant subspace of V and Ti = TIy, satisfies T, - Ai = 0, i.e., T(v) = Aiv for all v E V,. Then, by Example 4.26, if Vi has a basis Bi = (vi1, ... , vi,,), then [T;]8, = Ail,,,, and if B = U;=1Bi, then B is a basis of V by Equation (4.20) and [T]8 = so T is diagonalizable. (4.31) Corollary. Let T E EndF(V) be diagonalizable. Then the exponent of (X - Ai) in the characteristic polynomial cT(X) is equal to the number of times that Ai appears on the diagonal in any diagonal matrix (T]8.

Proof. If Ai appears ni times on the diagonal in a diagonal matrix representation of T, then

n, = dimF Ker(T - Ail,). This number depends only on T and not on any particular diagonalization of T. Now suppose that (T]8 = ®;=1Ai1,,,, let Vi = Ker(T - ally), and let Ti = T I v, as in the proof of Theorem 4.30. Then by Proposition 3.7.21 cT(X) = co(T) = co(T1)

co(Tt)

=CT'(X)...CT,(X) = (X - A1)"' ... (X - At)"' as claimed.

Since, by Proposition 4.23, we have an independent method for calculating the characteristic polynomial cT(X), the following result is a useful sufficient (but not necessary) criterion for diagonalizability.

(4.32) Corollary. Let V be an n-dimensional vector space and let T E EndF(V) be a linear transformation. If the characteristic polynomial cT(X)

is a product of distinct linear factors, then MT(X) = cT(X) and hence T is diagonalizable.

Proof. Suppose that cT(X) = rj; 1(X -,\i) where the Ai are distinct. Since every irreducible factor of cT(X) is also a factor of MTV), it follows that CT(X) divides mT(X). But since mT(X) always divides cT(X) (Corollary 4.10), it follows that mT(X) = cT(X ). The diagonalizability of T then follows from Theorem 4.30.


245

(4.33) Remark. It is interesting to determine the cyclic decomposition of VT when T is diagonalizable. By Theorem 4.30, we have

mT(X) _ (X - A1) ... (X - At) and

cT(X) = (X - A1)" ... (X - At)nk

+ nt. If k = where the A, are distinct and n = dim(V) = nl + Rvl ® ® Rvk, and the inmax{nl, ... , nt} then rank(VT) = k, VT variant factors of the torsion R-module VT are the polynomials

fk+1_i(X) = (X - A)e(+,nt) ... (X - At)e(i,nt) where 1

IE(i'j) =

0

for

1 j.

The following special case of Theorem 4.30 will be useful to us later.

(4.34) Corollary. Suppose that V is a finite-dimensional vector space over

a field F and let T : V -+ V be a linear transformation such that TIC = 1y. Suppose that F is a field in which the equation zk = 1 has k distinct solutions. Then T is diagonalizable.

Then X k - 1 = Proof. Let the solutions of zk -1 be 1 I [J k- (X - () is a product of distinct linear factors. By hypothesis, Tk ly =o0, so T satisfies the polynomial equation Xk - 1 = 0, and hence mT(X) divides Xk -1. But then mT(X) is also a product of distinct linear factors, and hence T is diagonalizable. (4.35) Remark. Note that the hypothesis on the field F is certainly satisfied for any k if the field F is the field C of complex numbers. From Theorem 4.30, we see that there are essentially two reasons why a linear transformation may fail to be diagonalizable. The first is that mT(X)

may factor into linear factors, but the factors may fail to be distinct; the second is that mT(X) may have an irreducible factor that is not linear. For example, consider the linear transformations T. : F2 by multiplication by the matrices

Al=[0 0]

r

and

F2, which are given 11

A2=[01 p1.

Note that mT, (X) = X2, so Tl illustrates the first problem, while mT, (X) _ X2 + 1. Then if F = R, the real numbers, X2+1 is irreducible, so T2 provides an example of the second problem. Of course, if F is algebraically closed (and in particular if F = C), then the second problem never arises.

246


We shall concentrate our attention on the first problem and deal with the second one later. The approach will be via the primary decomposition theorem for finitely generated torsion modules over a PID (Theorems 3.7.12 and 3.7.13). We will begin by concentrating our attention on a single primary cyclic R-module.

(4.36) Definition. Let A be in the field F and let n E N. An n x n Jordan block with value A is the matrix A

1

0

0

A

1

0 0

0

0

0

0

(4.21)

Note that Ja,n = AIn + Hn where n-1

Ei,i+1 E Mn(F)

Hn =

i-1

That is, Hn has a 1 directly above each diagonal element and 0 elsewhere. Calculation shows that n-k

(Jan - Aln)k = Hn =

Ei i+k

0

for

1

i=1 but

(Ja,n-AIn)n=Hn =0. Fn be the linear transformation obtained Therefore, if we let TA,n : F" by multiplying by Ja,n, we conclude that TTA,n (X) = (X - A)n = CT"..

(since deg(cT,,.,, (X)) = n) so that Lemma 4.11 (3) shows that the FIX]module (F")T,,,,, is cyclic.

(4.37) Proposition. Let V be a finite-dimensional vector space over the field

F, and suppose that T E EndF(V) is a linear transformation such that the R-module VT is a primary cyclic R-module. Suppose that Ann(VT) _ ((X - A)n), and let v E V be any element such that VT = Rv. Then

8={vk=(T-Aly)n-k(v):1
[T] e = Ja,n.


247

Proof. First we show that 13 is a basis of V. Since dim V = n by Lemma 4.11 (3), and since B has n elements, it is only necessary to show that B is linearly independent. To see this, suppose that Ek=1 akvk = 0 where

al,...,an E F. Then

Ek=1

ak(T -

AJv)n-k(v) = 0, i.e.,

n

g(X) = E ak(X - A)n-k E Ann(v) = Ann(V). k=1

But deg(g(X)) < n, so this can only occur if g(X) = 0, in which case a1 =

= an = 0. Thus B is linearly independent and hence a basis of V. Now we compute the matrix [T]B. To do this note that

T(vk) = T ((T - A)n-k(v)) = (T - A)(T - ))n-k(v) + A(T - A)n-k(v)

= (T =

A)n-(k-1)(v)

+ $T - $n-k(v)

Jvk_1+.vk ifk>2, ifk = 1.

1 Jivk

0

Therefore, [T]B = Ja,n, as required.

(4.38) Theorem. (Jordan canonical form) Let V be a vector space of dimension n over a field F and let T : V - V be a linear transformation. Assume that the minimal polynomial mT(X) of T factors into a product of (not necessarily distinct) linear factors. (Note that this hypothesis is automatically satisfied in case F is an algebraically closed field and, in particular

if F = C.) Then V has a basis B such that

[T]B=J=(Dii i=1

where each Ji is a Jordan block. Furthermore, J is unique up to the order of the blocks. (The matrix J is said to be in Jordan canonical form.) Proof. Let VT ®;=1Vi be the primary decomposition of the torsion F[X]module VT (see Theorem 3.7.12). According to the proof of Theorem 3.7.12,

each V is the pi (X)-primary component of VT for some irreducible polyno-

mial pi(X) dividing me(VT) = mT(X), so by the assumption on mT(X), each pi (X) is linear, i.e., pi (X) = X - Ai for some Ai E F. According to Theorem 3.7.13, each module V, has a decomposition into primary cyclic submodules

1;=Wit®

®Wi.,

for

1

By Proposition 4.37, there is a basis Bi; of Wig in which the restriction of T to Wig (recall that submodules of VT are T-invariant subspaces of V) is a Jordan block. Let B = Ui,3Bi,. Then B is a basis of V and [T]B is in Jordan canonical form.

248


It remains to show uniqueness of the Jordan canonical form, but this is immediate from the fact that the blocks are in one-to-one correspondence with the elementary divisors of the module VT-the elementary divisor corresponding to JA,4 is (X - A)9. We have already briefly encountered eigenvalues of linear transformations (see Definition 3.35 and Remark 4.29). We now consider this concept

in more detail and relate it to the canonical form theory just developed. Recall that A E F is an eigenvalue of T : V - V if T(v) = Av for some nonzero v E V. The nonzero element v is an eigenvector of T corresponding to A.

(4.39) Lemma. Let V be a finite-dimensional vector space over a field F and

let T E EndF(V) be a linear transformation. If A E F, then the following are equivalent.

(1) A is an eigenvalue of T. (2) X - A divides mT(X). (3) X - A divides cT(X). Proof. (1) =:> (2). Let v be an eigenvector of T corresponding to A. Then

(X - A) E Ann(v) D Ann(VT) = (mT(X)),

so X - A divides mT(X). (2) (1). Immediate from Theorem 3.7.1 and Lemma 3.7.17. (2) (3). mT(X) and cT(X) have the same irreducible factors (Corollary 4.10).

For convenience, we restate Corollary 3.34 and Theorem 4.30 in the current context.

(4.40) Theorem. Let T : V

V be a linear transformation of a finite-

dimensional vector space over the field F. Then the following are equivalent:

(1) T is diagonalizable. k2) mT(X) is a product of distinct linear factors. (3) V has a basis consisting of eigenvectors of T.

Proof. (1) a (2) is Theorem 4.30, while (1) q (3) is Corollary 3.34.

we see that if T is (4.41) Remark. Since diag(A1, diagonalizable, then the Jordan canonical form of T is diagonal. (4.42) Definition. Let T E EndF(V). Then a nonzero vector v E V is a generalized eigenvector of T corresponding to the eigenvalue A E F if p(T)(v) = 0 for p(X) = (X - A)k for some k > 0.


249

In other words, v $ 0 is a generalized eigenvector of T corresponding to the eigenvalue A if v is in the (X - A)-primary component Jf the F[X]module VT.

(4.43) Lemma. If {vi} 1 are generalized eigenvectors of T corresponding to distinct eigenvalues, then they are linearly independent. Proof. Decompose e

(4.22) i=1

by the primary decomposition theorem (Theorem 3.7.12). After reordering,

if necessary, we may assume that v; E V for 1 < i < r. If F,1 aivi = 0 then it follows that aivi = 0 since Equation (4.22) is a direct sum, so ai = 0

0

since vi 54 0.

(4.44) Theorem. The following are equivalent for a linear transformation

T:V-+V. (1) T has a Jordan canonical form. (2) V has a basis B consisting of generalized eigenvectors of T.

(3) MT(X) is a product of (not necessarily distinct) linear factors.

(2). If [T]p is in Jordan canonical form then the basis B Proof. (1) consists of generalized eigenvectors. be a basis of V, and assume (3). Let B = {vj, ... , (2) (T - ai)k' (vi) = 0, i.e., each vi is assumed to be a generalized eigenvector of T. Then

mT(X) = me(VT) = lcm{(X - A1)k', ... (X - \,a)k-} is a product of linear factors. (1). This is Theorem 4.38. (3)

0

Now we define some important invariants.

(4.45) Definition. Let T : V -. V be a linear transformation and let A E F be an eigenvalue of T. (1) (2)

The algebraic multiplicity of the eigenvalue A, denoted vj8(A), is the highest power of X - A dividing cT(X). The geometric multiplicity of the eigenvalue A, denoted vge,,,m(A) is the

dimension (as a vector space over F) of Ker(T - Alv).


250

(4.46) Remarks. (1) (2)

(3)

Ker(T - Alv) = {v E V : T(v) = Av} is called the elgenspace of A.

{v E V : (T - \)k(V) = 0 for some k E N} is called the generalized eigenspace of the eigenvalue A. Note that the generalized eigenspace corresponding to the eigenvalue A is nothing more than the (X - A)primary component of the torsion module VT. Moreover, it is clear from the definition of primary component of VT that the generalized eigenspace of T corresponding to A is Ker(T - Alv)' where r is the exponent of X - A in MT (X) = me(VT). Note that Lemma 4.43 implies that distinct (generalized) eigenspaces are linearly independent.

(4.47) Proposition. Let A be an eigenvalue of the linear transformation T E

EndF (V) where V is a finite-dimensional vector space over F. Then the geometric multiplicity of A is the number of elementary divisors of VT that are powers of (X - A). In particular, (4.23)

1 < vgeom(A) < vlg(A).

Proof. First note that 1 < veeom(A) since A is an eigenvalue. Now let VT ®;_1V, be the primary decomposition of the F[X]-module VT, and assume (by reordering, if necessary) that V1 is (X -A)-primary. Then T-A : V; -+ V; is an isomorphism for i > 1 because Ann([;) is relatively prime to X - A for i > 1. Now write V1 c--- W1® ... ( D

as a sum of cyclic submodules. Since V1 is (X - A)-primary, it follows that

Ann(Wk) = ((X - A)9;) for some qk > 1. The Jordan canonical form of Taw,, is J,\,4,, (by Proposition 4.37). Thus we see that Wk contributes 1 to the geometric multiplicity of A and qk to the algebraic multiplicity of A. 0 (4.48) Corollary.

The geometric multiplicity of A is the number of Jordan blocks J,\,, in the Jordan canonical form of T with value A. The algebraic multiplicity of A is the sum of the sizes of the Jordan blocks of T with value A. If mT(X) = (X - A)Qp(X) where (X - A) does not divide p(X), then q is the size of the largest Jordan block of T with value A. If A1, ..., Ak are the distinct eigenvalues of T, then .u(VT) = max{vgeom(Aj) : 1 < i < k}.

Proof. (1), (2), and (3) are immediate from the above proposition, while (4) follows from the algorithm of Theorem 3.7.15 for recovering the invariant factors from the elementary divisors of a torsion R-module. 0


251

(4.49) Corollary. The following are equivalent for a linear transformation T E EndF(V). (1) T is diagonalizable. (2) If,11, ... , A, are the distinct eigenvalues of T, then k

vgeom()i) = dimF(V). i=1

(3) mT(X) is a product of linear factors, and for each eigenvalue a, ugeom(A) = vaig(A).

Proof.

(4.50) Remark. The importance of the Jordan canonical form of a linear transformation can hardly be over-emphasized. All of the most important invariants of a linear transformation: its characteristic and minimal polynomials, its elementary divisors, its eigenvalues, and their algebraic and geometric multiplicities, can all be read off from the Jordan canonical form, and the (generalized) eigenspaces may be read off from the basis with respect to which it is in this form. We have seen that in order for T to have a Jordan canonical form, the minimal polynomial mT(X) must be a product of linear factors. We will conclude this section by developing a generalized Jordan canonical form to handle the situation when this is not the case. In addition, the important case F = R, the real numbers, will be developed in more detail, taking into the special relationship between the real numbers and the complex numbers. First we will do the case of a general field F. Paralleling our previous development, we begin by considering the case of a primary cyclic R-module (compare Definition 4.36 and Proposition 4.37).

(4.51) Definition. Let f (X) E F[X] be a monic polynomial of degree d. The

nd x nd generalized Jordan block corresponding to f(X) is the nd x nd matrix (given in blocks of d x d matrices)

C(f(X)) 0

N

0

C(f(X)) N ...

0

0

0 0

Jf(X),n = L

0 0 0

0 0

0 ....C(f(X)) cucX,> 0 0

0

N

cC xX

where N is the d x d matrix with entld(N) = 1 and all other entries 0. Jf(X),n is called irreducible if f (X) is an irreducible polynomial. Note, in

252


particular, that the Jordan matrix J),,, is the some as the irreducible Jordan block J(X_A),n.

(4.52) Proposition. Let T : V - V be a linear transformation, and assume that VT = Rv is a (primary) cyclic R-module such that Ann(v) = Ann(VT) = y(X)n), where f (X) is an (irreducible) polynomial of degree d. Define B as follows: Let vk = f(T)n-k(v) for 1 < k < n and let vkl = Tj-'(vk) for 1 < j < d. Then let 1

B = (vll,...,vlk, V21,...,V2d,.... un1) ...,vnd}.

Then B is a basis of V over F and [T]g is an (irreducible) nd x nd generalized Jordan block corresponding to f (X). Proof. This is a tedious computation, entirely paralleling the proof of Proposition 4.37; we shall leave it to the reader.

(4.53) Remark. Note that if f (X) = X - A, then Proposition 4.52 reduces to Propositon 4.37. (Of course, every linear polynomial is irreducible.)

(4.54) Theorem. (Generalized Jordan canonical form) Let V be a vector space of dimension n over a field F and let T : V - V be a linear transformation. Then V has a basis B such that

[T]a=J'=®J; i=1

where each J' is an irreducible generalized Jordan block. Furthermore, J'

is unique up to the order of the blocks. (The matrix J' is said to be in generalized Jordan canonical form.)

Proof. Almost identical to the proof of Theorem 4.38; we leave it for the reader.

0

Now let F = R be the field of real numbers. We will produce a version of the Jordan canonical form theorem which is valid for all matrices with entries in R. This canonical form will be somewhat different than the generalized Jordan canonical form of Theorem 4.54. Recall (Theorem 4.44) that a linear transformation T : V -4 V, where V is a finite-dimensional vector space over a field F, has a Jordan canonical form if and only if the minimal polynomial mT(X) is a product of (not necessarily distinct) linear factors. Of course, this says that all of the roots of mr(X ), i.e., all eigenvalues of T, are included in the field F. This condition

will be satisfied for all polynomials over F if and only if the field F is algebraically closed. The field C of complex numbers is algebraically closed (although we shall not present a proof of this fact in this book), but the field R of real numbers is not. In fact, X2+1 is a polynomial over R that does not


253

have any roots in R. Therefore, any linear transformation T E EndR(R2) such that mT(X) = X2 + 1 cannot have a Jordan canonical form. There is, however, a very simple variant of the Jordan canonical form which is valid over R and which takes into the special nature of R as it relates to the complex numbers. We will start by analyzing polynomials over R. Recall that if z E C then 'z denotes the complex conjugate of z, i.e., if

z = a+bi, then

a-bi. Then z(z+'') = a, while zi(z-z) = band z E R

if and only if z =1. The conjugation map on C extends in a natural way to

an involution a : C[XI - C(X] defined by a(f (X))= f(X) where 7W is the polynomial determined by conjugating all the coefficients of f (X), i.e.,

if f(X) = a

ao +a1X a ring homomorphism such that a2 = lc(XI, and

f (X) E R[X] if and only if 7(X) = f (X ). If f (X) E C[X] and z E C is a root of f (X), then a simple calculation shows that z is a root of f (X ). Thus if f (X) E R[X] and z E C is a root of f (X), then z is also a root of f(X). By Corollary 2.4.6 it follows that whenever z E C \ R is a root of f (X), then the polynomial (X - z)(X - -z) divides f (X) in C[X]. But if z= a+ bi, where a, b E R and b O 0, then

(X - z)(X - z) _ ((X - a) - bi)((X - a) + bi) _ ((X - a)2 + b2). Since the polynomial (X - a)2 + b2 has real coefficients and divides the real polynomial f (X) in the ring CIX], it follows from the uniqueness part of the division algorithm (Theorem 2.4.4) that the quotient of f (X) by (X - a)2 + b2 is in R[X]. From this observation and the fact that C is algebraically closed, we obtain the following factorization theorem for real polynomials. (4.55) Lemma. Let f (X) E R[X]. Then f (X) factors in R[X] as (4.24)

f(X)=a (h(X -ci )'"'

X-a )2+b2 "' j=1

whereci ER, aj,b3 ERwith bj #0, andaER. Proof. The real roots c1, ... ,c,. of f(X) give the first factor, while if zj = aj + ibj is a complex root, then (X - aj )2 + b? divides f (X), so we may argue by induction on the degree of f (X ).

13

Note that the factorization of Equation (4.24) implies that if zj is a nonreal root of f(X) of multiplicity nj, then zj is a root of the same multiplicity nj .

(4.56) Definition. Given z = a + bi E C with b # 0 and r E N, let JR E M2r(R) be defined by

254


(4.25)

JR =

A

12

0

0

0

0

A

12

0

0

A

12

0 . 0

A

.

.

.

0 0

0 0

0

...

where A = [ b -b]. JR is said to be a real Jordan block corresponding to

zEC\R. (4.57) Theorem. (real Jordan canonical form) Let V be a finite-dimensional

vector space over R and let T : V - V be a linear transformation. Then there is a basis B of V such that [T)8 = ® J, i=1

where J, =

is an ordinary Jordan block corresponding to .\, E R, or

J, = JR k, is a real Jordan block corresponding to some z, E C \ R. Proof. By the primary decomposition theorem (Theorem 3.7.13), we may suppose that the R[XJ-module VT is a primary cyclic R[XJ-module, say V7 = R[X]u. Thus, Ann(-,,) = (p(X)') where (by Lemma 4.55) (4.26)

p(X) = X - c

or (4.27)

p(X) = (X - a)2 + b2

with b -A 0.

In case (4.26), T has a Jordan canonical form J,,,.; in case (4.27) we will show that T has a real Jordan canonical form JR where z = a + bi. Thus, suppose that p(X) = (X -a)2+b2 and let z = a+bi. By hypothesis VT is isomorphic as an R[XI-module to the R[XJ-module R[X]/(p(X )'').

Recall that the module structure on V7 is given by Xu = T(u), while the module structure on R[X]/(p(X)') is given by polynomial multiplication. Thus we can analyze how T acts on the vector space V by studying how X acts on R[XJ/(p(X)') by multiplication. This will be the approach we will follow, without explicitly carrying over the basis to V. Consider the C[X]-module IV = C[XJ/(p(X)'). The annihilator of W

is p(X)', which factors in C[XJ as p(X)' = q(X)'q(X)' where q(X) ((X - a) + bi). Since q(X)' and q(X )' are relatively prime in CJX], 4(X)r92(X (4.28) 1 = q(X)r91(X) + ) for g, (X) and 92(X) in C[XJ. Averaging Equation (4.28) with it's complex conjugate gives


1 = q(X)r

(91(X) + 92(X) 1 + 2

J

4(X) I*

255

(92(X) + 91(X) ) 2

JJ

= f(X) +7(X). Thus, by Theorem 3.7.12, we may write (4.29)

W = C[X J/(p(X )r) = (v) ® (w),

where v = f (X) + (p(X )r), w = 7(X) + (p(X )r), Ann(v) = (4(X)') and Ann(w) = (q(X)'). Equation (4.29) provides a primary cyclic decomposition of C[XJ/(p(X)r) and, as in the proof of Theorem 4.37, C = {q(X )r-kw : 1 < k < r} U {q(X )r-kv : 1 < k < r}

is a basis of W = C(XJ/(p(X)r) over C. For 1 < k < r, let 1

Vk = and

Wk =

2

(q(X)r-kw + q(X )r-kV) E W

I (q(X )r-kw - q(X )r-kv)

E W.

Note that vk = vk and wk = wk since v = w, where the conjugation map on W = C[XJ/(p(X)r) is induced from that on C[X). Therefore vk and wk are in R[XJ/(p(X)r). Moreover, Vk + iWk = q(X)r-kw and vk - iWk = q(X )r-kv Therefore, the set B = {v1, Wi, v2, w21 ... vr, wr} C RIXJI(p(X)r)

spans C[XJ/(p(X)r), and hence, it is a basis. Since 8 is linearly independent over C, it is also linearly independent over R, and since

dimR R[X]/(p(X)r) = deg p(X )r = 2r,

it follows that 5 is a basis of R[XJ/(p(X)r) over R. Now we compute the effect of multiplication by X on the basis 8. Since

X =q(X)+(a-bi) and X =q(X)+(a+bi), Xvk = X

((q(xy_cw + q(X)r-kv)

= 2 (q(X) + (a - bi)) q(X )r-kw + 2 (q(X) + (a + bi)) q(X )r-kv = 2 (q(X)r-1k-1),w .+ +a

q(X)r-(k-1).vl

((q(xy_kw + q(X )r-kv)) + b ((q(x)y_kw - q(X )r-kv) Zi

= 1 vk_ I + avk + bwk lavk + bwk

if k > 2, if k = 1.

256


Similarly,

if k > 2, if k = 1.

Wk-1 - bvk +awk -bvk + awk

x Wk =

Thus, if S : R[X]/(p(X)r)

R(X]/(p(X)r) denotes multiplication by

X, then (S]e = JR , and since S corresponds to T under the isomorphism VT

R[XJ/(p(X)r), the proof is complete.

O

(4.58) Remark. It is worthwhile to compare the real Jordan canonical form derived in Theorem 4.57 with the generalized Jordan canonical form determined by Theorem 4.54. Suppose that f (X) = ((X - a)2 + b2)2. Then the real Jordan block determined by f (X) is a b 0 L0

R

-b

1

0

a 0

0 a

-b

0

b

a

1

while the generalized Jordan block determined by f (X) is 0

JI(x),2 =

-(a2 + b2) 0

1

0 0

2a 0 0

1

0

0 0

-(a2 + b2)

1

2a

(4.59) Corollary. If V is a finite-dimensional real vector space of dimension at least 2, then every linear transformation T : V -+ V has a 2-dimensional invariant subspace.

Proof. T has an elementary divisor of the form

(X - c)r

for r > 2,

((X-a)2+b2)r forr>1,

or T is diagonalizable. If T is diagonalizable, the result is clear, while if the real Jordan canonical form of T contains a block Jo,r (r > 2) the vectors corresponding to the first two columns of Jr,,. generate a two-dimensional invariant subspace. Similarly, if the real Jordan canonical form contains a block JR (r > 1), then the vectors corresponding to vl and w1 constructed in the proof of Theorem 4.57 generate a T-invariant subspace. 0

(4.60) Remark. If a linear transformation possesses an eigenvalue, then it has an invariant subspace of dimension 1. Corollary 4.59 is of interest in that it states that if a linear transformation of a real vector space does not have any eigenvalues, then at least there is the next best thing, namely, a two-dimensional invariant subspace. If F is a field that has an irreducible polynomial f (X) of degree n, then multiplication by A = C(f(X)) on F"

4.5 Computational Examples

257

is a linear transformation with no invariant subspace of dimension less than n. Thus Corollary 4.59 depends on the special nature of the real numbers.

At this point we would like to reemphasize the remark prior to Proposition 4.3.

(4.61) Remark. Our entire approach in this section has been to analyze a linear transformation T by analyzing the structure of the F[X]-module VT. Since Tl and T2 are similar precisely when VT, and VT, are isomorphic, each and every invariant we have derived in this section-characteristic and minimal polynomials, rational, Jordan, and generalized Jordan canonical forms, (generalized) eigenvalues and their algebraic and geometric multiplicities, etc.-is the same for similar linear transformations Ti and T2.

4.5 Computational Examples This section will be devoted to several numerical examples to illustrate the general canonical form theory developed in Section 4.4. In Chapter 5, further techniques will be presented using an analysis of equivalence of matrices over a PID. Throughout this section it will be assumed that the reader is familiar with the procedure for solving a system AX = B of linear equations over a field F by the process of row reduction of the matrix [A B]. This procedure is a standard topic in elementary linear algebra courses; it will be developed in the more general context of matrices with entries in a PID in Section 5.2.

(5.1) Example. Construct, up to similarity, all linear transformations T FB - Fe with minimal polynomial

MT(X) = (X - 5)2(X - 6)2. Solution. If mT(X) = (X - 5)2(X - 6)2, then we must have

CT(X) = (X - 5)'(X - 6)j

where i + j = 6 = dirn(F'6) and 2 <_ i, 2 < j. This is because mT(X) divides CT(X) (Corollary 4.10 (2)), and MAX) and cT(X) have the same irreducible factors (Corollary 4.10 (3)). Suppose i = 2 and j = 4. Then 5 has algebraic multiplicity 2, and 2 is the size of the largest Jordan block with eigenvalue 5, so there is exactly one such block. Also 6 has algebraic multiplicity 4, and 2 is the size of its largest Jordan block, so either it has two blocks of size 2, or one block of size 2 and two of size 1. Thus the possibilities for the Jordan canonical form of T are

Algebra

Linear

4.

Chapter

258

0 0 0 0 1 5 0 0 0 0 5 0

0 0 1 6 0 0 6 0

1 6 0 0

6 0 0 0

000 0 000 0 = 2, J6

0 0 0 0 1 5

0 0 0 0 0 0 0 1 6 0 6 0 50 0 0 0 0 =

I

,

J6 ED

I

E6

J6 (1

J6 G)

J5

,

2

2,

,

2,

J5

2

9 ,

J6

and

0 6 6 0 0 0 0 0 0 0 0 0

of roles

the

with

analysis,

similar

a is there

then

2, =

j and

4

=

i

If

are

T of form

canonical

Jordan

the

for

possibilities

The

reversed.

6

5

and

then

I. J5, EI)

®

J5,1

J5,2

(D J6.2

or

J5,2

Eli Js,2

4) J6,2

of block

one

have

must

we

eigenvalue

each

for

then

=

3,

j

and

3

=

i

If

possibility

single

the

gives

This

I. size

of one

and

2

size

00 00 00 0 1 0 0 0 6 0 05 0 1 50 0 5 00 0 = 1

,

,

2Ei)J6

IE0J6 ,

2®J5,

J5

0 6 6 0 0 0 0 0 00 0 0

0

T. of

form

canonical

Jordan

the

for

with

T

transformations

linear

all

similarity,

to

up

Construct,

Example.

(5.2)

_

mT(X)

polynomial

minimal

-8)7,

(X =

CT(X)

polynomial

characteristic

4.

multiplicity

geometric

of

eigenvalue

an

8

and

-

8)3,

(X

are

7-

a

we

dimensional

vector

space

V,

which

we

may

take

to

be

F7,

since

of

transformation

linear

a is

T

7,

=

deg(cT(X))

Since

Solution.

Jordan

degree

blocks

in

the

Jordan

canonical

form

of

T

are

Js,,..

Since

MT(X)

the has

Thus

T.

of

eigenvalue

only

the

is

8

and

similarity,

to

up

working

3,

or

equal

geometric

multiplicity

1, is 3, 8 3, of

be

must

sizes

block

the

or

1, 2, 3,

1,

or

3,

1. 1, 1, 1,

Since

so the

3,

size

has

block

Jordan

largest

ways

only

the

and

(1)),

4.48

(Corollary

blocks

Jordan

2+2+2+1,

=

=

3+2+1+1

=

4+1+1+1

7

are

parts

into

to

partition

7

44

be

must

there

the 2, 4, 3, 2, to

these

Comparing

(2)).

4.48

(Corollary

sizes

block

possible

the

being

these

possi-

one

only

is

there

thus

coincidence;

only

only

is

there

lists,

two

last

T:

for

bility


J8.3®J8,2®41 ®J8,1 =

8

1

0

0

0

0

0

0

8

1

0

0

0

0

0

0

8

0

0

0

0

0

8

1

0

0

0

0

8

0

0

0

0

0

0

0

0

0

0

0 0 0 8 0

0 0 0 0 8

259

0

In contrast to the first two examples, the problems which generally arise are not to construct matrices in Jordan canonical form, but rather to find the Jordan canonical form of a given linear transformation (or matrix). In this, Proposition 4.23 is our starting point, telling us how to find the characteristic polynomial. (We do not wish to address at this point the problem of practical methods for computing the characteristic polynomial,

but simply assume that this is done. Chapter 5 will contain further information, which will be of computational interest in computing canonical forms.)

V be the linear transformation

(5.3) Example. Let V = F3 and let T : V with matrix 1

[T]c =A= -2 -2

-1

1

1

2

-1

4

where C is the standard basis on F3. Find out "everything" about T. Solution. A calculation of the characteristic polynomial cT(X) shows that

CT(X) = det(XI3 - A) = X3 - 6X2 + 11X - 6

= (X - 1)(X - 2)(X - 3). Since this is a product of distinct linear factors, it follows that T is diagonalizable (Corollary 4.32), mT(X) = cT(X), and the Jordan canonical form of T is J = diag(1, 2, 3). Then the eigenvalues of T are 1, 2, and 3, each of which has algebraic and geometric multiplicity 1. Since MT(X) = cT(X), it follows that VT is a cyclic F[X]-module and the rational canonical form of T is 0

C(mT(X)) = R =

1

0 0

-11

6

0

1

6

We now compute the eigenspaces of T. This is a straightforward matter of solving systems of linear equations. The details are left to the reader. If VA = Ker(T - A1.. ), then we find


260

/

`

v,=(

1

v2=( o >,

},

\

\

V3=

(

0 1

1

Thus, if 1 1

C3

,

1

1

0

0

1

1

1

I I,

then [T]8 = J. Also, if

P=

1

1

0

1

0

1

1

1

1

is the matrix whose columns are the vectors of 5, then P-1 AP = J = diag(1, 2, 3). Finally, if we let 1

1

0

2

1

3

v= 1+ 0+ 1= 2 1

1

then

Ann(v) = ((X - 1)(X - 2)(X - 3)) = (mT(X)) = Ann(VT). Thus, v is a cyclic vector for the FIX]-module VT. If we let

1[21 5' _ {v,Tv,T2v} =

2

,

3

3 4

[511 ,

6

10 14

be the basis of V given by Corollary 4.16, then (TJ8- = R, and if Q is the matrix whose columns are the column vectors of 8', then Q-1AQ = R. (5.4) Example. Let V = F3 and let T : V -+ V be the linear transformation with matrix with respect to the standard basis C [T]c = A =

6 6

-2 -2 -1 -3

4

-2

0

Find out "almost everything" about T. Solution. We compute

CT(X) = det(XI3 - A)

=X3-5X2+8X-4 = (X - 1)(X - 2)2. We see at this point that we have two possibilities:


261

(a) valg(1) = vgem(1) = 1 and vatg(2) = vgeom(2) = 2.

In this case T is diagonalizable and the Jordan canonical form is J = diag(1, 2, 2). Also mT(X) = (X - 1)(X - 2), so the F[X]-module VT has invariant factors fl(X) = X - 2 and f2(X) = (X - 1)(X - 2), and VT has rank 2 as an FIX]-module. Thus, T has the rational canonical form R = C(f1(X )) ®C(f2(X )) =

2

0

0

0 0

1

3

0 -2

(b) valg(l) = ugeom(1) = 1 and valg(2) = 2, vgeom(2) = 1.

In this case T is not diagonalizable, and the eigenvalue 2 has a single Jordan block, so its Jordan canonical form is

J=J1,1®J2,2=

1

0

0

2

1

0

0

2

0

Also, mT(X) = (X- 1)(X-2)2 = cT(X) so that VT is a cyclic F[X]-module (Lemma 4.11 (3)) and T has the rational canonical form 0

R = C(mT(X)) =

1

0 0

-8

0

1

5

4

To decide between these two alternatives, we may proceed in either of two ways:

(1) Compute vg,,,,m(2). If it is 2, we are in case (a), and if it is (1), we are in case (b).

(2) Compute (T - 1)(T - 2) = (T - 1v)(T - 2. 1v). If it is 0 we are in case (a), if it is not, we are in case (b). Since we are in any event interested in (generalized) eigenvectors, we

choose the first method. We find

Ker(T - 2 lv) _

[]) = (v2, vs), ([i]

2

s o we are in case (a) here. Also,

[21 Ker(T - 1v) _ (

= (vl),

3 2

1 JJ

so in the basis B = {v1, v2i v3} (note the order), [T]8 = J.

0

262


(5.5) Example. Let V = F3 and let V -+ V be the linear transformation with the matrix

-5

5

[T]c = A = -4 -5

7

7

-6

7

-7

in the standard basis C. Find out "almost everything" about T.

Solution. We compute that cT(X) = (X - 1)(X - 2)2, exactly the same as in Example 5.4. Thus, we have the same two possibilities and we will proceed in the same way. Now we find that

Ker(T - 2. lv)

(Vs),

2

\

1

so in this example we are in case (b). To find the basis for the Jordan canonical form, we find a vector v2 E F3 with (T - 2)(V3) = v2. By solving a system of linear equations we find that we may take

-3 -2

v3=

0

Also,

-3

J

-1

Ker(T - 1v) = (

1

) _ (vi).

/

Thus, if B = {v1, v2, v3} (note the order), then B is a basis of F3 and [T]!3 = J.

As a practical matter, there is a second method for finding a suitable basis of F3. Note that dim(Ker(T - 2.1v )2) = 2. Pick any vector v3, which

is in Ker(T - 2 lv)2 but not in Ker(T - 2 1v), say v3 = [2 Then let v2 = (T - 2)(v3) _ [-1

0

-11'.

-2 -11'. If B' _ {v1, v3, v3}, then we

also have [T] g, = J.

O

(5.6) Example. Let V be a vector space over F of dimension 8, let B = {v;}8_1 be a basis of V, and suppose that T E EndF(V) is a linear transformation such that

[T]g=A =J2,2$J2.2ED J2,1 ED J1.2 (1) J1.1 0 0

0 0

0

0

2

1

0

0

0 0 0 0 0

2

0

0

2

0 0

0 0 0

2

1

0

2

0 0 0

0

0

0

0

0

0

0

0

0

0

1

0 0 0 0 0 0 0

0

0

1

0 0

0

0 0 0

0

1

1

0 0 0


263

That is, [TJg is already in Jordan canonical form. Compute p(VT), the cyclic decomposition of VT, and the rational canonical form of T.

3 and vgeom(1) = 2, it follows from Corollary Solution. Since 4.48 (4) that µ(VT) = 3. Moreover, the elementary divisors of T are (X-2)2,

(X - 2)2, (X - 2), (X - 1)2, and (X - 1), so the invariant factors are (see the proof of Theorem 3.7.15):

f3(X) = (X - 2)2(X - 1)2 f2(X) = (X - 2)2(X - 1) f, (X) = (X - 2). Since

Ann(v2) = Ann(v4) = ((X - 2)2) Ann(v5) = ((X - 2)) Ann(v7) = ((X - 1)2) Ann(v8) = ((X - 1)), it follows (by Lemma 3.7.18) that

Ann(w3) = (f3(X))

Ann(w2) = WW) Ann(wl) = (fl(X)) where W3 = V2 + v7, w2 = v4 + v8, and wi = v5. Thus VT a5 Rwl ® Rw2 ® Rw3

and the rational canonical form R of T is the matrix R = C(fi (X )) ® C(f2(X)) ® C(f3(X)).

Moreover, Q-'AQ = R if Q E M8(F) is the invertible matrix 0 0 0

0

0

0

1

4

12

0

0

1

2

4

8

1

4

0

2

4

1

0

1

2

3

0

0 0

0 0

0

0

0 0

0 0 0

0

1

0 0 0 0

0

0

1

1

1

1

0

1

1

1

0

0

0

0

0

Q= PBc =

0 0 0

0

0

0

where C = {w1, w21 Tw2, T2w2, W3, Tw3, T2w3, T3w3}. 11

264


(5.7) Remark. We have given information about the rational canonical form in the above examples in order to fully illustrate the situation. However, as we have remarked, it is the Jordan canonical form that is really of interest. In particular, we note that while the methods of the current section produce the Jordan canonical form via computation of generalized eigenvectors, and then the rational canonical form is computed (via Lemma 3.7.18) from the Jordan canonical form (e.g., Examples 5.3 and 5.6), it is the other direction that is of more interest. This is because the rational canonical form of T can be computed via elementary row and column operations from the matrix

XI,, - [71a. The Jordan canonical form can then be computed from this information. This approach to computations will be considered in Chapter 5.

We will conclude this survey of computations by sketching the algorithm that has been used in the previous examples to compute a basis of V in which a given linear transformation T E EndF(V) is in Jordan canonical form. The Jordan canonical form can be computed solely by solving systems of homogeneous linear equations, provided the eigenvalues of T are available. It should, however, be emphasized again that the computation of eigenvalues involves solving a polynomial equation. Suppose that T E EndF(V) has mT(X) = (X - A)'q(X) where q(A) 0. Then the primary decomposition theorem allows one to write

VT 5 Wl®...®Wt®V where W; = F[X]w; is cyclic with Ann(w;) = ((X - A)") and W=W1ED ...ED Wt

is the generalized eigenspace of the eigenvalue A. By rearranging the order of the r; if necessary we may assume that

r=r1 >r2>...>rt (and of course rl +

+ rt = va1g(A).) We will let

rl = rl > r2 > ... > r8 = rt denote the distinct r;. It is the vectors w; that are crucial in determining the Jordan blocks corresponding to the eigenvalue A in the Jordan canonical

form of T. We wish to see how these vectors can be picked out of the generalized eigenspace W corresponding to X. First observe that

B = {w1i Swl, ... ,Srl -lwl, W2, ... ,Sr2-1W2, .. . ,Sre-1Wd, where S = T - ally, is a basis of W. This is just the basis (in a different order) produced in Theorem 4.38. It is suggestive to write this basis in the following table:


265

Table 5.1. Jordan basis of T

wl Swl

... ...

Wk,

Swk,

Wkl+l Swk, +1

Wk3

Swk, Wt

Sr1-lwl

... Sr,-lwk, Sr2-lwk,+l ... Sr--lwk3 ...

Sr. -'WI

(Of course, W is an F[X]-module of rank t, and as we observe from Table 5.1, t = dimF Ker(T - Al v), as we expect from Corollary 4.48 (4).) Note that the linear span of each column is the cyclic submodule W; = Rw; while the last k rows form a basis of Ker(T-A1v)k. As a concrete example of this scheme, suppose that T : F13 - F13 is multiplication by the matrix A=J2,4eJ2,4®J2.2®J2,2ED J2,1.

Using the standard basis on F13 the above table becomes Table 5.2. Jordan basis of T : F13 -+ F13 e4

e8

e3

e7

e2

e6

elo

e12

el

e5

e9

ell

e,3

Thus, the bottom row is a basis of Ker(T - 2.1 v), the bottom two rows are a basis of Ker(T - 2 lv)2, etc. These tables suggest that the top vectors

of each column are vectors that are in Ker(T - \1v)k for some k, but they are not also (T - Alv)w for some other generalized eigenvector w. This observation can then be formalized into a computational scheme to produce the generators w; of the primary cyclic submodules W;. We first introduce some language to describe entries in Table 5.1. It is easiest to do this specifically in the example considered in Table 5.2, as otherwise the notation is quite cumbersome, but the generalization is clear. We will number the rows of Table 5.1 from the bottom up. We then say (in the specific case of Table 5.2) that {e13} is the tail of row 1, {elo, e,2} is

266


the tail of row 2, the tail of row 3 is empty, and {e4, es} is the tail of row 4. Thus, the tail of any row consists of those vectors in the row that are at the top of some column. Clearly, in order to determine the basis B of W it suffices to find the tails of each row. We do this as follows:

For i = 0, 1, 2,... , let Va') = Ker(S') (where, as above, we let S = T - Alv). Then

Let d; = dim(V(')) so that

0=do <

Note that r is the smallest value of i for which di = v,ig(A), where r is as above, i.e., where MT(X) = (X - A)rq(X) with q(A) 96 0. This observation is useful for determining r without first computing MAX) explicitly. In particular, we will have r = 1, i.e., d1 = v,jg(A), if and only if the generalized eigenspace of T corresponding to A is the eigenspace of T corresponding to A. Note that in any case Var) = W, so V(r+1) = V(r), and dr+i = dr.

Observe that for 1 < i < r, row i of Table 5.1 has length di - d;_1, and hence, its tail has length

(d: - d;-1) - (d:+1 - di). We make the following observation:

Let Vai+1) be any complement of V') in Va'+') i.e., any subspace of V('+i) such that VA(i+i) - Vai) ®Vai+1)

Then S(Va'+1)) and Va'-1) are both subspace of v') and (5.8) Claim. S(V('+')) n VA('-') = {0}.

To see this, let v E S(Vai+1)) n Va'-1). Then v = S(v) for some v E

andalso S'-1(v)=0. But 0 = S'-1(v) = S'-1(S(v)) = S'(v) implies that v = 0, by the definition of Va'+1)

Now for our algorithm. We begin at the top, with i = r, and work down. By Claim 5.8, we may choose a complement Va') of Vai-1) in V')

which contains the subspace S(VAi+1)). Let Va,) be any complement of


267

S(Vr') in V. Then a basis for Va gives the tail of row i of Table 5.1. W

(Note that Var+l) = {O}, so at the first stage of the process Va = Vary is any complement of V(' ' in V(r) = W. Also, at the last stage of the process 0a°) = {0}, so Vale = Vain and Vale is any complement of S(Va2)) in Vlll, the eigenspace of T corresponding to A.) (5.9) Example. We will illustrate this algorithm with the linear transformation T : F13 - F13 whose Jordan basis is presented in Table 5.2. Of course, this transformation is already in Jordan canonical form, so the purpose is

V\ , and V.\ relate to the just to illustrate how the various subspace basis B = {el, ... , e13}. Since there is only one eigenvalue, for simplicity, let V(') = Vt, with a similar convention for the other spaces. Then

V5=V4=F13 V4 = (el, ... , el3) V3 = (el, e2, e3, e5, e6, e7, e9, elo, ell, e,2, e,3) V2 = (el, e2, es, e6, e9, elo, ell, e12, e13)

V, = (el, e5, e9, ell, e,3) Vo = {0},

while, for the complementary spaces we may take

V5 = {0} V4 = (e4, es) V3 = (e3, e7)

V2 = (e2, e6, elo, e,2)

V, = (el, es, e9, ell, e13).

Since V4 = F13, we conclude that there are 4 rows in the Jordan table of T, and since V4 = V4, we conclude that the tail of row 4 is {e4, e8}. Since

S(e4) = e3 and S(es) = e7, we see that S(V4) = V3 and hence the tail of row 3 is empty. Now S(e3) = e2 and S(e7) = e6 so that we may take V2 = (elo, e,2), giving {elo, e12} as the tail of row 2. Also, S(e2) = el, S(e6) = e5, S(elo) = e9, and S(e,2) = ell, so we conclude that V, = (e,3), i.e., the tail of row 1 is {e13}. Examples 5.3 and 5.4 illustrate simple cases of the algorithm described above for producing the Jordan basis. We will present one more example, which is (slightly) more complicated.

(5.10) Example. Let V = F4 and let T : V -. V be the linear transformation with the matrix

268


-1

-3 [T]c - A - -8 2

1

1

3

1

2

5

0

-1

3

in the standard basis C. Find the Jordan canonical form of T and a basis B of V such that [TJB is in Jordan canonical form. Solution. We compute that CT(X) = CA(X) _ (X - 2)4. Thus, there is only one eigenvalue, namely, 2, and v,lg(2) = 4. Now find the eigenspace of 2, i.e.,

Vl')=Ker(T-2.1v)=`/

1

1

2

,

1

0

0

I

2

so dl = 2. We also find that V(2)

= Ker((T - 2. 1v)2) = V,

so d2 = 4. Hence, we choose {w1, w2} C V so that V(1)

® (w1, w2) = V

in order to obtain a Jordan basis of V. Take 0

10

w1 =

0

0

and

w2=

[1] 0

.

0

Then compute

V1 = (T - 2 lv)(w1) =

-3 -3 -8 2

and 1

vJ=(T-2.1v )(w2)=

1

2 0

Setting

v1 = (T - 2 ly)(w1),

v2 = wl, v4=W2,

\,

4.6 Inner Product Spaces and Normal Linear Transformations

269

we obtain a basis B = {v1, v2, v3, v4} such that [T]g = J. The table corresponding to Table 5.1 is V2

V4

V1

V3

We also note that VT = Rv2 ® Rv4 so that p(VT) = 2.

0

4.6 Inner Product Spaces and Normal Linear Transformations In this section we will study vector spaces possessing an additional structure, that of an inner product, as well as endomorphisms of these spaces which are "normal," normality being a property defined in of an inner product.

(6.1) Definition. Let V be a vector space over F = R or C. An inner product on V is a function and a E F,

F such that for all u, v, w E V

VxV

(1) (u + v : w) _ (u : w) + (v : w), (2) (au : v) = a u : v), (3) (u : v) = (v : u), and

(4) (u:u)>Oifu54 0. A vector space V, together with an inner product (:) is called an inner product space.

The bar in Definition 6.1 (3) refers to conjugation in the field F. Of course, if F = R, conjugation in F is trivial, but it is convenient to handle the real and complex cases simultaneously. (6.2) Examples.

(1) The standard inner product on Fn is defined by

=

(U: v)

n

F, ujvj j=1

(2) If A E M,n,n(F), then let A' = A` where q denotes the matrix ob-

tained from A by conjugating all the entries. A' is called the Hermitian transpose of A. If we define

(A : B) = Tr(AB')

270

(3)


then (: ) defines an inner product on M..... (F). If we identify F" with M1,,a(F), then this inner product agrees with the inner product in (1). Suppose that T : V -+ W is an injective linear transformation. If W is an inner product space, then T induces an inner product on V by the formula

(v1 : v2) = (T(vi) : T(v2)) In particular, every subspace of an inner product space inherits an inner product. (4) Let V = C((0, 1], F) be the F-vector space of continuous F-valued functions on the unit interval [0, 1]. Then an inner product on V is defined by

(f : 9) =

f

1

f (x)9(x) dx.

0

(6.3) Definition. If V is an inner product space, then llvll = (v: v)'"2 E R is called the norm of v. The norm of a vector v is well defined by Definition 6.2 (4). There are a number of standard inequalities related to the norm.

(6.4) Proposition. If V is an inner product space with inner product (: ), if u, v E V and a E F, then (1) Ilaull = la(Ilull, (2) (lull > 0 if u 0 0,

(3) (Cauchy-Schwartz) 1(u: v)l S IlulIlIvIl, and (4) (Triangle inequality) l(u + vll 5 (lull + (lvi( Proof. (1) and (2) are immediate from the definitions. (3) Let u, v E V be arbitrary and let x E R. By Definition 6.1 (4),

(u+xv : u+xv) > 0 (and the inequality is strict unless u is a multiple of v). Thus,

0<(u+xv:u+xv) = Ilul12 + ((u : v) + (v : u))x + (Ivl(2x2 = (lull2 + 2(Re(u : v))x + (Iv112x2 <_ IIuIl2 + 21(u: v)(x +

Ilv1l2x2.

Since this holds for all x E R, we conclude that f (x) = Ilul(2 + 21(u: v)(x + Ilv112x2

is a real quadratic function, which is always nonnegative. Thus, the discriminant of f (x) must be nonpositive, i.e.,


4(u: v)2

271

-4 IIUI1211VI12 < 0,

and the Cauchy-Schwartz inequality follows immediately from this. (4) By the Cauchy-Schwartz inequality, Re(u : v) < 1(u: v)I 5 IIvIllIv1I Hence,

llu+v112=(u+v:u+v) = IIuI12+(u:v)+(v:u)+IIv112 = IIuI12 + 2Re(u: v) + IIv112 IIui12 + 211ullIIvIl + IIv112

_ (IIuII + IIvII)2

Taking square roots gives the triangle inequality.

(6.5) Remark. The Cauchy-Schwartz inequality implies that, if u and v are nonzero, then (u : v) <

-1 5

(lull IIuII

1

Thus, we can define the angle between u and v by means of the equation

coso= (u:v)

Dull llvll

0<0<7r.

It is this equation that allows us to introduce the geometric idea of angle into an arbitrary real or complex vector space by means of the algebraic notion of inner product.

(6.6) Definition. If V is an inner product space, then vectors u, v E V are said to be orthogonal if (u : v) = 0. A subset S C V is said to be orthogonal if every pair of distinct vectors in S is orthogonal. S is said to be orthonormal if S is orthogonal and if every vector in S has norm 1.

(6.7) Proposition. Let V be an inner product space and let S C V be an orthogonal set of nonzero vectors. Then S is linearly independent.

Proof. Suppose that ul, ... , uk are distinct elements of S and suppose that there is an equation

alul + ... +akuk = 0

where a,,...,akEF.Then, for1
= ai(ut ui) + ... +ak(uk : ui) = ai(ui ui) But (ui : ui) > 0, so we conclude that ai = 0 for all i.

272


Proved in exactly the same manner as Proposition 6.7 is the following useful fact.

(6.8) Lemma. Let V be an inner product space and suppose that B =

... ,vn} is an orthonormal basis of V. Then for any u E V, the coordinate matrix of u is (u: v1) {vl,

lull; = (u : vn) Proof. Exercise.

11

The classical Gram-Schmidt orthogonalization process allows one to produce an orthonormal basis of any finite-dimensional inner product space; this is the content of the next result. (6.9) Theorem. Every finite-dimensional inner product space V has an orthonormal basis.

Proof. Let C = Jul, ... un} be any basis of V. We will produce an orthonormal basis inductively. Let vl = ul/IIu1II Now assume that an orthonormal set {vl, ... ,vk} has been chosen so that

(VI, ... , vt) = (ul ... , ut) for 1 < t < k. Then define k

Vk+1 = "uk+l - E(uk+l : Vj )Vj, j=1

and set vk+1 = vk+1/Ilvk+l II We leave to the reader the details of ing the validity of this construction and the fact that an orthonormal basis has 0 been produced.

(6.10) Definition. Suppose that V is an inner product space and W C V is a subspace. Let

W1 ={uE V: (u:w)=0 forallwEW}. The subspace W' of V is called the orthogonal complement of W. (6.11) Proposition. Let V be a finite-dimensional inner product space and let W C V be a subspace. Then (1) W n Wl = (0),


273

(W1)1 = W, (3) dim W + dim Wl = dim V, and (2)

(4)

Proof. Choose an orthonormal basis 5 = {vl, ... , vn} of V in which {v1, ... , vk} is a basis of W. This is possible by the algorithm of Theorem 6.9. Then W' =

This equation immediately implies (2) and (3). (1) is clear and (4) follows 0 from (1) and (3).

(6.12) Remark. If V is a vector space over F, its dual space V' is defined to be V* = HomF(V, F). If V is finite-dimensional, then, by Corollary 3.4.10, V and V' have the same dimension and so are isomorphic, but in general there is no natural isomorphism between the two. However, an inner product on V gives a canonical isomorphism 0: V -' V' defined as follows:

For y = V, 0(y) E V' is the homomorphism O(y)(x) = (x : y). To see that 0 is an isomorphism, one only needs to observe that 0 is injective since dim V = dim V. But if y 0 0 then 0(y)(y) = (V: y) > 0, so 0(y) 0 and Ker(4) _ (0). (6.13) Theorem. Let V be a finite-dimensional inner product space. Then for every T E EndF(V) there exists a unique T' E EndF(V) such that

(Tv:w)=(v:T'w) for all v, w E V. T' is called the adt of T. Proof. Let w E V. Then h,,, : V -+ F defined by h,,,(v) = (Tv : w) is an element of the dual space V. Thus (by Remark 6.12), there exists a unique w E V such that

(Tv:w)=(v:w) for all v E V. Let T'(w) = w. We leave it as an exercise to that T' E EndF(V). 0 (6.14) Lemma. Let V be an inner product space and let S, T E EndF(V).

Then (ST)' = T'S'.

0

Proof. Exercise.

(6.15) Lemma. Let V be a finite-dimensional inner product space and let 8 be an orthonormal basis of V. If T : V -+ V is a linear transformation and (T)e = A = [ai3], then

[T*]e=A'=A`

274


Proof. Let B = {vi, ...

If T'(vi) = F,k=lbkivk, then, according to

Lemma 6.8,

bji = (T* (vi) : vj) = (vj : TO (vi))

= (T(vj) : vi) n

_

\u akj Vk : k=1

Vi

ll

= Qij.

Thus, [T')B = T, as required.

0

(6.16) Definition. Let V be an inner product space, and let T : V - V be a linear transformation.

(1) T is normal if TT' = T'T, i.e., if T commutes with its adt. (2) T is self-adt if T = T', i.e., if T is its own adt. (3) T is unitary i f T' = T -' .

LetAEMn(F) be an n x n matrix. (1') A is normal if AN = A' A. (2') A is self-adt if A = N. (3') A is unitary if AN = I,,. (6.17) Remarks. (1) (2)

(3)

If T is self-adt or unitary, then it is normal. If F = C, then a self-adt linear transformation (or matrix) is called Hermitian, while if F = R, then a self-adt transformation (or matrix) is called symmetric. If F = R, then a unitary transformation (or matrix) is called orthogonal. Lemma 6.15 shows that the concept of normal is essentially the same for transformations on finite-dimensional vector spaces and for matrices. A similar comment applies for self-adt and unitary.

The importance of unitary transformations arises because of the following geometric property which characterizes them. (6.18) Proposition. Let T : V V be a linear transformation on the finitedimensional inner product space V. The following are equivalent:

(1) T is unitary. (2) (Tu : Tv) = (u: v) for all u, v E V. (3) IITvIJ = Jlvi) for all v E V.


275

Proof. (1) =: (2). Let u, v E V. Then

(Tu: Tv) = (u : T'Tv) = (u : T-'Tv)

=(u:v). (2) = (1). (u : v) _ (Tu : Tv) _ (u : T'Tv), so

(u:T'Tv-v)=0

for all

u,vEV.

Taking u = T'Tv - v shows that (u : u) = 0. Thus u = 0, i.e.,

T'Tv = v for all v E V. Therefore, T'T = 1v, and since V is finite dimensional, this shows that

T' = T-'. (2) * (3). Recall that IjTvjI2 = (Tv : Tv). (3) . (2). Let u, v E V. Then IIT(u + v)HH2 = (T(u + v) : T(u + v))

= (Tu : Tu) + (Tu : Tv) + (Tv : Tu) + (Tv : Tv) = (u : u) + 2Re ((Tu : Tv)) + (v : v), while similarly

(u+v:v+v)=(u: u) + 2Re ((u: v)) + (v: v). Thus, we conclude that

Re((Tu : Tv)) = Re(u : v). Applying the same argument to u + iv, we obtain

Re ((Tu : iTv)) = Re ((Tu : T(iv)) = Re(u : iv). But it is easy to check that Re((x : iy)) = Im((x : y)) for any x, y E V, so we have

(Tu: Tv) = (u: v)

forallu,vE V.

0

The following result collects some useful properties of normal transformations.

(6.19) Lemma. Let V be a finite-dimensional inner product space and let T : V -+ V be a normal transformation.

276

(1) (2)


If f(X) E FIX) then f(T) is normal.

IITvII=IIT'vIIforallvEV.

(3) Ker T = (Im T)1. (4) If T2v = 0 then Tv = 0. (5) v E V is an eigenvector for T with eigenvalue A if and only if v is an eigenvector for T' with eigenvalue A.

Proof. (1) This follows immediately from the fact that (aT")' = a(T')" and the definition of normality. (2)

IITvII2

= (Tv : Tv) = (v : T'Tv)

=(v:TT'v)=(T'v:T'v) = IIT'vII2.

(3) Suppose that (u : Tv) = 0 for all v E V. Then (T'u : v) = 0 for all v E V. Thus T'u = 0. By (2) this is true if and only if Tu = 0, i.e., u E (Im T)1 if and only if u E Ker T. (4) If T2v = 0 then Tv E Ker T n Im T = (0) by (3). (5) By (1), the linear transformation T - AI is normal. Then by (2), II(T - AI)vII = 0 if and only if II(T' - AI)vII = 0, i.e., v is an eigenvector of T with eigenvalue A if and only if v is an eigenvector of T' with eigenvalue

1

0

(6.20) Theorem. Suppose that V is a finite-dimensional inner product space

and T : V -. V is a normal linear transformation. (1)

The minimal polynomial mT(X) is a product of distinct irreducible

factors. (2) Eigenspaces of distinct eigenvalues of T are orthogonal. (3) If V is a complex inner product space, then T is diagonalizable.

Proof. (1) Let p(X) be an irreducible factor of mT(X). We need to show

that p2(X) does not divide mT(X). Suppose it did and let v E V with p2(T)(v) = 0 but p(T)(v) # 0 (such a v E V exists by Theorem 3.7.1 and Lemma 3.7.17, or see Exercise 43 in Chapter 3). By Lemma 6.19 (1), U = p(T) is normal, and U2(v) = 0 but U(v) # 0, contradicting Lemma 6.19 (4).

(2) Suppose that Tv1 = Alvl and Tv2 = A2v2 where v1 and v2 are nonzero and Al 96 A2. Then (v1

: V2) = ((.'1 - A2)-1(T - A21)Vi : v2) = (Al - A2)-1(vl:(T - A2)*v2) = 0,

since v2 is a eigenvector of T' with eigenvalue A2 by Lemma 6.19 (5).


277

(3) Every irreducible polynomial over C is linear, so by (1), MT(X) is 0 a product of distinct linear factors and so T is diagonalizable.

(6.21) Corollary. (Spectral theorem) If V is a finite-dimensional complex V is a normal transformation, then V inner product space and T : V has an orthonormal basis of eigenvectors of T. Proof. By Theorem 6.20 (3), T is diagonalizable and by (2) the eigenspaces are orthogonal. It is only necessary to choose an orthonormal basis of each 0 eigenspace.

(6.22) Remark. If V is a real vector space and T is normal, T may not be diagonalizable, but from Theorem 6.20 it follows that the real Jordan canonical form of T (cf. Theorem 4.57) will consist of 1-by-1 ordinary Jor-

dan blocks or 2-by-2 real Jordan blocks. For example, the second case occurs

The case of self-adt linear transformations (which are automatically normal) is of particular importance; such transformations are diagonalizable even in the real case.

(6.23) Theorem. (Spectral theorem, self-adt case) Suppose that V is an inner product space and T : V -+ V is a self adt linear transformation. (1) All of the eigenvalues of T are real. (2) V has an orthonormal basis of eigenvectors of T.

Proof. We consider first the case F = C. In this case (1) is immediate from Lemma 6.19 (5) and then (2) follows as in Corollary 6.21. Now consider the case F = R. In this case, (1) is true by definition. To prove part (2), we imbed V in a complex inner product space and apply part (1). Let Vc = V ® V and make VC into a complex vector space by defining the scalar multiplication (6.1)

(a + bi)(u, v) = (au - by, bu + av).

That is, V ®0 is the real part of VC and 0 ®V is the imaginary part. Define an inner product on Vc by (6.2) ((u1, vl) : (u2, v2)) = (u1 : u2) + (v1 : v2) + i((vi : u2) - (u1 : v2))

We leave it for the reader to check that Equations (6.1) and (6.2) make Vc into a complex inner product space. Now, extend the linear transformation T to a linear transformation Tc : Vc -' VC by

Tc(u, v) = (T(u), T(v)). (In the language to be introduced in Section 7.2, VC = C ®R V and TC = lc OR T.) It is easy to check that TC is a complex linear transformation of

278


VC, and in fact, Tc is self-adt. By part (1) applied to Tc we see that all the eigenvalues of Tc are real and mTQ(X) is a product of distinct (real) linear factors. Thus, mTa (X) E R(X ). If f (X) E R[X], then (6.3)

f(Tc)(u, v) = (f(T)(u), f(T)(v))

Equation (6.3) shows that MT(X) = mTo (X) and we conclude that T is diagonalizable. Part (2) is completed exactly as in Corollary 6.21. 0

4.7 Exercises 1.

2. 3. 4.

Suppose R is a finite ring with JR[ = s. Then show that Mm,n(R) is finite with JM,","(R)l = s". In particular, [M"(R)I = s"'. Prove Lemma I.I. Prove Lemma 1.2.

(a) Suppose that A = [al

AB=j:"1a;row{(B).

a", ] and B E M,..,, (R). Then show that

(b) Suppose

and B E Mm,n(R). Then show that BC =E", ci col,(B). This exercise shows that left multiplication of B by a row matrix pro-

duces a linear combination of the rows of B and right multiplication of B by a column matrix produces a linear combination of the columns of B. 5.

Let S C M2(R) be defined by

S={[a aa,bER}. that S is a subring of M2(R) and show that S is isomorphic to the 6.

field of complex numbers C. Let R be a commutative ring.

(a) If 1 < j < n prove that Ejj = P, 1 Ell P1t. Thus, the matrices E;; are all similar. (b) If A, B E M"(R) define [A, B] = AB - BA. The matrix [A, B] is called the commutator of A and B and we will say that a matrix C E Mn(R) is a commutator if C = [A, B] for some A, B E M,,(R). If i 54 j show

that E;, and E11 - E are commutators. (c) If C is a commutator, show that 7r(C) = 0. Conclude that In is not a commutator in any Mn(R) for which n is not a multiple of the characteristic of R. What about 12 E M2(Z2)?

7.

If S is a ring and a E S then the centralizer of a, denoted C(a), is the set

C(a) = {b E S : ab = ba}. That is, it is the subset of S consisting of elements which commute with a.

a that C(a) is a subring of S. b What is C(1).

4.7 Exercises

279

(c) Let R be a commutative ring and let S = MM(R). If A = D;(#) for 8.

,0 96 1 E R' then compute C(A). A matrix N E MM(R) is nilpotent if Nk = 0 for some k.

(a) If F is field of characteristic 0 and N E MnF) is nilpotent, show that

there is a matrix A E MM(F) such that A' = I,, + N for any natu-

ral number m. (Hint: The binomial series may be helpful.) Are there problems if we do not assume that char(F) = 0? (b) Let N = [o o] E M2(Q). Show that there is no matrix A E M2 (Q) such that A2 = N. 9. Show that there are infinitely many A E M2(R) such that A2 = 0. is idempotent if P2 = P. Give an example, other than 10. A matrix P E 0 or It,, of a diagonal idempotent matrix. Give an example of a nondiagonal 2 x 2 idempotent matrix. Show that if P is idempotent, then T-1 PT is also idempotent for all T E GL(n, R). 11. Let A E Mn(R). We say that A has constant row sums if the sum of the entries in each row is a constant a E R, i.e En_ a, = a for 1 < i < n. We define constant column sums similarly. (a) Show that A has constant row sums if and only if

for a E R and that A has constant column sums if and only if [1 1) for 0E R. 1]A =p[1

12. 13. 14.

(b) Prove that if A and B E Mn (R) both have constant row sums, then so does AB. (c) Prove that if A and B E Mn(R) both have constant column sums, then so does AB. Prove Proposition 1.13.

Prove Lemma 1.15. Let R be a commutative ring. the following formulas for the kronecker product of matrices:

(a) A®(B+C)=A®B+A®C.

(b) (A ® B)` = At ® Bt. Prove Lemma 1.20. Give an example of a function D : Mn(Z2) -# Z2 which is n-linear and satisfies D(P;,A) = -D(A) = D(A), but which is not alternating. 17. A matrix A E Mn(R) is symmetric if At = A and it is skew-symmetric if At = -A and all the diagonal entries of A are zero. 15. 16.

(a) Let V1 be the set of symmetric matrices and V2 the set of skew-symmetric

matrices. Show that V1 and V2 are both submodules of V = MM(R). If 2 is a unit in R show that V = V1 ®V2. (b) Let A E M,,(R) be skew-symmetric. If n is odd, show that det(A) = 0. If n is even, show det(A) is a square in R. 18. 19.

20.

If P,,, is a permutation matrix, show that P,; l = P. Let R and S be commutative rings and let f : R - S be a ring homomorA phism E Mn (S). Show that f

f . If Let R be a commutative ring and let H be a subgroup of the group of units

R' of R. Let N = A E GL(n, R) : det A E H}. Prove that N is a normal subgroup of GL(n, R) and that GL(n, R)/N 95 R'/H.

280

21.


(a) Suppose that A has the block decomposition A = [A. A, . Prove that det A = (det Al) (det A2 ).

(b) More generally, suppose that A = [Aij] is a block upper triangular (respectively, lower block triangular) matrix, i.e., Aii is square and Aij = 0 if i > j (respectively, i < j). Show that detA = fli(detAii). 22. If R is a commutative ring, then a derivation on R is a function b : R -> R such that 5(a + b) = b(a) + b(b) and b(ab) = ab(b) + b(a)b. (a) Prove that b(al ... an) = >2 1(a, ... ai-ib(ai)ai+1 ... an).

(b) If b is a derivation on R and A E Mn(R) let Ai be the n x n matrix obtained from A by applying b to the elements of the ith row. Show that

b(det A) = I:n 1 det Ai. 23.

If A E Mn(Q[X]) then detA E Q[X]. Use this observation and your knowledge of polynomials to calculate det A for each of the following matrices, without doing any calculation. 1

(a) A=

1

1

2-X2

2

2 1

3 3 1

1 1-X (b) A= 24.

25. 26. 27.

28.

29.

2

3

2

3 5

1

9-X2

1 1 1

1

1

2-X

1

1

1

..

1

..

1

1

... m-X

Let F be a field and consider the "generic" matrix [Xij] with entries in the polynomial ring F[Xij] in the n2 indeterminants Xs- (1 < i, i < n). Show that det[Xij] is an irreducible polynomial in F[Xijj. (Hint: Use Laplace's expansion to argue by induction on n.) Let A E Mn(Z) be the matrix with entii(A) = 2 (1 1. Compute det(A). Let An E Mn(Z) be the matrix with entii(An) = i for 1 < i < n and aij = 1 if i # j. Show that det(An) = (n - 1)!. Let A E Mn(Z) be a matrix such that entij(A) = ±1 for all i and j. Show that 2n-1 divides det(A). Let R be a commutative ring and let a, b E R. Define a matrix A(a, b) E Mn(R by entii(A(a, b)) = a for all i and entij(A(a, b)) = b if i j. Compute det(A a, b)). (Hint: First find the Jordan canonical form of A(a, b).) Let V(xi, ...,xn) be the Vandermonde determinant:

V(x1, ... ,xn) =

1

x1

X1

1

x X2

X2z

x1 .

.

xn-1

.

2 .

1

xn

x2n

xri-1 n J

(a) Prove that

detV(x1, ... xn) =

11 (xi -xj). 1

(b) Suppose that ti, ... , to+1 are n + 1 distinct elements of a field F. Let Pi (X) (1 < i < n + 1) be the Lagrange interpolation polynomials determined by t1, ... , to+1. Thus,

B = {Pi(X), ... ,Pn+1(X)}

4.7 Exercises

281

is a basis of the vector space A, (F) of polynomials in F[X] of degree at most n. But A = (1, X, ... , X "J is also a basis of P,.. Sh^w that

V(t1, ... ,t"+1) = P. 30. If R is an integral domain let A E GL(n, R) and let B E M"(R).

(a) Prove that there are at most n elements a E R such that det(aA+B) = 0. If R is a field, conclude that aA + B is invertible except for finitely many values of a E R. (b) If R = Zs X Z2 that every a E R satisfies the equation a2 - a = 0. If A = Is and B = (° ° ] show that part (a) is false without the assumption that R is an integral domain. 31. Recall that Qp,,, is the set of sequences a = (i1, ... ip) of p integers with

1 < i1 < is < . . . < ip < n. Thus there are (D) elements of Qp,n and we can order these elements lexicographically, i.e., (ii, ... ip) < (j1, ... , jp if i1 < j1 or i1 = j1, ..., i,._ 1 = j,._ 1 and i,. < if or some 1 < r p. For

example, the lexicographic ordering of Q2,4 is (1, 2) < (1, 3) < (1, 4) <

(2, 3) < (2, 4) < (3, 4). If A E M" (R) then the set of (n)2 elements {det A[a 1,61 : a, Q E Qp,, } can be arranged, using the lexicographic ordering

on Q n into a matrix (A) called the p° compound matrix of A. (a) If'A, B E M"(R) then (using the Cauchy-Binet theorem) that (AB) = (A)(B).

(b) Show that (In) = I(.). (c) Prove that if A is invertible, then (A) is also invertible, and give a 32.

formula for (A)-1. If A E Mm,n(R) then det(AA`) > 0. (Hint: Use the Cauchy-Binet theorem.)

If A = [%] E M"(C), then let A = Jai;] where aif denotes the complex conjugate of a;J and let A' = T. Show that det(AA') > 0. 34. Let F and K be fields with F C K. Then M"(F) C M"(K). If A E M"(F) and A is invertible in M"(K), prove that A is invertible in M"(F). That is, 33.

GL(n, K) f1 M"(F) = GL(n, F). (a) If A E Mm,n(R) then rank(A) = rank(AÀ). Give an example to show that this statement may be false for matrices A E Mm,n(C). (b) If A E Mn(C) show that rank(A) = rank(A'A). 36. If A E Mn(R) has a submatrix A[a y] = 01 where t > n/2, then det(A) = 0. 35.

37. 38.

Prove Lemma 2.24.

Let R be any subring of the complex numbers C and let A E Mm,n(R). Then show that the matrix equation AX = 0 has a nontrivial solution X E Mn,1(R) if and only if it has a nontrivial solution X E Mn,1(C).

39. Prove Corollary 2.29. 40. Let R be a commutative ring, let A E Mm,n(R), and let B E M,,,p(R). Prove

that

M-rank(AB) < min{M-rank(A), M-rank(B)}.

41. the claims made in Remark 2.30. 42. Let F be a field with a derivation D, i.e., D : F --+ F satisfies D(a+b) = a+b

and D(ab) = aD(b)+D(a)b. Let K = Ker(D), i.e., K = {a E F : D(a) = 0}. K is called the field of constants of D. (a) Show that K is a field.

282


(b) If ui, ..., un E F, define the Wronskian of u1, ..., un E F by UI

W(ui,.. ,un)=det

D(ul)

un ...

D(un)

pn-i(ul) ...

43.

44.

where D' = Do o D (i times). Show that ul, ..., un E F are linearly dependent over K if and only if W (ul, ... , u,,) = 0. E Mn(R) be a matrix such Let R be a commutative ring and let A = that all is not a zero divisor. If n > 2 prove that ai i 4 det(A) = det(B) where B E Mn_1(R is the matrix with ent,, = A[(1, i + 1) ] (1, j + 1)( for 1 < i, j < n - 1. (This formula is sometimes called Choi's pivotal condensation formula. It provides another inductive procedure for the computation of determinants.) Prove the following facts about the ads of a matrix A E Mn(R) (in part

c,) assume that R = C): a If A is diagonal, then Adj(A) is diagonal.

b If A is symmetric, then Adj(A) is symmetric. c) If A is Hermitian, then Adj(A) is Hermitian. d) If A is skew-symmetric, then Adj(A) is symmetric or skew-symmetric according as n is odd or even. 45. Let R be an arbitrary ring. (a) If A E M,n,n(R) and B E Mn,y(R), show that

(op(AB))t =op(Bt)op(A`). (b) If A E Mn(R) is invertible, show that

op(A-1)` = (op(A`))-' 46.

Let P3=If (X)EZ[X]:degf(X)<3}. LetA={1, X, X2, X3} and B = {l, X, X(2). X(3)}

(X - i + 1). (a) that A and B are bases of the Z-module P3. (b) Compute the change of basis matrices P8 and PAB. (c) Let D : P3 P3 be differentiation, i.e., D(f (X)) = f'(X); e.g., where X(') = X(X - 1)

D(X3 + 2X) = 3X2 +2. Compute [D]A and (D)B.

(d) LetO:P3-'P3 be defined by0(f(X))=f(X+1)-f(X),e.g.,

(X3+2X)=((X+1)3+2(X+1))-(X3+2X)=3X2+3X+3. that i is a Z-module homomorphism and compute (A)A and [A]B.

47.

Show that the vectors (1, 2, 1), (2, 3, 3), and (3, 2, 1) form a basis of R3 and that the vectors (3, 2, 4), (5, 2, 3), and (1, 1, -6) form a second basis. Calculate the matrix of transition from the first basis to the second.

48.

If A E Mn(R), then the columns of A form a basis of Rn if and only if det(A) is a unit in

4.7 Exercises 49.

283

Let R be a commutative ring. We will say that A and B E M8(R) are permutation similar if there is a permutation matrix P such tl,,.t P-1 AP = B. Show that A and B are permutation similar if and only if there is a free R-module M of rank n, a basis B = {v1, ... , vn} of M, and a permutation

v E Sn such that A = [f]2i and B = [f)c where f E EndR(M) and C = 50.

{v,(1), ... ,vv(n)}. Let R be a commutative ring, and let

B= {Eli, ... ,E1n, E21, ... ,E2n, ... ,Eml, ... ,E,nn} be the basis of M,n,n(R) consisting of the matrix units in the given order. Another basis of M,n,n(R) is given by the matrix units in the following order:

C={E11,...,Em1,E12,...,Em2,...,E1n...,Emn}. If A E Mm(R) then GA E EndR(Mm,n(R)) will denote left multiplication by A, while 1ZB will denote right multiplication by B, where B E Mn(R).

(a) Show that AE,, = Ek akiEki and E,,B = E 1 bleE{r. 1

(b) Show that (GA)B = A 0 In and [LAIC = In 0 A. Conclude that A 0 In and In ® A are permutation similar. (c) Show that [RB)B = I,,, ® B` and IRA IC = B' 0 Im. (d) Show that [LA 0 RB)B = A ® B`. This exercise provides an interpretation of the tensor product of matrices as the matrix of a particular R-module endomorphism. Another interpretation, using the tensor product of R-modules, will be presented in Section 7.2 (Proposition 7.2.35). 51. Give an example of a vector space V (necessarily of infinite dimension) over a field F and endomorphisms f and g of V such that a f has a left inverse but not a right inverse, and b g has a right inverse but not a left inverse. 52. Let F be a field and let A and B be matrices with entries in F. a) Show that rank(A ® B) ank A + (rank(B)). (( b)) Show that rank(A ® B) rank A (rank(B)). c Show that rank(AB) > r (A) + rank(B) - n if B E Mn,m(F). 53.

Let M be a free R-module of finite rank, let f E EndR(M), and let g E

EndR(M) be invertible. If v E M is an eigenvector of f with eigenvalue A, show that g(v) is an eigenvector of gfg-1 with eigenvalue A. 54. Let M be a finite rank free R-module over a PID R and let f E EndR(M). Suppose that f is diagonalizable and that M = N1 ® N2 where N1 and N2 are f-invariant submodules of M. Show that g = f IN. is diagonalizable for

i=1,2.

55.

Let A E Mn(R), and suppose

CA(X)=det(XIn-A)=Xn+a1Xn-1+ (a) Show that a1 = - Tr(A) and an = (-1)n det(A). b More generally, show that a,

detA[a I al.

1)r QEQ,.n

56.

Prove the Cayley-Hamilton theorem for matrices A E M8(R) where R is

any commutative ring. (Hint: Apply the noncommutative division algorithm (Exercise 44 of Chapter 2) and the adt formula (Theorem 2.17).) 57. Let A E Mn(F) where F is a field.

284


(a) Show that det(Adj(A)) _ (det(A))'-' if n > 2. (b) Show that Adj(Adj(A)) _ (det(A))n-2A if n > 2 and that Adj(Adj(A)) = A 58.

if n = 2.

Let F be a field and let A E Mn(F). Show that A and At have the same minimal polynomial.

59. Let K be a field and let F be a subfield. Let A E Mn(F).

Show that the minimal polynomial of A is the same whether A is considered

in M. (F) or Mn(K).

60. An algebraic integer is a complex number which is a root of a monic polynomial with integer coefficients. Show that every algebraic integer is an eigenvalue of a matrix A E Mn(Z) for some n. 61. Let A E Mn(R) be an invertible matrix.

(a) Show that det(X-'In - A-') = (-X)-ndet(A-')cA(X).

b If

CA(X) = Xn +a, Xn-1 + ... + an-IX + an and

CA-1(X) =Xn+b,X'-'+ +bn_,X+bn,

then show that bi = (-1)'

ao=1. 62.

for 1 < i < n where we set

If A E MM(F) (where F is a field) is nilpotent, i.e., Ak = 0 for some k, prove that A' = 0. Is the same result true if F is a general commutative ring rather than a field?

63.

Let F be an infinite field and let Y = {Aj}jEJ C Mn(F) be a commuting family of diagonalizable matrices. Show that there is a matrix B E Mn(F) and a family fj(X)}jEJ C F[X] of polynomials of degree < n-1 such that Aj = fj(B). (Hint: By Theorem 3.36 there is a matrix P E GL(n, F) such that P-'AjP = diag(Aij, ... , Ani). Let t,, ... , to be n distinct points in F, and let

B = Pdiag(tl, ... ,tn)P-'. Use Lagrange interpolation to get a polynomial fj (X) of degree _< n -1 such 64.

that fj(ti) = Aij for all i, j. Show that { fj(X)}jEJ works.) Let F be a field and V a finite-dimensional vector space over F. Let S E EndF(V) and define Ads : EndF(V) EndF(V) by Ads(T) = [S, T] = ST - TS.

(a) If S is nilpotent, show that Ads is nilpotent. b If S is diagonalizable, show that Ads is diagonalizable. 65. Let Ni, N2 E M,(F) be nilpotent matrices. Show that N, and N2 are similar if and only if rank(Ni) = rank(N2) for all k > 1. 66. Let F be an algebraically closed field and V a finite-dimensional vector space

over F.

(a) Suppose that T, S E EndF(V). Prove that T and S are similar if and only if

dim(Ker(T - A1V )k) = dim(Ker(S - Alv)k) for all A E F and k E N. (This result is known as Weyr's theorem.)

4.7 Exercises

285

If T (b) Suppose that T E Endp (V) has Jordan canonical form ®- 1 is invertible, show that T-1 has Jordan canonical form ®i 1JaT1 ,,, 67.

Let F be an algebraically closed field and V a finite-dimensional vector space over F. (a) If T E EndF(V), show that there is a basis B = {vl, ... ,v.,} of V such

that Vi = (vi, ... , vi) is a T-invariant subspace of V. Conclude that

T)a is an upper triangular matrix. (Hint: Since F is algebraically closed has an eigenvalue, say Al with associated eigenvector v1. Choose a complementary subspace W to (v1), define Ti E EndF (W) by T1 = ,r o T where it : V -. W is the projection. Now argue by induction.) (b) If T, S E EndF(V) commute then there is a basis B of V such that both [T]1s and [S)e are upper triangular. (c) Show that the converse of part (b) is false by finding two upper triangular matrices which do not commute. (d) While the converse of part (b) is not true, show that two upper triangular matrices are "almost commutative in the following sense. that

the commutator matrix [A, B] = AB - BA of two upper triangular

matrices is a nilpotent matrix. 68. Find matrices in M3(Q) with minimal polynomials X, X2, and X3. 69. Find the characteristic and minimal polynomial of each of the following matrices:

0 0 a (a)

1

1

70.

0

b

01

(b)

0

1

1

0 0

0

0 0

(c)

1

0

1

0

1

1

0

1

0

0

1

0

1

1

0

1

0

(F a field) has characteristic polynomial

If A E

CA(X) = X2(X - 1)2(X2 - 1), 71.

what are the possibilities for the minimal polynomial mA(X)? Let V be a vector space and T : V -+ V a linear transformation. Assume that MT(X) is a product of linear factors. Show that T can be written as a sum T = D + N where D is a diagonalizable linear transformation, N is a nilpotent linear transformation, and DN = ND. Note that the hypotheses are always satisfied for an algebraically closed field (e.g., C).

72. Show that the matrices 0

1

00 1

73.

0

0 1

and

r1 I0

1

0

1

1

0

0

1]

0

are similar in M3 Z3). For each of the following matrices with entries in Q, find III the characteristic polynomial; 2 the eigenvalues, their algebraic and geometric multiplicities; 3 bases for the eigenspaces and generalized eigenspaces; 4 Jordan canonical form (if it exists) and basis for V = Q° with respect to which the associated linear transformation has this form;

(5) rational canonical form and minimal generating set for VT as Q[X)module. 0

(a) 3

(c)

0 0

4]

-4

-2 -4 1 2

4

-1 2

(b) (d)

[4

91

-6

1

2

2 2

1

2 2

2

1

]

Chapte r 4. Lin ear Algebra

286

(e)

-2

0

1

1

0 0

2

0 0

1

-2

0

0

-1

1

1

1

4

0 3

2

0

(f)

-2

1

1

0

5

0

-2 0 -1

'

3

74.

In each case below, you are given some of the following information for a linear transformation T : V -. V, V a vector space over the complex numbers C: (1) characteristic polynomial for T; (2) minimal polynomial for T; (3) algebraic multiplicity of each eigenvalue; (4) geometric multiplicity of each eigenvalue; (5) rank(VT) as an C[X]-module; (6) the elementary divisors of the module VT. Find all possibilities for T consistent with the given data (up to similarity) and for each possibility give the rational and Jordan canonical forms and the rest of the data. (a) CT(X) = (X - 2)4(X - 3)2. (b) CT(X) = X2(X - 4)' and MT(X) = X(X - 4)3. (c) dim V = 6 and MT(X) = (X + 3)2(X + 1)2. d) CT X = X(X - 1)4(X - 2)5 , v m(1) = 2, and vèom(2) = 2. e cT(X = (X - 5)(X - 7)(X (X - 11).

75.

Recall that a matrix A E MM(F) is idempotent if A2 = A. (a) What are the possible minimal polynomials of an idempotent A?

f)

76.

dimV=4andmr(X)=X-1.

b If A is idempotent and rank A = r, show that A is similar to B = Ir ®On-r If T : Cn Cn denotes a linear transformation, find all possible Jordan

canonical forms of a T satisfying the given data: (a) cT(X) = (X - 4)3(X - 5)2.

(b) n=6andmT(X)=(X -9)3.

(c) n = 5 and mr(X) _ (X - 6)2(X - 7).

77.

78. 79.

80. 81. 82.

83.

d T has an eigenvalue 9 with algebraic multiplicity 6 and geometric multiplicity 3 (and no other eigenvalues). (e) T has an eigenvalue 6 with algebraic multiplicity 3 and geometric multiplicity 3, and eigenvalue 7 with algebraic multiplicity 3 and geometric multiplicity 1 (and no other eigenvalues). (a) Show that the matrix A E M3(F) (F a field) is uniquely determined up to similarity by CA(X) and mA(X). (b) Give an example of two matrices A, B E M4(F) with the same characteristic and minimal polynomials, but with A and B not similar. Let A E If all the roots of the characteristic polynomial CA(X) are real numbers, show that A is similar to a matrix B E MM(R). Let F be an algebraically closed field, and let A E Mn(F). (a) Show that A is nilpotent if and only if all the eigenvalues of A are zero. are the eigenvalues b Show that Tr(Ar) = ai + + an where of A counted with multiplicity. (c) If char(F) = 0 show that A is nilpotent if and only if Tr(A') = 0 for all r E N. (Hint: Use Newton's identities, Exercise 61 of Chapter 2.) Prove that every normal complex matrix has a normal square root, i.e., if A E M. (C) is normal, then there is a normal B E A. (C) such that B2 = A. Prove that every Hermitian matrix with nonnegative eigenvalues has a Hermitian square root. Show that the following are equivalent: a) U E M,,(C) is unitary. b) The columns of U are orthonormal. c) The rows of U are orthonormal.

(a) Show that every matrix A E M,(C) is unitarily similar to an upper triangular matrix T, i.e., UAU' = T, where U is unitary. (b) Show that a normal complex matrix is unitarily similar to a diagonal matrix.

4.7 Exercises

287

Show that a commuting family of normal matrices has a common basis of orthogonal eigenvectors, i.e., there is a unitary U such that UA3U` = D3 for all Aj in the commuting family. (D3 denotes a diagonal matrix.) 85. A complex matrix A E M,,(C) is normal if and only if there is a polynomial 84.

f (X) E C[X] of degree at most n - 1 such that A' = f (A). (Hint: Apply

Lagrange interpolation.) Show that B = ®A; is normal if and only if each Al is normal. Prove that a normal complex matrix is Hermitian if and only if all its eigenvalues are real. 88. Prove that a normal complex matrix is unitary if and only if all its eigenvalues have absolute value 1. 86. 87.

89.

Prove that a normal complex matrix is skew-Hermitian (A' = -A) if and

only if all its eigenvalues are purely imaginary. 90. If A E let H(A) = I (A + A') and let S(A) = (A - A'). H(A) is z called the Hermitian part of A and S(A) is called the skew-Hermitian part

of A. These should be thought of as analogous to the real and imaginary parts of a complex number. Show that A is normal if and only if H(A) and

S(A) commute. Let A and B be self-adt linear transformations. Then AB is self-adt if and only if A and B commute. 92. Give an example of an inner product space V and a linear transformation T : V -+ V with T'T = lv, but T not invertible. (Of course, V will necessarily be infinite dimensional.) 93. (a) If S is a skew-Hermitian matrix, show that I - S is nonsingular and the matrix 91.

U=(I+S)(I-S)-1

is unitary. (b) Every unitary matrix U which does not have -1 as an eigenvalue can be written as

U = (I + S)(I - S)-1 94.

for some skew-Hermitian matrix S. This exercise will develop the spectral theorem from the point of view of projections.

(a) Let V be a vector space. A linear transformation E : V -* V is called a projection if E2 = E. Show that there is a one-to-one correspondence between projections and ordered pairs of subspaces (V1, V2) of V with V1ED V2=V given by

E H (Ker(E), Im(E)). (b) If V is an inner product space, a projection E is called orthogonal if E _

E. Show that if E is an orthogonal projection, then Im(E)' = Ker(E) and Ker(E)1 = Im(E), and conversely. (c) A set of (orthogonal) projections {E1, ... , E,.} is called complete if

E;Ej=O ori$jand

lv Show that any set of (orthogonal) projections {E1, ... , E,.} with EiEE _ 0 for i 54 j is a subset of a complete set of (orthogonal) projections. (d) Prove the following result. Let T : V --> T be a diagonalizable (resp., normal) linear transformation on the finite-dimensional vector space V. Then there is a unique set of

288


distinct scalars {Al, ... , A,.} and a unique complete set of projections (reap., orthogonal projections) {E,, ... , Er} with

T=AiEi+...+ArEr. are the eigenvalues of T and {1m(E;)};al are Also, show that the associated eigenapaces.

(e) Let T and {E;) be as in part (d) Let U : V - ' V be an arbitrary

linear transformation. Show that U = UT if and only if E1U = UEt for 1 < i < r. (f) Let F be an infinite field. Show that there are polynomials pi(X) E FIX) for 1 < i < r with pi (T) = E. (Hint: See Exercise 63.)

Chapter 5

Matrices over PIDs

5.1 Equivalence and Similarity Recall that if T; : V V (i = 1, 2) are linear transformations on a finitedimensional vector space V over a field F, then T, and T2 are similar if and only if the F[XJ-modules VT, and VT, are isomorphic (Theorem 4.4.2). Since

the structure theorem for finitely generated torsion F[X]-modules gives a criterion for isomorphism in of the invariant factors (or elementary divisors), one has a powerful tool for studying linear transformations, up to similarity. Unfortunately, in general it is difficult to obtain the invariant factors or elementary divisors of a given linear transformation. We will approach the problem of computation of invariant factors in this chapter by studying a specific presentation of the F[X]-module VT. This presentation will be used to transform the search for invariant factors into performing elementary row and column operations on a matrix with polynomial entries. We begin with the following definition. (1.1) Definition. If R is a ring and M is a finitely generated R-module, then a finite free presentation of M is an exact sequence (1.1)

R"'

0

R^

M

0.

Note that M Coker(O) and the free presentation is essentially an explicit way to write M as a quotient of a finite-rank free R-module by a finite-rank submodule. While every finitely generated R-module has a free presentation (Definition 3.4.15), it need not be true that every finitely generated R-module has a finite free presentation as in Definition 1.1; however, if the ring R is a PID, then this is true. (1.2) Lemma. Let R be a PID and let M be a finitely generated R-module with p(M) = n. Then there is a finite free presentation (1.2)

withm

0- R' --04 R"_ M_0

290

Chapter 5. Matrices over PIDs

Proof. By Proposition 3.4.14, there is a free presentation

0-+K- R"- M-- 40. Since R is a PID, Theorem 3.6.2 implies that K is a free R-module of rank m < n, so Equation (1.2) is valid. Recall (Definition 4.3.14) that two R-module homomorphisms f and g from R' to Rn are equivalent if and only if there is a commutative diagram Imh,

T

2

Rn

Rrn

where h1 and h2 are R-module isomorphisms. That is, g = h2 f hi Also, matrices A, B E Mn,,,,(R) are equivalent if B = PAQ for some P E GL(n, R) and Q E GL(m, R). (1.3) Proposition. Let R be a commutative ring and let f, g : Rn -+ Rn be R-module homomorphisms. If f is equivalent to g, then Coker(f) a, Coker(g).

Proof. Let M = Coker f and N = Coker(g). Since f and g are equivalent there is a commutative diagram of R-modules and homomorphisms

M-0 Th Rm

a

Rn -# N -. 0

where the hi are isomorphisms and iri are the canonical projections. Define

0: M -i N by 0(x) = ir2(h2(y)) where irl(y) = x. It is necessary to check that this definition is consistent; i.e., if irl(yi) = irl(y2), then ir2(h2(yl)) = ir2(h2(y2)). But if 7rl(yi) = irl(y2), then yl - y2 E Ker(7rl) = Im(f), so y1 - y2 = f(z) for some z E R'n. Then h2(yi - 112) = h2f(z) = ghl(z) E Im(g) = Ker(7r2)

so a2h2(yl - y2) = 0 and the definition of 0 is consistent. We will leave it to the reader to check that 0 is an R-module isomorphism. In order to apply Proposition 1.3 to our F[X]-modules VT, it is necessary to produce a finite free presentation of VT. We now show how to use a basis of V to produce a finite free presentation of the F[X]-module VT. Thus, we assume that V is a finite-dimensional vector space over a field F and T : V V is a linear transformation. Define

5.1 Equivalence and Similarity

291

by,OB(ej) = vj for 1 < j < n where A = {e1, en} is the standard basis of the free F[X]-module F[X]n and B = {v1, ... , vn} is a given basis of V. Thus, (1.3)

f1(T)(vi)+...+fn(T)(vn)

VG13(f1(X), ... ,fn(X)) =

for all (f1(X), ... , f. (X)) E F[X]n. Let K = Ker(08). If A = [T]B then A = [aij] E Mn(F) where n

T(vj) _

aijvi. i=1

Let n

(1.5)

pj(X) = Xej - >aijei E F[X]n for 1 < j < n. i=1

(1.4) Lemma. K C F[X ]n is free of rank n and C = {p1(X ), ... , pn(X )} is a basis of K.

Proof. It is sufficient to show that C is a basis. First note that 13(pj(X)) = YOB

Xej -

aijei i=1 n

= X013(ej) - Y:aijVB(ei) i=1

n

= T (vj) - L aij vi i=1

=0 by Equation (1.4). Thus pj(X) E K = Ker(OB) for all j. By Equation (1.5) we can write n

Xej = pj(X) + >aijei.

(1.6)

i=1

By repeated uses of this equation, any n

H(X) = I: gj(X)ej E F[X]n j=1

can be written as n

(1.7)

n

n

H(X) = Egj(X)ej =Ehj(X)pj(X)+>bjej j=1

j=1

j=1


292

where bj E F. If H(X) E K then it follows from Equation (1.7) that E' 1 bjej E K, and applying the homomorphism ip we conclude that bjvj = 0 E V; but {v1i ... , vn} is a basis of V, so we must have 1

bi =

= bn = 0. Therefore,

C = {P1(X), ... ,Pn(X)} generates K as an F[X]-submodule. To show that C is linearly independent, suppose that n

E hj(X)pj(X) = 0j=1

Then

n

n

>hj(X)Xe, = E hj(X)aijei

j=1

i,j=1

and since {ei, ... , en} is a basis of F[X]n, it follows that n

hi(X)'X =Ehj(X)a;j.

(1.8)

j=1

If some hi (X) 96 0, choose i so that hi (X) has maximal degree, say,

deghi(X) = r > deghj(X)

(1 < j < n).

Then the left-hand side of Equation (1.8) has degree r + 1 while the righthand side has degree < r. Thus hi(X) = 0 for 1 < i < n and C is a basis of K. We can summarize the above discussion as follows:

(1.5) Proposition. Let V be a vector space of dimension n over the field F, let T : V -* V be a linear transformation, and let 8 be a basis of V. Then there is a finite free presentation of VT (1.9)

0 --

' F[X]n ±24 F[X]n -+ VT .0

in which the matrix of 08 in the standard basis of F[X]n is

XIn - [T]8. Proof. If O8(ej) = pj(X) as in Equation (1.5), then pj(X) = colj(XIn - A) where A = [T]8. Thus [O8]A = XIn - A, and sequence (1.9) is exact by Lemma 1.4.

According to our analysis, VT = Coker(O8). It is worthwhile to see this isomorphism in a simple explicit example.

5.1 Equivalence and Similarity

293

(1.6) Example. Let F be a field, let V = F2, and let T : V -+ V be defined

by T(ul, u2) = (u2i 0) (cf. Example 3.1.5 (13)). If B = (vi, v2) is the standard basis on F2, then '+,e(f (X), 9(X)) = f (T)(vi) + g(T)(v2) = aov1 + bov2 + blvl where

f(X)=ao+a1X+ g(X) = bo+b1X + +b,,,Xm.

Since A=[T]B= [oo],wehave X12-A= [° X) so that -013 L(1f(x)1) 9(X) l

Note that

(f(X)1 l g(X) J

- [xf(x)_9(X)1 X9(X)

- lao+bll + (Xf(X)-9'(X) Xg(X) bo

L

JJ

where

9'(X) =

9(X) - bo

X f(X) +9(X) - ao - bl f(X) = X Since f (X) and g(X) are arbitrary, we see that F[X]2/Im(OB) F2 (1.10) as F-modules, while as an F[X]-module, r

F[X]21 Im(s8) L- F[X]

11)

J

101.

Equation (1.11) follows from Equation (1.10), the observation

X [0] 1

and the fact that

0J+I

X1,

E Im(OB). It follows immediately that

I

XJ L

F[X]2/ Im(OB) ?` VT

as F[X]-modules.

(1.7) Theorem. Let F be a field and let A, B E

be matrices. Then

A is similar to B if and only if the polynomial matrices XI, - A and XI,, - B E M (F[X ]) are equivalent.

294


Proof. If A and B are similar, then B = P-'AP for some P E GL(n, F). Then

(XIn - A) (XIn - B) are equivalent in Mn(F(X]). Conversely, suppose that (XIn - A) and (XIn - B) are equivalent in Mn(F[X]). Thus there are matrices P(X), Q(X) E GL(n, F[X]) such that (1.12)

(XI, - B) = P(X)(XIn - A)Q(X).

Now, if V is a vector space of dimension n over F and B is a basis of V, then there are linear transformations T1 and T2 on V such that [T1]B = A and [T2]B = B. By Proposition 1.5, there are injective homomorphisms

Oi : F[X]n ~ F[X]n such that Coker(0i)

VT, and the matrix of ¢i with respect to the standard

basis on F[X]n is (XIn - A) and (XI,, - B) for i = 1, 2 respectively. By Equation (1.12), the F[X]-module homomorphisms 01 and 02 are equivalent. Therefore, Proposition 1.3 applies to give an isomorphism of VT, and VT, as F[X]-modules. By Theorem 4.4.2, the linear transformations T1 and T2 are similar, and taking matrices with respect to the basis B shows that A and B are also similar, and the theorem is proved. This theorem suggests that a careful analysis of the concept of equivalence of matrices with entries in F[X] is in order. Since it is no more difficult to study equivalence of matrices over any PID R, the next two sections will

be devoted to such a study, after which, the results will be applied to the computation of invariant factors and canonical forms for linear transformations. We will conclude this section by carefully studying the relationship between generating sets of a module M that has two different finite free presentations.

(1.8) Example. Let R be a commutative ring and let M be an R-module with two equivalent finite free presentations. That is, there is a commutative diagram of R-modules and homomorphisms Rm

(1.13)

jai

Rm g

M --'

0

11M

M -* 0

where h1 and h2 are isomorphisms. If A = {el, ... en} is the standard basis on Rn, then V = {vl, ... , vn} and W = {w1, ... , wn} are generating sets of the R-module M, where vi = 9r1(ei) and w3 = ir2(e,). Note that we are not assuming that these generating sets are minimal, i.e., we do not assume that µ(M) = n. From the diagram (1.13) we see that the generators v; and w3 are related by

5.1 Equivalence and Similarity (1.14)

295

wj = 7r2(ej) = ir1(h2 1(ej))

Let us analyze this situation in of matrix representations of the homomorphisms. All matrices are computed with respect to the standard bases

on R' and R. Thus, if A = [f], B = [g], Q = [hi], and P = [h2], then the commutativity of diagram (1.13) shows that B = PAQ-1. Furthermore, if P-1 = [pt], then Equation (1.14) becomes wj = ir1(h21(ej)) n

_ 11pijei) i=1

=

(1.15)

tPiv.

That is, wj is a linear combination of the vi where the scalars come from the jeh column of P-1 in the equivalence equation B = PAQ-1. (1.9) Example. We will apply the general analysis of Example 1.8 to a specific numerical example. Let M be an abelian group with three generators v1i v2, and v3, subject to the relations 6v1 + 4v2 + 2v3 = 0 -2v1 + 2v2 + 6v3 = 0.

That is, M = Z3/K where K is the subgroup of Z3 generated by yi = (6,4,2) and y2 = (-2,2,6), so there is a finite free presentation of M

0, Z2-

TA

(1.16)

+ Z3--

11

-+ M->0

where TA denotes multiplication by the matrix 6

A= 4

2

-2 2 6

We wish to find B = PAQ equivalent to A where B is as in Proposition 4.3.20. If

P=

0

0

1

0

1

1

-2

-2

and

1

_

1

Q=

[0

3

-1

'

then

2

B=PAQ= 0

0 10

0

0

(We will learn in Section 3 how to compute P and Q.) Then

296


P-1=

3 2

2

1

1

0

1

0

0

Therefore, w1 = 3v1 + 2v1 + v3, w2 = 2v1 + V2, and w3 = v3 are new generators of M, and the structure of the matrix B shows that 2w1 = 0 (since 2w1 = 6v1 + 4v2 + 2v3 = 0), lOw2 = 0, and w3 generates an infinite cyclic subgroup of M, i.e.,

MZ2®Z1o®Z.

5.2 Hermite Normal Form In this section and the next, we will be concerned with determining the simplest form to which an m x n matrix with entries in a PID can be reduced using multiplication by unimodular matrices. Recall that a unimodular matrix is just an invertible matrix. Thus, to say that A E M,, (R) is unimodular is the same as saying that A E GL(n, R). Left and right multiplications by unimodular matrices gives rise to some basic equivalence relations, which we now define.

(2.1) Definition. Let R be a commutative ring and let A, B E M"',n(R). (1)

We say that B is left equivalent to A, denoted B £ A, if there is a

unimodular matrix U E GL(m, R) such that B = UA. R (2) We say that B is right equivalent to A, denoted B A, if there is a unimodular matrix V E GL(n, R) such that B = AV. (3) We say that B is equivalent to A, denoted B N A, if there are unimodular matrices U E GL(m, R) and V E GL(n, R) such that B = UAV. Each of these relations is an equivalence relation, and we would like to compute a simple form for a representative of each equivalence class in an algorithmic manner. (Equivalence of matrices has been introduced in Definition 4.3.18, and, in fact, if R is a PID then we have described the equivalence classes of matrices under the relation of equivalence in Proposition 4.3.20. What we will concentrate on at the present time are algorithmic aspects of arriving at these representatives.) This will require the ability to

construct a number of unimodular matrices. The elementary matrices introduced in Definition 4.1.8 provide some examples of unimodular matrices over any commutative ring, and we will see that for a Euclidean domain, every unimodular, i.e., invertible, matrix is a product of elementary matrices. The following fundamental result describes an inductive procedure for constructing unimodular matrices with a prescribed row or column.

5.2 Hermite Normal Form

297

(2.2) Theorem. Let R be a PID, let a1, ..., a,a E R, and let d = gcd{a1, ... ,an}. Then there is a matrix A E Mn(R) such that (1) rowl(A)=[a1

an], and

(2) det(A) = d.

Prof. The proof is by induction on n. If n = 1 the theorem is trivially true. Suppose the theorem is true for n - 1 and let Al E Mn-1(R) be a an_1] and det(A1) = d1 = gcd{a1,..., matrix with rowl(A1) = [a1 an-11- Since d = gcd{al,...,an}

= gcd{gcd{a1i ... a, _1}, an} = gcd{dl, an},

it follows that there are u, v E R such that ud1 - van = d. Now define A by AI

an 0

A= 0 L

U

ai for 1 < i < n - 1, it follows that A E M. (R) and row, (A) = an ]. Now compute the determinant of A by cofactor expansion along the last column. Thus, Since d1 [a1

det(A) = udet(A1) + (-1)n+1an det(Ain) where A1n denotes the minor of A obtained by deleting row 1 and column n. an_1 ] Note that A1n is obtained from AI by moving row,(A1) = (a1 to the n -1 row, moving all other rows up by one row, and then multiplying the new row n -1 (= [ a1 an- I ]) by v/d1. That is, using the language of elementary matrices

A1n = Dn-1(v/d1)Pn-1,n-2 ... P3.2 P21A1. Thus,

det A1n = (i-) (-1)n-2 det Al = (-1)n-2V. Hence,

det A = u det A 1 + (-1)n+1 an det Ain (-1)n+lan(-1)n-2v

= ud1 +

= ud1 - Van = d.

Therefore, the theorem is true for all n by induction.

0


298

(2.3) Remark. Note that the proof of Theorem 2.2 is completely algorithmic, except perhaps for finding u, v with ud, - van = d; however, if R is a Euclidean domain, that is algorithmic as well. It is worth noting that the existence part of Theorem 2.2 follows easily from Theorem 3.6.16. The details are left as an exercise; however, that argument is not at all algorithmic.

(2.4) Corollary. Suppose that R is a PID, that a1,

..., a, are

relatively

prime elements of R, and 1 < i < n. Then there is a unimodular matrix A2 with rowi(A2) _ [a, an ] and a unimodular matrix B2 with col2(B2) _ [a,

...

an]

t

.

Proof. Let A E Mn(R) be a matrix with

...

row, (A) = [a,

an ]

and det A = 1, which is guaranteed by the Theorem 2.2. Then let Ai = P12A and let B2 = Al P1. where P12 is the elementary permutation matrix that (] interchanges rows 1 and i (or columns 1 and i).

(2.5) Example. We will carry through the construction of Theorem 2.2 in a specific numerical example. Thus let R = Z and construct a unimodular matrix A E GL(3, Z) with

row,(A) _ [25

7].

15

3 z Then 3 5 - 2 7 = 1,

Since 2.25-3-15 =5, we may take A, so the induction step will give 25

15

7

3

2

0

10

6

3

A=

Then det A = 1, so A is unimodular. Furthermore, if we wish to compute a unimodular matrix B E GL(4, Z) with

row,(B) = [25

15

7

9]

then we may use the matrix A in the induction step. Observe that 10-9 = 1, so the algorithm gives us a matrix

B-

25

15

7

9

3

2

0

10

6

3

0 0

25

15

7

10

that is unimodular and has the required first row.


299

(2.6) Lemma. Let R be a PID, let A # 0 E M,,,,1(R), and let d = gcd(A) (i.e., d is the gcd of all the entries of A). Then there is a unimodular matrix U E GL(m, R) such that d

UA=

0

0

Moreover, if R is a Euclidean domain, then U can be taken to be a product of elementary matrices. .. am it Proof We may write bla1 + + bmam = d 0 0 where A = [ al and b1, ..., bm E R. Then

bt\d/+.+bm(d)=1 so {b1, ... bm} is relatively prime. By Theorem 2.2 there is a matrix U1 E GL(m, R) such that rows (Ul) = [ b1 ... bm ]. Then

+ uimam so that d I ci for all i > 2. Hence ci = aid for ai E R. Now, if U is defined by

and ci = uilal +

U = T21(-a2)T31(-a3) ... Tml(-am)Ul then Proposition 4.1.12 shows that

This completes the proof in the case of a general PID. Now suppose that R is a Euclidean domain with Euclidean function v : R \ {0} - Z+. We shall present an argument, which is essentially a second proof of Lemma 2.6 in this case. This argument is more constructive in that only elementary row operations, i.e., left multiplications by elementary matrices, are used. Hence the U constructed will be a product of elementary matrices. Let v(A) = min{v(at) : 1 < j < m} and suppose that v(A) = Qm ]t has v(/31) < v(,0j) for j > 2. Each Qi v(ai). Then P1iA = 10, can be written as Qi = 7it1 + ri where ri = 0 or v(ri) < v(l31). Therefore, subtracting ryi row1(P1iA) from rowi(P1iA) gives a matrix

300

Chapter 5. Matrices over PII)s

al

Al = T2,(-72) ... T,.1(-7m)PjjA =

rl

rm

where ri = 0 or v(rj) < v(pl). If some rj # 0 then we have found a matrix Al left equivalent to A via a product of elementary matrices for which v(A1) < v(A). We can repeat the above process to find a sequence of matrices

amp

left equivalent to A via a product of elementary matrices such that (1) v(aii-ll) > v(a(')) for all i > 1, and v(a(')) for j > 2. 0 or (2)

Since condition (1) can only occur for finitely many A;, it follows that we must have a3(') = 0 for j > 2 for some A;, i.e.,

al') 1

UA=A,=

0 0

for some U E GL(m, R), which is a product of elementary matrices. It remains to observe that a(') = b is a gcd of the set {a1, ... ,am}. But the equation -b0

am

0

shows that b E (a 1, ... , am) and b

[ai] = U-1 am

U

in

shows that (al, ... am) C (b), i.e., (b) = (a1, proved.

, am) and the lemma is 0

Given an equivalence relation ti on a set X, a complete set of representatives of - is a subset P C X such that P has exactly one element from


301

each equivalence class. Thus, P is a complete set of representatives of - if each x E X is equivalent to a unique a E P. The cases we wish to consider concern some equivalence relations on a commutative ring R. A complete set of nonassociates of R is a subset P C R such that each element of R is an associate of a unique b E P, i.e., if a E R then there is a unique b E P and a unit u E R* such that b = au. Similarly, if I C R is an ideal, then a complete set of residues modulo I consists of a subset of R, which contains exactly one element from each coset a + I. If I is the principal ideal Ra for some a E R, then we speak of a complete set of residues modulo a. (2.7) Examples. (1)

(2)

(3)

If R = Z, then a complete set of nonassociates consists of the nonnegative integers; while if m E Z is a nonzero integer, then a complete set of residues modulo m consists of the m integers 0, 1, ... , Iml - 1. A complete set of residues modulo 0 consists of all of Z. If F is a field, then a complete set of nonassociates consists of {0, 1}; while if a E F \ {0}, a complete set of residues modulo a is {0}. If F is a field and R = F[X] then a complete set of nonassociates of R consists of the monic polynomials together with 0.

(2.8) Definition. Let R be a commutative ring, let P C_ R be a complete set

of nonassociates of R, and for each a E R let P(a) be a complete set of residues modulo a. Then a matrix A = [aij] E M,n,n(R) is said to be in Hermite normal form (following P, P(a)) if A = 0 or A 0 and there is an integer r with 1 < r < m such that 0 for 1 < i < r, rowi(A) = 0 for r + 1 < i < m; and (1) rowi(A) < n,. < m such there is a sequence of integers 1 < nl < n2 < (2)

that aij = 0 for j < ni (1 < i < r), ain, E P\101 (1 < i < r), and ajnj E P/lain;) for 1 < j < i. Thus, if the matrix A is in Hermite normal form, then A looks like the following matrix: Table 2.1. Hermite normal form 0 0

0

...

0

al.,

*

...

*

0

...

0

aln3 a2n3

* *

...

a2n,.

...

0

* *

alnr

0

al-2 a2n2

...

...

0

0

0

...

0

0

0

a3n3

*

...

a3nr

arn r 0

0 0

... ...

0 0

...

0

...

...

0

...

...

*

...

0

0

302


where * denotes an entry that can be any element of R. If R is a field and P = {0, 11 while P(a) = {0} for every a 96 0, then in the Hermite normal form we will have a,,,, = I while ajn, = 0 if j < i. The resulting matrix is what is usually called a reduced row echelon matrix and it is used to solve systems of linear equations Ax = b over a field. Our main result on left equivalence of matrices is that if R is a PID then every matrix is left equivalent to one in Hermite normal form. (2.9) Theorem. Let R be a PID, P C R a complete set of nonassociates, and P(a) (a E R) a complete set of residues modulo a. Then any A E M,a,n(R) is left equivalent to a matrix H in Hermite normal form. If R happens to be Euclidean, then H = UA where U is a product of elementary matrices.

Proof. The proof is by induction on the number of rows m. If m = 1 and A 34 0 let n1 be the first index with al,,, 0. Then let ualn, = bin, E P. Then B = uA is in Hermite normal form. Now suppose that m > 1 and that every matrix in Mm_1,,,(R) (for arbitrary n) is left equivalent (using a product of elementary matrices if R is Euclidean) to a matrix in Hermite normal form. Let n1 be the smallest integer such that coln, (A) $ 0. Let (cola, (A))' = [ a1 . a,a) and let d = ged{al, ... , a,,,}. Then d 36 0 and, by Lemma 2.6, there is an invertible matrix U1 E GL(m, R) (which may be taken as a product of elementary matrices if R is Euclidean) such

that d

Al=U1A=

*

0

B, 0

where B1 E M,_l,n_1(R). By the induction hypothesis, there is an invertible matrix V E GL(m - 1, R) (which may be taken as a product of elementary matrices if R is Euclidean) such that VB1 is in Hermite normal form. Let 1

U2

0

0 V

Then U2A1 = A2 is in Hermite normal form except that the entries a,,,,

(i > 1) may not be in P(a;a,). This can be arranged by first adding a multiple of row 2 to row 1 to arrange that a,,,, E P(a2n, ), then adding a multiple of row 3 to row 1 to arrange that a1n3 E P(a3n, ), etc. Since a;3 = 0 if j < n;, a later row operation does not change the columns before n;, so at the end of this sequence of operations A will have been reduced to Hermite normal form, and if R was Euclidean then only elementary row operations will have been used. 0 If we choose a complete set of nonassociates for R so that it contains 1 (as the unique representative for the units) and a complete set of representatives modulo 1 to be {0}, then the Hermite form of any U E GL(n, R) is


303

I,,. This is easy to see since a square matrix in Hermite normal form must be upper triangular and the determinant of such a matrix is the product of the diagonal elements. Thus, if a matrix in Hermite normal form is invertible, then it must have units on the diagonal, and by our choice of 1 as the representative of the units, the matrix must have all 1's on the diagonal. Since the only representative modulo 1 is 0, it follows that all entries above the diagonal must also be 0, i.e., the Hermite normal form of any U E GL(n, R) is I,,. If we apply this observation to the case of a unimodular matrix U with entries in a Euclidean domain R, it follows from Theorem 2.9 that U can be reduced to Hermite normal form, i.e., I,,, by a finite sequence of elementary row operations. That is,

El...E1U=I where each Ej is an elementary matrix. Hence, U = El 1 . . . ET' is itself a product of elementary matrices. Therefore, we have arrived at the following result. (2.10) Theorem. Let R be a Euclidean domain. Then every invertible matrix over R is a product of finitely many elementary matrices. Proof.

(2.11) Remark. If R is not Euclidean then the conclusion of Theorem 2.10 need not hold. Some explicit examples of matrices in GL(2, R) (R a PID), which cannot be written as a product of elementary matrices, have been given by P. M. Cohn in the paper On the structure of the GL2 of a ring, Institut des Hautes Etudes Scientifiques, Publication #30 (1966), pp. 554. A careful study of Lemma 2.6 shows that the crucial ingredient, which Euclidean domains have that general PIDs may not have, is the Euclidean algorithm for producing the gcd of a finite subset of elements.

(2.12) Example. Let R = Z, P = Z+, and for each m 54 0 E Z, let P(m) _ {0, 1, ... , Imp - 1}.

Let P(0) = Z. Thus we have chosen a complete set of nonassociates for Z and a complete set of residues modulo in for each m E Z. We will compute a Hermite normal form for the integral matrix 2

A= 6

3

9 4

8

4

1

4

5

3

-1

The left multiplications used in the reduction are U1, ..., U7 E GL(3, Z), while Al = U1A and A; = U;A;_1 for i > 1. Then

304


-1

1

0

0

1

0

0

0 0

1

UI =

1

U2= -3 -4

r1

U3=

U4=

U5 =

U8=

1

0

1

0 0

0

1

0

0

10

-9

-1

1

1

0

0

0

1

0

0

-2

1

1

0

0

0

1

0

0

0 -1

1

5

0 0

1

0 0

0

1

0

-2

1

0

0

1

U7= 10

Al = Ul A = A2 = U2A1 =

A3 = U3A2 =

A4 = U4A3 =

A5 = U5A4 =

1

6 8

3 4

2

1

-5 -2

0 0

0

19

9

0

21

7

2

1

-5 -2

0 0

0

1

27

0

2

2

1

-5

-2 -2

0 0

0

1

27

2

1

0 0

0 0

2

0

4

3

1

-1

-56 -5 -2 0

27 56

1

0

1

0

0 0

1

0

133 27 56

2

1

0

21

0 0

0 0

1

27 56

A6=UBA5= 0 0

A7=U7A6=

-5 -2

2

0

The matrix A7 is the Hermite normal form associated to the matrix A (using the system of representatives P and P(m)). A natural question to ask is whether the Hermite normal form of A guaranteed by Theorem 2.9 is unique. Certainly one can get a different Hermite normal form by changing the complete set of nonassociates or the complete set of residues modulo a E R, but if we fix these items then the Hermite normal form is uniquely determined, independent of the precise sequence of operations needed to achieve this form. This is the content of the next result. (2.13) Theorem. Let R be a PID, P C R a complete set of nonassociates and P(a) a complete set of residues modulo a for each a E R. If A E Mm,n(R) then the Hermite normal form of A is unique.

Proof. Without loss of generality we may assume that A 0 0. First note that the number of nonzero rows in any Hermite normal matrix H, which is left equivalent to A, is just rank(A). To see this, suppose the Hermite

normal matrix H has r nonzero rows, let 1 < nl < < n,. < n be the integers guaranteed by Definition 2.8, and let a = (1, 2, ... , r) and a,.,,, 0. Thus, Q = (ni, n2, ... , n,.). Then det H[a 10) = ain, a2n2 rank(H) > r and any submatrix of H with more than r rows will have a row of zeros. Hence, rank(H) = r, and since rank is preserved by equivalence


305

(Proposition 4.2.36), it follows that

r = rank(H) = rank(A).

Now suppose that A L H and A L K where H and K are both in Hermite normal form. Then there is a unimodular matrix U such that H = UK, and by the above paragraph, both H and K have r = rank(A) nonzero rows. Let 1 < nl < n2 < < nr < n be the column indices for K given by the definition of Hermite normal form and let 1 < t1 < t2 < < tr < n be the column indices for H. We claim that t; = ni for 1 < i < r. Indeed, n1 is the first nonzero column of K and ti is the first nonzero column of H; but col, (H) = U col, (K) and U is invertible, so n1 = t1. Then we conclude

that for j = ni = ti, hin, 54 0, kin, 0 0, and kln,

[hi ni

0

0

U

0

0

so kin,u,i = 0 for s > 1, and hence, u,1 = 0 for s > 1. Therefore, _

U-

u11

Ui

0

If H = [ro i(H)] and K = [ro i(K)] Hi KI where Hi, K1 E M,n_i,n(R) then H1 = Ui Ki and H1 and K1 are in Hermite normal form. By induction on the number of rows we can conclude that nj = tj for 2 < j < r. Moreover, by partitioning U in the block form U = [ u" u12 ] where Ul 1 E Mr (R) U2i Usz and by successively comparing cola, (H) = U Colnj (K) we conclude that U21 = 0 and that U1 i is upper triangular. Thus, U11

U12

0

U22

0

0

Uir u2r

U12

U= urr

0

and detU = u11

U22

urr(det(U22)) is a unit of R. Therefore, each uii is

a unit of R; but hin, = uiikin, for 1 < i < r so that hin, and kn, are associates. But hin, and kin, are both in the given complete system of nonassociates of R, so we conclude that hin, = kin, and hence that uii = 1 for 1 < i < r. Therefore, each diagonal element of U11 must be 1. Now

suppose that 1 < s < r - 1. Then m

hs,n.+, _ E us,Yk7,n.+j f=1

= ussks,n.+i + us,a+iks+i,n.+i

306

Chapter 5. Matrices over PIN

since u,.. = 0 for -y < s while k7n,+, = 0 if 'y > s + 1. Since u,, = 1, we conclude that h.,,n.+i = ks,n.+, + u,,,+1ks+1,n.+1 Therefore,

hs,n.+, = ks,n.+,

(mod k,+1,n.+, ),

and since h,,n,+, and k,.n,+, are both in P(k,+l,n,+, ), it follows that hs,n.+, = k,,n,+, and, hence, us,s+1 = 0 for I < s < r - 1. We now proceed by induction. Suppose that u,,,+j = 0 for 1 < j < r-1 (we just verified that this is true for j = 1) and consider

-f=l

= us,s+j+1 ks+j+1,n,+j+, + u,,k,,n,+,+, us,s+j+1ks+j+1,n,+,+1 + ks,n.+,+t

Therefore, h,,,,,+,+, = k,,n,+,+, since they both belong to the same residue Hence u,,,+j+I = 0. Therefore, we have shown class modulo that U has the block form k,+j+1,n.+,r,

U _ rIr 0

U12 U22

and, since the last m - r rows of H and K are zero, it follows that H = 0 UK = K and the uniqueness of the Hermite normal form is proved. We will conclude this section with the following simple application of the Hermite normal form. Recall that if R is any ring, then the (two-sided) ideals of the matrix ring M,, (R) are precisely the sets M,' (J) where J is an ideal of R (Theorem 2.2.26). In particular, if R is a division ring then the only ideals of M,, (R) are (0) and the full ring M,, (R). There are, however,

many left ideals of the ring M,,(R), and if R is a PID, then the Hermite normal form allows one to compute explicitly all the left ideals of M,(R), namely, they are all principal.

(2.14) Theorem. Let R be a PID and let J C Mn(R) be a left ideal. Then there is a matrix A E Mn(R) such that J = (A), i.e., J is a principal left ideal of Mn(R).

Proof. Mn(R) is finitely generated as an R-module (in fact, it is free of rankn2), and the left ideal J is an R-submodule of Mn(R). By Theorem 3.6.2, J is finitely generated as an R-module. Suppose that (as an R-module)

J=(B1,...,Bk) where B; E Mn(R). Consider the matrix

5.3 Smith Normal Form

307

B1

B=

E

Bk

There is an invertible matrix P E Mnk(R) such that PB is in Hermite normal form. Thus,

PB =

(2.1)

A 0

0

and if we partition P = [P,J) into blocks where P;j E Mn(R), it follows from Equation (2.1) that

A=P11B1+ -

+ P1kBk-

Therefore, A E J and hence the left ideal (A) C J. Since A-

B=Q

0 .

.0

where Q = P-1 = [Qt

,

it follows that B; = Qi1A. Therefore, J C (A),

and the proof is complete.

0

5.3 Smith Normal Form In contrast to the relatively complicated nature of the Hermite normal form, if multiplication by nonsingular matrices is allowed on both the left

and the right, then one can reduce a matrix A over a PID R to a particularly simple form. The existence of this simple diagonal form, known as the Smith normal form, has essentially already been proved in Proposition 4.3.20. Combining this with Theorem 2.10 will provide a reasonably efficient procedure for the computation of the invariant factors of a linear transformation. Additionally, this same computational procedure produces the change of basis map that puts a matrix (or linear transformation) in rational canonical form. We will also consider applications of the Smith normal form to the solution of linear diophantine equations.

(3.1) Theorem. Let R be a PID and let A E M,n,n(R). Then there is a U E GL(m, R) and a V E GL(n, R) such that


308

UAV =

(3.1)

Dr 0

0 0

where r = rank(A) and Dr = diag(sl, ... , s,.) with si 0 0 (1 < i < r) and Si I si+l for 1 < i < r - 1. Furthermore, if R is a Euclidean domain, then the matrices U and V can be taken to be a product of elementary matrices.

Proof. Consider the homomorphism TA : R" - Rm defined by multiplication by the matrix A. By Proposition 4.3.20, there is a basis B of R' and

a basis C of R' such that (3.2)

[TA]"

_

[Dr

0]

where Dr = diag(sl, ... , Sr) with si $ 0 (1 < i < r) and si

I

si+l for

1 < i < r - 1. If C' and B' denote the standard bases on R' and R" respectively, then the change of basis formula (Proposition 4.3.16) gives

]=

(3.3) [ D0r

[ZA]C = PC [TA]C, Phi

00

Since A = [TA]C, and since the change of basis matrices are invertible, Equation (3.3) implies Equation (3.1). The last statement is a consequence of Theorem 2.10.

(3.2) Remarks.

(1) The matrix

[r

00] is called the Smith normal form of A after H. J.

Smith, who studied matrices over Z. (2) Since the elements s1, ... , Sr are the invariant factors of the submodule

Im(TA), they are unique (up to multiplication by units of R). We shall

call these elements the invariant factors of the matrix A. Thus, two matrices A, B E M,,n(R) are equivalent if and only if they have the same invariant factors. This observation combined with Theorem 1.7 gives the following criterion for the similarity of matrices over fields.

(3.3) Theorem. Let F be a field and let A, B E Mn(F). Then A and B are similar if and only if the matrices XI7, - A and XI, - B E Mn(F[X]) have the same invariant factors. Proof.

(3.4) Remark. If R is a Euclidean domain then A can be transformed into Smith normal form by a finite sequence of elementary row and column operations since every unimodular matrix over R is a finite product of elementary matrices (Theorem 2.10). It is worthwhile to describe explicitly an algorithm by which the reduction to Smith normal form can be


309

accomplished in the case of a Euclidean domain R with degree function v : R \ 0 --+ Z+. The algorithm is an extension (to allow both row and column operations) of the second proof of Lemma 2.6. If A E M,,,,,,(R) is a nonzero matrix, let v(A) = min{v(ain) : 1 1, we may write ail = alibi + bt1 where bit = 0 or v(bil) < v(a11). By subtracting birow1(A) from rowi(A) we obtain a matrix All in which every element of coli(A(')) is divisible by all or v(A(11) < v(A). If we are not in the first case, repeat the process with A replaced by AM. Since v(A) is a positive integer, this process cannot go on indefinitely, so we must eventually arrive at a matrix B with b11 bil for 2 < i < m. By applying a similar process to the elements of the first I

row, we may also assume that elementary row and column operations have produced a matrix B in which b11 bit and b11 I bl, for 2 < i < m and I

2 < j < n. Then subtracting multiples of the first row and column of B produces an equivalent matrix B = [b11J ® C where C E Mm-l,,,-l. We may arrange that b11 divides every element of C. If this-is not the case already, then simply add a row of C to the first row of B, producing an equivalent matrix to which the previous process can be applied. Since each

repetition reduces v(B), only a finite number of repetitions are possible before we achieve a matrix B = [b11J ® C in which b11 divides every entry of C. If C is not zero, repeat the process with C. This process will end with the production, using only elementary row and column operations, of the Smith normal form.

(3.5) Example. It is worthwhile to see this algorithm in practice, so we will

do a complete example of the computation of the Smith normal form of an integral matrix. A simple method for producing U E GL(m, R) and V E GL(n, R) so that UAV is in Smith normal form is to keep track of the elementary row and column operations used in this reduction. This can be conveniently accomplished by simultaneouly applying to In each row operation done to A and to I each column operation performed on A. This process is best illustrated by a numerical example. Thus let 2

A=

1

4

1

-3

-1 -3

-4

0

-1 1

I E M3 4(Z). ,

16

We shall reduce A to Smith normal form by a sequence of row and column operations, and we shall keep track of the net effect of these operations by simultaneously performing them on 13 (for the row operations) and 14 (for the column operations). We will use the arrow to indicate the age from one operation to the next.

Chap ter 5. Matrices over PIDs

310

A

13 1

0 0

0 1

0 0

0 1

0 1

0 0 1

H 1 -2 0

-+

2

0

1

-1

1

4

1

0

1

0

2

0

1

0

0

1

16

-1

-3

1

0

-1 -3 3

3

12

0 1

0

0

0

0

3

3

-3

-4

1

0

0

12

12

0

1

0

1

0

0

1

-2 -4

0

0

3

0 12

0

0 0 12

0

1

0

1

0

0

0

1

-2

0

0

3

0

-4

1

0 0

U

1

1

0

0

0

1

0

0 0

0 0

0

1

0

0

0

1

0

0 1

0 3

1

0

0 0

1

12

0

0

0 0

1

0

0

0

0 0

-3

1

0

1

16

-2

1

0

0

-3 -1

1

0

0

0

1

0

-4

0

0

1

-1

-4 1

4

1

-3 -3

0

0

=

-4

0

14

0

12

0

S

0

0

1

1

-10 0 1

_

0-

1

2

1

-1

1

0

1

0

0

0 1

2

1

-2

1

-1

2

0

1

-1

0

0

1

V

Then UAV = S and S is the Smith normal form of A. (3.6) Remark. Theorem 3.3 and the algorithm of Remark 3.4 explain the origin of the adjective rational in rational canonical form. Specifically, the

invariant factors of a linear transformation can be computed by "rational" operations, i.e., addition, subtraction, multiplication, and division of polynomials. Contrast this with the determination of the Jordan canonical form, which requires the complete factorization of polynomials. This gives an indication of why the rational canonical form is of some interest, even though the Jordan canonical form gives greater insight into the geometry of linear transformations.

The ability to compute the invariant factors of a linear transformation by reduction of the characteristic matrix XIn - [T]13 to Smith normal form has the following interesting consequence.

(3.7) Proposition. Let F be a field and let A E to the transposed matrix At.

Then A is similar


311

Proof. Consider the matrix X In -A E M.(F[XI ). Then there are invertible matrices P(X), Q(X) E GL(n, F[X]) such that

P(X) (XI. - A)Q(X) = diag(s1(X), ... , sn(X))

(3.4)

where si(X) I si+1(X) for 1 < i < n - 1. Taking the transpose of Equation (3.4) shows that

Q(X)t(XIn -At)P(X)t =diag(sl(X), ... sn(X))Thus, A and At have the same invariant factors and hence are similar.

This result cannot be extended to matrices over arbitrary rings, even PIDs, as the following example shows:

(3.8) Example. Let R be a PID, which is not a field, and let p E R be a prime. Consider the matrix 0

A= 0

0

p 0 0

0

E M3(R).

1

0

Claim. A is not similar to At. Proof. Suppose that T = [tip] E M3(R) satisfies the matrix equation AT = TAt. Then (3.5)

0 p

0

t11

t12

t13

0

0

1

t21

t22

t23

0

0

0

t31

t32

t33

=

t11

t12

t13

0

0

0

t21

t22

t23

p

0

0

t31

t32

t33

0

1

0

,

which implies that

(3.6)

pt21

pt22

pt23

t31

t32

t33

0

0

0

=

pt12

t13

pt22

t23

0 0

pt32

t33

0

From Equation (3.6), we conclude that t32 = t23 = t33 = 0, t12 = t21, t13 = t31 = pt22 = ps for some s E R. Therefore, T must have the form T =

tll

t12

PS

t12

s

0

Ps

0

0

and hence, det(T) = -p2s3. Since this can never be a unit of the ring R, it follows that the matrix equation AT = TAt has no invertible solution T. Thus, A is not similar to At.

312


(3.9) Remark. Theorem 1.7 is valid for any commutative ring R. That is, if A, B E MM(R), then A and B are similar if and only if the polynomial matrices XI, - A and X In - B are equivalent in Mn (R[X ]) . The proof we have given for Theorem 1.7 goes through with no essential modifications. With this in mind, a consequence of Example 3.8 is that the polynomial matrix

X13-A=

X -p 0 X -1 0

E M3(R[X])

X is not equivalent to a diagonal matrix. This is clear since the proof of Proposition 3.7 would show that A and At were similar if XI3 - A was equivalent to a diagonal matrix. 0

0

What this suggests is that the theory of equivalence for matrices with

entries in a ring that is not a PID (e.g., R[X] when R is a PID that is not a field) is not so simple as the theory of invariant factors. Thus, while Theorem 1.7 (extended to A E M,,(R)) translates the problem of similarity of matrices in M,(R) into the problem of equivalence of matrices in M,(R[X]), this merely replaces one difficult problem with another that is equally difficult, except in the fortuitous case of R = F a field, in which case the problem of equivalence in M,,(F[X]) is relatively easy to handle.

The invariant factors of a matrix A E M,n,n (R) (R a PID) can be computed from the determinantal divisors of A. Recall (see the discussion prior to Definition 4.2.20) that Q,,,,,, denotes the set of all sequences a = (i1, ... , ip) of p integers with 1 < it < i2 < < ip m. If a E Qp,,,,,, ,3 E Qj,,,, and A E M,n...n(R), then A[a 10] denotes the submatrix of A whose row indices are in a and whose column indices are in 0. Also recall (Definition 4.2.20) that the determinantal rank of A, denoted D-rank(A), is the largest t such that there is a submatrix A[a 1,31 (where a E Qt,,,, and /3 E Qt,,,) with det A[a 10] 0. Since R is an integral domain, all the ranks of a matrix are the same, so we will write rank(A) for this common number. For convenience, we will repeat the following definition (see Definition 4.2.21):

:3.10) Definition. Let R be a PID, let A E M,n,,, (R), and let k be an integer

such that 1 < k _< min{m, n}. If det A[a

1

a] = 0 for all a E Qk,m,

l3 E Qk,n, then we set dk (A) = 0. Otherwise, we set dk(A) = gcd{det A[a 10] : a E Qk,m., 0 E Qk,n}. dk (A) is called the kth determinantal divisor of A. For convenience in some formulas, we set do(A) = 1.

(3.11) Lemma. Let R be a PID and let A, B E Mm,n(R). Suppose that A is equivalent ofdk(B) to B and that 1 < k < min{m, n}. Then dk(A) is an associate


313

Proof. Suppose UAV = B where U E GL(m, R) and V E GL(n, R). If a E Qk,,n and 0 E Qk,n, then the Cauchy-Binet theorem (Thei.rem 4.2.34) shows that (3.7)

det B[a I p] _ E det U[a I w] det A[w I r] det V [r I p]. WEQk,,, TEQk.n

Thus, if dk(A) = 0 then det B[a I /3] = 0 for all a E Qk,n, Q E Qk,n and hence dk(B) = 0. If dk(A) 34 0 then dk(A) I detA[w I r] for all w E Qk,,n, r E Qk.n, so Equation (3.7) shows that dk(A) I det B[a I ,o] for all a E Qk,m, (3 E Qk,n Therefore, dk(A) I dk(B).

Since it is also true that A = U-'BV-1, we conclude that dk(A) = 0 if and only if dk(B) = 0 and if dk(A) 54 0 then dk(A) I dk(B)

and

dk(B) I dk(A)

0

so dk(A) and dk(B) are associates.

Now suppose that R is a PID and that A E Mm,n(R) is in Smith normal form. That is, we suppose that rank(A) = r and

A[0

Dr

0

0

where Dr = diag(al, ... ,Sr) with s; 36 0 (1 < i < r) and s,

I

s;+l for

1
while det A[(3 I ry] = 0 for all other /3 E Qk,,n, -y E Qk,n. Then, since si 18 +1

for 1 < i < r - 1, it follows that (3.8)

dk(A) =

at Sk if1
0

From Equation (3.8) we see that the diagonal entries of A, i.e., sl, ..., sr, can be computed from the determinantal divisors of A. Specifically, al = d1(A) 8 =

Sr __

d2(A)

di(A)

4(A) 4-1(A).

By Lemma 3.11, Equations (3.9) are valid for computing the invariant factors of any matrix A E M,n,n(R).


314

(3.12) Examples. (1) Let

-2 A=

0

10

2

-1

-3 -4

0 1

E M3(Z).

Then the Smith normal form of A is diag(1, 1, 8). To see this note that ent31(A) = 1, so d1(A) = 1;

detA[(1,2)I(1,2)]=6 and detA[(2,3)l(2,3)]=11, so d2(A) = 1, while det A = 8, so d3(A) = 8. Thus, s1(A) = 1, s2(A) _ 1, and s3(A) = 8. (2) Let

X(X - 1)3 B

0 0

0

0

0

X

X-1

0

E M3(Q[X]).

Then d1(B) = 1, d2(B) = X(X - 1), and d3(X) = X2(X - 1)4. Therefore, the Smith normal form of B is diag(1, X (X -1), X (X -1)3 ).

Let M be a finitely generated R-module (where R is a PID) and choose a finite free presentation of M

0--.R"TA R'-M--.0

where TA denotes multiplication by the matrix A E Mm,n(R). If A is put in Smith normal form

UAV=I

j

]=B

where Dr = diag(s1, ... ,Sr) with s, i 0 for all i and s; I s;+1 for 1 < i < r - 1, then by Proposition 1.3 (3.10)

M Coker(TB) ?° (RI (81)) ®. ® (R/(Sr)) ® Rm

Therefore, we see that the si 54 1 are precisely the invariant factors of the torsion submodule Mr of M. This observation combined with Equation (3.9) provides a determinantal formula for the invariant factors of M. We record the results in the following theorem.

(3.13) Theorem. Let R be a PID and let A E Mm,n(R). Suppose that the Smith normal form of A is [ ° o]. Suppose that s, = 1 for 1 < i < k (take

k = 0 if s1 # 1) and s; 96 1 fork < i < r. If M = Coker(TA) where TA : Rn --y Rm is multiplication by A, then

(1) µ(M) = m - k;


315

(2) rank(M/M,) = m - r; (3) the invariant factors of M, are t; = sk+i for 1 < i < r - A*,; and dk+1(A) for l < i < r - k. (4)

ti = dk+i-1(A)

Proof. All parts follow from the observations prior to the theorem; details 0 are left as an exercise.

If we apply this theorem to the presentation of VT from Proposition 1.5, we arrive at a classical description of the minimal polynomial mT(X) due to Frobenius. (3.14) Theorem. (Frobenius) Let V be a finite dimensional vector space over a field F and let T : V V be a linear transformation. Let 8 be any basis

of V, let [T]8 = A E Mn(F), and let d(X) = 4_1(XIn - A). Then (3.11)

MAX) = CT(X) d(X)

Proof. CT(X) = det(XIn - A) = dn(X). Since mT(X) is the highest degree invariant factor of VT (Definition 4.4.6 (2)), formula (3.11) follows immediately from Theorem 3.13 (4). 0 The determinantal criterion for invariant factors also allows one to prove the following fact.

(3.15) Theorem. Let F be a field and let A, B E Mn(F). If K is field that contains F as a subfield, then A and B are also in Mn(K). If A and B are similar in Mn(K), then they are similar in Mn(F). Proof. This follows immediately from Theorem 3.3 and the following observations. (1)

If f(X) and g(X) 34 0 are in F[X], then the quotient and remainder upon division of f(X) by g(X) in K[X] are, in fact, in FIX]. To see this, divide f(X) by g(X) in F[X] to get

f (X) = g(X)q(X) + r(X) where q(X), r(X) E F[X] and degr(X) < degg(X). The uniqueness of division in K[X] shows that this is also the division of f (X) by g(X) in K[X].

(2) Let f1(X), ..., fk(X) E F[X]. Then the greatest common divisor of these polynomials is the same whether they are considered in FIX] or in K[X]. This follows from (1) because the greatest common divisor can be computed by the Euclidean algorithm, which only uses the division algorithm.


316

If A and B E

then the polynomial ma-

are similar in

trices XI,, - A and XI - B have the same invariant factors in K[XJ. But since the invariant factors are computed as quotients of determinantal

divisors, and since XI, - A and XI,, - B are in M,a(F[X]), we conclude from items (1) and (2) above that the invariant factors of both polynomial matrices are in F[X]. Hence A and B are similar in (3.16) Remark. The content of Theorem 3.15 is that in order to determine if two matrices are similar, we may, without loss of generality, assume that they are matrices over a large (e.g., algebraically closed) field. This observation is useful, for example, in Theorem 5.13 (see Remark 5.14).

Let A E M,,,,,,(R) (R a PID) be a matrix of rank r and let sl, ..., s, be the invariant factors of A. Then, by definition, si # 0, si s2+1 for 1 < i < r - 1 and A is equivalent to f °r of where Dr = diag(sl, ... , s,.). Let pi, ... , A be a complete set of nonassociate primes that occur as prime I

divisors of some invariant factor. Then for appropriate nonnegative integers eij and units ui of R, we have

(3.12)

ulpil l p212 ... pelk Sl = s2 = u2P1 p22' ... pk2k

Sr = urpellpe,2 ... perk. Since si I s;+l, it follows that (3.13)

0 < el.i G e2j <

< erg

(1 < j < k).

The prime power factors eiJ > 0}, counted according to the number of times each occurs in the Equation (3.12), are called the elementary divisors of the matrix A. Of course, the elementary divisors of A are nothing more than the elementary divisors of the torsion submodule of Coker(TA), where, as usual, TA : R" R' is multiplication by the matrix A (Theorem 3.13 (3)). For example, let A = diag(12, 36, 360, 0, 0).

A is already in Smith normal form, so the invariant factors of A are 12 = 3, 36 = 22 . 32, and 360 = 23 . 32 . 5. Hence the elementary divisors of A are 22

{22, 3,2 2 , 32, 23.32, 5}.

(3.17) Theorem. Let R be a PID and let A, B E M,,,,n(R). Then A and B are equivalent if and only if they have the same rank and the same set of elementary divisors (up to multiplication by a unit).


317

Proof. According to Remark 3.2 (2) (uniqueness of the invariant factors), A and B are equivalent if and only if they have the same invar,ant factors. Since the invariant factors determine the elementary divisors, it follows that equivalent matrices have the same set of elementary divisors. Conversely, we will show that the set of elementary divisors and the rank determine the invariant factors. Indeed, if

ej= maxeij 1

1<j

then sr is an associate of pi' ... p','. Delete {pj', ... , pk' } from the set of elementary divisors and repeat the process with the set of remaining elementary divisors to obtain sr_l. Continue this process until all the elementary divisors have been used. At this point the remaining si are 1 and we have recovered the invariant factors from the set of elementary divisors. 0 (3.18) Remark. The argument of Theorem 3.17 is essentially a reproduction of the proof of Theorem 3.7.15.

(3.19) Example. It is worthwhile to present a complete example illustrating the process of recovering the invariant factors from the elementary divisors. Thus, suppose that A E M7,6(Z) is a rank 5 matrix with elementary divisors {2, 22, 22, 23,3 2 , 32, 32, 7, 11, 112}.

Then ss = 23 .3 2 7. 112 = 60984. Deleting 23, 32, 7, 112 from the set of elementary divisors leaves the set 12,2 2 , 22, 32, 32, 11}.

Thus, s4 = 22 32 . 11 = 396. Deleting 22, 32, 11 leaves the set {2, 22, 32) so that 83 = 22.32 = 36. Deleting 22 and 32 gives a set {2} so that 32 = 2. Since the set obtained by deleting 2 from {2} is empty, we must have that sl = 1. Therefore, A is equivalent to the matrix 0 0 0

0 2 0 0

0 0 36 0

0 0 0 396

0

0

0

0 0

0 0

0 0

0 0 0

1

0 0 0 0 60984 0 0

00 0 0 0 0 0

The next result is useful if an equivalent diagonal matrix (not necessarily in Smith normal form) is known.

318


(3.20) Proposition. Suppose that R is a PID and A E Mm,n(R) is a matrix of rank r, which is equivalent to the matrix 0

1Dr 0

0

where Dr = diag(t1, ... , tr). Then the prime power factors of the t; (1 < i < r) are the elementary divisors of A. Proof. Let p be any prime that divides some t; and arrange the tj according to ascending powers of p, i.e., ti3=Pe'91

tir = Pe,gr

where (p, q;) = 1 for 1 < i < r and 0 < e1 < e2 <

- - -

< er. Then the

pej+-.-+e,, exact power of p that divides the determinantal divisor dk(A) is for 1 <_ k <_ r and hence the exact power of p that divides the kth invariant factor sk(A) = dk(A)/dk_I(A) is pek for 1 < k < r. (Recall that do(A) is defined to be 1.) Thus pek is an elementary divisor for 1 < k < r. Applying this process to all the primes that divide some t; completes the proof. 0

(3.21) Remark. Proposition 3.20 is a matrix theoretic version of Proposition 3.7.19. The proof presented above is simpler than the proof of Proposition 3.7.19, because we now have the determinantal divisor description of the invariant factors. The following result is a consequence of Proposition 3.20 in exactly the same way that Corollary 3.7.20 is a consequence of Proposition 3.7.19.

(3.22) Corollary. Let B E Mm,n(R), C E Mp,g(R), and let

A=BEC=

B

0

0

C

Then the elementary divisors of A are the union of the elementary divisors of B and C.

Proof. If U1 BV1 and U2CV2 are in Smith normal form, then setting U = U1ED U2 and V=V1®V2 we see that

UAV-

Dr

0

0

0

0

0 0 0

0 E. 0

0

0

0

0

0

where Dr = diag(di, ... , d,.) and E, = diag(t1, ... , t,). Therefore, A is equivalent to the block matrix


Dr

0

0

0

E, 0

0

0

319

,

0

0

and Proposition 3.20 applies to complete the proof.

(3.23) Example. Let F be a field, and let

A = diag(X2(X - 1)2, X(X - 1)3, X - 1, X) E M4(F[X]). Then by Proposition 3.20, the elementary divisors of A are

X2, (X - 1)2,

X,

(X - 1)3,

(X - 1),

X

so the invariant factors are given by

s4(A) = X2(X - 1)3 s3(A) = X(X - 1)2 s2(A) = X(X - 1) sl(A) = 1. Therefore, A is equivalent to the matrix

diag(1, X(X - 1), X(X -1)2, X2(X -1)3).

5.4 Computational Examples This section will be devoted to some computational examples related to the Smith normal form. Specific computations to be considered include the reduction of matrices to rational and Jordan canonical form, generators and relations for finitely generated abelian groups, and linear diophantine equations. We will start with some examples of reduction of matrices to canonical form. These calculations are supplemental to those of Section 4.5, and it is recommended that the reader review the discussion there. In particular, note that the use of elementary row and column operations to produce the Smith normal form is also a particularly efficient technique if one is only interested in producing the characteristic polynomial of a linear transformation, which was the starting point for the calculations in Section 4.5.

(4.1) Example. Let V be a four-dimensional vector space over Q with basis

13 = {vl, V2, v3, v4} and let T : V -' V be a linear transformation with matrix [T)g given by

320


2

-4 -3

0

-

1

1

-1

2

[T]e=A=

0

3 2

1

2

-1

0

1

E M4 (Q) .

Compute the rational canonical form of T, and if it exists, compute the Jordan canonical form.

Solution. To compute the rational canonical form of T we compute the Smith normal form of the matrix XI4 - A E M4(Q[X]). Since Q[X] is a Euclidean domain, we may compute the Smith normal form by means of a finite sequence of elementary row and column operations on XI4 - A. We will use the symbol -- to indicate the age from one matrix to another by means of finitely many row or column operations. We will write the row operations to the left of the matrix, and the column operations to the right to keep a record of the row and column operations performed. They will be recorded by means of the elementary matrices, which were left (or right) multiplied to obtain the given matrix. Thus, the symbol Tij (a)

'--' An

D2(a')

D1(3) P23 (7')

indicates that A7, is obtained from A,,-, by multiplying on the right by A by multiplying on the left Tij(a)D2(a'), while by D1((3)P23(y). (See Propositions 4.1.12 and 4.1.13.) With these preliminaries out of the way, our calculations are as follows:

X-2 XI4-A=

X+3

0

1

-1

-1) DiP14

-1

-2

X+3

0

1

X-2

4

-1

T21(2)

T41(-(X - 2))

0X+1

`0

[10

1

-3 -2

1

X

X-1 -2

1

1

0

-1

4

-2

-1 0

-X -2

X-1 -2 -1 -3 -1 -X -2

X-1

-2X - 2 -2

X+2 X-3 X2-2X-3

1

0

0

1

0X+1

0

-2

X-1

0

-2X - 2 -2

0 X+2 X-3 X2-2X-3 P23

r-+

1

0

0

1

0 X+1

0

X-1 -2

0

-2

-2X-2

0 X+2 X-3 X2-2X-3

T12( Ti3 1 T14(

)


T42(-(X +2)) T32 - X + 1

0

0

1

0

0 0 0

1

X-1 -2 0 0 -X2 - 1 0 -X2-1 X2+1

1

0

0 0 0

1

0 X2+1 0 X2+1 X2+1

1

0

0

1

0 0

0 0

0X

0 0 0

0 0 2

T23(-(X - 1)) T24(2)

D3(-1)

0 0 0

+1

T43(-1).

X2+1

0

0

321

Therefore, XI4 - A is equivalent to diag(1, 1, X2 + 1, X2 + 1) so that the nonunit invariant factors of A are X2 + 1 and X2 + 1. The minimal polynomial of T is s4(X) = X2 + 1 while cT(X) = (X2 + 1)2. Since the companion matrix of X2 + 1 is r

1

C(X2 + 1) = I 0

_11,

it follows that the rational canonical form of T is (4.1)

R=C(X 2 +1)ED C(X 2 +1) =

0

-1

0

1

0

0

0

0

0 0

0

-1

1

0

0

0

Since MT (X) does not split into linear factors over the field Q, it follows that T does not possess a Jordan canonical form. Our next goal is to produce a basis B' of V such that [71s, = R. Our calculation will be based on Example 1.8, particularly Equation (1.15). That is, supposing that

P(X)(XI4 - A)Q(X) = diag(1, 1, X2 + 1, X2 + 1) then

VT °-` Q(XJwI

Q[x]w2

where w3 = F 1Pij+zv; if P(X)-' _ [p,,]. But from our caclulations above (and Lemma 4.1.11), we conclude that

P(X)-1 = P14Di(-1)T21(-2)T41(X - 2)P23T32(X + 1)T42(X + 2). Therefore,

X-2 X+2 0 (4.2)

P(X)-1 .

-2

1

X+1

1

0

1

0

0 0

-1

0

0

0

Thus, w1 = v2 and W2 = vi each have annihilator (X2 + 1), and hence,

322

Chapter 5. Matrices over PI1)s

8' = {w1, T(w1) = -4v1 - 3v2 - v3 - v4, w2, T(w2) = 2v1 + 2v2 + v4} is a basis of V such that [T]n- = R. Moreover, S-IAS = R where

S

0

-4

1

2

0 0

-1 -1

0 0

0

-P8.

1

O

(4.2) Remark. Continuing with the above example, if the field in Example

4.1 is Q[i], rather than Q, then mT(X) = X2 + 1 = (X + i)(X - i), so T is diagonalizable. A basis of each eigenspace can be read off from the caclulations done above. In fact, it follows immediately from Lemma 3.7.17 that Ann((X - i)wj) = (X + i) for j = 1, 2. Thus, the eigenspace corresponding to the eigenvalue -i, that is, Ker(T + i), has a basis

{(X - i)w1i (X - i)w2} and similarly for the eigenvalue i. Therefore, diag(i, i, -i, -i) = S1'AS1, where

2+i

-4

Si =

-3+i

2

-1 -1

-4

-3-i -1 -1

0 1

2-i 2 0 1

(4.3) Example. Let V be a vector space over Q of dimension 4, let B = {v1i v2, v3, v4} be a basis of V, and let T : V - V be a linear transformation such that 0

_ [T]g

1

2 1

2

1

-1

-1 4

-1

1

-1 -2 -2 -1 2

Compute the Jordan canonical form of T. Solution. As in Example 4.1, the procedure is to compute the Smith normal form of X I4 - A by means of elementary row and column operations. We will leave it as an exercise for the reader to perform the actual calculations, and we will be content to record what is needed for the remainder of our computations. Thus, (4.3)

P(X)(X14 - A)Q(X) = diag(1, 1, X - 1, (X - 1)(X + 1)2)

where

P(X) =T43(X - I)D3(1/4)T34(1)T42(-(X - 1))T32(X - 1)D2(1/4) P23T41(1)T31(2)T21(-X)D1(-1)P12


323

and

Q(X) = T12(X + 1)T13(1)T14(-1)P24D3(4)D4(2)

T23(-(X - 1))T24(X + 3)D4(2)T34(-(X + 1)).

Therefore, we can immediately read off the rational and Jordan canonical forms of T:

R=

1

0

0

0

1

0

0

0

0

1

0

1

0 0

0

1

0

1

0

0

-1

1

0

0

1

-1

0

0

0

-1

and J =

0 0

It remains to compute the change of basis matrices which transform A into R and J, respectively. As in Example 4.1, the computation of these matrices is based upon Equation (1.15) and Lemma 3.7.17. We start by computing P(X)-1:

P(X)-1 = P12D1(-1)T21(X)T31(-2)T41(-1)P23D2(4) T32(-(X - 1))T42(X - 1)T34(-1)D3(4)T43(-(X - 1))14

X -(X - 1) _

X+3

-1

-1 -2

0

0

0

4

0

0

-1

X-1

-(X - 1)

1

Therefore, we see from Equation (1.15) that the vector

v = (X + 3)vl - (X - ON is a cyclic vector with annihilator (X -1), i.e., v is an eigenvector of T with eigenvalue 1. We calculate

v=(T+3)(vl)-(T-1)(v4) = (3v1 + V2 + 2v3 + v4) - (-vl + V2 - 2v3 + v4) = 4(vl + v3). Let w1 = (1/4)v = vl + v4.

Also, the vector w2 = -v1 + V4 is a cyclic vector with annihilator ((X-1)(X+1)2) (again by Equation (1.15)). If W3 =T(w2) = -vi -4V3 +V4 and w4 = T2(w2) = -5v1 + 4v2 + 5v4i then {w2, w3, w4} is a basis for Q [X I w2 i and hence B' = {w1, w2, w3, W41 is a basis of V in which T is in rational canonical form, i.e., [T] 13- = R=

1

0

0

0

0

0

0

1

0

1

0

1

0

0

1

-1

324


Moreover, S-1AS = R where 11

-1 -1 -5 0 0

0

4

1

-4

0

0

1

1

5

0

S_

_ e. = P8

Now for the Jordan canonical form. From Lemma 3.7.17, we see that Ann((X - 1)w2) = ((X + 1)2)

(4.4)

and

Ann((X + 1)2w2) = ((X - 1)).

(4.5)

Equation (4.5) implies that w2

= (X + 1)2w2 = -8v1 + 4v2 - 8v3 + 8v4

is an eigenvector of T with eigenvalue 1. Let w4 = (X - 1)w2 = -4v3 and let w3 = (X + 1)w4 = -4v1 + 4v2 + 4v4 (see Proposition 4.4.37). Then

B"={w1,w;, W3,' W4)

is a basis of V in which T is in Jordan canonical form. In particular, U-'AU = J = [TJB where

U = PB 8 =

1

-8

4

0

4

-4

0 0

1

-8

0

-4

0

8

-4

0

(4.4) Remark. The theory developed in this chapter, as illustrated by the above examples, allows one to compute the rational canonical form of any matrix in (and hence any linear transformation T : V --+ V) using only the operations of the field F. Nothing more involved than the division algorithm for polynomials is needed to be able to reduce the characteristic

matrix XI,, - A E M (F[X J) to Smith normal form. Once one has the rational canonical form (and the transforming matrix), there are two steps to computing the Jordan canonical form. First, one must be able to factor each invariant factor into irreducible factors (which must be linear if A has a Jordan normal form). This step is the difficult one; factorization of polynomials is known to be difficult. To get an appreciation for this, see any book on Galois theory. Assuming the first step has been completed, the second step in computing the Jordan canonical form is the application of Lemma 3.7.17 and Proposition 4.4.37, as in the above example.

The main applications of the Smith normal form and the description of equivalence of matrices via invariant factors and elementary divisors are


325

to the similarity theory of matrices over fields, as illustrated by the above examples. There are, however, some other applications of the Smith normal form. Two of these applications are to the computation of the structure of a finitely generated abelian group given by generators and relations and the use of the Smith normal form in the problem of solving systems of linear equations over PIDs. We will consider examples of both of these problems, starting with abelian groups defined by generators and relations. (4.5) Example. One explicit way to describe an abelian group is by giving generators and relations. This is expressed by saying that M = (xl, ... , x.) where the generators x; are subject to the relations

allxl + ... + aInxn = O

amlxl+ - - +annxn=0 -

where A = [a.;,,] E Mm,n(Z). We can express this more formally by saying

that M = Zn/K where K is the subgroup of Zn generated by y, = a11e1 + ... + alnen

ym = amlel + ... + amnen,

Here, {el, ... ,en} is the standard basis of Zn. If 7r : Zn -' M is the natural projection map, then w(e;) = x;. We can compute the structure of an abelian group given by generators x1, ... , xn subject to relations (4.6) by using the invariant factor theorem for submodules of a free module (Theorem 3.6.23). That is, we find a basis {v1, ... ,vn} of Zn and natural

numbers a1 i ... , s, such that {s1v1, ... , srVr} is a basis of the relation subgroup K C Zn. Then

M = Zn/K

Z,, ®... ®Z,,, ®Zn-r.

Note that some of the factors Z,, may be 0. This will occur precisely when

s; = 1 and it is a reflection of the fact that it may be possible to use fewer generators than were originally presented. The elements sl, ... , Sr are precisely the invariant factors of the integral matrix A E Mm,n(Z). To see this, note that there is an exact sequence

Z'"- Zn --"-+ M-O where 0(c) = Atc. This follows from Equation (4.7). Then the conclusion follows from the analysis of Example 1.8, and Example 1.9 provides a numerical example of computing the new generators of M.

326


(4.6) Example. By a system of m linear equations over Z in n unknowns we mean a system

a11x1 + a12x2 + ... + alnxn = bl + a2nxn = b2 a21x1 + a22x2 + amlxl + am2x2 +

. + amnxn = bm

where ail E Z and b; E Z for all i, j. System (4.8) is also called a linear diophantine system. We let A = [a;3] E Mm,n(Z), X = [xl . . . xn ]', and B = [b1 bm ]t. Then the system of Equations (4.8) can be written in matrix form as

AX = B.

(4.9)

Now transform A to Smith normal form UAV

(4.10)

A.

0

0

0

where U E GL(m, Z), V E GL(n, Z), and DT = diag(s1.... , s,.) with si 0 0 and si I s;+1 for 1 < i < r - 1. Thus, Equation (4.9) becomes cl (4.11)

UAV(V-'X) =UB=C= Cm

Setting Y = V -1 X = [y,

.

. .

y,n ] 1 gives the equivalent system of equa-

tions

slyl = Cl s2y2 = C2

(4.12)

sryr = Cr

0 = Cr+1

0=cn,. The solution of the system (4.12) can be easily read off; there is a solution

if and only if s;Icifor1

327

We will illustrate the method just described with a numerical example.

(4.7) Example. Consider the linear diophantine system AX = B where 2

A=

-3 -1

1

-1 -3 -4 0

1

4

8

and B=

1

1

.

16

16

We leave it as an exercise for the reader to (via elementary row and column operations) that if 0

U=

1

0

1

0

-2 0 -4

and

V=

1

-2

1

1

2

0

1

-1

2

0

0

1

-1

0

0

0

1

then

UAV =

1

0

0 0

3 0

0 0

0 0 0

12

= S.

Let 1

C=UB=

6 12

Then the system AX = B is transformed into the system SY = C, i.e.,

yl = 1 3112 = 6

12y3 = 12.

This system has the solutions

where t = y4 is an arbitrary integer. We conclude that the solutions of the original equation AX = B are given by X = VY, i.e.,

X =VY=

5-2t

1

1

2

-2

1

0

1

-1

2

2

0

0 0

1

-1

1

1-t

0

1

t

t

0

-

1+2t

where t is an arbitrary integer.

(4.8) Remark. The method just described will work equally well to solve systems of linear equations with coefficients from any PID.

328


5.5 A Rank Criterion for Similarity Given two matrices A and B E M,,(F) (F a field) one way to determine if A and B are similar is to determine the invariant factors of both A and B, e.g., by reducing both X In - A and X In - B to Smith normal form. If the invariant factors are the same, then the matrices are similar. This approach, however, is not particularly amenable to providing an explicit description of the set similar to A}

OA

by means of polynomial equations and inequations in the entries of the matrix B. Note that OA is just the orbit of A under the group action (P, A)

PAP-'

of GL(n, F) on Mn(F). Another approach is via Weyr's theorem (Chapter 4, Exercise 66), which states that if F is algebraically closed, then A is similar to B if and only if rank((A - \In)k) = rank((B - \In)k)

(5.1)

for all A E F and k E N. This can be reduced to a finite number of rank conditions if the eigenvalues of A are known. But knowledge of the eigenvalues involves solving polynomial equations, which is intrinsically difficult.

(5.1) Example. As a simple example of the type of equations and inequations that can be derived from Equation (5.1) to describe an orbit under similarity, one can show that A = o I is similar to B = [ - d] if and only i ing sets of matrices: if B is in one of the rfollow]:bo}

S1={[' S2={[1 ?]:ci40} S3=

[0

d]

:bc#O,a+d=2,ad-be=1}.

We leave the verification of this description of the orbit of Al as an exercise.

In this section we will present a very simple criterion for the similarity of two matrices A and B (linear transformations), which depends only on the computation of three matrices formed from A and B. This has the effect of providing explicit (albeit complicated) equations and inequations for the orbit OA of A under similarity. Unlike the invariant factor and elementary divisor theory for linear transformations, which was developed in

5.5 A Rank Criterion for Similarity

329

the nineteenth century, the result we present now is of quite recent vintage. The original condition (somewhat more complicated than the one we present) was proved by C. I. Byrnes and M. A. Gauger in a paper published in 1977 (Decidability criteria for the similarity problem, with applications to the moduli of linear dynamical systems, Advances in Mathematics, Vol. 25, pp 59-90). The approach we will follow is due to J. D. Dixon (An is(>morphism criterion for modules over a principal ideal domain, Linear and Multilinear Algebra, Vol. 8, pp. 69-72 (1979)) and is based on a numerical

criterion for two finitely generated torsion modules over a PID R to be isomorphic. This result is then applied to the F[X]-modules VT and VS, where S, T E EndF(V), to get the similarity criterion. (5.2) Lemma. Let R be a PID and let a and b be nonzero elements of R. If d = (a, b) = gcd{a, b}, then

HomR(R/(a), R/(b)) L, R/(d). Proof. This is essentially the same calculation as Example 3.3.11. We leave it to the reader.

(5.3) Lemma. Let R be a PID, and let ...ED R/(sn) and

N ?` R/(ti) ®... ® R/(tm) be two finitely generated torsion R-modules. Then n

m

HomR(M, N) . ®® R/(si, tj). i=1 j=1

Proof. This follows immediately from Lemma 5.2 and Proposition 3.3.15.

(5.4) Definition. If R is a PID and M is a finitely generated torsion Rmodule, then let

t(M)

= > ki i=1

where {pi' , ... ,p} is the set of elementary divisors of M.

In the language of Section 7.1, t(M) is the length of the R-module M.

330


(5.5) Definition. Let M and N be finitely generated torsion R-modules (R a PID). Then let (M : N) = P(HomR(M, N)).

The notation (M : N) is suggestive of an inner product, and it is precisely for this reason that the notation was chosen. The following result gives some facts concerning (M : N), which are reminiscent of basic properties of inner products, and the main theorem (Theorem 5.7) is analogous to the Cauchy-Schwartz inequality. (5.6) Proposition. Let R be a PID and let M, N, and P be finitely generated torsion R-modules. Then

(1) (M®N: P)=(M:P)+(N:P); (2) (M: M) > 0 with equality if and only if M = {0};

(3) (M:N)=(N:M).

Proof. All three parts are immediate consequences of Lemma 5.3.

(5.7) Theorem. Let M and N be finitely generated torsion modules over a PID R. Then

(M : N)2 < (M: M)(N : N). Equality holds if and only if M9 = N' for some relatively prime integers 8 and t. Proof. We may write (5.2)

111 = RI(s I)

(D RI (s,)

and (5.3)

N

R/(tl) (D ... ©? R/(tm).

Let pi, ..., p,. be the distinct primes of R that divide some elementary divisor of either M or N; by Proposition 3.7.19, these are the distinct prime divisors of the s, and tj. Let

r = lcm{Ann(M), Ann(N)} = pi'

prr

+ k,.. Let Rk be identified with the vector space of 1 x k and let k = k1 + matrices over R, and consider each A E Rk as a block matrix

A= IAl

... A,)

where A, E Rk' . Given any divisor a of c, write

a=pi' ... pr', and define v(a) E Rk by

7l(a) = [v(al)

... v(ar)t

t


331

where

v(ai) _ [1

1

0

1

0J

with ei ones. Then define v(M)

v(si)

i=1 and

m

v(N) = 1: v(tj). j=1

Notice that the matrix v(M) determines M up to isomorphism since one can recover the elementary divisors of M from v(M). To see this, choose the largest ti > 0 such that v(M) - v(p2') has nonnegative entries. Then pt` is an elementary divisor of M. Subtract v(pi') from v(M) and repeat the process until the zero vector is obtained. (See the proof of Proposition 3.7.19.)

Let si = pi" ... pe"' t r, j= for 1 <1
pIf''

(Si,

.

"prf''' and define dijl = minffe l al, fji} +

d,3,

tj) = (p1 ... pr'" ). If (:) denotes the standard inner product on Rk, then d

r (5.4)

(v(si) : v(tj)) =

dijl 1=1

= £(R/(si, tj)). Therefore, n

(v(M) : v(N)) =

m

E(v(si) : v(tj)) i=1 j=1 n

m

_

e(R/(si, tj)) i=1 j=1

=(M:N). Similarly, (M : M) = (v(M) : v(M)) and (N : N) = (v(N) : v(N)). By the Cauchy-Schwartz inequality in Rk we conclude that (M : N)2 = (v(M) : v(N) )2 < (v(M) : v(M))(v(N) : v(N)) _ (M : M) (N : N),

as required. Moreover, equality holds if and only if v(M) and v(N) are linearly dependent over R, and since the vectors have integral coordinates, it follows that we must have v(M) and v(N) linearly dependent over Q, i.e.,

sv(M) = tv(N)

332


where s and t are relatively prime natural numbers. But

v(M') = sv(M) = tv(N) = v(N') so that M' = N' since we observed above that v(W) determines the elementary divisors, and hence, the isomorphism class of a finitely generated torsion R-module W. O (5.8) Corollary. Let M and N be finitely generated torsion R-modules. Then

M N if and only if (1) e(M) = I(N), and (2) (M: N)2 = (M : M)(N : N). Proof. ML, N certainly implies (1) and (2). Conversely, suppose that (1) and (2) are satisfied. Then by Theorem 5.7, M' L" N' for relatively prime integers a and t. But

se(M) = e(M')

=e(N') = te(N) = te(M) so s = t. Since s and t are relatively prime, it follows that s = t = 1, and

0

hence, M 2.1 N.

(5.9) Remark. If M' N' for relatively prime integers s and t, then it is an easy consequence of the uniqueness of the primary cyclic decomposition of a finitely generated torsion R-module (Theorem 3.7.15) that M ?t P' and N P' where P is a finitely generated torsion R-module.

We now wish to apply Corollary 5.8 to derive a simply stated rank criterion for the similarity of two linear transformations. We will start by computing matrix representations for some basic linear transformations.

Let V be a finite-dimensional vector space over a field F and let B = {v1, ... be a basis of V. Then a basis of EndF(V) is given by C = where fii(vk) = hikvi

Under the F-algebra isomorphism 4) : EndF(V) -. M,,(F) given by 4)(f) _ [f ]8, we have 4)(f i) is the matrix unit Ei,. Given a linear transformation T E EndF(V), define two linear transformations GT E EndF(EndF(V)) and RT E EndF(EndF(V)) by (5.5)

GT(U) = TU

and (5.6)

RT(U) = UT.

That is, CT is left multiplication by T and 7ZT is right multiplication by T in the F-algebra EndF(V). Let us now order the basis C of EndF(V) as follows:


C= {fl 1, 121,

333

, fn 1, f12, ... , f22, ... , fn2, .. , fnn}

With these notations there is the following result. (Recall that the tensor product (or kronecker product) of matrices A and B was defined in Definition 4.1.16 as a block matrix [Cij] where Cij = aijB.) (5.10) Lemma. Let T E EndF(V) with [T]f; = A = [aij]. Then [LT]C = A ® In

(5.7)

and

[RT]C = In ® At.

(5.8)

Proof. Note that

Tfij(vk) =T(bikvj) = bikT(vj) n

atjvi

= bik l=1

n ai j bik vt

t=1 n

Eaijfi!

(Vk).

I=1

This equation immediately gives [CT]c = A ® In. 1 ail f1j, so Equation (5.8) is also A similar calculation gives fijT = satisfied.

(5.11) Corollary. Let S, T E EndF(V), and define TS,T E EndF(EndF(V)) by

TS,T(U) = SU - UT. If [SIB = A and [T]8 = B, then (5.9)

(TS.T]C = A ® In - In ® Bt.

Proof. Since TS,T = LS -1T, the result is immediate from Lemma 5.10.

If S, T E EndF(V), we will let (as usual) VS and VT denote the F[X]-module structures on V determined by S and T respectively. Then by Proposition 4.4.1, we know that


334

HomF[x[(VT, Vs) = {U E EndF(V) : UT = SU}. Thus, we have an identification (5.10)

Homp[x](VT, Vs) = Ker(Ts,T),

and hence:

(5.12) Lemma. dimF HomF[xl(VT, Vs) = n2 - rank(A ® In - I (9 Bt). Proof. By Proposition 3.8.8, we have

dimF (Ker(Ts, T)) + dimF (Im(Ts,T)) = dimF EndF(V) = n2. The result then follows from Equation (5.10).

0

We can now give the proof of Dixon's theorem.

(5.13) Theorem. (Dixon) Let V be a vector space of finite dimension n over an algebraically closed field F, and let S, T E EndF(V) be linear transformations. Let B be a basis of V and let A = [S]B, B = [T]B. Then S and T are similar if and only if (rank(A(9

Bt))2 =

Proof. To simplify the notation, we will let

rAA = rank(A ®I - In (9 A'), with a similar definition for rAB and rBB. Since VT

Vs as F[X]-modules

if and only if S and T are similar (Proposition 4.4.2), it follows that if S and T are similar then HomF[x[(VT, Vs)

HomF[x)(VT, VT)

HomF[x)(VS, Vs)

as F[X]-modules. Hence, they have the same rank as F-modules, and thus rAA = rAB = rBB follows immediately from Lemma 5.12. Conversely, assume that 2

(5.11)

rAB = rAArBB

Since F is assumed to be algebraically closed, the elementary divisors of any finitely generated torsion F[X]-module W are of the form (X - Xt)k-. Since

W .`

t

F[X]/((X - a:)k'), it follows that


335

e

£(W) _

(5.12)

k; = dimF W.

In particular, Lemma 5.12 shows that

(VT:VS)=n2-TAB (VT:VT)=n2-rBB and

(Vs:VV)=n2-rAA. Equation (5.11) then gives (VT : Vs)2 - (VT : VT)(VS

VS) = n2(rAA + rBB - 2rAB)

+ (rAB - rAArBB)

= n2( rAA - rBB 2 >0. By Theorem 5.7 (VT : Vs)2 < (VT : VT)(VS

VS)

Thus,

(VT:Vs)2=(VT:VT)(VS:V5) Since £(VT) = £(V5) = n, Corollary 5.8 then shows that VT modules. Hence T and S are similar.

Vs as F[X]0

(5.14) Remark. The restriction that F be algebraically closed in Theorem 5.13 is not necessary. Indeed, let K be an algebraically closed field containing F (see Remark 2.4.18 (3)) and consider A = [SIB and B = [T]B E Then A is similar to B in M,,(F) if and only if A is similar to B in M,,(K) (Theorem 3.15) and the rank condition in Theorem 5.13 does not depend upon which field we are using.

The computation of HomR(M, N) where M and N are finitely generated torsion R-modules over a PID R (Lemma 5.3) is also useful for some applications other than Theorem 5.13. We will give one such example. Suppose that V is a finite-dimensional vector space over a field F and T E EndF (V ). The centralizer of T in the ring EndF (V) is

C(T) = {U E EndF(V) : TU = UT}. Note that, according to Proposition 4.4.1, C(T) = EndF(x)(VT).

The F-algebra generated by T, namely, F[T], is certainly contained in the centralizer C(T). There is a theorem of Frobenius, which computes the dimension of C(T) over F. This result is an easy corollary of Lemma 5.3.

336


(5.15) Theorem. (Frobenius) Let F be a field, V a finite-dimensional vector

space over F, andT E EndF(V). Iff1(X), , fk(X) (where fi(X) divides f,+1(X) for 1 < i < k - 1) are the invariant factors of T, then k

dimFC(T) = E(2k - 2i + 1) deg(fi(X )).

(5.13)

,=1

Proof. By Lemma 5.3, 25

k

k

®®F[X]/(f,(X), fj(X)).

C(T) = EndF(XI(VT)

i=1 j=1

But (fi(X ), fj (X )) = (fmin{,, j} (X )), so k

dimF C(T) = E deg fmin{,,j}(X ) +.j=1

But

{(i, j) : 1 < i, j < k and

min{i, j} = t}] = 2k - 2t + 1,

so k

dimF C(T) = E(2k - 2t + 1) deg ft(X) t=1

as required.

We have observed above that F[T]C C(T). As a corollary of Frobenius's theorem, there is a simple criterion for when they are equal, i.e., a criterion for when every linear transformation that commutes with T is a polynomial in T.

(5.16) Corollary. Let T E EndF(V). Then F[T] = C(T) if and only if mT(X) = cT(X), i.e., if and only if VT is a cyclic F[X]-module.

Proof. First note that dimFF[T] = degmT(X) and if {fi(X)} 1 are the invariant factors of T, then mT(X) = fk(X). By Equation (5.13) k

dimF C(T) =

(2k - 2i + 1) deg fi(X)

,cl k-1

= dimF F[T] + E(2k - 2i + 1) deg fi(X ). ,=1

From this we see that C(T) = F[T] if and only if k = 1, i.e., if and only if VT is a cyclic F[X]-module.

5.6 Exercises

337

(5.17) Corollary. If T E EndF(V) then dimF C(T) > n = dimF(V).

Proof. Since f 1 f;(X) = cT(X), it follows that k

k

dimF C(T) = 1:(2k - 2i + 1) deg f; (X) >

deg f; (X) = n. 11

(5.18) Example. The Jordan matrix Ja,

n is cyclic, so

C(JA, n) = F[JA, n].

It is easily checked that a basis of F(JA, n] consists of the n matrices A; _ Jo, n for 0 < i < n - 1. That is, a matrix A = (a.;j] commutes with J,\, n if and only if it is upper triangular and constant on the lines parallel to the main diagonal.

5.6 Exercises 1.

Compute a finite free presentation of the Z-module

M=Z2®Z3o®Z. 2.

3.

Compute two distinct finite free presentations of the R(X -module (P3)D, where P3 denotes the real vector space of polynomials of degree at most 3 and D E EndR(P3) is the differentiation map. Let M be an abelian group with three generators v1, v2i and v3, subject to the relations

2v1 - 4v2 - 2v3 = 0 10v1 - 6v2 + 4v3 = 0 6v1 - 12v2 - 6v3 = 0. Assuming the matrix identity 10 11 l

1 21

4. 5.

°1

[2

6, [0 0 1 ] _ [0 0 0]

show that M ?5 Z2 ® Z14 ® Z, and find new generators Wi, w2, and W3 such that 2w1 = 0, 14w2 = 0, and W3 has infinite order. Use Theorem 3.6.16 to give an alternative proof of Theorem 2.2.

Construct a matrix A E GL(4, Z) with

rowl(A)=(12 -10 6.

Construct a matrix A E GL(3, Q(X)) with

9

8).


338

coll(A) = [X(X - 1) X2 7.

X+1]'.

Construct a matrix A E M3(Z[i]) with

row2(A) _ [1-2i 1+3i 3-i] 8.

and with det A = 2 + i. Reduce each of the following matrices to Hermite normal form: 2 6 91 (a) A -2 0 4 1 E M3(Z). 2

1

1

(b) B = [a 0X 1 + X 22X

E M3(Q[X]).

L

(c) C= 9.

3+i] EM2(Z[i]).

17-i

Write the unimodular matrix

[3

A= 3 6

1

31 3 EJY13(Z)

3

5

3

as a product of elementary matrices. 10.

Let R = Z[/]. Show that no matrix [ a d ] with a = 2 and c = 1-

is left equivalent to a matrix in Hermite normal form. Same as Exercise 10 with R = Q[X2, X3] and a = X2, C = X3. Let R be a PID. Show that there is a one-to-one correspondence between the left ideals of M (R) and the R-submodules of Ml,,,(R). 13. Find the Smith normal form for each of the following matrices: 11. 12.

2

(a)

0 2

6

142

14

8

(d) 14.

0 0

E M3(Z).

6 8

X(X - 1)3 (c)

E M3(Z).

2

1

(b)

10

0

-1 -3 -4

0

0l

0

XJ

(X - 1)

1+8i -23 + 2i

-5+i

13i

E

0

E M 2(Z[i]).

Find the invariant factors and elementary divisors of each of the following matrices: (a)

0 0

2+4X

0

-4+X 5

7-6X 2X

E 1113(ZS[X]).

0

(b) diag(20, 18, 75, 42) E 1i14(Z). 15.

(c) diag (X(X - 1)2, X(X - 1)3, (X - 1), X). Let R be a PID and let A E Mm,,,(R) with m < n. Extend Theorem 2.2 by proving that there is a matrix B = [A, ] E (so that Al E

5.6 Exercises

339

(R)) such that det B is an associate of d..(A), the m`h determinantal M,, divisor of A. (Hint: First put A in Smith normal form.) 16.

Let S = {vi, ... , vk } C M",1(R where R is a PID. Show that S can be extended to a basis of M",1(R) if and only if dk(A) = 1, where A= [vi ... Vk ].

17.

Suppose that A E M3(Z) and det A = 210. Compute the Smith normal form of A. More generally, suppose that R is a PID and A E M"(R) is a matrix such that det A is square-free. Then compute the Smith normal form of A. Let A E M" (Z) and assume that det A y6 0. Then the inverse of A exists in M"(Q), and by multiplying by a common denominator t of all the nonzero

18.

entries of A-, we find that to-1 E M"(Z). Show that the least positive integer t such that tA'1 E M"(Z) is t = Is"(A)l where s"(A) is the n1h invariant factor of A. 19.

Let A, B E M"(Z) such that AB = kI" for some k 54 0. Show that the

invariant factors of A are divisors of k. Let R be a PID and let A E M"(R), B E M,"(R). Show that the elementary divisors of A®B are the product of elementary divisors of A and of B. More precisely, if p' is an elementary divisor of A® B where p E R is a prime, then pr = pkpl where pk is an elementary divisor of A and p' is an elementary divisor of B, and conversely, if pk is an elementary divisor of A and p' is an elementary divisor of B, then pk+i is an elementary divisor of A 0 B. 21. Let A E M4(F) where F is a field. If A has an invariant factor s(X) of degree 20.

2 show that the Smith normal form of XI4 - A is diag(1, 1, s( X), s(X)). Conclude that CA(X) is a perfect square in F[X). 22. Find all integral solutions to the following systems AX = B of equations:

23.

0

2

-1

1

1

2

0

0 -1

(c) A= [68

19

(b) A=

14

[51

B

,

1

7

B= [7]'

22]'

Show that the matrices A = [ 8

i ] and B = [ 232 -15 ] in M2 (Q) are

similar.

24. Show that the matrices

25.

(0)

10

3

4

021

and

0

2

4

5

0

1

0

01

1

[0

0

are similar in M3(Z7). Show that the matrices 1]

and

1

[1

26.

0

0,

1

are similar in M3(Z3). Find the characteristic polynomial, invariant factors, elementary divisors, rational canonical form, and Jordan canonical form (when possible) of each of the matrices from Exercise 73 of Chapter 4. Additionally, find bases of Q" with respect to which the matrix (or linear transformation) is in rational

or Jordan canonical form. Do this exercise by reducing XI" - A to Smith

Chapter 5. Matrices over PIN

340

canonical form and compare your results with the same calculations done in Chapter 4. 27.

Find an example of a unimodular matrix A E M3(Z) such that A is not similar to A'. (Compare with Example 3.8.)

28.

Show that the matrix A =

2X X I is not equivalent in M2(Z(XJ) to a

2

diagonal matrix. (Hint: Use `Fitting ideals.) 29.

Let Z" have the standard basis lei, ... ,e"} and let K C Z" be the submodule generated by f, _ j a,jej where a,j E Z and 1 < i < n. Let A = [a,j] E MM(Z) and let d = det A. Show that Z/K is torsion if and only

ifdetA=d00andifd#0showthatIZ/Kl=l 1.

Suppose that an abelian group G has generators xl, x2, and x3 subject to the relations xl - 3x3 = 0 and xl + 2x2 + 5x3 = 0. Determine the invariant factors of G and Cl iif G is finite. 31. Suppose that an abelian group G has generators xi, x2, and x3 subject to the relations 2x1 - x2 = 0, xl - 3x2 = 0, and x1 + x2 + x3 = 0. Determine the invariant factors of G and Cl lif G is finite. 30.

32. 33.

the claim of Example 5.1.

Let F be a field and let A E M"(F), B E Mm(F). Show that the matrix equation

AX - XB = 0 for X E M",,"(F) has only the trivial solution X = 0 if and only if the

characteristic polynomials cA(X) and CB(X) are relatively prime in FIX]. In particular, if F is algebraically closed, this equation has only the trivial solution if and only if A and B have no eigenvalues in common. 34. Let F be a field. Suppose that A = Al ® A2 E M"(F) where Al E Mk(F) and A2 E Mm(F) and assume that cA,(X and CA3(.X) are relatively prime.

Prove that if B E M"(F) commutes with A, then B is also a direct sum

35.

B = B1 ® B2 where B1 E Mk(F) and B2 E M,,,(F). Let F be a field. Recall that C(f(X)) denotes the companion matrix of the monic polynomial f (X) E FIX]. If deg(f (X)) = n and deg(g(X)) = n, show

that

rank(C(f (X)) ® In - I" ® C(9(X ))) = deg(lcm{ f (X), 9(X)}). 36.

Let V be a finite-dimensional vector space over a field F and let T E EndF(V). Prove that the center of C(T) is FIT].

Chapter 6

Bilinear and Quadratic Forms

6.1 Duality Recall that if R is a commutative ring, then HomR(M, N) denotes the set of all R-module homomorphisms from M to N. It has the structure of an Rmodule by means of the operations (f + g) (x) = f (x) + g(x) and (a!) (x) = a(f (x)) for all x E M, a E R. Moreover, if M = N then HomR(M, M) =

EndR(M) is a ring under the multiplication (fg)(x) = f (g(x)). An Rmodule A, which is also a ring, is called an R-algebra if it satisfies the extra axiom a(xy) = (ax)y = x(ay) for all x, y E A and a E R. Thus EndR(M) is an R-algebra. Recall (Theorem 3.4.11) that if M and N are finitely generated free R-modules (R a commutative ring) of rank m and n respectively, then HomR(M, N) is a free R-module of rank mn.

In this section R will always denote a commutative ring so that HomR(M, N) will always have the structure of an R-module. (1.1) Definition. If M is an R-module, then HomR(M, R) is called the dual module of M and is denoted M*.

(1.2) Remark. If M if free of rank n then M' = HomR(M, R) R^ °5 M by Corollary 3.4.10. Note, however, that this isomorphism is obtained by choosing a basis of M' and M. One particular choice of basis for M* is the following, which is that described in the proof of Theorem 3.4.11 if the basis 111 is chosen for R. (1.3) Definition. If M is a free R-module and B = {v1, ... , vn} is a basis

of M, then the dual basis of M' is defined by B' = {vi, ... , v;} where vi E M' is defined by

v,(vi)=bij=

1

0

ifi=j

ifi#j.

(1.4) Example. Let R = Z and M = V. Consider the basis B = {vl = (1, 0), v2 = (0,1)}. Then vi (a, b) = a and vz (a, b) = b. Now consider the

Chapter 6. Bilinear and Quadratic Forms

342

basis C = {wl = (1, 1), w2 = (1, 2)}. Then (a, b) = (2a - b)wl + (b - a)w2 so that wi (a, b) = 2a - b and w2 (a, b) = b - a. Therefore, vl 34 w' and v2 # w2. Moreover, if D = Jul = (1,0), u2 = (1,1)} then ui(a,b) = a - b and u2 (a, b) = b so that ui 54 vl even though ul = vl. The point is that an element v, in a dual basis depends on the entire basis and not just the single element vi.

(1.5) Proposition. Let M be a free R-module of finite rank n and let B =

{v1,...,v } be a basis of M. Then B'={vl,...,vn} is a basis of M'. Proof. B' is the basis produced in the proof of Theorem 3.4.11.

0

(1.6) Corollary. Let M be a free R-module of finite rank n and let B = {vl, ... vn} be a basis of M. Then the map w : M --i M' defined by n

n

waivi) _ i.1 aivi i=1 is an R-module isomorphism. Proof.

O

The isomorphism given in Corollary 1.6 depends upon the choice of a basis of M. However, if we consider the double dual of M, the situation is much more intrinsic, i.e., it does not depend upon a choice of basis. Let M be any R-module. Then define the double dual of M, denoted M",

by M" = (M')' = HomR(M', R). There is a natural homomorphism n: M M" = HomR(M', R) defined by 17(v)(w) = w(v)

for allvEMandWEM'=HomR(M,R). (1.7) Theorem. If M is a free R-module, then the map r, : M -y M" is injective. If rank(M) < oo then r) is an isomorphism.

Proof. Suppose that v # 0 E M. Let B be a basis of M. If v = alvl + + anvn where al # 0 and {v1i ... , vn} C B, then we can define an element w E M' by w(vi) = 1 and w(w) = 0 for all w 34 vl E B. Then + anvn) = al # 0. Hence, n is injective. Now suppose that rank(M) < oo and let B = {vl, ... ,vn} be a basis

7](v)(w) = w(alvl +

of M. Let B' = {vi, ... ,v;a} be the dual basis of M' and let B" = {vi', ... , vn } be the basis of M" dual to the basis B' of M'. We claim that i(vi) = v1' for 1 < i < n. To see this, note that rl(vi)(vv) = viNvi) = bi j

= v,' (vj*)

6.1 Duality

343

Since 77(v1) and v;' agree on a basis of M', they are equal. Hence, q(M) (vl', ... vn) = M" so that q is surjective, and hence, is an isomorphism. 0

(1.8) Remark. For general R-modules M, the map q : M - M" need not be either injective or surjective. (See Example 1.9 below.) When q happens to be an isomorphism, the R-module M is said to be reflexive. According to Theorem 1.7, free R-modules of finite rank are reflexive. We shall prove below that finitely generated projective modules are also reflexive, but first some examples of nonrefiexive modules are presented. (1.9) Examples.

(1) Let R be a PID that is not a field and let M be any finitely generated nonzero torsion module over R. Then according to Exercise 9 of Chap-

ter 3, M' = HomR(M, R) = (0). Thus, M" = (0) and the natural map q : M -' M" is clearly not injective. (2) Let R = Q, and let M = ®nENQ be a vector space over Q of countably infinite dimension. Then M' °-' fl EN Q. Since

M=®QC fQ, nEN

nEN

we see that M' 5 M ® M' where M' is a vector space complement of M = ®nENQ in M. Then

M" = M. ®(M')' so that M" contains a subspace isomorphic to M. But ®nENQ is countably infinite, while the infinite product fl EN Q " M' is un-

countable (the decimal representation identifies every real number with an element of 11nEN Q). Therefore, q : M - M" cannot be surjective by cardinality consideration and we conclude that M is not a reflexive Q-module. Let Ml and M2 be R-modules and let M = Ml ®M2. Then according to Corollary 3.3.13, M" °-' Ml** ED M2**. In order to study reflexivity for direct sums and summands, it is necessary to identify carefully this isomorphism. To this end, define W : (MI ®M2)" - Ml** ED M2

by 'Y(w) = (w;, w2), where w, E M;' is defined by w,(9;) = w(9, o ir;) for each 0, E M, = HomR(M;, R). 7r; : Ml ® M2 -+ M, is the canonical projection map. Similarly, define OD: M'

M2' -i (Ml

®M2).'


344 by

4)(wi, w2)(8) =w1(Oot1)+w2(Oot2)

where 0 E (M1 ® M2)' and i, : Mi -+ M1 ® M2 is the canonical injection. (1.10) Lemma. 1Y and 4) are inverse R-module homomorphisms.

Proof. Let w E (M1 ®M2)" and let 0 E (M1 (D M2)'. Then 0 o T(w)(0) = 4)(wi, w2)(0)

=w1(00ll)+w2(00t2) =w(9ot1 0711)+w(B01.207r2) =w(Ootl o7r1 +001.20712) =w(Oo(t1o7r1+1.20712)) =w(Oo1M1®M2)

= w(e), and

('I+ 0 4)(wl, w2))(01, 02) ='I'(Nwl, w2))(01, 02) = (4)(wl,w2)(01 0 711), 4)(wl,w2)(02 0 7r2)) = (w1(01 0711 0ti),w2(02071201.2)) _ (w1(0i),w2(02))

_ (wl,w2)(01,02).

Therefore, 4)o 41 = 1(MI(DM2).. and

W o 4) = 1Mi ®Mr,

0

and the lemma is proved.

Now let 71i:Mi-'M;' (i = 1, 2) andq :M1ED M2- (M1(D M2)" be the natural maps into the double duals. (1.11) Lemma. Using the notation introduced above, there is a commutative diagram M1 ®M2

InM2)..

(M1 (D

w

That is, W 077 = (711, 712)

M1"eMM'

6.1 Duality

345

Proof. ((Y

rl) (v1, v2)) (wl, W2) =' (77(v1, v2)) (w1, w2)

= (rl(vl, v2)(w1 o rrl), rl(vl, v2)(w2 o l2)) = ((w1 o 7r1)(vl, v2) , (w2 o rr2)(vl, v2)) = (w1(vI), w2(v2))

= ((rll, r72)(v1, V2)) (w1, L02)-

That is,Worl=(771,772). (1.12) Lemma. (1) 11 : Ml ®M2 -p (MI ® M2)** is injective if and only if rli

: M; -+ Mi** is injective for each i = 1, 2. (2) MI ® M2 is reflexive if and only if MI and M2 are reflexive.

Proof. Both results are immediate from Lemma 1.11 and the fact that IF is an isomorphism.

(1.13) Proposition. If P is a projective R-module, then 77 : P -> P** is injective. If P is also finitely generated, then P is reflexive. Proof. Since P is projective, there is an R-module P' such that P ® P' = F where F is a free R-module (Theorem 3.5.1), and furthermore, if P is finitely generated, then F may be taken to have finite rank (Corollary 3.5.5). The result now follows from Lemma 1.12 and Theorem 1.7.

The remainder of this section will be concerned with the relationship between submodules of an R-module M and submodules of the dual module

M*. The best results are obtained when the ring R is a PID, and the module M is a finite rank free R-module. Thus, we will make the following convention for the rest of the current section.

Convention. For the remainder of this section R will denote a PID and M will denote a free R-module of finite rank unless explicitly stated otherwise. (1.14) Definition. (1) If N is a submodule of M, then we define the hull of N, denoted

Hull(N)={x'EM:rx'EN for somer540ER}. If A is a subset of M, then we define Hull(A) = Hull((A)). (2) If A is a subset of M then define the annihilator of A to be the following

subset of the dual module M*:

K(A) = Ann(A)

={wEM*:Ker(w)DA}

={wEM*w(x)=0 for all xEAl C M*.

346

(3)


If B is a subset of M' then define the annihilator of B to be the following subset of M:

K'(B) = Ann(B)

={xEM:w(x)=O forallwEB} C M.

(1.15) Remarks. (1) If N is a submodule of M, then M/ Hull(N) is torsion-free, so Hull(N)

is always a complemented submodule (see Proposition 3.8.2); further-

more, Hull(N) = N if and only M/N is torsion-free, i.e., N itself is complemented. In particular, if R is a field then Hull(N) = N for all subspaces of the vector space M. (2) If A is a subset of M, then the annihilator of A in the current context

of duality, should not be confused with the annihilator of A as an ideal of R (see Definition 3.2.13). In fact, since M is a free R-module and R is a PID, the ideal theoretic annihilator of any subset of M is automatically (0).

(3) Note that Ann(A) = Ann(Hull(A)). To see this note that w(ax') _ 0 r* aw(x') = 0. But R has no zero divisors, so aw(x') = 0 if and only if w(x') = 0. Also note that Ann(A) is a complemented submodule

of M' for the same reason. Namely, aw(x) = 0 for all x E A and

a#OER *w(x)=0 forallxEA.

(4)

Similarly, Ann(B) = Ann(Hull(B)) and Ann(B) is a complemented submodule of M for any subset B C M*.

The concepts of annihilators of subsets of M and M' will be used to get a duality between submodules of M and M'. But since annihilators of subsets are complemented submodules, we see immediately that it is necessary to restrict any correspondence between submodules of M and M* to the set of complemented submodules. Thus, if M if a free R-module, then we will denote the set of all complemented submodules by C(M). The following result is a collection of straightforward properties of annihilators. The verifications are left as an exercise.

(1.16) Proposition. Let M be a free R-module of finite rank, let A, A1, and

A2 be subsets of M, and let B, B1, and B2 be subsets of M. Then the following properties of annihilators are valid: (1) If Al C A2, then K(A1) J K(A2).

(2) K(A) = K(Hull(A)). (3) K(A) E C(M'). (4) K({0}) = M' and K(M) = {0}.

6.1 Duality

(5) K'(K(A))

347

A.

(1') If B1 C B2, then K'(B1) D K'(B2).

(2') K'(B) = K'(Hull(B)). (3') KO(B) E C(M).

(4') K'({O}) = M and K'(M') = {0}. (5') K(K'(B)) B. 0

Proof. Exercise.

The following result is true for any reflexive R-module (and not just finite rank free modules). Since the work is the same, we will state it in that context:

(1.17) Lemma. Let M be a reflexive R-module and let n : M - M" be the natural isomorphism. Then for every submodule T of M', we have

r)(K'(T)) = K(T) C M. Proof. Let w E K(T). Then w = rl(x) for some x E M. For any t E T,

t(x) = zl(x)(t) = w(t) = 0 because w E K(T). Therefore, x E K'(T) by definition and hence K(T) C i7(K*(T)) Conversely, if x E K' (T) then 0 = t(x) = 77(x)(t)

for any t E T so r)(x) E K(T) by definition. Thus, rt(K'(T)) C K(T), and the lemma is proved.

(1.18) Theorem. Let M be a free R-module of finite rank, let S be a complemented submodule of M, and let T be a complemented submodule of M*. Then

rank(M) = rank(S) + rank(K(S)), and

rank(M') = rank(T) +rank(K'(T)). Proof. Let B1 = {v1i ... ,vk} be a basis of S. Since S is complemented, it follows (Corollary 3.8.4) that 81 extends to a basis

B = {vl, ... , Vk) Vk+l, ... , vm} of M. Let B' = {v,*, ... , vm } be the basis of M dual to B. If i < k and j > k then vj*(vi) = 0. Therefore,

348


(vk+1, ... , v.n) C K(S). If w E K(S), then we may write w =

1 ajv,*, and if 1 < i < k, then

m

0 = W(v8) = E ajvj* (vi) = a,. j=1

Therefore, w = F jk+1 aj vj*, and hence, K(S) = (vb+1, ... , v;R)

so that rank(K(S)) = m - k = rank(M) - rank(S). Similarly,

rank(M') = rank(T) + rank(K(T)) = rank(T) + rank(ri(K'(T))) = rank(T) + rank(K'(T))

where the last equality is valid because q : M M" is an isomorphism, and hence, it preserves ranks of submodules. (1.19) Theorem. Let M be a free R-module of finite rank. Then the function

K : C(M) - C(M') is a one-to-one correspondence with inverse K*.

Proof. We claim that for every complemented submodule S C M and T C M', we have

K'(K(S)) = S and

K(K'(T)) = T. We will prove the first of these equalities; the second is similar. First note the K'(K(S)) 2 S for every complemented submodule S C M by Proposition 1.16 (5), so Corollary 3.8.5 implies that it suffices to show

that rank(K'(K(S))) = rank(S). But rank(S) = rank(M) - rank(K(S)) and

rank(K(S)) = rank(M') - rank(K'(K(S))) by Theorem 1.18. Since rank(M) = rank(M'), the result follows.

6.1 Duality

349

(1.20) Definition. If M and N are R-modules and f E HomR(M, N) then

the adt off is the function f' : N' - M' defined by f' (w) = w o f , that is, (f'(w))(x) = w(f(x))

forallxEM. (1.21) Remarks. (1)

f' : N'

M' is an R-module homomorphism.

(2) Ad : HomR(M, N) - HomR(N', M'), defined by Ad(f) = f', is an R-module homomorphism. (3) If M and N are free, Ker(f) is always a complemented submodule of

M, but Im(f) need not be complemented. (See Proposition 3.8.7.)

(1.22) Theorem. Let M and N be free R-modules of finite rank and let f E HomR(M, N). Then

(1) Ann(Im(f )) = Ker(f') C N', (2) rank(Im(f')) = rank(Im(f))), and (3) Im(f') = Ann(Ker(f)) C M' if Im(f') is a complemented submodule of M*.

Proof. (1) Let w E N'. Then

w E Ker(f') p f'(w) = 0

qwo f =0 aw(f(x))=0 tlxEM aw(y)=0 VyEIm(f) a w E Ann(Im(f)). (2) Since f' : N' -+ M', Proposition 3.8.8 gives rank(N') = rank(Im(f')) + rank(Ker(f')) while Theorem 1.18 shows

rank(N) = rank(Im(f)) + rank(Ann(Im(f))). Since rank(N) = rank(N'), (2) follows from (1). (3) Now let r E M'. Then r E Im(f') a r= f' (w) for some w E N'

a r(x) = w(f (x)) Vx E M. If x E Ker(f) then f (x) = 0, so w(f (x)) = 0. Therefore, r(x) = 0, and we conclude that r E Ann(Ker(f )). Hence, Im(f') C Ann(Ker(f )). By Theorem 1.18 and part (2), rank(Ann(Ker(f))) = rank(M) - rank(Ker(f)) = rank(Im(f)) = rank(Im(f')). Since Im(f') is assumed to be complemented, we conclude that Im(f') _ Ann(Ker(f)). 0


350

(1.23) Corollary. Let F be a field, let V and W be finite-dimensional vector spaces over F, and let f E HomF(V, W). Then

(1) f is injective if and only if f * is surjective; (2) f is surjective if and only if f * is injective; and (3) f is an isomorphism if and only if f * is an isomorphism. Proof.

(1.24) Proposition. Let M and N be free R-modules of finite rank with bases B and C, respectively, and let f E HomR(M, N). Then [f"]C _ ([.f]CC)t

Proof. Let B = {v}1 and C = {wj}T 1. If A = [aid]

[f]' and B =

[bid] _ [f*]C., then by definition m

f (vj) = > akj wk k=1

and

f*

R

(wi) _ E bkivk k=1

But then aii = wi (.f(vj)) = (w: o f)(vj) = (.f*(w2M(vj) = bpi.

0

6.2 Bilinear and Sesquilinear Forms In this section we present an introduction to an important branch of mathematics that is the subject of much study. Throughout this section R will be a commutative ring with 1 and all R-modules will be free. (2.1) Definition. A conjugation on R is a function c : R -> R satisfying (1) (2) (3)

c(c(r)) = r for all r E R; c(rl + r2) = c(ri) + c(r2) for all ri, r2 E R; and c(rlr2) = c(rl)c(r2) for all r1, r2 E R.

6.2 Bilinear and Sesquilinear Forms

351

That is, a nontrivial conjugation of R is a ring automorphism, which has order 2 as an element of the group Aut(R). (2.2) Examples.

(1) Every ring has the trivial conjugation c(r) = r. Since Aut(Q) = {1Q), it follows that the trivial conjugation is the only one on Q. The same is true for the ring Z. (2) The field C has the conjugation c(z) = z, where the right-hand side is complex conjugation. (This is where the name "conjugation" for a function c as above comes from.)

(where d is not a square) both have the conjugation c(a + bVd-) = a - bf .

(3) The field Q[/] and the ring

Because of Example 2.2 (2), we will write 'r, instead of c(r), to denote conjugation.

(2.3) Definition. Let M be a free R-module. A bilinear form on M is a

function 0: M x M R satisfying (1) 4 (rlxl + r2x2, y) = r14 (xl, y) + r20(x2, y), and (2) O(x, rlyl +r2y2) = ri (x, yl) +r2j(x, y2) for all x1, x2, y1, y2 E M, and r1i r2 E R. A sesquilinear form on M is a function 0: M x M - R satisfying (1) and (2) O(x, r1y1 + r2y2) = f1 (x, yl) + r2O(x, y2)

for a nontrivial conjugation r

T on R.

Observe that this notion is a generalization of the notion of inner product space that we considered in Section 4.6. Some (but not all) authors use

the term "inner product space" to refer to this more general situation. (Strictly speaking, in the second part of the definition we should say that 0 is sesquilinear with respect to the given conjugation, but we shall assume that we have chosen a particular conjugation and use it throughout.)

(2.4) Definition. Let R be a ring with conjugation and let M and N be R-modules. A map f : M -+ N is called an antihomomorphism if f(r1m1 +r2m2) = T1f(m1) + T2f(m2) for all r1, r2

R, m1, m2

M.

We observed in Section 6.1 that there is no canonical isomorphism from M to its dual module M'; however, a bilinear form produces for us a canonical map, and, conversely, a map produces a canonical form. (Here we

352


do not necessarily have isomorphisms, but we shall investigate this point shortly.) (2.5) Proposition.

(1) Let 0 be a bilinear (reap., sesquilinear) form on M. Then am M - M', defined by am(y)(x) = O(x, y)

is an R-homomorphism (reap., R-antihomomorphism).

(2) Let a : M --+ M' be an R-homomorphism (neap., R-antihomomorphism). Then 0Q : M x M - R, defined by

ma(x, y) = a(y)(x) is a bilinear (reap., sesquilinear) form on M. Proof. Exercise.

(2.6) Examples.

(1) Fix s E R. Then ¢(rl, r2) = rlsr2 (reap., = rl02) is a bi- (reap., sesqui-) linear form on R.

(2) O(x, y) = xty is a bilinear form on Mn,l(R), and O(x, y) = x1V is a sesquilinear form on Mn,1(R). Note that ji is obtained from y by entry by entry conjugation. (3) More generally, for any A E Mn(R), O(x, y) = xtAy is a bilinear form, and O(x, y) = xtA'y is a sesquilinear form on Mn,l (R).

(4) Let M = Mn,,n(R). Then 4,(A, B) = Tr(AtB) (reap., ¢(A, B) _ TT(AtB)) is a bi- (reap., sesqui-) linear form on M. (5) Let M be the space of continuous real- (reap., complex-) valued functions on [0, 1]. Then

Of, 9) =

f

1

.f (x)9(x) dx

0

is a bilinear form on the R- (reap., C-) module M. If M is the space of continuous complex-valued functions on [0, 1], then

O(,f, 9) = fo f(x)9(x) dx 1

is a sesquilinear form on the C-module M.

We will often have occasion to state theorems that apply to both bilinear and sesquilinear forms. We thus, for convenience, adopt the language that 0 is a b/s-linear form means 0 is a bilinear or sesquilinear form. Also,


353

the theorems will often have a common proof for both cases. We will then write the proof for the sesquilinear case, from which the proof for the bilinear case follows by taking the conjugation to be trivial (i.e., r = r for all r E R). We will start our analysis by introducing the appropriate equivalence relation on b/s-linear forms.

(2.7) Definition. Let 01 and 02 be b/s-linear forms on free R-modules M1 and M2 respectively. Then ¢1 and 02 are isometric if there is an R-module isomorphism f : M1 --. M2 with 02 (f (X), f(y)) _ 01(x, y)

forall

x,yEM1.

(If there is no danger of confusion, we will call M1 and M2 isometric.) The map f is called an isometry.

Our object in this section will be to derive some general facts about b/s-linear forms, to derive canonical forms for them, and to classify them up to isometry in favorable cases. Later on we will introduce the related notion of a quadratic form and investigate it. We begin by considering the matrix representation of a b/s-linear form with respect to a given basis.

(2.8) Definition. Let M be a free R-module of rank n with basis B = {vi, ... , V.) and 0 a b/s-linear form on M. Define the matrix of ¢ with respect to the basis B, denoted [01g, by entii (1018) = 0(vi, vi)

1

(2.9) Proposition. (1) Let M be a free R-module of rank n with basis B and let 0 be a bilinear form on M. Then for any x, y E M, (2.1)

O(x, y) = [x]B[01a[y18.

(2) If 0 is a sesquilinear form on M and x, y E M, then (2.2)

0(x, y) = [x]B[m18[y1 B.

Proof. Just as a linear transformation is determined by its values on a basis,

a b/s-linear form is determined by its values on pairs of basis elements. According to Example 2.6 (3), the right-hand sides of equations (2.1) and (2.2) define such forms, and the two sides clearly agree on each pair (vi, v,). 0

(2.10) Definition. If M is a free R-module, then we will denote the set of all bilinear forms on M by Bilin(M), and if R has a conjugation, then we will denote the set of all sesquilinear forms on M by Seslin(M). Each of

354


these sets is an R-module in a natural way, i.e., via addition and scalar multiplication of R-valued functions.

(2.11) Corollary. Let M be a free R-module of rank n. Then there are Rmodule isomurphisms

Bilin(M) = Mn(R)

(2.3)

and (2.4)

Seslin(M) = M, (R)

given by

where B is any basis of M.

Proof. Proposition 2.9 gives a bijection, and it is easy to check that it is a homomorphism (in both cases). 0 (2.12) Remarks.

(1) Note that this corollary says that, in the case of a free module of finite rank, all forms arise as in Example 2.6 (3). (2) We have now seen matrices arise in several ways: as the matrices of linear transformations, as the matrices of bilinear forms, and as the matrices of sesquilinear forms. It is important to keep these different roles distinct, though as we shall see below, they are closely related. One obvious question is how the matrices of a given form with respect to different bases are related. This is easy to answer.

(2.13) Theorem. Let 0 be a b/s-linear form on the fire R-module M of rank n. Let B and C be two bases for M. If P = PP is the change of basis matrix from C to B, then (2.5)

[01c =

Proof. By definition, [0Jc is the unique matrix with O(x, y) = [x]c[0)c[v]c

But, also, O(x, y) = [x1B[01e[v1e,

and if P = PP, then Proposition 4.3.1 gives [x)g = P[x}c

Thus,

and

[y] B = P[y]c.


355

0(x, y) = (P[xl c)` 101B (P[y]c) = [x]`c (Pl[0]BP) [y]c

yielding the theorem.

(2.14) Remark. Note that the matrix P is nonsingular (Proposition 4.3.1) and that every nonsingular matrix arises in this way (Proposition 4.3.2). The relation A - B if A and B are matrices of the same b/s-linear form with respect to different bases is clearly an equivalence relation, and the above theorem states that this equivalence relation is given by A - B if and only if A = P1 BP for some invertible matrix P. In the case of a bilinear form, this relation becomes A = P° BP and is known as congruence. There is no generally accepted name in the sesquilinear case; we shall refer to it as conjugate congruence. Note that this relation is completely different from the relation of similarity.

(2.15) Remark. We shall often have occasion to speak of det(0) in this section. By this, we mean det(A), where A is the matrix of 0 in some basis. Note that this is well defined up to multiplication by a unit of R of the form rf, for if B is the matrix of 0 in a different basis, then B = PtAP, for some

invertible matrix P. Then det(B) = rir det(A) where det(P) = r is a unit, and for any r we may find such a matrix P, e.g., P = diag(r, 1, 1, ...). On the other hand, this observation gives an invariant of a b/s-linear form: If 01 and 02 are two forms with the equation det(01) = rfdet(02) having no solution for r a unit in R, then 01 and 02 are not isometric.

(2.16) Lemma. Let M be a free R-module of rank n and D a b/s-linear form on M. Let B be a basis of M and B' the dual basis of M*. Then [0113 = 1a01e.

Proof. Let B = {v1, ... , vn}. We claim that am(vj)

_

0(vi, vi)v: i=1

In order to see that this is true we need only check that am(vj)(vk) = 0(vk, vj), which is immediate from the definition of ao (see Proposition 2.5). Then from the definition of A = [ao18. (Definition 4.3.3), we see that A is the matrix with entij (A) = 0(vi, vj), and this is precisely the definition of [0113 (Definition 2.8).

We shall restrict the forms we wish to consider. (2.17) Definition. Let M be a free R-module.

356


(1) A bilinear form 0 on M is said to be symmetric if 4)(x, y) = O(y, x) for all x, y E M. (2) A bilinear form 0 on M is said to be skew-symmetric if 4)(x, y) _ -4)(y, x) for every x, y E M, and 4)(x, x) = 0 for every x E M. (3) A sesquiiinear form.0 on M is said to be Hermitian if O(x, y) = 4)(y, x) for every x, y E M. (4)

If 2 is not a zero divisor in the ring R, then a sesquilinear form ¢ on M is said to be skew-Hermitian if 4)(x, y) = -4)(y, x) for every x,

yEM. (2.18) Remarks.

(1) We do not define skew-Hermitian if 2 divides 0 in R.

(2) Let 0 be a b/s-linear form on M and let A be the matrix of ¢ (with respect to any basis). Then the conditions on m in Definition 2.17 correspond to the following conditions on A: (a) 0 is symmetric if and only if At = A;

(b) 0 is skew-symmetric if and only if At = -A and all the diagonal entries of A are zero; (c) 0 is Hermitian if and only if A 54 A and At = A; and (d) 0 is skew-Hermitian if and only if A # A and At = -A (and hence every diagonal entry of A satisfies a = -a). (3) In practice, most forms that arise are one of these four types. We introduce a bit of terminological shorthand. A symmetric bilinear form will be called (+1)-symmetric and a skew-symmetric bilinear form will be called (-1)-symmetric; when we wish to consider both possibilities simultaneously we will refer to the form as c-symmetric. Similar language

applies with c-Hermitian. When we wish to consider a form that is either symmetric (bilinear) or Hermitian (sesquilinear) we will refer to it as (+1)-symmetric b/s-linear, with a similar definition for (-1)-symmetric b/s-linear. When we wish to consider all four cases at once we will refer to an c-symmetric b/s-linear form. (2.19) Definition. The c-symmetric b/s-linear form 0 is called non-singular if the map ao : M - M' is bijective. It is called non-degenerate if am : M -+ M' is injective. Note that if R is a field and M has finite rank, then these notions are equivalent. (2.20) Proposition. Let V be a finite-dimensional vector space over a field F,

and let 0 be an c-symmetric b/s-linear form on V. Then 0 is non-singular if and only if it is non-degenerate, which is the case if and only if, for every y O O E V, there is an x E V with ao(y)(x) _ O(x, y) 0 0.


357

Proof. Since am is a homomorphism (or antihomomorphism) between vector spaces of the same dimension, in order to show that it is an isomorphism (or antiisomorphism) we need only show that it is injective, i.e., that Ker(am) 0 (0). But

Ker(a,0)_{yEV:a0(y)(x)_0(z, y)=0 foreveryxEV}. 0 To give a criterion for non-singularity over a ring, we need to use the matrix representation of a form.

(2.21) Theorem. Let M be a free R-module of finite rank, and let 0 be an c-symmetric b/8-linear form on M. (1) The form 0 is non-singular if and only if in some (and hence in every)

basis B of M, det([-0]B) is a unit in R. (2)

The form 0 is non-degenerate if and only if in some (and hence in every) basis B of M, det([0]e) is not a zero divisor in R.

Proof. This follows immediately from Lemma 2.16 and Proposition 4.3.17.

0 Note that if N is any submodule of M, the restriction 46N = 'IN of any c-symmetric b/s-linear form on M to N is an c-symmetric b/s-linear form on N. However, the restriction of a non-singular b/s-linear form is not necessarily non-singular. For example, let 0 be the b/s-linear form on M2,1(R) with matrix [ i a ] . If N1 = ([ o ]) and N2 = ([0]), then 01N, is non-singular, but cIN2 is singular and, indeed, degenerate. The following is standard terminology: (2.22) Definition. Let 0 be an c-symmetric b/s-linear form on M. A submodule N C M is totally isotropic if 01 N is identically zero. Thus, in the above example, N2 is a totally isotropic subspace of M. Recall that in studying free modules, we found it useful to decompose them into direct sums, and in studying a vector space with a linear transformation, we found it useful to decompose it into a direct sum of invariant subspaces. There is an analogous, and similarly useful notion in our present context, which we now introduce. (2.23) Definition. Let 0 be an c-symmetric b/s-linear form on M.

(1) Two submodules N1 and N2 of M are orthogonal if q(n1, n2) = 0 for

every n1EN1in2EN2. (2) M is the orthogonal direct sum of two submodules N1 and N2, written M = N1 1 N2i if M = N1 9 N2 and N1 and N2 are orthogonal.


358

(2.24) Remark. Let N1 have a basis B1, N2 a basis B2, and let M = N1®N2, in which case B = Bl U B2 is a basis of M. Then M = N1 1 N2 if and only if A

0

0B

[0Jg = Conversely, if [O]B is of this form, and if N1 (reap., N2) denotes the span of B1 (reap., B2)1 then M = N1 1 N2. In this case we will also say 0 _ 01 102 where 4i = 01,y, .

(2.25) Definition. Let 0 be an e-symmetric b/s-linear form on M. The kernel of 0, denoted M° = M°(O) C M, is defined to be

M° = Ker(am) _ {y E M : 41(x, y) = 0 for all x E M}. As a first step in studying forms, we have the following decomposition:

(2.26) Proposition. Let M be a finite rank free module over a PID R, and let 0 be an e-symmetric b/s-linear form on M. Then 0 is isometric to 00 1 01 defined on M° 1 M1, where 41o is identically zero on M° and 01 is nondegenerate on M1. Furthermore, 00 and 01 are uniquely determined up to isometry.

Proof. Note that M° is a pure submodule of M (since it is the kernel of a homomorphism (Proposition 3.8.7)) and so it is complemented. Choose a complement M1. Then M1 is free and M °_f M°®M1. We let db = 411M0 and

01 = 01 M, . Of course M° and M1 are orthogonal since M° is orthogonal

to all of M, so we have M = M° 1 Ml with 0 = 00 1 01. Also, if mlEM1with 01(m'1,ml)=Ofor all m'EM1,then 41(m,ml)=0 for all m E M, i.e., m1 E M°. Since M = M° ® M1, M° n M1 = (0) and so 01 is non-degenerate. The construction in the above paragraph is well defined except for the choice of M1. We now show that different choices of M1 produce isometric

forms. Let it : M -+ M/M° = M'. Then M' has a form 0' defined as follows: If x', y' E M', choose x, y E M with w(x) = x' and 7r(y) = y'. Set 41'(x', y') = 41(x, y), and note that this is independent of the choice of x and y. But now note that regardless of the choice of M1, not only is RIM, : M1 - M' an isomorphism, but is in fact an isometry between 01 and 0'. 0 The effect of this proposition is to reduce the problem of classifying e-symmetric b/s-linear forms to that of classifying non-degenerate ones. It also says that the following definition does indeed give an invariant of such a form. (2.27) Definition. Let 0 be an e-symmetric b/s-linear form on a finite rank free module M over a PID R. If 0 is isometric to (0° 1 01 with 41° identically


359

zero and 01 non-degenerate on M1, then 01 is called the non-degenerate part of 0 and we set rank(O) = rank(M1).

(2.28) Example. Let ¢ be the symmetric bilinear form on M = M3,1(Z) with matrix (with respect to the standard basis)

-1 A=

1

-1

1

-1

-1

-3

-3 -1

Then det(A) = 0, so 0 is degenerate. Routine computation shows that Ker(am) = {X E M : AX = 0} is the rank 1 subspace spanned by v1 = (2, 1, -1)t. Note that det

0

2

1

1

0

1

-1

0

0

= 1

so that vl, v2 = [ 1 0 0]t, and v3 = [0

1

0]t form a basis B for M.

Then we let M1 be the submodule spanned by v2 and v3. The matrix of 01 in this basis is Ov2, V2)

4(v2, V3)

O(v3, V2)

O(v3, V3)

_-

and

0

0 0

-1

0 1

0

1

-3

(2.29) Definition. Let ¢ be an c-symmetric b/s-linear form on M and let N be a submodule of M. Then N1, the orthogonal complement of N, is defined to be

N1={xEM:0(x,y)=0 forallyEN}. (2.30) Examples.

(1) Ml = M°. (2)

If N C M°, then N1 = M.

(3) Let 0 be the b/s-linear form on M2,1(R) whose matrix with respect to the standard basis is (a)

[o

1J

(b) [1 OJ;

rChapter 6. Bilinear and Quadratic Forms

360

(c)

(d)

l0

0l ; and

1

OJJ

0

0

Let N1 = ([ o ]) and N2 = ([ ]). Then in the above cases we have

(a) N1' =N2,N2 =N1,MN11NIL, '= M=N21N2; (b) N1 = ([ 12J ), Ny = M, M = Hull(N1 1 Ni ); (c) N1 = N1, Nz = N2;

(d) N11- =N2iN2 =M,M=N11NIL.

(2.31) Lemma. Let N be a submodule of M and suppose that IN is nondegenerate. Then N fl N1 = (0). Conversely, if N fl N1 = (0), then q5I N is non-degenerate.

Proof. If 0 = 4IN, then (2.6)

N fl N1 = Ker(ap).

The result follows immediately from Equation (2.6).

(2.32) Proposition. Let R be a PID, 0 an c-symmetric b/s-linear form on a free R-module M, and N C M a submodule of finite rank. If 'I N is non-degenerate, then M = Hull(N I N1). If 0I N is non-singular, then

M=NIN1.

Proof. Let 10 = IN. If M E M, then fm : N -+ R, defined by fm(n) _ 0(n, m), is an R-module homomorphism, i.e., f E N. Since ap is assumed

to be injective, it follows that Im(ap) is a submodule of N' of rank n where n = rank(N) = rank(N'). Thus, N'/Im(ay) is a torsion R-module by Proposition 4.3.11, and hence there are r E R and no E N such that rf n = ao(na); in case ap is an isomorphism, we may take r = 1. If m1 = rm - no, then for every n E N, we have

m(n, mi) = O(n, rm - no) = rO(n, m) - i(n, no) = rf,n(n) - ao(no)(n) = 0,

i.e., ml E N1. Thus, rm. = no + ml, where no E N, and m1 E N. Thus,

M = Hull(N + N1) (or M = N + N1 if r = 1). But N fl N1 = (0) by Lemma 2.31, yielding the proposition.

(2.33) Remark. Note that in Lemma 2.31 and Proposition 2.32 there is no restriction on 0, just on GIN. The reader should reexamine Examples 2.30 in light of the above lemma and proposition.


361

(2.34) Corollary. Let R be a PID and M a free R-module of finite rank. Let N be a pure submodule of M with 41N and OINJ. both non-singular. Then

(N')' = N. Proof. We have M=NIN' = (N')' _L N-L. But it is easy to check 0

that (N1)1 D N, so they are equal.

Now, a bit of obvious notation: no denotes 0 1 ... 1 0 (where there are n summands). We now come to the first classification result. Note that it suffices to classify non-degenerate forms.

(2.35) Theorem. Let R be a PID, and let ¢ be a non-degenerate skewsymmetric bilinear form on a free R-module M of finite rank. Then ¢ is classified up to isometry by M'/Im(am), a torsion R-module. Both M and

M'/Im(am) have even rank, say 2n and 2k respectively. The invariant factors of M'/Im(am) are of the form el, el, e2, e2, ... , ek, ek.

Furthermore, 0 is isometric to

r0 (n - k) I

-1

1

0

ell

0

-el

0J

0 1[ -e2

e2

0

1 ...

0

-ek

ek 0

Proof. Let rank(M) = rank(M') = in. Then Q = M*/Im(am) is a torsion R-module of rank q, and hence it is determined by its invariant factors. We will write the invariant factors of the submodule lm(am) as fl, f2, ..., fm where f1 = = fm-q = 1, fm-q+1 # 1 and fi I fi+l for m-q+l < i < m. (See Proposition 3.6.23.) It is evident then that Q ®;_1R/ f;R. Clearly, M*/ Im(am) is an invariant of the isometry class of 0. We need

to show, conversely, that this determines ¢, or in other words, that the sequence f1, f2, ..., fm determines 0. In fact, we will show this by showing that 0 is isometric to the form given in the statement of the theorem, with m = 2n, q = 2k, and (fm-q+1, fm-q+2, ... , fm) = (el, el,

e2, e2,

... , ek, ek)

The proof is by induction on the rank of M. If rank(M) = 1, then the only possible skew-symmetric form on M is [0], which is degenerate, and hence this case is excluded. If rank(M) = 2, then 0 must have a matrix in some (and, hence, in any) basis of the form [ ° a], and so the theorem is true in this case also. Now for the inductive step. Assume the theorem is true for all free R-modules of rank less than in. By Proposition 3.6.23, we may choose a


362

basis {w1, ... ,w,,,} of M' such that {f1w1, ... , fmwm} is a basis of the submodule Im(am). Since fl I f2 I I fm, we see that lm(am) C f1 M', i.e., O(v1, v2) is divisible by fl for every v1, v2 E M. Let x1, ..., x,,, be the dual basis of M, that is, wi(xj) = b,j. Let yl E M with am(yl) = f1w1. If ax1 + by, = 0, then 0 = O(xi, ax, + by,)

= a¢(xi, x1) + bm(xi, yl)

= bfl. Thus, b = 0 and hence a = 0 also. Thus, {x1, yl } is linearly independent. Let N be the submodule of M with basis 8 = {x1, yl}. If tli = OIN, then has the matrix

)s = [

f0 .

l 0f ] Note that N is a pure submodule. To see this, suppose az = bx1 +cy1 where a, b, c E R; by cancelling common factors we can assume that gcd{a, b, c} =

1. Then O(xl, ax) = .(xl, bxl + cyl) = cfl, while O(xl, az) = a4(x, z) = adfi since O(v1, v2) is divisible by fl for all v1, v2 E M. Thus, ad = c, i.e.,

arc. A similar computation with 0(yi, az) shows that a I b, so a is a common

divisor of a, b, and c, i.e., a is a unit and z E N. By Proposition 2.32,

M = Hull(N 1 N'). But, in fact, M = N 1 N'. To see this, let us consider the form 0' defined by O'(vi, v2) = fl 10(Vl, V2)

Then N' is also the orthogonal complement of N with respect to ,b' and O'IN is non-singular, so M = N 1 N', i.e., io = ip 1 ¢1 with ml = OINI Note that N ' / Im(ay,) has "invariant factors" (in the above sense) f1

and fl, so we see that f2 = fl. Then the "invariant factors" of ..., f,,,, and the theorem follows by induction.

(N-L)'/ Im(am,) are f3,

(2.36) Corollary. Let R be a PID, and let m be a non-singular skewsymmetric bilinear form on a free R-module M of finite rank. Then rank(M) = 2n is even, and 0 is isometric to

n[0101' Proof.

Remark. Recall that over a field, 0 is non-degenerate if and only if it is non-singular, so Corollary 2.36 classifies skew symmetric forms over a field.


363


Consider the skew-symmetric bilinear form 0 over Z with matrix 0

A=

-2 0 2

2 0 2 8

0

-2

0

4

-4

0

-2 -8

According to the theory, to classify 0 we need to find the "invariant factors" of Z4/AZ'. To do this, we apply elementary row operations: 0

2

0

-2

-2

0

-2

-8

0 2

2

0

8

-4

4 0

0

2

0

-2

-2 0 -2 -8 0 0

2 8

0

2

-2

0

0 0

0

0

4

-6 -8 0

-2

-2 -8 0

0 -6

6 0

from which we see that the invariant factors are (2, 2, 6, 6) and, hence, that 0 is isometric to [ -2

0, 1 [ -6 0]

(2) Let A be any invertible n x n skew-symmetric matrix over a field F,

i.e., At = -A and the diagonal entries are 0. Then A is the matrix of a non-singular skew-symmetric form over F, and hence, PAP = mJ where m = n/2 and J = [ 01 0]. Then det(A)(det(P) )2 = det(nJ) = 1.

In particular, det(A) is a square in F. Now let R = Z[Y] where Y = {Xij : 1 < i < j < n}, that is, R is the polynomial ring over Z in the (2) indeterminates X;j for 1 < i < j < n. Let F = Q(Y) be the

quotient field of R and let A E Mn(F) be the skew-symmetric matrix

with entij = X;j for 1 < i < j < n, ent;, = -X.- for 1 < j < i < n, and entii = 0. Then det(A) = P(Xij)2 for some element P(Xij) E F. But R is a UFD, and hence, the equation Z2 = det(A) has a solution in F if and only if it has a solution in R. Thus, P(Xij) E R, and since P(Xi3) is a root of a quadratic equation, there are two possibilities for the solution. We choose the solution as follows. Choose integers xij so

364


that the evaluation of the matrix A at the integers xij gives the matrix mJ. Then choose P(Xij) so that the polynomial evaluation P(xij) = det(mJ) = 1. Then we will call the polynomial P(Xi3) E Z[Xi)J the generic Pfaffian and we will denote it Pf(A). If S is any commutative ring with identity, then the evaluation Xij bi3 induces a ring homomorphism -+ S.

71:

Under this homomorphism, the generic skew-symmetric matrix A is

sent to the skew-symmetric matrix B = [bid[. Since determinants commute with ring homomorphisms, we find

det(B) = det(i (A)) = n(det(A)) = 11((Pf(Xi,)2) = Pf(b,)'. We conclude that the determinant of every skew-symmetric matrix over any commutative ring is a square in the ring, and moreover, the square root can be chosen in a canonical manner via the Pfaffian. (2.38) Remark. The case of skew-Hermitian sesquilinear forms is quite different. For example, the form [ii (i refers to the complex number i) is a

non-singular form over C (or over the PID Z[i)), a module of odd rank. Also, consider the following two non-singular forms:

_ t 12

0

0 -i]

We leave it to the reader to check the following facts: (1) 01 and 02 are not isometric.

(2) 01 is not isometric to any form with matrix []. (3) 02 is not isometric to [

01 0]

over Z[i], but is isometric to it over C.

We shall not discuss these forms any further. Now we come to the case of symmetric bilinear or Hermitian sesquilinear forms.

(2.39) Definition. A(+1)-symmetric b/s-linear form 0 on an R-module M is called diagonalizable if 0 is isometric to the form [al] 1 [a2J 1 ... 1 [an]

for some elements a1, a2, ..., an E R. Here [ail denotes the form on R whose matrix is [ail, i.e., the form O(rl, r2) = rlair2. (The terminology "diagonalizable" is used because in the obvious basis 0 has the matrix

diag(al, a2, ... an)-)


365

Note that the notion of diagonalizability for a form is quite different than the notion of diagonalizability for a linear transformation; in of matrices, a form 4i is diagonalizable if its matrix is (conjugate)-congruent to a diagonal matrix, while a linear transformation is diagonalizable if its matrix is similar to a diagonal matrix. (2.40) Definition A symmetric bilinear form 45 on a module M over a ring

R is called even if O(x, X) E 2R for every x E M. If ¢ is not even, it is called odd.

Here, 2R denotes the principal ideal of R generated by 2. Note that if 2 is a unit in R, then every form over R is even. (2.41) Lemma. A symmetric bilinear form 0 on a module M is even if and only if for some (and hence for every) basis B = {v,}SEI of M, 0(v1, v,) E 2R for every i E I. Proof. Since 45 is symmetric, it follows that 4'(x + y, x + y) = 45(x, x) + 45(x, y) + 45(y, x) + ¢(y, y)

= O(x, x) + 20(x, y) + 0(y, y) for any x, y E M. Given this observation, the rest of the proof is left to the reader. 0 (2.42) Theorem. Let 0 be a non-singular (+1)-symmetric b/s-linear form on a module M of finite rank over a field R. If char(R) = 2 and 45 is symmetric bilinear, assume also that 0 is odd. Then 45 is diagonalizable.

Proof. We prove this by induction on rank(M). If rank(M) = 1, then 0 _ [a], and there is nothing to prove. Thus, suppose that the theorem is true for rank(M) < n and let rank(M) = n. First, we claim that there is an x E M with 0(x, x) 96 0. If char(R) = 2 and 0 is symmetric bilinear, this is true by hypothesis. Otherwise, pick y E M. If 45(y, y) 0 0, set x = y. If 45(y, y) = 0, pick z E M with 45(y, z) # 0. Such a z exists because 45 is assumed to be non-singular. If 45(z, z) 96 0, set

x = z. Otherwise, note that for any r E R, 45(ry + z, ry + z) = 45(ry, ry) + 0(ry, z) + 0(z, ry) + 45(z, z) = rc6(y, z) + rO(y, z).

Ifchar(R)02setx=ry+zwith r=45(y,z)-1. If char(R) = 2 and 0 is Hermitian, let a E R with a 0 a, and set z = ry + z with r = a45(y, z)-1. Then, in any case, we have 45(x, x) 0 0. Let N be the subspace of M spanned by x. By construction, OWN is

non-singular, so M = N I NJ- by Proposition 2.32. But then rank(Nl) =

366


rank(M) -1 and t/i = 0IN is non-singular (since O(x, x) det(>[i) = det(O) # 0). Thus, unless char(R) = 2 and 0 is symmetric bilinear, we are done. If char(R) = 2 and 0 is symmetric bilinear, we cannot apply induction yet as we do not know that the form t/, is odd. Indeed, it is possible that t/i is even and so there is more work to be done. First, consider the case rank(M) = 2. Let a = O(x, x) and choose a basis {x, x2} of M. Then, in this basis has the matrix

[b c, with a 54 0. Let e = b/a. Then [ab

1

c] [0 1]

Ie

- [0

d,

1) (with d = ae2 + c) and 0 is diagonalized. Now suppose that rank(M) > 3. Find x as above with 0(x, x) = a 0 0

and write M = N I N1 as above. Pick y E N1. If 0(y, y) 0 0, then OINl is odd and we are done (because we can apply induction). Thus, suppose 0(y, y) = 0. Since VP = OIN1 is non-singular, there is z E N1 with

0(y, z) = b # 0. If 0(z, z) 0 0, then ik is odd and we are again done by induction. Thus suppose 0(z, z) = 0. Let M1 be the subspace of M with basis {x, y, z} and let 01 = OIM, . Then, in this basis 01 has the matrix

A=

a

0

0

0

0

0

b

b 0

1

e

e

1

1

0

1

0

1

Let e = b/a and

P= and note that det(P) = 1. Then a

PAP =

0

0 be

(b + a)e

0

(b + a)e

be

0

in a basis, which we will simply denote by {x', y', z'}. Now let N be the subspace of M spanned by x', and so, as above, M = N 1 N1. But now 0 = OI N1 is odd. This is because y' E N1 and t,i(y', y') = be 0 0. Thus, we may apply induction and we are done. 0 (2.43) Example. We will diagonalize the symmetric bilinear form with matrix


A=

0

1

1

0

2 3

2

3

0

367

over Q. The procedure is to apply a sequence of elementary row/column operations. If A is symmetric and E is any elementary matrix, then EtAE is also symmetric. We indicate the matrices and record the results. Al = T12(1)AT21(1) =

2

1

5

1

0

3

5

3

0

0

2

A2 = T21(-1/2)A1T12(-1/2) =

1/2

5

A3 = T31(-5/2)A2T13(-5/2) =

A4 = T32(1)A3T23(1) =

5

-1/2 -1/2

0

0

2

0

0

0 0

-1/2

-1/2

1/2

-25/2

2

0

0

-1/2

0 0

0

0

-12

The reader should not be under the impression that, just because we have been able to diagonalize symmetric or Hermitian forms, we have been able to classify them. However, there are a number of important cases where we can achieve a classification. (2.44) Corollary. Let 0 be a non-degenerate symmetric bilinear form on a module M of finite rank over a field R of characteristic not equal to 2 in which every element is a square. (Note that R = C satisfies this hypothesis.) Then 0 is determined up to isometry by rank(M). If rank(M) = n then ¢ is isometric to nil]. Proof. This follows immediately from Theorem 2.42 and Remark 2.15.

0

(2.45) Corollary. Let 0 be a non-degenerate symmetric bilinear form on a module M of finite rank over a finite field R of odd characteristic. Then 0 is determined up to isometry by rank(M) and det(O), the latter being well defined up to multiplication by a square in R. Let x E R be any element

that is not a square, and let rank(M) = n. If det(4) is a square, then 0 is isometric to nil]. If det(O) is not a square, then 0 is isometric to (n - 1)(1] 1 (x]. Proof. By Theorem 2.42 we know that 0 is isometric to the form (r1) 1 (r2] 1 ... 1 (rn]

368

Chapter 6. Bilinear and Quadratic Fbrms

for some ri E R. Note that the multiplicative group R' has even order, so the squares form a subgroup of index 2. Then det(4)) is a square or a nonsquare accordingly as there are an even or an odd number of nonsquares

among the {r,}. Thus, the theorem will be proved once we show that the form [ri] I [rj] with r, and rj both nonsquares is equivalent to the form [1] 1 (s] for some a (necessarily a square). ] in a basis B = {v1, V2} of M. R has an odd Thus, let [0]g = [

o°

number of elements, say 2k + 1, of which k + 1 are squares. Let A = {a2r1 : a E R} and B = {1-b2r2 : b E R}. Then A and B both have k+1 elements,

soAf1B00.Thus, forsome ao,b0ER, ar 1

= 1 - ba re,

4r, +b 0 r2 = 1.

Let N be the subspace of M spanned by a0v1 + bov2. Then 4)]N = [1] and

M = N 1 N- L, so, in an appropriate basis, 0 has the matrix [ o , J, as 0

claimed.

(2.46) Corollary. Let 0 be a non-degenerate symmetric bilinear form on a module M of finite rank over a finite field R of characteristic 2. Then 0

is determined up to isometry by n = rank(M) and whether ¢ is even or odd. If n is odd, then 0 is odd and is isometric to n[l]. If n is even, then either 0 is odd and isometric to n[1], or 0 is even and 0 is isometric to

(n/2) l0

11.

Proof. Since R' has odd order, every element of this multiplicative group is a square. If 0 is odd then, by Theorem 2.42, 0 is diagonalizable and then 0 is isometric to n[1] as in Corollary 2.44. Suppose that 0 is even. Note that an even symmetric form over a field of characteristic 2 may be regarded as a skew-symmetric form. Then, by 0 Corollary 2.36, n is even and 0 is isometric to (n/2) [ ° o ] .

(2.47) Theorem. (Witt) Let 0 be a (+1)-symmetric b/s-linear form on a module M of finite rank over a field R. If char(R) = 2, assume also that 0 is Hermitian. Let N1 and N2 be submodules of M with 4)]N, and 4)IN2 non-

singular and isometric to each other. Then OiN; and 4)IN; are isometric.

Proof. If N1 = N2, then Nl = N21, and there is nothing to prove; so we

assume N1 0 N2. Let 4i = ON,, and let f : N1 - N2 be an isometry between 01 and 02. We prove the theorem by induction on n = rank(Ni) = rank(N2). Let n = 1. Let m = rank(M). If m = 1 the theorem is trivial. Let m = 2. Let v1 generate N1 and v2 = f (v1) generate N2, so


369

((vl, v1) = 0(v2, v2) = a # 0.

Then M has bases Bi = {vi, v; } with vi E Nil, i = 1, 2, and hence JOJ,g, = diag(a, bi) for i = 1, 2, with b1 and b2 both nonzero. By Theorem 2.13, there is an invertible matrix P with Pt[0Js1P = 101s21

and taking determinants shows that ab2 = ccab1 (where c = det(P)); so g : NIL -+ N2' defined by 9(vi) = c-1vs gives an isometry between cIN. and 0IN.L.

Next let m > 3 and consider the submodule N12 of M with basis {v1, v2}, where v1 generates N1, and v2 = f(v1) generates N2. Then 4,(v1, vi) = O(v2, v2) # 0.

(Since we are assuming Ni $ N2, v1 and v2 are linearly independent in M and so N12 has rank 2.) Consider 01 N, .. Either it is non-singular or it is not. First suppose that 4,IN1, is non-singular. Then, by the case m = 2 we have that 01N , IN12 and OIN2 nN1, are isometric, and hence, 0I(N;nN12)1Ns

and '0I(N1nN12)1N, L

are also isometric. But in this case M = N12 1 N12, from which it readily follows that

(N: nN12)1N 2=N; , yielding the theorem. Now suppose WI N1, is singular. Then there is a 0 0 w E N12 with ¢(v, w) = 0 for all v E N12. Suppose there is a v3 E M with q5(v3i w) 0 0. (Such an element v3 certainly exists if 0 is non-singular on M.) Of course, v3 0 N12, so {v1, v2, v3} form a basis for a submodule N123 of M of rank

3 and 0IN1 is non-singular. (To see this, consider the matrix of 0 in the basis {v1, w, v3} of N123.) Now for i = 1, 2,

N123=Ni1(Ni nN123) and w E Ni n N12 3 with 4,(w, w) = 0, so there is a basis {w, w1 } of Ni n N123 with 0 having matrix [° aj in that basis (with ai = di). We claim that any two such forms are isometric, and this will follow if, given

anyaERwitha=Z, there is abERwith Lb

1J 11

a110 1J - 11 0J

If char(R) 96 2, take b = -a/2 (and note that b = 6). If char(R) = 2, let c E R with c 34 c (which exists as we are assuming that 0 is Hermitian)

370


and let b = ac/(c+c). Hence, tINi nN,23 and 0IN2 nN,23 are isometric, and M = N123 1 N12 3 (as OI N, 2 3 is non-singular); so, as in the case 01 N, 2 is non-singular, it follows that 0IN. and OINz are isometric. It remains to deal with the case that O(v, w) = 0 for every v E M. We claim that in this case N1 = N2, so the theorem is true. To see this, let B1 2 = {V1, V21 and extend B1 2 to a basis B = {v1 i ... , vm } of M. Let A = [O]B. Then A[(1, 2)

1

(1, 2)] _

La

a,

with a 0 0 and a2 - bb = 0, and we may assume w E N12 is given by

[w]B=[a -b

0

...

0]t

(as w is well defined up to a scalar factor). Then 0(vi, w) = 0 for i = 3, m implies that there exist scalars ci so that [entli(A)1 = { (v1, vi)J = ci a ent2i(A) JJ

'(v2, vi)

,

i = 3, .

.

, m.

LbJ

If we let P be the matrix defined by

entii(P) = 1 ent2Z(P) = -Ci

i = 3, ... , m

entij(P) = 0

otherwise,

then

PtAP = rA L0

0 *

(the right-hand side being a block matrix). This is [¢]B' in the basis

13 ={v1i712,7/3,...,vm}

and then Nl = N2 is the subspace with basis {w, v3,

...

, vm}

This concludes the proof of the theorem in case n = 1. Now we apply induction on n. Assume the theorem is true for

rank(N1) = rank(N2) < n, and let rank(N1) = rank(N2) = n.

Let v1 E N1 with O(v1, v1) 0 0, and let v2 = &1) E N2; so cb(v2, v2) = O(711, 711) 54 0.


371

Note that such an element exists by the proof of Theorem 2.42. Let N11 be the subspace of M generated by v1 and let N21 be the subspace of M generated by v2. Then

M=N111(NinN1)1NIL =N211(N,nN2)1N2. Then the case n = 1 of the theorem implies that

(NL1nN1)1Ni and (N21nN2)1N2 are isometric, and then the inductive hypothesis implies that NIL and N2 are isometric, proving the theorem.

This theorem is often known as Witt's cancellation theorem because of the following reformulation:

(2.48) Corollary. Let 01, 02, and 13 be forms on modules of finite rank over a field R, all three of which are either symmetric or Hermitian. If char(R) = 2, assume all three are Hermitian. If 01 is non-singular and 01 1 02 and 01 1 03 are isometric, then 02 and 03 are isometric. Proof.

(2.49) Remark. Witt's theorem is false in the case we have excluded. Note

that 1

0

0

0

0

1

0

1

0

and

1

0

0

0

1

0

0

0

1

are isometric forms on (Fr2)3, as they are both odd and non-singular, but I

0

1J

and

[01

1]

are not isometric on (F2)2, as the first is even and the second is odd. Now we come to the very important case of symmetric bilinear forms over R and Hermitian forms over C. (2.50) Definition. If 46 is a symmetric bilinear form over R, or a Hermitian

form over C, on a module M, then ¢ is said to be positive definite if ¢(v, v) > 0 for every v 34 0 E M. If q(v, v) < 0 for every v 36 0 E M, then 0 is said to be negative definite. (2.51) Theorem. (Sylvester's law of inertia) Let 0 be a non-degenerate sym-

metric bilinear form over R = R, or Hermitian form over R = C, on a module M of finite rank over R. Then 0 is isometric to

372


r[1] 1 s[-1] with r+s = n = rank(M). Furthermore, the integers r ands are well defined and 45 is determined up to isometry by rank(o) = n, and signature(O) = r-s

Proof. Except for the fact that r and s are well defined, this is all a direct corollary of Theorem 2.42. (Any two of n, r, and s determine the third, and we could use any two of these to classify 0. However, these determine and are determined by the rank and signature, which are the usual invariants that are used.) Thus, we need to show that r and s are well defined by 0. To this end, let M+ be a subspace of M of largest dimension with OPM+ positive definite. We claim that r = rank(M+). Let B = {v1, ... be a basis of M with [0113 = Ir ®-I,.

If Ml = (v1, ... , vr), then OIM, is positive definite. Thus, rank(M+) > r. This argument also shows that if M_ is a subspace of M of largest possible dimension with 01M_ negative definite, then rank(M_) > s.

We claim that r = rank(M+) and s = rank(M_). If not, then the above two inequalities imply

rank(M+) + rank(M_) > r + s = n, so M+f1M_ # {0}. Let x 96 0 E M+f1M_. Then 4(x, x) > 0 since x E M+, while 0(x, x) < 0 since x E M_, and this contradiction completes the proof. We present an alternative proof as an application of Witt's theorem. Suppose rl[1] I sl[-1] and r2[1] 1 a2[- 1]

are isometric. We may assume rl c r2. Then by Witt's theorem, s1 [-l) and (r2 - rl)[1] 1 82[-l] are isometric. As the first of these is negative-definite, 0 so is the second, and so rl = r2 (and hence si = s2).

(2.52) Remark. If 0 is not assumed to be non-degenerate then applying Theorem 2.51 to the non-degenerate part of 0 (Definition 2.27) shows that a symmetric bilinear form over R = R or a Hermitian form over R = C is isometric to

r[1] 1 s[-1] 1 k[0]

with r + s = rank(0) = rank(M) - k, and with r and a well defined by 0. Again, we let signature(46) =,r - s.

Of course, we have a procedure for determining the signature of 0, namely, diagonalize 0 and inspect the result. In view of the importance of this case, we give an easier method.


373

(2.53) Proposition. Let 0 be a symmetric bilinear form over R = R, or a Hermitian form over R = C, on a module M of finite rank over R, and let A = [0]B for some basis B of M. Then

(1) rank(O) = rank(A); (2) all of the eigenvalues of A are real; and

(3) signature(O) = r-s, where r (reap., s) is the number of positive (reap., negative) eigenvalues of A.

Proof. It is tempting, but wrong, to try to prove this as follows: The form 0 is diagonalizable, so just diagonalize it and inspect the diagonal entries. The mistake here is that to diagonalize the form 0 we take PAP, whereas to diagonalize A we take PAP-', and these will usually be quite different. For an arbitrary matrix P there is no reason to suppose that the diagonal entries of P=AP are the eigenvalues of A, which are the diagonal entries of

PAP`. On the other hand, this false argument points the way to a correct argument: First note that we may write similarity as (P)-'AP. Thus if P

is a matrix with P' = (P)'', then the matrix B = PtAP will have the same eigenvalues as A. Let us regard A as the matrix of a linear transformation a on R" where

R = R or R = C. Then A is either real symmetric or complex Hermitian. In other words, a is self-adt in the language of Definition 4.6.16. But then by Theorem 4.6.23, there is an orthonormal basis B = {v,, ... , v"} of R" with B = [a]B diagonal, i.e., P-'AP = B where P is the matrix whose columns are v1, ..., v,,, and furthermore B E M"(R), where n = rank(A). But then the condition that B is orthonormal is exactly the condition that Pt = (15)-1, and we are done. 0 There is an even handier method for computing the signature. Note that it does not apply to all cases, but when it does apply it is easy to use.

(2.54) Proposition. Let 0 be a symmetric bilinear form over R = R or a Hermitian form over R = C, on an R-module M of finite rank n. Let B be any basis for M and let A = [']B. Set bo(A) = 1, and for 1 < i < n, let 6i(A) = det(Ai), where Ai = A[(1, 2, ..., i) (1, 2, ..., i)] is the it' principal submatrix of A, i.e., the i x i submatrix in the upper left-hand corner.

(1) Ifbi(A)#0 foralli, then (2.7)

signature(O) = I{i : b;(A) and bi_,(A) have the same sign}l - I{i : bi(A) and bi_,(A) have opposite signs) .

(2) 0 is postive definite if and only if b; (A) > 0 for all i. (3) 0 is negative definite if and only if (-1)ibi(A) > 0 for all i.


374

Proof We first prove part (1) by induction on n. The proposition is trivially

true if n = 1. Assume it is true if rank(M) < n, and consider M with rank(M) = n. Write Equation (2.7) as

signature(o) = r'(A) - s'(A).

(2.8)

Of course, r'(A) + s'(A) = n. Let B = {v1, ... , vn}, let N C M be the subspace with basis B' = (v1i ... ,vn_1), and let t = GIN. Set B = [,0jfj,. Then B = An-1, so det(B) 54 0 by hypothesis, and thus ik is non-singular. Let C' = {w1 i ... , wn_1 } be a basis of N with [01c' diagonal, say,

['G]c' = diag(c1, ... cn-1) Then, by definition,

signature(o)_ I { i : c i > 0 } 1 - {i:ci <0}l = r1 - 81

with r1 + s1 = n - 1. By induction we have

signature(O) = r'(B) - s'(B),

with r'(B) + s'(B) = n - 1, so r'(B) = r1 and s'(B) = s1. Now, since 0k' = V) is non-singular, we have M = N I Nl. Hence dim(N1) = 1; say Nl = (wn). Then C = {wi, ... wn} is a basis of M, and [O]c = diag(c1i ... , cn) with cn = 0(wn, wn). Also, by definition,

signature(O)=I{i:ci>0}I-J{i:ci<0}1 =r-s with r + s = n. Note also that det(B) and c1

cn_1 have the same sign, as do det(A) and c1 c,,, as they are determinants of matrices of the same form with respect to different bases. Now there are two possibilities:

(a) cn > 0. In this case, signature(o) = signature( s) + 1, so r = ri + 1 and s = s1. But also and det(A) have the same sign, so r'(A) = r'(B) + 1 and s'(A) = s'(B); hence, r'(A) = r and s'(A) = s, so

signature(4) = r'(A) - s'(A). (b) c,, < 0. Here the situation is reversed, with r = ri, s = si + 1, and also

r'(A) = r'(B), s'(A) = s'(B) + 1, again yielding signature(O) = r'(A) - s'(A). Thus, by induction, part (1) is proved. (2) In light of (1), we need only prove that 0 positive definite implies

that all bi(A) are nonzero. But this is immediate as bi(A) = det(OIN,), Ni = (vi, ... , vi), and ON, is non-singular since 0(v, v) 0 0 for every v E M (so Ker(a.j,) = (0), where i/i = 0[N,).


375

(3) Observe that m is negative definite if and only if -0 is positive

0

definite; thus (3) follows from (2).

(2.55) Example. Diagonalize the symmetric bilinear form over R with matrix

A-

-2

3

1

0

3

-6

0

1

1

0

-3

1

0

1

1

-1

L

To do this, calculate that the determinants of the principal minors are bo(A) = 1,

61(A) = -2,

62(A) = 3,

63(A) = -3,

64(A) = -59

giving 3 sign changes, so this form diagonalizes to

-1

0

0

-1

0 0

0 0

0 0

-1

0 0 0

0

1

Although we do not use it here, we wish to remark on a standard notion. In the following definition 0 may be arbitrary (i.e., 0 need not be c-symmetric):

(2.56) Definition. Let 0 be a bilinear form on M such that ao : M --+ M*

is an isomorphism. Let f E EndR(M). Then fT : M -' M, the adt of f with respect to 0, is defined by (2.9)

.0(f (X)'

y) = O(x, fT (y))

for all x, y E M.

To see that this definition makes sense, note that it is equivalent to the equation (am(fT (y))) (x) = (am (y)) (f (x)) = (f'(ao(y))) (x),

or in other words,

ao o fT = f' o am,

i.e., fT = am 1 o f' o a0 where f' is the adt of f as given in Definition 1.20.

Note that fT and f' are quite distinct (although they both have the name "adt"). First, f' : M' - M', while fT : M -4 M. Second, f' is always defined, while fT is only defined once we have a form 0 as above, and it depends on 0. On the other hand, while distinct, they are certainly closely related. Now when it comes to finding a matrix representative for f T, there is a subtlety we wish to caution the reader about. There are two natural

376


choices. Choose a basis B for M. Then we have the dual basis B' of M*. Recall that [f']B. = ([f]a)t (Proposition 1.24), so

[f']s =

[a0']e* ([f]e)`

On the other hand, if B = {v;}, then we have a basis C' of M' given by C' = (am(v1)}. Then there is also a basis C dual to C' (using the canonical isomorphism between M and M**). By definition, I,,, the identity matrix, where n = rank(M), so we have more simply [IT ]8 = If *]C = ([I]C)` .

The point we wish to make is that in general the bases B' and C' (or equivalently, B and C) are distinct, so care must be taken. There is, however, one happy (and important) case where B' = C' and B = C. As the reader may check, this is true if and only if the basis B is orthonormal with respect to 0, i.e., if ¢(vi, vi) = b;,. In this case [cb]8 = I,,, so we see that 0 has an orthonormal basis if and only if ¢ is isometric to n(1], and we have seen situations when this is and is not possible.

6.3 Quadratic Forms This section will be devoted to some aspects of the theory of quadratic forms. As in the previous section, R will denote a commutative ring with 1 and all modules are assumed to be free. (3.1) Definition. Let M be a free R-module. A quadratic form

on M is a

function 4, : M - R satisfying

r24(x) for any (1) (2) the function 0: M x M

r E R, x E M; and R defined by

0(x, y) = Cx + y) - 4'(x) - Cy) is a (necessarily symmetric) bilinear form on M. In this situation we will say that 4 and 0 are associated. The quadratic form 0 is called non-singular or non-degenerate if the symmetric bilinear form 0 is. A basic method of obtaining quadratic forms is as follows: Let t/i be any

(not necessarily symmetric) bilinear form on M and define the symmetric bilinear form 0 on M by O(x, y) = '+,b(x, y) + V '(Y' x)

(Of course, if 0 is symmetric then 4,(x, y) = 2tp(x, y).) The function

6.3 Quadratic Forms

377

4':M-+R defined by

4i(x) = t'(x, x) is a quadratic form on M with associated bilinear form (3.2) Theorem. Let M be a free module over the ring R. (1) Let 4i be a quadratic form on M. Then the associated bilinear form 45

is uniquely determined by 4'. (2) Let 0 be a symmetric bilinear form on M. Then 45 is associated to a

quadratic form 4) if and only if QS is even (see Definition 2.40). (3)

If 2 is not a zero divisor in R and ' is an even symmetric bilinear form, then the associated quadratic form 4i is uniquely determined by 0.

Proof. Part (1) is obvious from Definition 3.1 (2) and is merely stated for emphasis. To prove (2), suppose that ' is associated to 4;. Then for every x E M, 44'(x) = 4'(2x) = 4'(x + x) = 24'(x) + 45(x, x). Thus, 45(x, x) = 24'(x)

(3.1)

and 0 is even. Conversely, suppose that 0 is even. Let B = {vi}gEJ be a basis for M and choose an ordering of the index set I. (Of course, if rank(M) = n is finite, then I = 11, ... , n} certainly has an order, but it is a consequence of Zorn's lemma that every set has an order.) To define O(x, y) it suffices to define ii(vi, vj) for i, j E I. Define i/i(vi, v3) as follows: ?P(vi, vj) = 0(vi, vi) if i < j, tI'(vi, vj) = 0 if j < i, and ?P(vi, vi) is any solution of the equation 2:fi(vi, vi) = QS(vi, vi). Since 0 is assumed to be even, this equation has a solution. Then 4'(x) = 1li(x, x) is a quadratic form with associated bilinear form 4'.

0

(3) This is a direct consequence of Equation (3.1).

(3.3) Lemma. (1) Let 4ii be a quadratic form on a module M1 over R with associated bilinear form 0j, for i = 1, 2. Then 4i : M1 (D M2 -+ R, defined by

4'(x1,x2) = 01(x1) +'02(x2), is a quadratic form on M1 ® M2 with associated bilinear form 451 102 (In this situation we write 4i = 4'1 1 4'2.) (2) Let 45i be an even symmetric bilinear form on Mi, i = 1, 2, and let 4i be a quadratic form associated to 451 1 452. Then 4i = 4i1 I. 4i2 where 4'i is a quadratic form associated to QSi, i = 1, 2. Also, 4'1 and 4'2 are unique. -

378


Proof. (1) First we check that 4'(r(xl, x2)) = 4'(rx1, rx2) = 4'1(rx1) + 4'2(rx2) = r24'1(xl) + r24'2(x2) = r2(4'1(x1) +'02(X2)) = r24'(x1, x2)

Next we need to check that (3.2) 4'(x1 + y1, x2 + 1/2) = 4'(x1, x2) + 4'(y1, Y2) + O((xl, x2), (y1, y2))

where 0 = 01 102. Since 0 = 01 1 02, we have 0((xl, x2), (y1, y2)) = O((xl, 0), (y1, 0)) + 0((x1, 0), (0, y2)) (3.3)

+ OR x2), (y1, 0)) + OR X2), (0, y2)) = 01(xl,y1) + 02(x2,y2)

On the other hand, 4'(x1 + Y1, X2 + y2) = 4'1(x1 + y1) + 4'2(x2 + y2)

4'1(x1) + (DI (y]) +01(xt,y1) + 4'2(x2) + 4'2(1/2) + 02(x2,1/2)

= 4'(x1. x2) + 4'(y1,y2) + 41(x1,1/1) + 02(x2,1/2),

which together with Equation (3.3) gives Equation (3.2), as desired. (2) Set 4'1 = 4'(,u, and 4'2 = 4'(tif,. Uniqueness is trivial, as 4' certainly determines 4'1 and 4'2.

We call 4' the orthogonal direct sum of 4'1 and 4'2. Thus Lemma 3.3 tells us that the procedures of forming orthogonal direct sums of associated bilinear and quadratic forms are compatible.

It follows from Theorem 3.2 that if 2 is not a zero divisor in R, the classification problem for quadratic forms over R reduces to that for even symmetric bilinear forms. (Recall that if 2 is a unit of R, then every symmetric bilinear form is even.) Thus we have already dealt with a number of important cases-R = R, R = C, or R a finite field of odd characteristic. We will now study the case of quadratic forms over the field F2 of 2 elements. In this situation it is common to call a quadratic form 4' associated to the even symmetric bilinear form 0 a quadratic refinement of 0, and we shall use this terminology. Note that in this case condition (1) of Definition 3.1 simply reduces to the condition that 4'(0) = 0, and this is implied by condition (2)---set y = 0. Thus, we may neglect condition (1). (3.4) Proposition. Let 0 be an even symmetric bilinear form on a module M over the field R = F2.

6.3 Quadratic Forms

379

(1) If'F1 and 4i2 are two quadratic refinements of 0, then

f(x) _'F1(x)+4'2(x) is an R-linear function f : M R, i.e., f E M*. (2) If 'F1 is any quadratic refinement of ¢ and f E M' is arbitrary, then

t2 ='F1 + f is also a quadratic refinement of 0. Proof. These are routine computations, which are left to the reader.

Suppose now that M is a module of finite rank n over R = F2. Then any even symmetric bilinear form ¢ on M has JM'1 = IMO = 2" quadratic refinements. We have already classified these forms in Corollary 2.46. If rank(m) = 2m (necessarily even), then 0 is isometric to

(n-2m)[0]1mI0 111

11

0

We will now see how to classify quadratic forms on M. By Lemma 3.3, we may handle the cases 0 identically zero and ¢ non-singular separately. (Of course, we are interested in classifying 0 up to isometry. An isometry has the analogous definition for quadratic forms as for bilinear forms: Two quadratic forms'Fi on R-modules Mi, i = 1, 2, are isometric if there is an R-isomorphism f : Ml -. M2 with 'F2(f (x)) _'F1(x) for every x E M,.) First we will deal with the case where ¢ is identically zero.

(3.5) Proposition. Let M be a free F2-module of rank n and let 0 be the identically zero bilinear form on M. Then a quadratic refinement 4> of 0 is simply an element 4V E M*. There are two isometry classes of these. One, containing one element, consists of 0 E M' . The other, containing 2" - 1 elements, consists of all'F E M' \ {0}.

Proof. In this case, Definition 3.1 (2) says 'F(x + y) = 4,(x) +'F(y), i.e., 'F : M -+ F2 is a homomorphism, so 'F E M. Clearly, the zero homomorphism is not isometric to any nonzero homomorphism. On the other

hand, a nonzero homomorphism 'F is uniquely determined by 'F ' (0), a subspace of M of rank n - 1, and there is an automorphism of M taking any one of these to any other. Of course, the case we are really interested in is the classification of quadratic refinements of a non-singular form. We prepare for this classification with the following elementary lemma. (3.6) Lemma. Define two function e : N the following recursions:

N and o : N

N by means of

380

Chapter 6. Bilinear and Quadratic Forms e(1) = 3,

o(1) = 1

e(m) = 3e(m - 1) + o(m - 1),

o(m) = 3o(m - 1) + e(m - 1)

form > 1. Then for every m E N,

e(m) = 2'-1(2'" + 1)

and

o(m)=2'-1(2m-l).

Proof. The proof is a routine induction.

0

(3.7) Proposition. Let 4) be a quadratic refinement of a non-singular even symmetric bilinear form on a module M of (necessarily) even rank n = 2m over F2. Then either (a) It-1(0)I = e(m) and I4D -1(1)I = o(m), or (b) I4 '(0)I = o(m) and I4-1(1)I = e(m).

Furthermore, among the 22m such quadratic forms, there are e(m) forms 4) with It-'(0)I = e(m) and o(m) forms $ with I$-1(0)I = o(m).

Proof. The proof is by induction on m. Let m = 1. Then M has a basis B = {x, y} in which [0IB = [° a], where 0 is the bilinear form associated to 4). Then M = {0, x, y, x + y}. Of course, $(0) = 0, and then from Definition 3.1 (2),

I = O(x, y) = 4)(x + y) +t(X) +t(y)

so we see that 4)(z) = 1 for z either exactly one of x, y, and x + y (3 possibilities) or for all three of them (1 possibility). Since, in this case, the above equation is the only condition on 4), all possibilities indeed occur. This is precisely the statement of the proposition in case m = 1.

Now suppose the proposition is true for rank(M) < 2m, and let rank(M) = 2m. Choose an element x 0 of M, let y E M with O(x, y) = 1 (which exists since ¢ is non-singular), let M1 be the subspace of M with

basis 8={x,y}, and let M2=Mi .Then M=M11M2,&=011m2, and 4) = t1 1 42, where O; = OIM, and t; = $IM,. Now 0 = +(x1, x2) = 4)(x1) + 4)(x2)

implies that either 4)(x1) = 4)(x2) = 0 or 4)(x1) = 4)(x2) = 1. If 41 and 4)2 are both as in part (a) (resp., both as in part (b)), the first (reap., second) case arises for e(1)e(m - 1) = 3e(m - 1) values of (xl, x2), and the other for o(1)o(m - 1) = o(m - 1) cases. Thus,

It' (0)I = 3e(m - 1) + o(m - 1) = e(m)

by Lemma 3.6. If $1 is as in (a) and t2 as in (b) (resp., vice-versa), then the first (resp., second) case arises for e(1)o(m - 1) = 3o(m - 1) values of (x1i x2), and the other for o(1)e(m - 1) cases. Thus, Lemma 3.6 again gives

6.3 Quadratic Forms

381

J--'(0)J = 3o(m - 1) + e(m - 1) = o(m). This proves the first claim of the proposition, but also the second, as we

see that 4' is in case (a) if either b1 and 4'2 are both in (a) or both in (b), and in case (b) if one of them is in (a) and the other in (b), and this gives exactly the same recursion. (We need not check o(m) separately since

e(m) + o(m) = 2" = (MI.)

0

Given this proposition we may define a "democratic" invariant of 4': We let each x E M "vote" for 4'(x) and go along with the majority. This is formalized in the following definition.

(3.8) Definition. Let 4' be a quadratic refinement of a non-singular even symmetric bilinear form on a module M of rank 2m over F2. The Arf invariant Arf(4') E F2 is defined by Arf(O) = 0 if l4-1(0)I = e(m) and I4' '(1)l = o(m) and

Arf(4') = 1

if 14-1(0)l = o(m) and I0-1(1)1 = e(m).

The form 0 is called even or odd according as Arf(0) = 0 or 1. Note that the proof of Proposition 3.7 yields the following:

(3.9) Corollary. Arf(4'1 14)2) = Arf(4'1) + Arf(4'2).

Proof

0

(3.10) Theorem. Let 4' be a quadratic refinement of a non-singular even symmetric bilinear form on a module M of rank 2m over F2. Then {4'} falls into two isometry classes. One, containing e(m) elements, consists of all 0 with Arf(O) = 0, and the other, containing o(m) elements, consists of all 0 with Arf(0) = 1. Proof. 10-1(0)1 is certainly an invariant of the isometry class of 4', so two forms with unequal Arf invariants cannot be isometric. It remains to prove

that all forms 0 with Arf(O) = 0 are isometric, as are all forms 0 with Arf(0) = 1. First let m = 1, and let 0 be the bilinear form associated to 0. Since is non-singular, there are x, y E M with O (x, y) = 1. Then O(x, x + y) = ¢(x, x) + O(x, y) = 1,

and similarly, 0(y, x + y) = 1. Thus, 0 is completely symmetric in x, y, and z = x + y. In the proof of Proposition 3.7, we saw that there are four possibilities for 4':


382

4,(x) = 1 4(x) = 0 4(x) = 0 O(x) = 1

0

4,(y) = 0

Cy) = 1

0

4(z) = I $(z) = 1.

4,(y) = 0

Cy) =1

The first three have Arf(4b) = 0; the last one has Arf($) = 1. But then it is easy to check that AutF, (M) S3, the permutation group on 3 elements, acting by permuting x, y, and z. We observed that 0 is completely symmetric in x, y, and z, so S3 leaves 0 invariant and permutes the first three possibilites for 4' transitively; hence, they are all equivalent. (As there is only one 4' with Arf(4') = 1, it certainly forms an equivalence class by itself.)

Now let m > 1. We have that 4' is isometric to 41 1 . . . 1 ',,, (by Lemma 3.3) and

Arf(4') = Arf(41) + -

+ Arf(4'm)

(by Corollary 3.9), so Arf(4') = 0 or 1 accordingly as there are an even number or an odd number of the forms 4'i with Arf(4'i) = 1. Each 4i has rank 2, and we have just seen that all rank 2 forms 4i with Arf(4'i) = 0 are isometric. Thus to complete the proof we need only show that if ' is the unique rank 2 form with 1, then T 1'P is isometric to +i 1 4D2 with Arf(4,) = 0 for i = 1, 2. Let 0 be the bilinear form associated to 4' _' 1 41 and let M have a basis B = {21, y1, x2, 112} in which [Ole _

0

1

1

0 0 0

0 0

0

0 0

0

0

1

0

1

(Such a basis is called symplectic.) Let f : M - M be defined by

f(xi) = x;,

f(yt) = yi,

where xi = Xi + x2,

Y = Yl,

X2'=X2,

1/2=111+112.

It is easy to check that f is an isometry of 0, i.e., that O(f (u), f (v)) _ ¢(u, v) for all u, v E M. (It suffices to check this for basis elements.) This implies that f is invertible, but in any case, direct computation shows that f2 = IM. Also, it is easy to check that

B' = {a4, yi, xz, yz} is a symplectic basis as well.

Then if Mi is the subspace of M spanned by {x,, y,}, i = 1, 2, then 0 = 0', 1 42' with 0, non-singular for i = 1, 2, and hence, 4, = 4 1 4Z, where V = 4'1 M, . But now

6.3 Quadratic Forms

383

1(xl) ='t(xl) = 4(xl +XAA2) = 4'(xl) + (x2) +W(xl, x2)

=1+1+0=0 VI(yi) _ Cyi) = 4(yl) = 1 (xi + yi) _ 4D(xi + yi)

_ ' (xi) + Cyi) + O(xi, yi )

=0+1+1=0, so 4 is indeed a form with

0. Similarly,

z(xx) = 1,

(Dz(xz + yz) = 0

0,

so that Arf(4Z) = 0, and we are done.

0

We have been careful to give an intrinsic definition of Arf (4))-one that does not depend on any choices. However, there is an alternate extrinsic definition, which is useful for calculations.

(3.11) Proposition. Let 0 be a non-singular even symmetric bilinear form on M, a module of rank 2m over F2, and let

B={x1,yl,...,xm,ym} be a 8ymplectic basis for M, i.e., a basis in which O(xi, y3) = bij

O(xi, x,) = O(yi, y,) = 0,

(1) If ' is a quadratic refinement of 0, then

i=1

(2) Let c = (a1 i b1, ... , a., bm) E (F2)2m be arbitrary. If v E M, then v = Ei ` 1(rix; + siyi), for ri, si E F2, and we define m m m -tc(v) _ risi + bisi. airi + i=1

i=1

i=1

The function 4 : M -- F2 is a quadratic refinement of 0, and m

Arf(4) _

aibi. i=1

Proof. Exercise.

384


(3.12) Remark. The reader should not get the impression that there is a canonical 4?, obtained by taking c = (0, ... , 0) of which the others are modifications; the formula for Lc depends on the choice of symplectic basis, and this is certainly not canonical.

We will now give a brief discussion of the concept of isometry groups. The definitions will be given in complete generality, and then we will spe-

cialize to the case of fields of characteristic # 2 and prove a theorem of Cartan and Dieudonne concerning the generation of isometry groups by reflections. The theory of reflections will also allow us to give a second (much easier) proof of Witt's theorem for quadratic forms over fields of characteristic # 2. (3.13) Definition. (1) Let 0 be an arbitrary b/s-linear form on a free module M over a ring

R. Then the isometry group Isom(q) is defined by Isom(0) =

If E AutR(M) : 0(f (x), f (y)) = OX, y) for all x, y E M}. (2) Let 4P be an arbitrary quadratic form on a free module M over a ring

R. Then the isometry group Isom(4)) is defined by

Isom(') = If E AutR(M) : 4?(f (x)) = 4i(x) for all x E M}.

Consider the situation where (D and 0 are associated. In many cases 1 and 0 determine each other, so Isom(4)) = Isom(¢). (In particular, this happens if R is a field of characteristic not equal to 2.) In the other cases, 1 determines 0, so Isom(4?) C Isom(0). There is a simple matrix criterion for an R-module isomorphism of M to be an isometry: (3.14) Proposition. Let 0 be an arbitrary b/s-linear form on a free module M of finite rank over a ring R, and let B be a basis of M. If f E AutR(M), then f E Isom(0) if and only if (3.4)

[f1e[01s[7]B = Ms.

Proof. If x, y E M, then we have Of W, f(y)) = [f(x))e(0)13 [f(y)]B = ([f]B[x]B)` [0113 ([f18[9113) (3.5)

But

= [X1 BI 01 Is [0]B[71 B)

[y)B.

6.3 Quadratic Forms (3.6)

385

¢(x, y) = [x]g[O]B[Y]13

for all x, y E M. Comparing Equation (3.5) and Equation (3.6) gives the result.

O

(3.15) Corollary. Let ¢ be a non-degenerate b/s-linear form on a free module

M over a ring R, and let f E EndR(M) be such that O(f (x), f (y)) = ¢(x, y)

for all

x, y E M.

Then f is an injection. Furthermore, if M has finite rank, then f is an isomorphism (and hence f E Isom(q5)).

Proof. Let 0 96 y E M. Since 0 is non-degenerate, there is an x E M with 4'(x, y) 36 0. Then 4'(f(x), f(y)) 0 0, so f(y) # 0. In case M has finite rank, the proof of Proposition 3.14 applies to show that Equation (3.4) holds. Since, by Proposition 2.21, det(4') is not a zero 0 divisor, det(f) is a unit, and so f is an isomorphism. (3.16) Remarks. (1)

If M has infinite rank then f need not be an isomorphism in the situation of Corollary 3.15. For an example, let M = Q°° = with ¢ the bilinear form on M given by o°

xiyi

O((xl, x2, ... ), (y1, y2, ... )) _ i=1

Then f : M

M, defined by M X1, x2, ... )) _ (0, x1, x2, ... ),

satisfies the hypothesis of Corollary 3.15, but it is not an isomorphism. (2) Some authors use the term "the orthogonal group of 0" for what we are calling the isometry group of 0, while other call the isometry group of 0 "the orthogonal/unitary/symplectic group of 0" when 0 is symmetric bilinear/Hermitian/skew-symmetric, so beware!

(3.17) Example. Let M = R2, and let 4'(x, y) = (x : y) be the standard inner product on M. Thus, if we use the standard basis on R2, then [4'] = 12, so f E EndR,(R2) is an isometry if and only if

[f]`[f]=12. Geometrically, this means that an isometry of R2 is determined by a pair of orthonormal vectors of R2, namely, the first and second columns of [f]. Hence, the isometries of R2 (with respect to the standard inner product) are one of the two types:


386

_ cos8 -sing Pe

sin 8

cos 0

sin 0 - cos 8

re/2 _ (cos 0 l sin 9

The isometry Pe is the counterclockwise rotation through an angle of 8, while

re/2 =Peoro is the orthogonal reflection of R2 through a line through the origin making an angle of 8/2 with the x-axis. (Check that this geometric description of re/2 is valid.) In particular, Equations (3.7) and (3.8) show that an isometry of R2 is a reflection if and only if its determinant is -1, while any isometry is a product of at most 2 reflections. The theorem of Cartan-Dieudonne is a far reaching generalization of this simple example. (3.18) Definition. Let R be a field with char(R) 54 2 and let 4) be a quadratic

form on an R-module M. Let M1 be a finite rank submodule of M with 4iM, non-singular. Set M2 = M', so M = Ml 1 M2. The reflection of M determined by MI is the element fm, E AutR(M) defined as follows:

Let x E M and write x uniquely as x = xl + x2, where xl E M1, X2 E M2. Then

far, (x) = -XI + x2.

If Ml = (y) with 4)(y) 0 0, then we write fm, = fy and call fy the hyperplane reflection determined by y.

(3.19) Lemma. Let M, 4), M1, and y be as in Definition 3.18, and let 0 be the symmetric bilinear form associated to it.

(1) fm, E Isom(4)), and (fM,)2 = 'Al. (2) If g E Isom(4'), then 9fM,9-1 = f9(nf,. (3) fy is given by the formula

fy(x) = x - 4)(y)) y

for all

x E M.

0

Proof. Exercise.

Since a hyperplane reflection f,,, is an isometry, it is certainly true that

+(x) = 4)(f,,,(x)). The following lemma gives a simple criterion for the existence of a hyperplane reflection that interchanges two given points.

(3.20) Lemma. Let 4) be a quadratic form on M over a field R with char(R) 96 2. Let x, y, and w = x - y E M satisfy 4)(x) = 4)(y) 96 0 Then fw(x) = y.

and +(w)

0.

6.3 Quadratic Forms

387

0

Proof. Exercise.

As an easy application of reflections, we will present another proof of Witt's theorem (Theorem 2.47). The difficult part of the proof of Theorem 2.47 was the n = 1 step in the induction. When reflections are available, this step is very easy.

(3.21) Theorem. (Witt) Let 1 be a quadratic form on a module M of finite rank over a field R of characteristic 54 2. Let N1 and N2 be submodules of M with $IN1 and DIN, non-singular and isometric to each other. Then 4DIN,. and tIN. are isometric.

Proof. As in the case of Theorem 2.47, the proof is by induction on n = rank(N1) = rank(N2). Thus, suppose that n = 1 and let v1 generate N1 and v2 generate N2. Since N1 and N2 are non-singular and isometric, we may assume that 4i(v1) = 4P(v2) # 0. Let 0 be the symmetric bilinear form associated to I . Then (D(v1 + v2) = 24'(v1) + ¢(v1, v2) and 4'(v1 - v2) =

2,D(v1) - 0(v1, v2). If both '(v1 + v2) and (v1 - v2) are zero, then it follows that t(v1) = 0. Thus, either 4(v1 +v2) -A 0 or 4 (v1-v2) # 0. Since N2 is also generated by -v2i we may thus assume that t(v1 - v2) 96 0. If w = v1 - v2 then Lemma 3.20 shows that the reflection f,,, takes v1 to v2, and hence it takes Ni to N2, and the theorem is proved in case n = 1.

The inductive step is identical with that presented in the proof of 0

Theorem 2.47 and, hence, will not be repeated.

The next two lemmas are technical results needed in the proof of the Cartan-Dieudonn6 theorem.

(3.22) Lemma. Let 0 be a non-singular symmetric bilinear form on the finite

rank R-module M, where R is a field of characteristic different from 2. If f E Isom(O), then

Ker(f - 1M) = (Im(f - 1M))1 . Proof. Suppose x E Ker(f - 1M) and y E M. Then f(x) = x and O(x, (f - 1M)(y)) =,O(x, f(y) - y) = O(x, f (y)) - q (x, y) = Of (x), f (y)) - O(x, y)

=0. Thus, x E (Im(f - 1M))l. Conversely, if x E (Im(f - 1M))1, then

388


-Xf (x) - x, f (y)) = O(f (x), f (y)) - O(x, f (y) = O(x, Y) - O(x, f (y)) = O(x, y - f(y))

=0.

Since f is invertible, this implies that f (x) - x E M1 = {0}, so x E Ker(f - 1M). (3.23) Corollary. With the notation of Lemma 3.22: (1) (Ker(f - 1M))1 = Im(f - 1M). (2) (f -1M)2 = 0 if and only if Im(f -1M) is a totally isotropic subspace.

Proof. (1) is immediate from Lemma 3.22. (2) Note that a subspace N C M is totally isotropic (Definition 2.22) if and only if N C N1. Then Im(f - 1M) is totally isotropic if and only if (3.9)

Im(f - 1M) S (Im(f - 1M))1 = Ker(f - 1M).

But Equation (3.9) is valid if and only if (f - 1M)2 = 0.

(3.24) Theorem. (Cartan-Dieudonne) Let 4D be a nonsingular quadratic form on a vector space M of dimension n over a field R with char(R) # 2. Then any g E Isom(c) is a product of < n hyperplane reflections. Proof. Let 0 be the associated bilinear form. The proof is by induction on n.

If n = 1, then Isom(4) = {±1}, where -1 is the unique reflection. Since 1 is a product of 0 reflections, the result is clear for n = 1. So assume the theorem holds for nonsingular quadratic forms on vector spaces of dimension < n, and let g E Isom(4D) = Isom(0). We will consider several cases.

Case 1. There exists x E Ker(g - 1) such that (D(x) 36 0.

Let N = (x). If y E Nl, then we have 0(g(y), x) = 0(g(y), g(x)) = O(y, x) = 0.

Thus, 91N-L E Isom(OINJ) and by induction 91N-L is a product of < n - 1 reflections of N-L. Each reflection of N' can be extended (by x H x) to a reflection of M and hence g is a product of < n - 1 reflections of M.

Case 2. There is x E M with D(x) $ 0 and D(x - g(x)) 54 0. By Lemma 3.20, if w = x - g(x), then there is a hyperplane reflection

f,,, E Isom(f) such that f,,,(g(x)) = x. Thus x E Ker(f o g - 1), and hence, by Case 1, f o g is a product of < n - 1 reflections. Therefore, g is a product of < n reflections. Case 3. dim(M) = 2.

6.3 Quadratic Forms

389

If O (x) 36 0 for all x 3& 0 E M, then this follows from the first two cases. Thus, suppose there exists x 0 0 E M with 4i(x) = 0. Choose y E M such that 4(x, y) 54 0, which is possible since 0 is non-singular. Since O(x, rx) = rq$(x, x), it is clear that B = {x, y} is a linearly independent subset of M and, hence, a basis. Replacing y by a multiple of y, we may assume that O(x, y) = 1. Furthermore, if r E R, then .O(y + rx, y + rx) = 0(y, y) + 2r4(x, y)

so by replacing x with a multiple of x we can also assume that ¢(y, y) = 0. That is, we have produced a basis B of M such that 0

1

1

0

If g E Isom(0), then [g]g = [: d] satisfies (Proposition 3.14)

[0

ab

[0

1

1

01 Lc

0J

or

[9)s = I b ,

This equation implies that [9)s = I 0 ao l ] Since

a [0

01] =

-0

a-[.0,

a

0

b]

1

1 0'

0] is the matrix of a reflection, for all b 34 0 E R. But if g(x) = by and g(y) = b-lx, then x + by E Ker(g - 1M) and Case 1 implies that g is a reflection. we are finished with Case 3 when we observe that

[ b_ 1

Case 4. dim(M) > 3, 4'(x) = 0 for all x E Ker(g - 1M), and whenever 4(x) 54 0, 4(x - 9(x)) = 0. Note that Case 4 simply incorporates all the situations not covered by Cases 1, 2, and 3. Our first goal is to show that in this situation, Im(g-IM) is a totally isotropic subspace. We already know (by hypothesis) that

45(x - g(x)) = 0

whenever

t(x) 54 0.

Thus, suppose that y 0 0 and '(y) = 0. We want to show that

4(y - 9(y)) = 0. Choose z E M with 0(y, z) # 0 and consider the two-dimensional subspace N = (y, z) with basis B = {y, z}. If i,b = GIN, then ('01B =

[

0(y, z)1 0 0(y, z) O(z, z) J

390


so V) is non-singular. Hence, we may write 0 = I >1" where tp' is nonsingular on N1 and dim N1 = dim M - 2 > 0. Thus, there exists x 0 0 E N1 with F(x) = 0 0, but ¢(x, y) = 0. Then O(y ± x, y) = 0 so that x) = F(x) 96 0.

The hypotheses of Case 4 then imply that

Cx - 9(x)) = 1((y + x) - 9(y + x)) = 4((y - x) - 9(y - x)) = 0, and from this we conclude that 4i(y - g(y)) = 0 and we have verified that Im(g - 1M) is a totally isotropic subspace of M. By Lemma 3.23, it follows that (g - IM)2 = 0. Hence, the minimum polynomial of g is (X - 1)2 (since g = 1M is a trivial case), and therefore, detg = 1 (Chapter 4, Exercise 55). Now Ker(g - 1M) is assumed to be totally isotropic (one of our hypotheses for Case 4). Thus,

Ker(g - 1M) C (Ker(g - 1M))1 = Im(9 - 1M) Since (g - 1)2 = 0, we have Im(g - 1,%f) C Ker(g - 1m) and conclude that Im(g - I m) = Ker(g - 1m). By the dimension formula (Propostion 3.8.8),

n = dim(Im(g - 1)) + dim(Ker(g - 1)) = 2dim(Ker(g - 1)), and hence, n is even. To complete the proof, suppose that g E satisfies the hypotheses of Case 4, and let f= be any hyperplane reflection. Then g' = fZ og is an isometry with det(g') = -1, and hence g' does not fall under Case 4, so it must be covered by Case I or Case 2. Thus, it follows that g' is a product of m < n reflections. Since m must be odd (in order to get a determinant

of -1) and n is even, it follows that m < n - 1. Therefore, g is a product of < n reflections and the proof is complete.

0

We conclude with some corollaries of this theorem.

(3.25) Corollary. Suppose dim M = n. (1) If g E Isom(4') is the product of r < n reflections, then

dim (Ker(g - 1M)) > n - r. (2) If Ker(g-1) = (0), then g cannot be written as a product of fewer than

n reflections.

Proof. (1) Let g = fy,

fy. and let Nj = Ker(fy, - 1). Then

Since dim N) = n - 1, it follows that (Proposition 3.8.10)

6.4 Exercises

391

n - r.

dim(Ni (2) This follows immediately from (1). (3.26) Corollary.

(1) If dim M = 2 then every isometry of determinant -1 is a reflection. (2) If dim M = 3 and g is an isometry of determinant 1, then g is the

product of 2 reflections. Proof. Exercise.

6.4 Exercises 1.

2.

Give an example of a field F with no/exactly one/more than one nontrivial conjugation. Let 0 be a b/s-linear form (or 4i a quadratic form) on a free R-module N and

let f : M - N be an R-module homomorphism. Show that f'(0) defined by

(f * (4,))(x, y) = Of (x), f (y)),

or f' (4i), defined by

(f*(4))(x) = 4V W), 3. 4.

is a b/s-linear (or quadratic) form on M. Let 0 be an even symmetric bilinear form over a PID R in which 2 is prime (e.g., R = Z) and suppose that det(4,) 2R. Show that rank(O) is even. Note that n 10 of is a form satisfying the conditions of Exercise 3, for any n. It is far from obvious that there are any other even symmetric non-singular forms over Z, but there are. Here is a famous example. Consider the form 0 over Z with matrix 1

0

0

0

0

2

1

0

0

0 0

0

1

0

0

1

2

1

0

0

0

0 0

0

0

1

2

1

0

0

0

0

0

0

1

2

1

0

0

0

0

0

1

2

1

0 0

0

0

0

1

2

0

0 0

0 0

0

1

0

0

2

2

E8 =

5.

1

a Show that det(4,) = 1. b Show that signature(4,) = 8, where 0 is regarded as a form over R. a) Let 0 be a b /s-linear form on a module M over a ring R. Suppose is an integral domain. If there is a subspace H C M with rank(H) > (1/2) rank(M), which is totally isotropic for 0, show that 0 is degenerate. b Suppose that 0 is either an even symmetric bilinear form on a module M over a ring R or a Hermitian form on a module M over a field R. If 0 is non-sin lar and there is a totally isotropic subspace H C M with rank(H) = (1/2) rank(M), show that 45 is isometric to

(rank(M)/2) [0

1 1


392 6.

Let 4; be a quadratic form associated to a bilinear form 0. Derive the following identities:

(a) 24D(x + y) = 24)(x, y) + 4)(x, x) + m(y, y)

(b) 4(x + y) + b(x - y) = 2($(x) + 4'(y)). The latter equation is known as the polarization identity. 7. Prove Proposition 2.5. 8. The following illustrates some of the behavior that occurs for forms that are not e-symmetric. For simplicity, we take modules over a field F. Thus 4) will

denote a form on a module M over F. (a) Define the right/left/two-sided kernel of 45 by

M°={yEM:0(x,y)=0 forallxEM} MM ={yEM:4)(y,x)=0 forallxEM} M°={yEM:4)(x,y)=m(y,x)=0 forallxEM}. Show that if rank(M) < oc, then M° = {0} if and only if M° = {0}.

Give a counterexample if rank(M) = oo. (b) Of course, M° = M° n Mi . Give an example of a form on M (with rank(M) < oo) where M° # {0}, Ml {0}, but M° = {0}. We say that 0 is right/left non-singular if Mr = 0}/M1 = {0} and non-singular if it is both right and left non-singular. (c) If N C M, define

NN ={yEM:4)(x,y)=0 forallxEN} with an analogous definition for N1. Let N1 = Nr n NIL. Give an

example where these three subspaces of M are all distinct. (d) Suppose that N C M is a subspace and is non-singular. Show that M = N ® N; and M = N ® N, L. Give an example where M # N ® N1. Indeed, give an example where N is a proper subspace of M but N1 = 101-

9.

Let R be a PID, M a free R-module of rank 4, and let 0 be a non-degenerate

skew-symmetric form on M. Show that 0 is classified up to isometry by det(m) and by

gcd{4)(v, w) : v, w E M}.

Let R be a PID, and let 0 be a skew-symmetric form on a free R-module M. If det(4)) is square-free, show that 0 is classified up to isometry by its rank. 11. Classify the following forms over Z (as in Theorem 2.35): 10.

(a)

(b)

12.

0

1

1

-1

2

1

-1 -1

0

5

-3

2

-3

-5

0

3

-7

1

3

0

6

4 5

-2

-2

7

1

3

-4

-6 -5

0

-1

-1

0

0 0 6 6 6 8

0

0 6 7 7

9

-3

-6 -6 -6 -8 -6 -7 -7 -9 0 -1 -5 -7 1

5 7

0 6 8

-6 -8 0 0

0 0

Find the signature of each of the following forms over R. Note that this also gives their diagonalization over R.

6.4 Exercises

40

40 50

9

12

5

(a)

2

(c)

-4 6

9 12 3

-4 13

-12

(b)

-11 3

-12

3

6

1

-12

(d)

3 4

18

(e)

13. 14.

5

-11

2

1

1

2

1

0 0

0 0 0 0

1

2

1

0 0 0

1

0

0 0

1

-2

0

0

3 9 5

0

0

0 0 1

0 0 0

1

-2

393

3

-12 0

4 5

0

1

Also, diagonalize each of these forms over Q. Carry out the details of the proof of Lemma 2.41. Analogous to the definition of even, we could make the following definition:

Let R be a PID and p a prime not dividing 2 (e.g., R = Z and p an odd

prime). A form 0 is p-ary if O(x, x) E pR for every x E M. Show that if 0 is p-ary, then O(x, y) E pR for every x, y E M. 15. Prove Proposition 3.4. 16. Prove Lemma 3.6. 17. Prove Proposition 3.11. 18. Prove Lemma 3.19. 19. Prove Lemma 3.20. 20.

Let f E AUtR(M) where R = R or C and where rank(M) < oo. If fk = Im

21.

for some k, prove that there is a non-singular form 0 on M with f E Isom(O). Diagonalize the following forms over the indicated fields: (a)

13

4

6

4 6

7

8

8

2

1

(b)

4

-1

4

over F2, F3, F5, and Q 1

-2

10

10

4

over F2, F3, F5, and Q

22. A symmetric matrix A = (a;2 J E M (R) is called diagonally dominant if laii l

a1+ ?

i#l for 1 < i < n. If the inequality is strict, then A is called strictly diagonally dominant. Let 0 be the bilinear form on R" whose matrix (in the standard basis) is A.

(a) If A is diagonally dominant, show that 0 is positive semidefinite, i.e., O(x, x) >0 for all x E R". (b) If A is strictly diagonally dominant, show that 0 is positive definite. 23. Let 0 be an arbitrary positive (or negative) semidefinite form. Show that is non-degenerate if and only if it is positive (negative) definite. 24. Let R be a ring with a (possibly trivial) conjugation. Show that

{P E GL(n, R) : Pt = (F)_'} is a subgroup of GL n, R). If R = R, with trivial conjugation, this group is called the orthogonal group and denoted O(n), and if R = C, with complex conjugation, it is called the unitary group and denoted U(n).

394 25.


Let A E M2(C) be the matrix of a Hermitian form, so

A= ab

b

c

with a, c E R. Find an explicit matrix P with P` = (P)-' such that PÀP is diagonal. 26.

Let F be an arbitrary subfield of C and 0 a Hermitian form on an F-module

M of finite rank. Show how to modify the Gram-Schmidt procedure to produce an orthogonal (but not in general orthonormal) basis of M, i.e., a

basis B={vl,...,vn} in which O(v;,v,)=0ifi54 j. 27.

Let R be a commutative ring and let A E MM(R) be a skew-symmetric matrix. If P E Mn(R) is any matrix, then show that PAP is also skewsymmetric and

Pf(PÀP) = det(P)Pf(A). 28.

Let A E be a Hermitian matrix. Show that A is positive definite if and only if A-1 is positive definite. More generally, show that the signature of A-' is the signature of A. (We say that A is positive definite if the associated Hermitian form 0(x, x) = x1Ax is positive definite.)

29.

If V is a real vector space, then a nonempty subset C C V is a cone if a, bEC=a+bECand C. Prove that the set of positive definite Hermitian matrices in Mn (C) is a cone.

30.

If A is a real symmetric matrix prove that there is a E R such that A+aIn is positive definite.

If A E M,n,n(C), show that A'A and AA' are positive semidefinite Hermitian matrices. 32. Let A, B E Mn(C) be Hermitian and assume that B is positive definite. Prove that the two Hermitian forms determined by A and B can be simultaneously diagonalized. That is, prove that there is a nonsingular matrix P 31.

such that P`BP = In and PAP = diag(A1, ... ,,\n). (Hint: B determines an inner product on C".)

Chapter 7

Topics in Module Theory

This chapter will be concerned with collecting a number of results and constructions concerning modules over (primarily) noncommutative rings that will be needed to study group representation theory in Chapter 8.

7.1 Simple and Semisimple Rings and Modules In this section we investigate the question of decomposing modules into "simpler" modules.

(1.1) Definition. If R is a ring (not necessarily commutative) and M 54 (0) is a nonzero R-module, then we say that M is a simple or irreducible Rmodule if (0) and M are the only submodules of M.

(1.2) Proposition. If an R-module M is simple, then it is cyclic.

Proof. Let x be a nonzero element of M and let N = (x) be the cyclic submodule generated by x. Since M is simple and N # (0), it follows that

M=N.

0

(1.3) Proposition. If R is a ring, then a cyclic R-module M = (m) is simple if and only if Ann(m) is a maximal left ideal. Proof. By Proposition 3.2.15, M 91 R/ Ann(m), so the correspondence theorem (Theorem 3.2.7) shows that M has no submodules other than M and (0) if and only if R has no submodules (i.e., left ideals) containing Ann(m) other than R and Ann(m). But this is precisely the condition for Ann(m) to be a maximal left ideal. 0 (1.4) Examples.

(1) An abelian group A is a simple Z-module if and only if A is a cyclic group of prime order.

Chapter 7. Topics in Module Theory

396

(2) The hypothesis in Proposition 1.3 that M be cyclic is necessary. The Z-module A = Z2 has annihilator 2Z but the module A is not simple. (3) Consider the vector space F2 (where F is any field) as an F(x]-module via the linear transformation T(ul, u2) = (u2i 0). Then F2 is a cyclic F[X]-module, but it is not a simple F[X]-module. Indeed,

but N = {(u, 0) : u E F} is an F(X]-submodule of F2. Thus the converse of Proposition 1.2 is not true. (4) Let V = R2 and consider the linear transformation T : V V defined by T(u, v) (-v, u). Then the R[X]-module VT is simple. To see

this let w = (ul, v1) # 0 E V and let N be the R[X]-submodule of VT generated by w. Then w E N and Xw = T(w) = (-vi, ul) E N. Since any (x, y) E V can be written as (x, y) = aw + QXw where a = (xul + yvl)/(ui + vi) and Q = (yul - xvl)/(ui + vi), it follows that N = VT and hence VT is simple. (5) Now let W = C2 and consider the linear transformation T : W -+ W defined by T(u, v) = (-v, u). Note that T is defined by the same formula used in Example 1.4 (4). However, in this case the C(X]-module

WT is not simple. Indeed, the C-subspace C (i, 1) is a T-invariant subspace of W, and hence, it is a C[X]-submodule of WT different from W and from (0). The following lemma is very easy, but it turns out to be extremely useful:

(1.5) Proposition. (Schur's lemma) (1) Let M be a simple R-module. Then the ring EndR(M) is a division

ring. (2)

If M and N are simple R-modules, then HomR(M, N) $ (0) if and only if M and N are isomorphic.

Proof. (1) Let f 96 0 E EndR(M). Then Im(f) is a nonzero submodule of M and Ker(f) is a submodule of M different from M. Since M is simple, it follows that Im(f) = M and Ker(f) = (0), so f is an R-module isomorphism and hence is invertible as an element of the ring EndR(M). (2) The same argument as in (1) shows that any nonzero homomor-

phism f : M N is an isomorphism. We have a second concept of decomposition of modules into simpler pieces, with simple modules again being the building blocks.

(1.6) Definition. If R is a ring (not necessarily commutative), then an Rmodule M is said to be indecomposable if it has no proper nontrivial com-

7.1 Simple and Semisimple Rings and Modules

397

plemented submodule M1, i.e., if M = M1 ® M2 implies that M1 = (0) or M1 = M.

If M is a simple R-module, then M is also indecomposable, but the converse is false. For example, Z is an indecomposable Z-module, but Z is not a simple Z-module; note that Z contains the proper submodule 2Z. One of the major classes of modules we wish to study is the following: (1.7) Definition. An R-module M is said to be semisimple if it is a direct sum of simple R-modules.

The idea of semisimple modules is to study modules by decomposing them into a direct sum of simple submodules. In our study of groups there was also another way to construct groups from simpler groups, namely, the extension of one group by another, of which a special case was the semidirect

product. Recall from Definition 1.6.6 that a group G is an extension of a group N by a group H if there is an exact sequence of groups 1

If this exact sequence is a split exact sequence, then G is a semidirect product of N and H. In the case of abelian groups, semidirect and direct products coincide, but extension of N by H is still a distinct concept. If G is an abelian group and N is a subgroup, then the exact sequence

(0)---N-pG-iH--,(0) is completely determined by the chain of subgroups (0) C N C G. By allowing longer chains of subgroups, we can consider a group as obtained by multiple extensions. We will consider this concept within the class of R-modules.

(1.8) Definition. (1) If R is a ring (not necessarily commutative) and M is an R-module, then a chain of submodules of M is a sequence {M;} o of submodules

of M such that (1.1)

(0)=M0 5M15M2

The length of the chain is n. (2)

We say that a chain (Njo is a refinement of the chain {M;}

o

if each M, is equal to N3 for some j. Refinement of chains defines a partial order on the set C of all chains of submodules of M. (3) A maximal element of C (if it exists) is called a composition series of M.

398


(1.9) Remarks.

(1) Note that the chain (1.1) is a composition series if and only if each of the modules M;/M;_1 (1 < i < n) is a simple module. (2) Our primary interest will be in decomposing a module as a direct sum of simple modules. Note that if M = ® 1 M; where M; is a simple R-module, then M has a composition series n

(0) 511115Ml®M25...5®M:=M. ;=1

On the other hand, if M = ®°_M;, then M does not have a composition series. In a moment (Example 1.10 (2)) we shall see an example of a module that is not semisimple but does have a composition se-

ries. Thus, while these two properties-semisimplicity and having a composition series-are related, neither implies the other. However, our main interest in composition series is as a tool in deriving results about semisimple modules.


Let D be a division ring and let M be a D-module with a basis

{xl,...,xm}. LetMo=(0)andfor1

Then {M;} o is a chain of submodules of length n, and since

1 i/Mi-l = (x1, ... ,xi)/(x1, ... ,x1_1) Dx, D,

we conclude that this chain is a composition series because D is a simple D-module. (2) If p is a prime, the chain

(0) 5 pzn' 5 Zpz is a composition series for the Z-module Zp2. Note that Z,2 is not semisimple as a Z-module since it has no proper complemented submodules.

(3) The Z-module Z does not have a composition series. Indeed, if {I,) !'o is any chain of submodules of length n, then writing I, = (a,), we can properly refine the chain by putting the ideal (2a1) between I, and to = (0). (4) If R is a PID which is not a field, then essentially the same argument as Example 1.10 (3) shows that R does not have a composition series as an R-module.


399

(1.11) Definition. Let M be an R-module. If M has a composition series let 1(M) denote the minimum length of a composition series for M. If M does not have a composition series, let 1(M) = oo. 1(M) is called the length of the R-module M. If 1(M) < oo, we say that M has finite length.

Note that isomorphic R-modules have the same length, since if f : M -+ N is an R-module isomorphism, the image under f of a composition series for M is a composition series for N. (1.12) Lemma. Let M be an R-module of finite length and let N be a proper

submodule (i.e., N # M). Then 1(N) < 1(M). Proof. Let (1.2)

be a composition series of M of length n = 1(M) and let Ni = NnM1 C N. Let 0 : Ni -+ Ms/M;_1 be the inclusion map Ni -+ M; followed by the projection map M; -4 M;/Mi_1. Since Ker(O) = N;_1, it follows from the first isomorphism theorem that N;/N;_1 is isomorphic to a submodule of M,/M;_1. But (1.2) is a composition series, so M;/Mi_1 is a simple Rmodule. Hence Ni = N;_1 or N;/N;_1 = Mt/M;_1 for i = 1, 2, ... ,n. By deleting the repeated of the sequence { Ni J !'O we obtain a composition series for the module N of length < n =1(M). Suppose that this composition series for N has length n. Then we must have N;/Ni_1 = M,/Mc_1

for all i = 1, 2, ... , n. Thus N1 = M1, N2 = M2,

..., N = M, i.e.,

N = M. Since we have assumed that N is a proper submodule, we conclude that the chain {N;} o has repeated , and hence, after deleting repeated we find that N has a composition series of length < 1(M), that is, 1(N) < 1(M). (1.13) Proposition. Let M be an R-module of finite length. Then every composition series of M has length n = 1(M). Moreover, every chain of submodules can be refined to a composition series. Proof. We first show that any chain of submodules of M has length < 1(M). Let

(0)=Ma5M1 be a chain of submodules of M of length k. By Lemma 1.12,

0 = 1(Ma) < 1(M1) < ... < 1(Mk) = 1(M). Thus, k:5 1(M). Now consider a composition series of M of length k. By the definition of composition series, k > 1(M) and we just proved that k < 1(M). Thus, k = 1(M). If a chain has length 1(M), then it must be maximal and, hence, is a composition series. If the chain has length < 1(M), then it is not a

400


composition series and hence it may be refined until its length is e(M), at which time it will be a composition series.

According to Example 1.10 (1), if D is a division ring and M is a D-module, then a basis S = {x1, ... , x,,} with n elements determines a composition series of M of length n. Since all composition series of M must have the same length, we conclude that any two finite bases of M must have

the same length n. Moreover, if M had also an infinite basis T, then M would have a linearly independent set consisting of more than n elements. Call this set {yl, ... , yk} with k > n. Then (0)5(yi)5(y1,Y2)5...5(yi,...,yk)5M

is a chain of length > n, which contradicts Proposition 1.13. Thus, every basis of M is finite and has n elements. We have arrived at the following result.

(1.14) Proposition. Let D be a division ring and let M be a D-module with a finite basis. Then every basis of M is finite and all bases have the same number of elements. Proof.

An (almost) equivalent way to state the same result is the following. It can be made equivalent by the convention that D°° refers to any infinite direct sum of copies of D, without regard to the cardinality of the index set.

(1.15) Corollary. If D is a division ring and D"` = D' then m = n. Proof.

We conclude our treatment of composition series with the following result, which is frequently useful in constructing induction arguments.

M L -p 0 be a short exact sequence of R-modules. If K and L are of finite length then so is M, and

(1.16) Proposition. Let 0 -+ K

e(M) = e(K) + e(L). Proof. Let

(0)=Ko5K15...5Kn=K

be a composition series of K, and let

(0)=Lo

5L15...5L,n=L


401

be a composition series for L. For 0 _< i < n, let Mi = q(Ki), and for n + 1 < i < n + m, let M. = O-1(Li_n) Then {Mi}n o'" is a chain of submodules of M and Mi/Mi_1

for 1 < i < n

Ki1Ki_1

for n+ l< i< n+ m

so that {M}°on is a composition series of M. Thus, £(M) = n + m.

O

(1.17) Example. Let R be a PID and let M be a finitely generated torsion R-module. We may write M as a finite direct sum of primary cyclic torsion modules: A;

M a` ®Rl(p:') i=1

Then it is an easy exercise to check that M is of finite length and k

C(M) _

ei. i=1

We now return to our consideration of semisimple modules. For this purpose we introduce the following convenient notation. If M is an R-module and s is a positive integer, then sM will denote the direct sum M ® ®M (s summands). More generally, if I' is any index set then FM will denote the R-module I'M = ®.yErM.y where M. = M for all y E F. Of course, if jF = s < oo then FM = sM, and we will prefer the latter notation. This notation is convenient for describing semisimple modules as direct sums of simple R-modules. If M is a semisimple R-module, then (1.3)

MQ5 ®Mi iEI

where Mi is simple for each i E I. If we collect all the simple modules in Equation (1.3) that are isomorphic, then we obtain (1.4)

M ®(I'aMa) aEA

MP if a ,l3) where {Ma}aEA is a set of pairwise distinct (i.e., M. simple modules. Equation (1.4) is said to be a simple factorization of the semisimple module M. Notice that this is analogous to the prime factorization of elements in a PID. This analogy is made even more compelling by the following uniqueness result for the simple factorization. (1.18) Theorem. Suppose that M and N are semisimple R-modules with simple factorizations

402


M

(1.5)

® (raMa)

aEA

and

N ® (AONO)

(1.6)

OEB

where {Ma}aEA and (N0}OEB are the distinct simple factors of M and N, respectively. If M is isomorphic to N, then there is a bijection V) : A B

such that Ma ^_' No(,,,) for all a E A. Moreover, Iral < oo if and only if IA,p(a)I < oo and in this case Iral = IAw(a)I Proof. Let ' : M - N be an isomorphism and let a E A be given. We may write M Ma®M' with M' = ®yEA\{a} (ryM.y)®raMa where rQ is ra

with one element deleted. Then by Proposition 3.3.15,

HomR(M, N) = HomR(Ma, N) ® HomR(M', N) (1.7)

= (®AO HomR(Ma, NO)) ® HomR(M', N). 8EB

By Schur's lemma, HomR(Ma, NO) = (0) unless M. ^_' NO. Therefore, in Equation (1.7) we will have HomR(Ma, N) = 0 or HomR(M0, N) AO HomR(M0, NO) for a unique 3 E B. The first alternative cannot occur since the isomorphism 0 : M - N is identified with (0 0 1-1, 10 0 1.2) where M is the injection). tl : M. --+ M is the canonical injection (and 12: M' If HomR(M,,, N) = 0 then 0 o t1 = 0, which means that 01m. = 0. This is impossible since 0 is injective. Thus the second case occurs and we define

0(a) _ 3 where HorR(M0, NO) # (0). Thus we have defined a function 0:A B, which is one-to-one by Schur's lemma. It remains to check that 0 is surjective. But given 0 E B, we may write N NO ® N'. Then

HomR(M, N) = HomR(M, NO) ® HomR(M, N') and

HomR(M, NO) = [ (11 HomR(Ma, NO)). aEA ra Since 0 is surjective, we must have HomR(M, NO) # (0), and thus, Schur's lemma implies that

HomR(M, NO) H Hom(M0, NO) ra

for a unique a E A. Then i (a) = /3, so ip is surjective. According to Proposition 3.3.15 and Schur's lemma, HomR(M, N)

fl ((DHomR(raM0, AONO)) aEA OEB

H HomR(r0M0, A,1,(a)N,p(a)). aEA


403

Therefore, 0 E HomR(M, N) is an isomorphism if and only if

0a = 0Ir,M0 : ram.

AG(a)Np(Q)

is an isomorphism for all a E A. But by the definition of ' and Schur's lemma, MQ is isomorphic to No(,,,). Also, I'QMQ has length Ir0I, and A,p(Q)N ,(.) has length IA ,(Q)I, and since isomorphic modules have the same 0 length, Il'a) I = IAO(Q) I, completing the proof.

(1.19) Corollary. Let M be a semisimple R-module and suppose that M has two simple factorizations

M - (D (I'QMQ) ® (ApNN) aEA

(3EB

with distinct simple factors {MQ}QEA and {Nfl}pEB. Then there is a bijec-

tion tli : A - B such that MQ N p(Q) for all a E A. Moreover, jr,,, I < 00 if and only if IA,(Q) I < oo and in this case II',, I = IAy(Q) I.

0

Proof. Take 0 = 1M in Theorem 1.18. (1.20) Remarks. (1)

While it is true in Corollary 1.19 that MQ - N,,(,) (isomorphism as R-modules), it is not necessarily true that MQ = N,,(0). For example, let R = F be a field and let M be a vector space over F of dimension s. Then for any choice of basis {ml, ... , m,} of M, we obtain a direct sum decomposition

Rm,. (2) In Theorem 1.18 we have been content to distinguish between finite and

infinite index sets rQ, but we are not distinguishing between infinite sets of different cardinality. Using the theory of cardinal arithmetic, one can refine Theorem 1.18 to conclude that II'QI = IA>G(a)I for all a E A, where ISI denotes the cardinality of the set S. We will now present some alternative characterizations of semisimple modules. The following notation, which will be used only in this section, will be convenient for this purpose. Let {Mi}iEI be a set of submodules of a module M. Then let

MI=I: M. iEI

be the sum of the submodules {Mil$E,. (1.21) Lemma. Let M be an R-module that is a sum of simple submodules {Mi}iEI, and let N be an arbitrary submodule of M. Then there is a subset

J C I such that

404


M'=N®(®Mi). iEJ

Proof. The proof is an application of Zorn's lemma. Let

S={PcI:Mp2, ®MiandMpnN=(0)}. iEP

Partially order S by inclusion and let C = {PQ}QEA be an arbitrary chain

in S. If P = UQEAPQ, we claim that P E S. Suppose that P 0 S. Since it is clear that Mp n N = (0), we must have that Mp ®iEPMI. Then Theorem 3.3.2 shows that there is some po E P, such that Mpo nMp' 0 (0), where P' = P\ {po}. Suppose that 0 96 x E Mpo nMp'. Then we may write (1.8)

x=xp,

where xp, 0 0 E Mp, for {pi, ... ,pk} C P. Since C is a chain, there is an index ct E A such that {po, pi, ... ,pie} C PQ. Equation (1.8) shows that Mpo 9t EIEP, MI, which contradicts the fact that PQ E S. Therefore, we must have P E S, and Zorn's lemma applies to conclude that S has a maximal element J. Claim.

N®((DiEJM1)

If this were not true, then there would be an index io E I such that Mio 0 N + Mj. This implies that Mio ¢ N and Mio ¢ M. Since Mio n N and M,, n Mj are proper submodules of Mio, it follows that M,o n N = (0) and MionMJ = (0) because Mio is a simple R-module. Therefore, {io}UJ E 0 S, contradicting the maximality of J. Hence, the claim is proved.

(1.22) Corollary. If an R-module M is a sum of simple submodules, then M is semisimple. Proof. Take N = (0) in Theorem 1.21.

0

(1.23) Theorem. If M is an R-module, then the following are equivalent:

(1) M is a semisimple module. (2) Every submodule of M is complemented. (3) Every submodule of M is a sum of simple R-modules.

(1) is immediate Proof. (1) . (2) follows from Lemma 1.21, and (3) from Corollary 1.22. It remains to prove (2) = (3). Let Ml be a submodule of M. First we observe that every submodule of Ml is complemented in Ml. To see this, suppose that N is any submodule of MI. Then N is complemented in M, so there is a submodule N' of M such


405

that N®N' M. But then N+(N'nM1) = M1 so that N® (N'nM1) 2e M1, and hence N is complemented in M1.

Next we claim that every nonzero submodule M2 of M contains a nonzero simple submodule. Let m E M2, m -A 0. Then Rm C M2 and, furthermore, R/ Ann(m) S5 Rm where Ann(m) = {a E R : am = 01 is a left ideal of R. A simple Zorn's lemma argument (see the proof of Theorem 2.2.16) shows that there is a maximal left ideal I of R containing Ann(m). Then Im is a maximal submodule of Rm by the correspondence theorem. By the previous paragraph, Im is a complemented submodule of Rm, so

there is a submodule N of Rm with N ® Im L Rm, and since Im is a maximal submodule of Rm, it follows that the submodule N is simple. Therefore, we have produced a simple submodule of M2.

Now consider an arbitrary submodule N of M, and let N1 C N be the sum of all the simple submodules of N. We claim that N1 = N. N1 is complemented in N, so we may write N S5 N1 ® N2. If N2 96 (0) then N2 has a nonzero simple submodule N', and since N' C N, it follows that N' C N1. But N1 n N2 = (0). This contradiction shows that N2 = (0), i.e., N = N1, and the proof is complete. (1.24) Corollary. Sums, submodules, and quotient modules of semisimple modules are semisimple. Proof. Sums: This follows immediately from Corollary 1.22. Submodules: Any submodule of a semisimple module satisfies condition (3) of Theorem 1.23. Quotient modules: If M is a semisimple module, N C M is a submod-

ule, and Q = M/N, then N has a complement N' in M, i.e., M ^_' N ®N'. But then Q N', so Q is isomorphic to a submodule of M, and hence, is semisimple.

(1.25) Corollary. Let M be a semisimple R-module and let N C M be a submodule. Then N is irreducible (simple) if and only if N is indecomposable.

Proof. Since every irreducible module is indecomposable, we need to show

that if N is not irreducible, then N is not indecomposable. Let N1 be a nontrivial proper submodule of N. Then N is semisimple by Corollary 1.24, so N1 has a complement by Theorem 1.23, and N is not indecomposable. 0

(1.26) Remark. The fact that every submodule of a semisimple R-module M is complemented is equivalent (by Theorem 3.3.9) to the statement that whenever M is a semisimple R-module, every short exact sequence

O -+ N -+ M ---. K ---p 0 of R-modules splits.

406


(1.27) Definition. A ring R is called semisimple if R is semisimple as a left R-module.

Remark. The proper terminology should be "left semisimple," with an analogous definition of "right semisimple," but we shall see below that the two notions coincide.

(1.28) Theorem. The following are equivalent for a ring R:

(1) R is a semisimple ring. (2) Every R-module is semisimple. (3) Every R-module is projective.

Proof. (1) . (2). Let M be an R-module. By Proposition 3.4.14, M has a free presentation

0-+K

F--yM---40

so that M is a quotient of the free R-module F. Since F is a direct sum of copies of R and R is assumed to be semisimple, it follows that F is semisimple, and hence M is also (Corollary 1.24). (2) (3). Assume that every R-module is semisimple, and let P be an arbitrary R-module. Suppose that (1.9)

is a short exact sequence. Since M is an R-module, our assumption is that it is semisimple and then Remark 1.26 implies that sequence (1.9) is split exact. Since (1.9) is an arbitrary short exact sequence with P on the right, it follows from Theorem 3.5.1 that P is projective. (1). Let M be an arbitrary submodule of R (i.e., an arbitrary (3) left ideal). Then we have a short exact sequence

Since all R-modules are assumed projective, we have that RIM is projective, and hence (by Theorem 3.5.1) this sequence splits. Therefore, M ® N for some submodule N C R, which is isomorphic (as an R R-module) to RIM. Then by Theorem 1.23, R is semisimple.

0

(1.29) Corollary. Let R be a semisimple ring and let M be an R-module. Then M is irreducible (simple) if and only if M is indecomposable. Proof.

11

(1.30) Theorem. Let R be a semisimple ring. Then every simple R-module is isomorphic to a submodule of R.

Proof. Let N be a simple R-module, and let R = ®IE!Mi be a simple factorization of the semisimple R-module R. We must show that at least


407

one of the simple R-modules M; is isomorphic to N. If this is not the case, then

HomR ((DM,, N) = II HomR(Mi, N) = (0)

HOMR(R, N)

iEI

iEI

where the last equality is because HomR(Mi, N) = (0) if Mi is not isomorphic to N (Schur's lemma). But HomR(R, N) N 0 (0), and this contradiction shows that we must have N isomorphic to one of the simple

0

submodules Mi of R. (1.31) Corollary. Let R be a semisimple ring.

(1) There are only finitely many isomorphism classes of simple R-modules. (2) If {Ma}QEA is the set of isomorphism classes of simple R-modules and

R = ®(I'aMQ), oEA

then each ra is finite. Proof. Since R is semisimple, we may write

R= ®Np OEB

where each No is simple. We will show that B is finite, and then both finiteness statements in the corollary are immediate from Theorem 1.30. Consider the identity element 1 E R. By the definition of direct sum, we have

1= 1: rpnp OE B

for some elements rp E R, np E N0, with all but finitely many rp equal to zero. Of course, each Np is a left R-submodule of R, i.e., a left ideal.

Now suppose that B is infinite. Then there is a go E B for which r00 = 0. Let n be any nonzero element of Npo. Then

n = n 1 = n(E rpnp) _

F,

(nrp)np,

pEB\{po}

pEB so

nE ® Np. pEB\{po}

Thus,

n E No. n( @ No) _ {0}, pEB\{po}


408

by the definition of direct sum again, which is a contradiction. Hence, B is finite.

We now come to the basic structure theorem for semisimple rings.

(1.32) Theorem. (Wedderburn) Let R be a semisimple ring. Then there is a finite collection of integers n1, ... , nk, and division rings D1, ... , Dk such that k

R

(DEndD,(D°`). i=1

Proof. By Corollary 1.31, we may write k

R®niMi i=1

where {M,}k 1 are the distinct simple R-modules and nl, ... , nk are positive integers. Then R is anti-isomorphic to R°P, and R°P = EndR(R) k

k

(

HomR (® niMi, ® niMil/ i=1

i=1

k

®HomR(niMM, niMi) i=1 k

^_' ®EndR(niM,), i=1

by Schur's lemma. Also, by Schur's lemma, EndR(MM) is a division ring,

which we denote by Ei, for each i = 1, ..., k. Then it is easy to check (compare the proof of Theorem 1.18) that EndR(n1Mi)

EndE.(E;')

Setting Di = E,'P, the proof is completed by observing that EndE, (E; ") is anti-isomorphic to EndD, (Dr ). Remark. Note that by Corollary 4.3.9, EndD(D") is isomorphic to M,, (DP). Thus, Wedderburn's theorem is often stated as, Every semisimple ring is isomorphic to a finite direct sum of matrix rings over division rings.

(1.33) Lemma. Let D be a division ring and n a positive integer. Then R = EndD(D") is semisimple as a left R-module and also as a right Rmodule. Fhrthermore, R is semisimple as a left D-module and as a right D-module.


409

Proof. Write D' = Dl ®D2 ® ® D,, where Di = D. Let

Mi = If E EndD(D") : Ker(f) 2 (DD, k#i

Nj = {f E EndD(D") : Im(f) C Dj}, and let

Pij=MinN.. Note that Pij =' D°p. Then

EndD(D") `= Ml ®... ®M,, as a left R-module, and EndD(D")2?5

as a right R-module. We leave it to the reader to check that each Mi (resp., Nj) is a simple left (resp., right) R-module. Also,

EndD(D") = ® Pij as a right (resp., left) D-module, and each Pij is certainly simple (on either side).

(1.34) Corollary. A ring R is semisimple as a left R-module if and only if it is semisimple as a right R-module. Proof. This follows immediately from Theorem 1.32 and Lemma 1.33.

Observe that R is a simple left R-module (resp., right R-module) if and only if R has no nontrivial proper left (resp., right) ideals, which is the case if and only if R is a division algebra. Thus, to define simplicity of R in this way would bring nothing new. Instead we make the following definition:

(1.35) Definition. A ring R with identity is simple if it has no nontrivial proper (two-sided) ideals.

Remark. In the language of the next section, this definition becomes "A ring R with identity is simple if it is simple as an (R, R)-bimodule." (1.36) Corollary. Let D be a division ring and n a positive integer. Then EndD(D") is a simple ring that is semisimple as a left Conversely, if R is a simple ring that is semisimple as a left R-module, or, equivalently, as a right R-module, then R

EndD(D")

for some division ring D and positive integer n.

410


Proof. We leave it to the reader to check that Endo(D's) is simple (compare Theorem 2.2.26 and Corollary 2.2.27), and then the first part of the corollary follows from Lemma 1.33. Conversely, if R is semisimple we have the decomposition given by Wedderburn's theorem (Theorem 1.32), and then the condition of simplicity forces k = 1.

Our main interest in semisimple rings and modules is in connection with our investigation of group representation theory, but it is also of interest to reconsider modules over a PID from this point of view. Thus let R be a PID. We wish to give a criterion for R-modules to be semisimple. The following easy lemma is left as an exercise. (1.37) Lemma. Let R be an integral domain. Then R is a semisimple ring if and only if R is a field. If R is a field, R is simple. Proof. Exercise.

From this lemma and Theorem 1.28, we see that if R is a field, then every R-module (i.e., vector space) is semisimple and there is nothing more to say. For the remainder of this section, we will assume that R is a PID that is not a field. Let M be a finitely generated R-module. Then by Corollary 3.6.9, we

have that M F ® M where F is free (of finite rank) and Mr is the torsion submodule of M. If F 76 (0) then Lemma 1.37 shows that M is not semisimple. It remains to consider the case where M = M i.e., where M is a finitely generated torsion module. Recall from Theorem 3.7.13 that each such M is a direct sum of primary cyclic R-modules. (1.38) Proposition. Let M be a primary cyclic R-module (where R is a PID is not a field) and assume that Ann(M) = (pe) where p E R is a prime. If e = 1 then M is simple. If e > 1, then M is not semisimple.

Proof. First suppose that e = 1, so that M°5 R/(p). Then M is a simple R-module because (p) is a prime ideal in the PID R, and hence, it is a maximal ideal.

Next suppose that e > 1. Then (0) -A p`-1 M 5 M,

and pe-! M is a proper submodule of M, which is not complemented; hence, M is not semisimple by Theorem 1.23 (2).

(1.39) Theorem. Let M be a finitely generated torsion R-module (where R is a PID that is not a field). Then M is semisimple if and only if me(M)


411

(see Definition 3.7.8) is a product of distinct prime factors. M is a simple R-module if and only if

me(M) = co(M) = (p) where p E R is a prime.

Proof. First suppose that M is cyclic, and me(M) = (pi' ... pr). Then the primary decomposition of M is given by MSM (R/(pi'))ED ...(D (Rl

k'')),

and M is semisimple if and only if each of the summands is semisimple, which by Proposition 1.38, is true if and only if ei=e2=...=ek=1.

Now let M be general. Then by Theorem 3.7.1, there is a cyclic decomposition

Rw such that Ann(wi) = (si) and si si+l for 1 < i < n - 1. Then M is [

semisimple if and only if each of the cyclic submodules Rwi is semisimple, which occurs (by the previous paragraph) if and only if si is a product of distinct prime factors. Since si I si+i, this occurs if and only if s = me(M) is a product of distinct prime factors. The second assertion is then easy to .

0

(1.40) Remark. In the two special cases of finite abelian groups and linear transformations that we considered in some detail in Chapters 3 and 4, Theorem 1.39 takes the following form: (1) A finite abelian group is semisimple if and only if it is the direct product of cyclic groups of prime order, and it is simple if and only if it is cyclic of prime order.

(2) Let V be a finite-dimensional vector space over a field F and let T : V -+ V be a linear transformation. Then VT is a semisimple F[XJmodule if and only if the minimal polynomial mT(X) of T is a product of distinct irreducible factors and is simple if and only if its characteristic polynomial cT(X) is equal to its minimal polynomial mT(X), this polynomial being irreducible (see Lemma 4.4.11.) If F is algebraically closed (so that the only irreducible polynomials are linear ones) then VT is semisimple if and only if T is diagonalizable and simple if and only if V is one-dimensional (see Corollary 4.4.32).

412


7.2 Multilinear Algebra We have three goals in this section: to introduce the notion of a bimodule, to further our investigation of "Hom," and to introduce and investigate tensor

products. The level of generality of the material presented in this section is dictated by the applications to the theory of group representations. For this reason, most of the results will be concerned with modules over rings that are not commutative; frequently there will be more than one module structure on the same abelian group, and many of the results are concerned with the interaction of these various module structures. We start with the concept of bimodule.

(2.1) Definition. Let R and S be rings. An abelian group M is an (R, S)bimodule if M is both a left R-module and a right S-module, and the compatibility condition

r(ms) = (rm)s

(2.1)

is satisfied for every r E R, m E M, and s E S. (2.2) Examples. (1)

(2)

Every left R-module is an (R, Z)-bimodule, and every right S-module is a (Z, S)-bimodule.

If R is a commutative ring, then every left or right R-module is an (R, R)-bimodule in a natural way. Indeed, if M is a left R-module, then according to Remark 3.1.2 (1), M is also a right R-module by means of the operation r,ir = rm. Then Equation (2.1) is

r(ms) = r(sm) = (rs)m = (sr)m = s(rm) = (rm)s. (3) If T is a ring and R and S are subrings of T (possibly with R = S = T),

then T is an (R, S)-bimodule. Note that Equation (2.1) is simply the associative law in T. (4) If M and N are left R-modules, then the abelian group HomR(M, N)

has the structure of an (EndR(N), EndR(M))-bimodule, as follows. If

f E HomR(M, N), ¢ E EndR(M), and V) E EndR(N), then define f ¢ = f o 0 and f = o f . These definitions provide a left EndR(N)module and a right EndR(M)-module structure on HomR(M, N), and Equation (2.1) follows from the associativity of composition of functions. (5)

Recall that a ring T is an R-algebra, if T is an R-module and the Rmodule structure on T and the ring structure of T are compatible, i.e.,

r(tit2) = (rtl)t2 = ti(rt2) for all r E R and t1, t2 E T. If T happens to be an (R,S)-bimodule, such that r(tlt2) = (rtl)t2 = tl(rt2) and (tlt2)s = tl(t2s) = (tls)t2 for all r E R, s E S, and t1, t2 E T, then we

7.2 Multilinear Algebra

413

say that T is an (R, S)-blalgebra. For example, if R and S are subrings of a commutative ring T, then T is an (R, S)-bialgebra. Suppose that M is an (R, S)-bimodule and N C M is a subgroup of the additive abelian group of M. Then N is said to be an (R, S)-bisubmodule of M if N is both a left R-submodule and a right S-submodule of M. If Ml and M2 are (R, S)-bimodules, then a function f : Ml M2 is an (R, S)-bimodule homomorphism if it is both a left R-module homomorphism and a right S-module homomorphism. The set of (R, S)-bimodule homomorphisms will be denoted Hom(R,S)(M1, M2). Since bimodule homomorphisms can be added, this has the structure of an abelian group,

but, a priori, nothing more. If f : Ml -+ M2 is an (R, S)-bimodule homomorphism, then it is a simple exercise to check that Ker(f) C Ml and Im(f) C M2 are (R, S)-bisubmodules. Furthermore, if N C M is an (R, S)-bisubmodule, then the quotient abelian group is easily seen to have the structure of an (R, S)-bimodule. We leave it as an exercise for the reader to formulate and the noether isomorphism theorems (see Theorems 3.2.3 to 3.2.6) in the context of (R, S)bimodules. It is worth pointing out that if M is an (R, S)-bimodule, then there are three distinct concepts of submodule of M, namely, R-submodule, S-submodule, and (R, S)-bisubmodule. Thus, if X C M, then one has three

concepts of submodule of M generated by the set X. To appreciate the difference, suppose that X = {x} consists of a single element x E M. Then the R-submodule generated by X is the set

Rx = {rx : r E R},

(2.2)

the S-submodule generated by X is the set XS = {X8: s E S},

(2.3)

while the (R, S)-bisubmodule generated by X is the set (2.4)

RxS={>r;xsj:nENandr1ER,siESfor1
(2.3) Examples. (1)

If R is a ring, then a left R-submodule of R is a left ideal, a right R-submodule is a right ideal, and an (R, R)-bisubmodule of R is a (two-sided) ideal.

(2) As a specific example, let R = M2(Q) and let x = [a R-submodule of R generated by {x} is

Rx= l[b 0] :a,bEQ},

o]. Then the left


414

the right R-submodule of R generated by {x} is

xR = { [a0

]:abEQ},

b 01

while the (R, R)-bisubmodule of R generated by {x} is R itself (see Theorem 2.2.26).

When considering bimodules, there are (at least) three distinct types of homomorphisms that can be considered. In order to keep them straight, we will adopt the following notational conventions. If M and N are left R-modules (in particular, both could be (R, S)-bimodules, or one could be an (R, S)-bimodule and the other a (R, T)-bimodule), then HomR(11/, N) will denote the set of (left) R-module homomorphisms from M to N. If M and N are right S-modules, then Hom_S(M, N) will denote the set of all (right) S-module homomorphisms. If M and N are (R, S)-bimodules, then Hom(R,s)(M, N) will denote the set of all (R, S)-bimodule homomorphisms from M to N. With no additional hypotheses, the only algebraic structure that can be placed upon these sets of homomorphisms is that of abelian groups, i.e., addition of homomorphisms is a homomorphism in each situation described. The first thing to be considered is what additional structure is available.

(2.4) Proposition. Suppose that Al is an (R, S)-bimodule and N is an (R, T)-bimodule. Then HomR(IVI, N) can be given the structure of an (S, T) -bimodule.

Proof. We must define compatible left S-module and right T-module struc-

tures on HomR(M, N). Thus, let f E HomR(M, N), s E S, and t E T. Define s f and ft as follows: (2.5)

s f (m) = f(ms) for all m E M

and (2.6)

f t(m) = f (m)t for all m E M.

We must show that Equation (2.5) defines a left S-module structure on HomR(M, N) and that Equation (2.6) defines a right T-module structure on HomR(M, N), and we must the compatibility condition s(ft) _ (s f )t.

We first that s f t, is an R-module homomorphism. To see this, suppose that rl, r2 E R, m1i m2 E M and note that

sf t(rimr + r2m2) = f((rlrnl +.r2m'2)s)t = f ((rime)s + (r2m2)s)t = f (rl (ml s) + r2(m2s))t = (ref (mis) + ref (m2s)) t


415

= (rif(mts))t+ (r2f(m2s))t = rj (f (mis)t) + r2(f (mis)t) = ri (sf t)(mi) + r2(sft)(m2), where the third equality follows from the (R, S)-bimodule structure on M, while the next to last equality is a consequence of the (R, T)-bimodule structure on N. Thus, s f t is an R-module homomorphism for all s E S, t E T, and f E HomR(M, N). Now observe that, if s1i 32 E S and m E M, then (s1(s2f )) (m) = (s2f)(ms1) = f((ms1)s2) = f(m(s1s2))

= ((s1s2)f) (m)

so that HomR(M, N) satisfies axiom (ci) of Definition 3.1.1. The other axioms are automatic, so HomR(M, N) is a left S-module. Similarly, if t1i

t2 ETandmEM, then ((ft1)t2) (m) = ((ft1)(m))t2 = (f (m)t1) t2

= f(m)(t1t2) = (f(t1t2)) (m) Thus, HomR(M, N) is a right T-module by Definition 3.1.1 (2). We have only checked axiom (Cr), the others being automatic. It remains to check the compatibility of the left S-module and right T-module structures. But, if s E S, t E T, f E HomR(M, N), and m E M, then ((s f )t) (m) = (s f)(m)t = f (ms)t = (f t)(ms) = s(f t)(m).

Thus, (s f )t = s(f t) and HomR(M, N) is an (S, T)-bimodule, which completes the proof of the proposition. 0 Proved in exactly the same way is the following result concerning the bimodule structure on the set of right R-module homomorphisms. (2.5) Proposition. Suppose that M is an (S, R)-bimodule and N is a (T, R)bimodule. Then Hom_R(M, N) has the structure of a (T, S)-bimodule, via the module operations

(t f)(m) = t(f (m)) and (f s)(m) = f (sm) where s E S, t E T, f EHom_R(M,N), and in E M. Proof. Exercise.

0


416

Some familiar results are corollaries of these propositions. (Also see Example 3.1.5 (10).) (2.6) Corollary. (1) If M is a left R-module, then M' = HomR(M, R) is a right R-module. (2) If M and N are (R, R)-bimodules, then HomR(M, N) is an (R, R)-

bimodule, and EndR(M) is an (R, R)-bialgebra. In particular, this is the case when the ring R is commutative.

0

Proof. Exercise.

Remark. If M and N are both (R, S)-bimodules, then the set of bimodule homomorphisms Hom(R,S)(M, N) has only the structure of an abelian group.

Theorem 3.3.10 generalizes to the following result in the context of bimodules. The proof is identical, and hence it will be omitted. (2.7) Theorem. Let

0-iM1 0+M 0-+M2

(2.7)

be a sequence of (R, S)-bimodules and (R,S)-bimodule homomorphisms. Then the sequence (2.7) is exact if and only if the sequence

0 -+ HomR(N, Ml)

(2.8)

HomR(N, M) V.+ HomR(N, M2)

is an exact sequence of (T, S)-bimodules for all (R, T)-bimodules N.

If (2.9)

Ml 0-+ M -i M2 -i 0

is a sequence of (R, S)-bimodules and (R, S)-bimodule homomorphisms, then the sequence (2.9) is exact if and only if the sequence (2.10)

0 -i HomR(M2i N)

HomR(M, N)

HomR(Ml, N)

is an exact sequence of (S, T)-bimodules for all (R, T) -bimodules N.

0

Proof.

Similarly, the proof of the following result is identical to the proof of Theorem 3.3.12. (2.8) Theorem. Let N be a fixed (R, T) -bimodule. If (2.11)

miM_M2-+0


417

is a split short exact sequence of (R, S)-bimodules, then

(2.12) 0 - HomR(N, M1) - HomR(N, M) 'P'+ HomR(N, M2)

0

is a split short exact sequence of (T, S)-bimodules, and

(2.13) 0 -+ HomR(M2i N)

HomR(M, N) m-+ HomR(Ml, N)

0

is a split short exact sequence of (S, T) -bimodules.

0

Proof

This concludes our brief introduction to bimodules and module structures on spaces of homomorphisms; we turn our attention now to the concept of tensor product of modules. As we shall see, Hom and tensor products are closely related, but unfortunately, there is no particularly easy definition of tensor products. On the positive side, the use of the tensor product in practice does not usually require an application of the definition, but rather fundamental properties (easier than the definition) are used.

Let M be an (R, S)-bimodule and let N be an (S, T)-bimodule. Let F be the free abelian group on the index set M x N (Remark 3.4.5). Recall that this means that F = ®(m,n)EMxNZ(m,n) where Z(m,n) = Z for all (m, n) E M x N, and that a basis of F is given by S = {e(m,n)}(m,n)EMxN where e(m,n) _ (bmkbnt)(k,t)EMxN, that is, e(m,n) = 1 in the component of F corresponding to the element (m, n) E M x N and e(m,n) = 0 in all other components. As is conventional, we will identify the basis element e(m,n)

with the element (m, n) E M x N. Thus a typical element of F is a linear combination

E

C(m.n)('m,n)

(m,n)EMxN

where c(m,n) E Z and all but finitely many of the integers c(m,n) are 0. Note

that F can be given the structure of an (R, T)-bimodule via the multiplication k

(2.14)

r E ci(mi, ni)

k

t = > ce(rmei net)

e=1

i_1

where rER,tET,andc1,...,ckEZ. Let K C F be the subgroup of F generated by the subset Hl U H2 U H3 where the three subsets H1, H2, and H3 are defined by

H1={(ml+m2,n)-(m1,n)-(m2,n)m1,m2EM,nEN} H2={(m,ni+n2)-(m,ni)-(m,n2):mEM,n1,n2EN} H3={(ms,n)-(m,sn):mEM, nEN, aES}.


418

Note that K is an (R, T)-submodule of F using the bimodule structure given by Equation (2.14). With these preliminaries out of the way, we can define the tensor prod-

uct of M and N. (2.9) Definition. With the notation introduced above, the tensor product of the (R, S)-bimodule M and the (S, T)-bimodule N, denoted M ®s N, is the quotient (R,T)-bimodule

M®sN=F/K. If 7r

: F - F/K is the canonical projection map, then we let m ®s n =

ir((m, n)) for each (m, n) E M x N C F. When S is clear from the context we will frequently write m ®n in place of m ®s n.

Note that the set (2.15)

{m ®s n : (m, n) E M x N}

generates M ®s N as an (R, T)-bimodule, but it is important to recognize that M ®s N is not (in general) equal to the set in (2.15). Also important to recognize is the fact that m ®s n = (m, n) + K is an equivalence class, so that m ® n = m' O n' does not necessarily imply that m = m' and n = n'. As motivation for this rather complicated definition, we have the following proposition. The proof is left as an exercise.

(2.10) Proposition. Let M be an (R,S)-bimodule, N an (S,T)-bimodule, and let m, m; E M, n, n; E N, and s E S. Then the following identities hold in M ®s N. (2.17)

(m1+m2)®n=m1®n+m2®n m®(nl +n2) = m®nl +m®n2

(2.18)

ms ®n = m ®sn.

(2.16)

Proof. Exercise.

0

Indeed, the tensor product M ®s N is obtained from the cartesian product M x N by "forcing" the relations (2.16)-(2.18), but no others, to hold. This idea is formalized in Theorem 2.12, the statement of which requires the following definition.

(2.11) Definition. Let M be an (R, S) -bimodule, N an (S,T)-bimodule, and

let M x N be the cartesian product of M and N as sets. Let Q be any (R,T)-bimodule. A map g : M x N --- Q is said to be S-middle linear if it satisfies the following properties (where r E R, s E S, t E T, m, mj E M and n, n; E N):


419

(1) g(rm, nt) = rg(m, n)t, (2) 9(mi + m2, n) = 9(ml, n) + 9(m2, n),

(3) 9(m, nl + n2) = 9(m, n1) + 9(m, n2), and (4) g(ms, n) = g(m, sn). Note that conditions (1), (2), and (3) simply state that for each m E M the function gm : N -- Q defined by gm (n) = 9(m, n) is in Hom-T(N, Q) and for each n E N the function g" : M -' Q defined by g"(m) = g(m, n) is in HomR(M, Q). Condition (4) is compatibility with the S-module structures on M and N. If it : F -+ M ®S N = F/K is the canonical projection map and e : M x N F is the inclusion map that sends (m, n) to the basis element

(m, n) E F, then we obtain a map B : M x N - M ® N. According to Proposition 2.10, the function 0 is S-middle linear. The content of the following theorem is that every S-middle linear map "factors" through 0. This can, in fact, be taken as the fundamental defining property of the tensor product.

(2.12) Theorem. Let M be an (R, S)-bimodule, N an (S, T)-bimodule, Q an (R, T) -bimodule, and g : M x N Q an S-middle linear map. Then there exists a unique (R,T)-bimodule homomorphism g` : M ®S N -, Q with g = g o 0. Furthermore, this property characterizes M ®S N up to isomorphism.

Proof. If F denotes the free Z-module on the index set M x N, which is used to define the tensor product M ®S N, then Equation (2.14) gives an (R, T)-

bimodule structure on F. Since F is a free Z-module with basis M x N and g : M x N Q is a function, Proposition 3.4.9 shows that there is a unique Z-module homomorphism g' : F -+ Q such that g' o t = g where t : M x N -' F is the inclusion map. The definition of the (R, T)-bimodule structure on F and the fact that g is S-middle linear implies that g' is in fact an (R, T)-bimodule homomorphism. Let K' = Ker(g'), so the first isomorphism theorem provides an injective (R, T)-bimodule homomorphism g" : F/K' Q such that g' = g"o7r' where ir' : F -+ F/K' is the canonical

projection map. Recall that K C F is the subgroup of F generated by the sets H1, H2, and H3 defined prior to Definition 2.9. Since g is an Smiddle linear map, it follows that K C Ker(g') = K', so there is a map 7r2 : F/K -i F/K' such that 7r2 0 7r = 7r'. Thus, g : M x N Q can be factored as follows: (2.19)

MxN

+F/K'-4Q.

Recalling that F/K = M Os N, we define g" 0 72. Since 8 = 7r o t, Equation (2.19) shows that g = g` o 8. It remains to consider uniqueness of g`. But M ®S N is generated by the set {m ®S n = 0(m, n) : m E M, n E N}, and any function g such

420


that g o 0 = g satisfies g(m (9 n) = g(8(m, n)) = g(m, n), so g` is uniquely specified on a generating set and, hence, is uniquely determined. Now suppose that we have (R, T)-bimodules P, and S-middle linear maps 0, : M x N - P; such that, for any (R, T)-bimodule Q and any S-middle linear map g : M x N - Q, there exist unique (R, T)-bimodule homomorphisms g; : P1 - Q with g = g'; 0 8i for i = 1, 2. We will show that P1 and P2 are isomorphic, and indeed that there is a unique (R, T)bimodule isomorphism 0 : P1 - P2 with the property that 02 = 0 0 01.

Let Q = P2 and g = 82. Then by the above property of P1 there is a unique (R, T)-bimodule homomorphism 0 : P1 - P2 with 82 = 0 0 01. We need only show that 0 is an isomorphism. To this end, let Q = P1 and g = 01 to obtain 0 : P2 -. P1 with 01 = 1/i o 02. Then 81 ='+Go02 =VLo(4o81) = ('Po4)o81.

Now apply the above property of P1 again with Q = P1 and g = 81. Then

there is a unique g' with g = g o 81, i.e., a unique g with 01 = g o 81. Obviously, g` = 1p, satisfies this condition but so does conclude that o 0 = 1 p, . Similarly, zJi=0- 1, and we are done.

>[i o 0, so we

(2.13) Remarks. (1) If M is a right R-module and N is a left R-module, then M OR N is

an abelian group. (2) If M and N are both (R, R)-bimodules, then M OR N is an (R, R)-

bimodule. A particular (important) case of this occurs when R is a commutative ring. In this case every left R-module is automatically a right R-module, and vice-versa. Thus, over a commutative ring R, it is meaningful to speak of the tensor product of R-modules, without explicit attention to the subtleties of bimodule structures. (3) Suppose that M is a left R-module and S is a ring that contains R as a subring. Then we can form the tensor product S®R M which has the structure of an (S, Z)-bimodule, i.e, S OR M is a left S-module. This construction is called change of rings and it is useful when one would like to be able to multiply elements of M by scalars from a bigger ring. For example, if V is any vector space over R, then C ®R V is a vector space over the complex numbers. This construction has been implicitly used in the proof of Theorem 4.6.23. (4) If R is a commutative ring, M a free R-module, and 0 a bilinear form on M, then 0: M x M - R is certainly middle linear, and so 0 induces an R-module homomorphism

4:M®RM-.R.


421

(2.14) Corollary. (1) Let M and M' be (R, S)-bimodules, let N and N' be (S,T)-bimodules,

and suppose that f : M -+ M' and g : N -+ N' are bimodule homomorphisms. Then there is a unique (R,T)-bimodule homomorphism (2.20)

f ®g=f Os g: M®sN -+M'®sN'

satisfying (f ®g) (m (9 n) = f (m) ®g(n) for all m E M, n E N. (2) If M" is another (R, S)-bimodule, N" is an (S, T)-bimodule, and f" :

M' -. M", g" : N' -. N" are bimodule homomorphisme, then letting

f®g:M®N-+M'®N' and f'®g':M'®N'---WON" be defined as in part (1), we have (f'®9)(f®g)=(f'f)0(9g):M®N-+M"®N".

Proof. (1) Let F be the free abelian group on M x N used in the definition of M®s N' be the unique Z-module homomorphism M ®s N, and let h : F such that h(m, n) = f (m) Os g(n). Since f and g are bimodule homomorphisms, it is easy to check that h is an S-middle linear map, so by Theorem

2.12, there is a unique bimodule homomorphism h : M 0 N -+ M' ® N' such that h = h o 0 where 0 : M x N -+ M 0 N is the canonical map. Let f ®g = h. Then (f 0 g)(m (9 n) = h(m (9 n) = h o 0(m, n) = h(m, n) = f (m) 0 g(n) as claimed. (2) is a routine calculation, which is left as an exercise.

0

We will now consider some of the standard canonical isomorphisms relating various tensor product modules. The verifications are, for the most part, straightforward applications of Theorem 2.12. A few representative calculations will be presented; the others are left as exercises.

(2.15) Proposition. Let M be an (R, S)-bimodule. Then there are (R, S)bimodule isomorphisms

R®RM5M and

M®sS:L, M.

Proof. We check the first isomorphism; the second is similar. Let f : R x M -' M be defined by f (r, m) = rm. It is easy to check that f is an R-middle linear map, and thus Theorem 2.12 gives an (R, S)-bimodule

homomorphism f : R OR M -. M such that A r 0 m) = rm. Define g : M --+ R OR M by g(m) = 1 ® m. Then g is an (R, S)-bimodule homomorphism, and it is immediate that f and g are inverses of each other. 0

422


(2.16) Proposition. Let R be a commutative ring and let M and N be Rmodules. Then

M®IIN=N®RM. Proof. The isomorphism is given (via an application of Theorem 2.12) by

0 (2.17) Proposition. Let M be an (R,S)-bimodule, N an (S, T)-bimodule, and P a (T, U)-bimodule. Then there is an isomorphism of (R, U)-bimodules

(M ®s N) OT

MOs (N®TP).

Proof. Fix an element p E P and define a function

fp:MXN--IM®s(N®TP) by

fp(m, n) = m ®s (n ®T p)-

fp is easily checked to be S-middle linear, so Theorem 2.12 applies to give an (R, T)-bimodule homomorphism fp : M ®s N -a M ®s (N ®T P). Then we have a map f : (M ®s N) x P --+ lvi ®s (N ®T P) defined by

f ((m ®n), p) = fp(m ®n) = m ®(n (& p)

But f is T-middle linear, and hence there is a map of (R, U)-bimodules

f: (M os N) OT P -

M®s(N®TP)

satisfying f ((m®n)®p) = m®(n®p). Similarly, there is an (R, U)-bimodule homomorphism

g:M®s(NOT P) (M®sN)®TP satisfying gg(m (9 (n ®p)) = (m (9 n) ®p. Clearly, §1 (respectively j g-) is the identity on elements of the form (m ®n) ®p (respectively, m ®(n (&p)), and since these elements generate the respective tensor products, we conclude

that f and g are isomorphisms.

O

(2.18) Proposition. Let M = ®iEIMi be a direct sum of (R, S)-bimodules, and let N = ®)EJNj be a direct sum of (S,T)-bimodules. Then there is an isomorphism

M®sN®®(MM®sN3) iEI jEJ of (R, T) -bimodules.

Proof. Exercise.

O


423

(2.19) Remark. When one is taking Horn and tensor product of various bimodules, it can be somewhat difficult to keep track of precisely what type of module structure is present on the given Hom or tensor product. The following is a useful mnemonic device for keeping track of the various module structures when forming Hom and tensor products. We shall write RMS to indicate that M is an (R, S)-birnodule. When we form the tensor product of an (R, S)-birnodule and an (S, T)-birnodule, then the resulting module has the structure of an (R, T)-bimodule (Definition 2.9). This can be indicated mnemonically by (2.21)

RMS ®S SNT = RPT

Note that the two subscripts "S" on the bimodules appear adjacent to the subscript "S" on the tensor product sign, and after forming the tensor product they all disappear leaving the outside subscripts to denote the bimodule type of the answer (= tensor product).

A similar situation holds for Horn, but with one important difference. Recall from Proposition 2.4 that if M is an (R, S)-bimodule and N is an (R, T)-bimodule, then HornR(M, N) has the structure of an (S,T)bimodule. (Recall that HomR(M, N) denotes the left R-module homomorphisms.) In order to create a simple mnemonic device similar to that of Equation (2.21), we make the following definition. If M and N are left Rmodules, then we will write M fiR N for HomR(M, N). Using mR in place of OR, we obtain the same convention about matching subscripts disappearing, leaving the outer subscripts to give the bimodule type, provided that the order of the subscripts of the module on the left of the OR sign are reversed. Thus, Proposition 2.4 is encoded in this context as the statement

RMS and

RNT . SMR 'I'R RNT = SPT A similar convention holds for homomorphisms of right T-modules. This is illustrated by

Hom_T(RMT, SNT) = SNT 'I'-T TMR = SPR, AA,,

the result being an (S, R)-birnodule (see Proposition 2.5). Note that we must reverse the subscripts on M and interchange the position of M and N. We shall now investigate the connection between Horn and tensor product. This relationship will allow us to deduce the effect of tensor products on exact sequences, using the known results for Hom (Theorems 2.7 and 2.8 in the current section, which are generalizations of Theorems 3.3.10 and 3.3.12).

(2.20) Theorem. (Adt associativity of Hom and tensor product) Let Ml and M2 be (S, R)-bimodules, N a (T, S)-bimodule, and P a (T, U)bimodule. If 0 : M2 , Ml is an (S, R)-bimodule homomorphism, then


424

there are (R, U)-bimodule isomorphisms

Homs(Mi, HomT(N, P))

HomT(N (&s Mi, P)

such that the following diagram commutes:

Homs(Mi, HomT(N, P))

f

Homs(M2, HomT(N, P))

(2.22)

1 02

HOMT(N (&s Ml, P) Proof. Define

(IN

O).

____+

HOMT(N OS M2, P)

Homs(Mi, HomT(N, P)) -+ HomT(N (&s Mi, P) by I{(f)(n (& m) = (f (m)) (n)

where f E Homs(Mi, HorT(N, P)), m E Mi, and n E N. It is easy to check that 4i (f) E HomT (N (&s M, P) and that $ is a homomorphism of (R, U)-bimodules. The inverse map is given by ('Ii(g)(m)) (n) = g('m ® n)

where g E HomT(N ®s M, P), m E M, and n E N. To check the commutativity of the diagram, suppose that f E Homs(M, HomT(N, P), n E N, and M2 E M2. Then

((4'2oP*)(f))(n(9 m2) _ (42(f o,')) (n(9 m2) _ ((f oV)(m2)) (n) = f (IG(m2))(n)

_ ('1(f )) (n ®?P(ms))

_ (4(f)) ((In (9 ',)(n ® m2)) _ (1n (9 1,)' (Òi (f )) (n ® m2)

_ ((1N (9'G)* o-61(f)) (n(9 m2).

Thus, 02

(IN

o 4i1 and diagram (2.22) is commutative.

0

There is an analogous result concerning homomorphisms of right modules. In general we shall not state results explicitly for right modules; they can usually be obtained by obvious modifications of the left module results. However, the present result is somewhat complicated, so it will be stated precisely.

(2.21) Theorem. Let M1 and M2 be (R, S)-bimodules, N an (S, T)-bimodule, and P a (U, T)-bimodule. If ip : M2 -+ M1 is an (R, S) -bimodule homomorphism, then there are (U, R)-bimodule isomorphisms

,ti : Hom_s(M;, Hom_T(N, P)) -i Hom_T(Mi ®s N, P)


425

such that the following diagram commutes:

Hom_s(Ml, Hom_T(N, P)) (2.23)

j.

Hom_s(M2, Hom_T(N, P)) t'D2

1

(1NuS 1W

Hom_T(M1 ®s N, P)

Hom_T(M2 (&s N, P)

Proof. The proof is the same as that of Theorem 2.20.

Remark. Note that Theorems 2.20 and 2.21 are already important results

in case M1=M2=Mandt=1M. As a simple application of adt associativity, there is the following result. (2.22) Corollary. Let M be an (R, S) -bimodule, N an (S, T) -bimodule, and

let P = M ®s N (which is an (R, T) -bimodule). If M is projective as a left R-module (resp., as a right S-module) and N is projective as a left Smodule (resp., as a right T-module), then P is projective as a left R-module (resp., as a right T-module).

Proof. To show that P is projective as a left R-module, we must show that, given any surjection f : A B of R-modules, the induced map

f.: HomR(P, A) -> HomR(P, B) is also surjective. By hypothesis, M is projective as a left R-module so that

f.: HomR(M, A)

HomR(M, B)

is surjective. Also, N is assumed to be projective as a left S-module, so the map

(f.).: Homs(N, HomR(M, A)) -> Homs(N, HomR(M, A)) is also surjective. But, by Theorem 2.20, if C = A or B, then

Homs(N,HomR(M, C)) = HomR(P, C). It is simple to check that in fact there is a commutative diagram Homs(N, HomR(M, A)) 1

411

HomR(P, A)

and this completes the proof.

(

Homs(N, HomR(M, B))

I

'2

HomR(P, B)


426

One of the most important consequences of the adt associativity property relating Horn and tensor product is the ability to prove theorems concerning the exactness of sequences of tensor product modules by appealing to the theorems on exactness of Horn sequences, namely, Theorems 2.7 and 2.8. (2.23) Theorem. Let N be a fixed (R, T)-bimodule. If

M1-00-.M- - M2 -+0

(2.24)

is an exact sequence of (S, R)-bimodules, then

Ml®RN --+ M®RNg, 01N M2®RN-+0 "Iri

(2.25)

is an exact sequence of (S, T)-bimodules, while if (2.24) is an exact sequence of (T, S) -bimodules, then

NOT Ml'N®mN®TM'-i N®TM2 -+0

(2.26)

is an exact sequence of (R, S)-bimodules. Proof. We will prove the exactness of sequence (2.26); exactness of sequence (2.25) is similar and it is left as an exercise. According to Theorem 2.7, in order to check the exactness of sequence (2.26), it is sufficient to check that the induced sequence (2.27)

HomR(N ®T M2, P)

0

HomR(N ®T M, P)

-+ HomR(N ®T M1, P) is exact for every (R, U)-bimodule P. But Theorem 2.20 identifies sequence (2.27) with the following sequence, which is induced from sequence (2.24) by the (T, U)-bimodule HomR(N, P): (2.28)

0 - HomT(M2, HomR(N, P)) - HomT(M, HomR(N, P)) - HomT(Ml, HomR(N, P)).

Since (2.24) is assumed to be exact, Theorem 2.7 shows that sequence (2.28)

is exact for any (R, U)-bimodule P. Thus sequence (2.27) is exact for all 0 P, and the proof is complete. (2.24) Examples.

(1) Consider the following short exact sequence of Z-modules: (2.29)

0

where 0(i) = mi and i is the canonical projection map. If we take N = Z,,, then exact sequence (2.25) becomes


427

Z®Zn +Z®Zn -® Zm®Zn -+ 0.

(2.30)

By Proposition 2.15, exact sequence (2.30) becomes the exact sequence

Zn -- 4Zn '1'+ Zm®Zn-'0

(2.31)

mi. Thus Zm ® Zn

where

Coker(QS). Now let d = gcd(m, n) is the composite

and write m = m'd, n = n'd. Then the map Zn -01 + Zn

02

+ Zn

where 01(i) = m'i and 02(i) = di. Since gcd(m', n) = 1, it follows that 01 is an isomorphism (Proposition 1.4.11), while IM(02) = dZn. Hence, Coker(q) Zn/dZn Zd, i.e.,

Zm®Zn

Zd.

(2) Suppose that M is any finite abelian group. Then M ®z Q = (0). To see this, consider a typical generator x ®r of M ®z Q, where x E M

and r E Q. Let n = IMI. Then nx = 0 and, according to Equation (2.18),

x ® r = x 0 n(r/n) = xn ® (r/n) = 0 ® (r/n) = 0. Since x E M and r E Q are arbitrary, it follows that every generator

ofM®Q is 0. (3)

Let R be a commutative ring, I C R an ideal, and M any R-module. Then (2.32)

(R/I) OR M M/IM.

To see this consider the exact sequence of R-modules

0-+I `+R-+R/I-+0. Tensor this sequence of R-modules with M to obtain an exact sequence

I®RM `®+RORM-+(R/I) ®RM--+0. But according to Proposition 2.15, R®R M = M (via the isomorphism

4i(r (9 m) = rm), and under this identification it is easy to see that Im(c (9 1) = IM. Thus, (R/I) OR M M/IM, as we wished to . Example 2.32 (1) shows that even if a sequence

0 -+ M1 -+ M ' M2

0

428


is short exact, the tensored sequence (2.25) need not be part of a short exact sequence, i.e., the initial map need not be injective. For a simple situation where this occurs, take m = n in Example 2.32 (1). Then exact sequence (2.30) becomes

Zn001- ' Zn - Zn -+ 0. 1 is the zero map, so it is certainly not an injection. The map This example, plus our experience with Hom, suggests that we consider criteria to ensure that tensoring a short exact sequence with a fixed module produces a short exact sequence. We start with the following result, which is exactly analogous to Theorem 2.8 for Hom. (2.25) Theorem. Let N be a fixed (R, T)-bimodule. If

0-pM1-.M-.M2-+0

(2.33)

is a split short exact sequence of (S, R) -bimodules, then (2.34)

0- Ml®RNO 1 M®RN4

M2®RN-+0

is a split short exact sequence of (S, T)-bimodules, while if (2.33) is a split short exact sequence of (T, S)-bimodules, then (2.35)

0

N ®T M1

14 N ®T M 1- N ®T M2 --i 0

is a split short exact sequence of (R, S)-bimodules. Proof. We will do sequence (2.34); (2.35) is similar and is left as an exercise.

Let of : M - M1 split 0, and consider the map

a®1:Al OR N-'M1OR N. Then

((a®1)(O®1))(m(9 n) = (aO(9 1)(m®n) = (1(9 1)(m(& n) = m®n so that 0®1 is an injection, which is split by a®1. The rest of the exactness 0 is covered by Theorem 2.23.

(2.26) Remark. Theorems 2.7 and 2.23 show that given a short exact sequence, applying Hom or tensor product will give a sequence that is exact on one end or the other, but in general not on both. Thus Horn and tensor product are both called half exact, and more precisely, Hom is called left exact and tensor product is called right exact. We will now investigate some conditions under which the tensor product of a module with a short exact sequence always produces a short exact sequence. It was precisely this type of consideration for Hom that led us to the concept of projective module. In fact, Theorem 3.5.1 (4) shows that if P is a projective R-module and


429

0- M1 0+M &M2-i0 is a short exact sequence of R-modules, then the sequence

0 -+ HomR(P, M1) -i HomR(P, M) 0.+ HomR(P, M2) -+ 0 is short exact. According to Theorem 3.3.10, the crucial ingredient needed

is the surjectivity of t/i. and this is what projectivity of P provides. For the case of tensor products, the crucial fact needed to obtain a short exact sequence will be the injectivity of the initial map of the sequence.

(2.27) Proposition. Let N be an (R, T)-bimodule that is projective as a left R-module. Then for any injection t : M1 -' M of (S, R)-bimodules,

t®1:M1®RN -+M®RN is an injection of (S, T)-bimodules. If N is projective as a right T-module and t : M1 --+ M is an injection of (T, S)-bimodules, then

1®t:N®TMi -+ N®TM is an injection of (R, S)-bimodules.

Proof. First suppose that as a left R-module N is free with a basis {nj}jEJ.

Then N ®jEJRj where each summand Rj = Rnj is isomorphic to R as a left R-module. Then by Proposition 2.18

MI OR N°-(D(MlOR Rj)=$Mlj jEJ

jEJ

where each MI j is isomorphic to M1 as a left S-module, and similarly M®RN ®jEJMj, where each Mj is isomoprhic to M as a left S-module. Furthermore, the map t ®1 : M1®R N -+ M OR N is given as a direct sum

®(tj:Mij--+Mj) jEJ

where each tj agrees with i under the above identifications. But then, since t is an injection, so is each tj, and hence so is t ®1. Now suppose that N is projective as a left R-module. Then there is a left R-module N' such that N ® N' = F where F is a free left R-module. We have already shown that

t®1:Mi®RF-+M®RF is an injection. But using Proposition 2.18 again,

MI OR F= (M1 OR N)®(MI OR N')

430


so we may write t(91 = c1® t2 where (in particular) c1 = L(91 : M1®R N --+ M OR F. Since c ® 1 is an injection, so is c1, as claimed. Thus the proof is

complete in the case that N is projective as a left R-module. The proof in case N is projective as a right T-module is identical. Note that we have not used the right T-module structures in the above proof. This is legitimate, since if a homomorphism is injective as a map of left S-modules, and it is an (S, T)-bimodule map, then it is injective as an (S, T)-bimodule map.

(2.28) Corollary. Let N be a fixed (R,T)-bimodule that is projective as a left R-module. If (2.36)

0--.M1

is a short exact sequence of (S, R)-bimodules, then (2.37)

0-+ M1 ORNMORN"0

M2®RN--+0

is a short exact sequence of (S, T)-bimodules; while if (2.36) is an exact sequence of (T, S)-bimodules and N is projective as a right T-module, then (2.38)

0--+N®TM1'- N®TM' - N®TM2-+0

is a short exact sequence of (R, S)-bimodules. Proof. This follows immediately from Theorem 2.23 and Proposition 2.27. 0

(2.29) Remark. A module satisfying the conclusion of Proposition 2.27 is said to be flat. That is, a left R-module N is flat if tensoring with all short exact sequences of right R-modules produces a short exact sequence, with a similar definition for right R-modules. Given Theorem 2.23, in order to prove that a left R-module N is flat, it is sufficient to prove that for all right R-modules M and submodules K, the inclusion map e : K -+ M induces an injective map

t®1:K®RN- M®RN.

Thus, what we have proven is that projective modules are flat. In Section 6.1 we discussed duality for free modules over commutative rings. Using the theory developed in the current section, we will extend portions of our discussion of duality to the context of projective (bi-)modules.

(2.30) Definition. Let M be an (R, S)-bimodule. The dual module of M is the (S, R)-bimodule M' defined by HomR(M, R).


431

In particular, if M is a left R-module, i.e., take S = Z, then the dual module M* is a right R-module. The double dual of M is defined to be

M" = Hom_R(M', R). As in Section 6.1, there is a homomorphism rl

: M - M" of (R,S)-

bimodules defined by

(rl(v))(w) = w(v) for all v E M, w E M' and if r) is an isomorphism, then we will say that M is reflexive.

If M is an (R, S)-bimodule, which is finitely generated and free as a left R-module, then given any basis B of M, one may construct a basis of M* (as a right R-module) exactly as in Definition 6.1.3 and the proof of Theorem 6.1.7 goes through verbatim to show that finitely generated free Rmodules are reflexive, even when R need not be commutative. Furthermore, the proofs of Theorems 3.5.8 and 6.1.13 go through without difficulty if one keeps track of the types of modules under consideration. We will simply state the following result and leave the details of tracing through the module types as an exercise. (2.31) Proposition. Let M be an (R, S)-bimodule, which is finitely generated

and projective as a left R-module. Then the dual module M' is finitely generated and projective as a right R-module. Furthermore, M is reflexive as an (R, S)-bimodule. Proof. Exercise. See the comments above.

0

If M is an (R, S)-bimodule and P is an (R, T)-bimodule, then define

(:M'xP-+HomR(M,P) by

(((w, p)) (m) = w(m)p

for

w E M', p E P, and m E M.

Then ( is S-middle linear and hence it induces an (S, T)-bimodule homomorphism

t;:M'®RP-kHomR(M,P)

given by

(2.39)

(((W'& p))(m) = w(m)p

for all w E M*, p E P, and m E M. (2.32) Proposition. Let M be an (R, S)-bimodule, which is finitely generated and projective as a left R-module, and let P be an arbitrary (R, T) -bimodule. Then the map


432

( : M' OR P - HomR(M, P) defined by Equation (2.39) is an (S, T) -bimodule isomorphism.

Proof. Since S is an (S, T)-bimodule homomorphism, it is only necessary to prove that it is bijective. To achieve this first suppose that M is free of finite rank k as a left R-module. Let B = {vl, ... , vk} be a basis of M and let {vi, ... ,v;} be the basis of M' dual to B. Note that every element of M' OR P can be written as x = Fk 1 of ®pi for pt, ... , Pk E P. Suppose

that ((x) = 0, i.e., (((x))(m) = 0 for every m E M. But ((x)(vi) = pi so that pi = 0 for 1 2v1 ®f(vi) Then (((xf))(vi) = f (vi) for 1 < i < k, i.e., S(xf) and f agree on a basis of M; hence, ((xf) = f and C is a surjection, and the proof is complete in case M is free of rank k. Now suppose that M is finitely generated and projective, and let N be a left R-module such that F = M ® N is finitely generated and free. Then S : F' OR P - HomR(F, P) is a Z-module isomorphism, and

F'®RP= (M®N)'®RPQ (M'(DN')ORP°` (M* OR P)®(N'®R P) while

HomR(F, P) = HomR(M ® N, P)

HomR(M, P) ® HomR(N, P)

where all isomorphisms are Z-module isomorphisms. Under these isomorphisms,

CF=CM®(N CM : M' OR P

HomR(M, P)

CN : N' OR P -' HomR(N, P). Since (F is an isomorphism, it follows that CM and (N are isomorphisms as well. In particular, CM is bijective and the proof is complete. 0

(2.33) Corollary. Let M be an (R, S)-bimodule, which is finitely generated and projective as a left R-module, and let P be an arbitrary (T, R)-bimodule. Then

(P®RM)' as (S,T)-bimodules.

Proof. From Proposition 2.32, there is an isomorphism


M' OR P'

433

HomR(M, P') = HomR(M, HomR(P, R)) (by adt associativity)

LY HomR(P OR M, R)

= (P OR M). 0

(2.34) Remark. The isomorphism of Corollary 2.33 is given explicitly by

0(f ®9)(P ®m) = f (m)9(p) E R

where f EM',gEP',pEP,andmEM. We will conclude this section by studying the matrix representation of the tensor product of R-module homomorphisms. Thus, let R be a commutative ring, let M1, M2, N1, and N2 be finite rank free R-modules, and let f; : M1 Ni be R-module homomorphisms for i = 1, 2. Let m; be the rank of M1 and n; the rank of Ni for i = 1, 2. If M = Ml 0 M2 and N = N10 N2, then it follows from Proposition 2.18 that M and N are free R-modules of

rank m1n1 and m2n2i respectively. Let f = ft 0 f2 E HomR(M, N). We will compute a matrix representation for f from that for f1 and f2. To do this, suppose that A = {a1 i ... , am, } 5 = {b1, ... , bn, } C = {c1, ... , Cm2 }

D = {d1, ...

, cn2 }

are bases of M1, N1, M2, and N2, respectively. Let E _ {al ®cl, a1 ®c2,... a2 ® C1, a2 ® C2, ...

a1 ®c

,

a2 0 Cm2,

am, ®C1, am, ®C2, ... , amt ® Cm2 } and .2 = {b1 ®dl, b1 ®d2,...

b1 ® dn2,

b2®d1, b2®d2, ... ,b2®42, bn1 ®d1, bn1 ®d2, ... , bn1 ®d2}.

Then E is a basis for M and F is a basis for N. With respect to these bases, there is the following result:

434


(2.35) Proposition. With the notation introduced above,

[fl ®f2].y = [fl]B ®[f2]D Proof. Exercise.

l7

Recall that the notion of tensor product of matrices was introduced in Definition 4.1.16 and has been used subsequently in Section 5.5. If [fl]14 = A = [ai3] and (f2]) = B = [,33j], then Proposition 2.35 states that (in block matrix notation)

I (fl ®f2]Y _ an11 B

an12B

There is another possible ordering for the bases C and F. If we set

E'={ai®cj :1 and

i<m1, 1<j<m2}

1"={bi0 dj :1
where the elements are ordered by first fixing j and letting i increase (lex-

icographic ordering with j the dominant letter), then we leave it to the reader to that the matrix of fl ® f2 is given by ... a1m2A 311A 012A [fl ® f2 ]F'

Nn21A /n22A

Qnzm2A

7.3 Exercises 1.

2.

3.

4.

Let M be a simple R-module, and let N be any R-module. a Show that every nonzero homomorphism f : M -. N is injective. (b) Show that every nonzero homomorphism f : N - M is surjective. Let F be a field and let R = { [ o b] : a, b, c E F} be the ring of upper triangular matrices over F. Let M = F2 and make M into a (left) R-module by matrix multiplication. Show that EndR(M) °_° F. Conclude that the converse of Schur's lemma is false, i.e., EndR(M) can be a division ring without M being a simple R-module. Suppose that R is a D-algebra, where D is a division ring, and let M be an R-module which is of finite rank as a D-module. Show that as an R-module, f(M) < rankD(M). An R-module M is said to satisfy the decending chain condition (DCC) on submodules if any strictly decreasing chain of submodules of M is of finite length.

7.3 Exercises

5.

6. 7.

8.

435

(a) Show that if M satisfies the DCC, then any nonempty set of submodules of M contains a minimal element. (b) Show that e(M) < oo if and only if M satisfies both the ACC (ascending chain condition) and DCC. Let R = { [o b] a, b E R; c E Q}. R is a ring under matrix addition and multiplication. Show that R satisfies the ACC and DCC on left ideals, but neither chain condition is valid for right ideals. Thus R is of finite length as a left R-module, but e(R) = oo as a right R-module. Let R be a ring without zero divisors. If R is not a division ring, prove that R does not have a composition series. Let f : Ml --, M2 be an R-module homomorphism. a If f is injective, prove that e(M1) < t(M2)b) If f is surjective, prove that £(M2) < e(Ml). Let M be an R-module of finite length and let K and N be submodules of M. Prove the following length formula: :

e(K + N) + e(K n N) = e(K) + e(N). 9.

a) Compute e(Z,-). b) Compute e(Z (G3 ® Zq c) Compute e where

.

is any finite abelian group. d) More generally, compute e(M) for any finitely generated torsion module over a PID R. 10.

Compute the length of M = F[X]/(f(X)) as an F[X]-module if f(X) is

a polynomial of degree n with two distinct irreducible factors. What is the length of M as an F-module? 11. Let F be a field, let V be a finite-dimensional vector space over F, and let T E EndF(V). We shall say that T is semisimple if the F[XI-module VT is semisimple. If A E we shall say that A is semisimple if the linear transformation TA : F' -, F' (multiplication by A) is semisimple. Let F2 be the field with 2 elements and let F = F2 (Y) be the rational function field in the indeterminate Y, and let K = F[XJ/(X2 +Y). Since X2 + Y E FIX] is irreducible, K is a field containing F as a subfield. Now let

A=C(X2+Y)= L0 0Y ] EM2(F). Show that A is semisimple when considered in M2(F) but A is not semisimple

12.

13. 14. 15. 16. 17.

when considered in M2(K). Thus, semisimplicity of a matrix is not necessarily preserved when one es to a larger field. However, prove that if L is a subfield of the complex numbers C, then A E M,,(L) is semisimple if and only if it is also semisimple as a complex matrix. Let V be a vector space over R and let T E EndR(V) be a linear transformation. Show that T = S + N where S is a semisimple linear transformation, N is nilpotent, and SN = NS. Prove that the modules Mi and N3 in the proof of Lemma 1.33 are simple, as claimed. Prove Lemma 1.37. If D is a division ring and n is a positive integer, prove that EndD(D") is a simple ring. Give an example of a semisimple commutative ring that is not a field.

(a) Prove that if R is a semisimple ring and I is an ideal, then R/I is semisimple.

(b) Show (by example) that a subring of a semisimple ring need not be semisimple.

436


Let R be a ring that is semisimple as a left R-module. Show that R is simple if and only if all simple left R-modules are isomorphic. 19. Let M be a finitely generated abelian group. Compute each of the following 18.

pups:

a Homy M, Q/Z . b Homz(Q/Z, M). c) M Oz Q/Z. 20.

Let M be an (R, S)-bimodule and N an (S, T)-bimodule. Suppose that E x; ®y; = 0 in M Os N. Prove that there exists a finitely generated (R, S)-bisubmodule Mo of M and a finitely generated (S, T)-bisubmodule No of N such that E x; 0 y; = 0 in Mo Os No.

21.

Let R be an integral domain and let M be an R-module. Let Q be the quotient field of R and define m : M - Q OR M by 0(x) = 1®x. Show that

Ker(4') = M, = torsion submodule of M. (Hint: If 10 x = 0 E Q OR M then 1(9 x = 0 in (Re`) OR M M for some c 54 0 E R. Then show that

cx=0.) 22.

Let R be a PID and let M be a free R-module with N a submodule. Let Q be

the quotient field and let 4': M Q OR M be the map 4'(x) = 1®x. Show

that N is a pure submodule of M if and only if Q Q. (0(N ) fl Im(4') = O(N). 23.

Let R be a PID and let M be a finitely generated R-module. If Q is the

quotient field of R, show that M OR Q is a vector space over Q of dimension 24.

25.

27.

28.

equal to rankR(M/M,). Let R be a commutative ring and S a multiplicatively closed subset of R containing no zero divisors. Let Rs be the localization of R at S. If M is an R-module, then the Rs-module Ms was defined in Exercise 6 of Chapter 3. Show that Ms Rs OR M where the isomorphism is an isomorphism of Rs-modules. If S is an R-algebra, show that S OR M and N be finitely generated R-modules over a PID R. Compute M ®R N. As a special case, if M is a finite abelian group with invariant factors st, ..., st (where as usual we assume that s; divides s;+1), show that M®z M is a finite group of order n;=1 aie-2)+1

Let F be a field and K a field containing F. Suppose that V is a finitedimensional vector space over F and let T E EndF(V). If B = {v;} is a basis of V, then C = {1 ®B = 110 vi I is a basis of K ®F V. Show that )10T]c = IT)s If S E ndF(V), show that 1®T is similar to 1®S if and only if S is similar to T. V a normal linear Let V be a complex inner product space and T : V transformation. Prove that T is self-adt if and only if there is a real inner W, and an product space W, a self-adt linear transformation S : W isomorphism 0: C OR W -+ V making the following diagram commute. C ®R W

29.

®s

C OR W

Let R be a commutative ring. (a) If I and J are ideals of R, prove that

R/I OR R/J

R/(I + J).

(b) If S and T are R-algebras, I is an ideal of S, and J is an ideal of T, prove that

S11 OR T/J °_° (S OR T)/(l, J),

7.3 Exercises

30.

437

where (I, J) denotes the ideal of S®RT generated by I ®RT and S®R J. (a) Let F be a field and K a field containing F. If f (X) E F[X ), show that there is an isomorphism of K-algebras:

K®F (F[X]/(f(X))) ` K[Xll(f(X)). (b) By choosing F, f (X), and K appropriately, find an example of two fields

K and L containing F such that the F-algebra K ®F L has nilpotent

elements. Let F be a field. Show that F[X, Y) ?f F[X]®F F[Y] where the isomorphism is an isomorphism of F-algebras. 32. Let G1 and Gs be groups, and let F be a field. Show that 31.

F(GI X Gz) 33.

F(G1) or F(G2).

Let R and S be rings and let f : R -+ S be a ring homomorphism. If N is an S-module, then we may make N into an R-module by restriction of scalars, i.e., a x = f (a) x for all a E R and x E N. Now form the S-module

Ns=S®RN and define 9:N-'Ns by

9(y) = I ®y. Show that g is injective and g(N) is a direct summand of Ns. 34. Let F be a field, V and W finite-dimensional vector spaces over F, and let T E EndF(V), S E EndF(W). (a) If a is an eigenvalue of S and Q is an eigenvalue of T, show that the product a# is an eigenvalue of S ® T. (b) If S and T are diagonalizable, show that S ® T is diagonalizable. 35. Let R be a semisimple ring, M an (R, S)-bimodule that is simple as a left Rmodule, and let P be an (R, T)-bimodule that is simple as a left R-module. Prove that

M ®R P= J EndR(M) if P L' M as left R-modules otherwise.

0 36.

Let R be a commutative ring and M an R-module. Let

M®k=M®...®M, where there are k copies of M, and let S be the submodule of M®k generated

by all elements of the form ml ®- - 0 Mk where m; = mi for some i 54 j. Then Ak(M) = M®k/S is called an exterior algebra. (a) Show that if M is free of rank n, then A``k(M) is free and rank(Ak(M)) _

(k!

0

if k < n

ifk>n.

(b) As a special case of part (a),

rank(A"(M)) = 1. Show that HomR(A"(M), R) may be regarded as the space of determinant functions on M.

Chapter 8

Group Representations

8.1 Examples and General Results We begin by defining the objects that we are interested in studying. Recall that if R is a ring and G is a group, then R(G) denotes the group ring of G with coefficients from R. The multiplication on R(G) is the convolution product (see Example 2.1.10 (15)).

(1.1) Definition. Let G be a group and F a field. A (left) F-representation of G is a (left) F(G)-module M.

In other words, M is an F vector space, and for each g E G we have

a linear transformation a(g) : M M given by the action of g regarded as lg E F(G) on M. These linear transformations satisfy a(e) = 1M and a(9291)(m) = a(92)(a(91)(m)) for all m E M and g E G. Note then that a(9-1)a(9) = a(e) = 'Al

so that o(g-1) = a(g)-1. In particular, each a(g) is invertible, i.e., a(g) E Aut(M). Conversely, we may view an F-representation of G on M, where M is an F vector space, as being given by a homomorphism a : G - Aut(M). Then we define an F(G)-module structure on M by

a99 9EG

aga(9)(m)-

(m) = gEG

In this situation, we say that the representation is defined by a. We will denote a representation as above by M, when we wish to emphasize the underlying vector space, or (more often) by a, when we wish to emphasize the homomorphism, and we will use the term representation instead of F-representation, when F is understood. Occasionally (as is often done) we shall omit a when it is understood and write g(m) for o(g)(m).

8.1 Examples and General Results

439

(1.2) Definition. The degree deg(M) of a representation M is dimF(M) E {0, 1, 2, ... } U {oo}.

Two F-representations Ml and M2 of G, defined by v; : G - Aut(M;) for i = 1, 2, are said to be isomorphic (or equivalent) if they are isomorphic

as F(G)-modules. Concretely, this is the case if and only if there is an invertible F-linear transformation f : Ml f (ol (g)(m)) = 0`2(g)(f (m))

M2 with

for every

m E Ml, g E G.

We will be considering general groups G and fields F, though our strongest results will be in the case G is finite (and F satisfies certain restrictions). Accordingly, we will adopt the following notational convention throughout this chapter: n will always denote the order of G. (The reader should note that many variables in this chapter will range over the set (0, 1, 2, ... } U {oo}, and the comment of Remark 7.1.20 (2) is relevant here. Namely, distinction is made between finite and infinite, but no distinction is made between sets of different infinite cardinality.) Since we are considering F(G)-modules, we make some elementary remarks about F(G) itself.

(1) F(G) is an F-algebra.

(2) F(G) is an (F(G), F(G))-bimodule, as well as an (F(K), F(H))bimodule for any pair of subgroups K and H of G. (3) F(G) is commutative if and only if G is abelian. (4) If G has torsion (i.e., elements of finite order), then F(G) has zero divisors. For let g E G with gm = 1. Then

(1 -g)(1 +g+ (5)

+g'"-') = 0 E F(G).

Any F(G)-module is an (F(G), F)-bimodule. It will be useful to us to single out the following class of fields F:

(1.3) Definition. Let G be a finite group. A field F is called good for G (or simply good) if the following conditions are satisfied. (1) The characteristic of F is relatively prime to the order of G. (2) If m denotes the exponent of G, then the equation X1 - 1 = 0 has m

distinct roots in F. Remark. Actually, (2) implies (1), and also, (2) implies that the equation Xk - 1 = 0 has k distinct roots in F, for every k dividing m (Exercise 34, Chapter 2). Furthermore, since the roots of Xk - 1 = 0 form a subgroup of the multiplicative group F', it follows from Theorem 3.7.24 that there is (at least) one root C = tk such that these roots are

Chapter 8. Group Representations

440

I = S°, S, S2, ... ,(,k-1.

We shall reserve the use of the symbol ((or (k) to denote this. We further assume that these roots have been consistently chosen, in the following sense: If k1 divides k2, then ((k2)k2/k, = (k,

Note that the field C (or, in fact, any algebraically closed field of characteristic zero) is good for every finite group, and in C we may simply choose (k = exp(21ri/k) for every positive integer k.

In this chapter we will have occasion to consider HomR, EndR, or OR for various rings of the form R = F(H), where H is a subgroup of G (or R = F, in which case we may identify R with F((1))). We adopt the notational convention that Hom, End, and 0 mean HomF, EndF, and OF, respectively, and HomH, EndH, and OH mean HomR, EndR, and OR, respectively, for R = F(H). (1.4) Examples.

Aut(M) by a(g) = lm for all g E G. We call M the trivial representation of degree 1, and we denote it by r. (2) M = F(G) as an F(G)-module. This is a representation of degree n. As an F vector space, M has a basis {g : g E G}, and an element (1) Let M be an F vector space of dimension 1 and define a : G

go E G acts on M by a(go) E ag9

=

gEC

(3)

a99o9 9EC

M is called the (left) regular representation of G and it plays a crucial role in the theory. We denote it by R = R(G). Permutation representations. Let P be a set, P = {pi}sEI, and suppose we have a homomorphism a : G Sp = Aut(P) (see Remark 1.4.2).

Let M = F(P) be the free F-module with basis P. Then G acts on F(P) by the formula (aiPi) _

a\9) sEl

ai(a(9)(pi)), iE1

giving a representation of degree jPJ. Note that the regular representation is a special case of this construction, obtained by taking P = G.

As an important variant, we could take P = G/H = {gH}, the set of left cosets of some subgroup H of G. (This concept was used in a preliminary way in Section 1.4.)

8.1 Examples and General Results (4)

441

If M1 and M2 are two representations of G, defined by al and a2,

then M1 ® M2 is a representation of G defined by al ® a2. Note that deg(Mj (D M2) = deg(Mi) + deg(M2). (5) If M1 and M2 are two representations of G, defined by 01 and a2, then M1 ® M2 is a representation of G defined by a1 ® 02. Note that deg(M1 (9 M2) _ (deg(M1))(deg(M2)).

(6) Let G = Z,, _ (g : g" = 1) be cyclic of order n and let F be a field that is good for G. We have the one-dimensional representations Ok : G -p Aut(F) F' defined by 9k(g) = (k for k = 0, ... , n - 1. These are all distinct (i.e., pairwise nonisomorphic) and 90 = T.

(7) Let

G=D2m=(x,y:xm=1,y2=1,xy=yx-1)

be the dihedral group of order 2m, and let F be a field that is good for G. The representations of D2m are described in Tables 1.1 and 1.2, using the following matrices: r

(1.1)

k

S'k

Ak = I o

B= 0

and

0

(Note that it is only necessary to give the value of the representation on the two generators x and y.)

Table 1.1. Representations of D2m (m odd)

Representation

x

y

degree

T

1

1

1

1

-1

1

Ak

B

2

4'k

1 <(m-1)/2

Table 1.2. Representations of D2m (m even)

Representation

Ok

x

y

degree

1

1

1

1

-1

1

-1 -1 Ak

1

1

-1

1

B

2

1< k< (m/2) - 1

442


(8) Suppose that M has a basis 8 such that for every g E G, [0'(9)113 is a matrix with exactly one nonzero entry in every row and column. Then M is called a monomial representation of G. For example, the representations of D2,,, given above are monomial. If all nonzero entries of [v(g)] g are 1, then the monomial representation is called a permutation representation. (The reader should check that this definition agrees with the definition of permutation representation given in Example 1.4 (3).)

(9) Let X = {1, x, x2, ... } with the multiplication x'xi = x'+'. Then X is a monoid, i.e., it satisfies all of the group axioms except for the existence of inverses. Then one can define the monoid ring exactly as

in the case of the group ring. If this is done, then F(X) is just the polynomial ring F[x]. Let M be an F vector space and let T : M -- M be a linear transformation. We have already seen that M becomes an

F(X)-module via x'(m) = T'(m), for m E M and i > 0. Thus, we have an example of a monoid representation. Now we may identify Z with -2

-1

2

If the linear transformation T is invertible, then 1YI becomes an F(Z)module via the action x'(m) = Ti(m) for m E M, i E Z. (10) As an example of (9), let T F2 -> F2 have matrix [ o i t (in the standard basis). Then we obtain a representation of Z of degree 2. If char(F) = p > 0, then Tn = lF2, so in this case we obtain a representation of Z P of degree 2. (11) Let e : R(G) -> F be defined by E

(agg 9EG

=

ag.

gEG

If we let G act trivially on F, we may regard E as a map of Frepresentations

E:7 (G)->T. We let Ro(G) = Ker(E). The homomorphism E is known as the augmen-

tation map and Ro(G) is known as the augmentation ideal of R(G). It is then, of course, an F-representation of G. (12) If F is a subfield of F' and 111 is an F-representation of G, then M' _ F' ®F M is an F'-representation of G. An F'-representation arising in this way is said to be defined over F. (13) If Mi is an F-representation of Gi, defined by of for i = 1, 2, then 1v11 ® M2 is a representation of G1 x G2i defined by (o1 ® (72)(91, 92) = oi(91) ®x2(92)

(Compare with Example 1.4 (4).)

for

gi E Gi.


443

(14) If M1 is an F-representation of Gi, defined by ai for i = 1, 2, then M1 ® M2 is a representation of G1 x G2, defined by (a1 (9 a2)(g1, 92) = a1(91) ® a2(92)

for

gi E Gi.

(Compare with Example 1.4 (5).) (15) If a : G - Aut(M) is injective, then or is called faithful. If not, let K =

Ker(a). Then a determines a homomorphism a' : G/K - Aut(M), which is a faithful representation of the group G/K. (16) Let f : G1 -+ G2 be a group homomorphism and let a : G2 - Aut(M)

be a representation of G2. The pullback of a by f , denoted f *(a), is the representation of G1, defined by f *(or) = a o f, i.e., f *(or) G1 - Aut(M) by for

f*(a)(9) = a(f(9))

g E G1.

(17) Let M be an F-representation of G, so M is an F(G)-module. Then

for any subgroup H of G, M is also an F(H)-module, i.e., an Frepresentation of H.

(18) Let H be a subgroup of G and let M be an F-representation of H, i.e., an F(H)-module. Then F(G) ®F(H) M is an F(G)-module, i.e., an F-representation of G. (1.5) Definition.

(1) M is an irreducible representation of G if M is an irreducible F(G)module. (See Definition 7.1.1.) (2) M is an indecomposable representation of G if M is an indecomposable F(G)-module. (See Definition 7.1.6.)

One of our principal objectives will be to find irreducible representations of a group G and to show how to express a representation as a sum of irreducible representations, when possible. The following examples illustrate this theme. (1.6) Example. Let F be good for Zn. Then the regular representation R(Zn) is isomorphic to 00 ®61 ® ... ®9n_1

Proof. Consider the F-basis {1, g, ... ,gn-1} of 7Z(Zn). In this basis, a(g) has the matrix 0

0

0

1

1

0

...

0

0

0

1

...

0

0

0

0

...

0

0

0

0

...

1

0

444


which we recognize as C(Xn - 1), the companion matrix of the polynomial

X' - 1. Hence, ma(9)(X) = C, (9)(X) = Xn - 1.

By our assumption on F, this polynomial factors into distinct linear factors n-1

Xn-1=H (X-(k). k=0

Hence, v(g) is diagonalizable, and indeed, in an appropriate basis B, it has matrix

diag(1, (, (2, ..., (n-1). Each of the eigenspaces is an F(Zn)-submodule, so we see immediately that R(Zn)L00®91®... ®9n-1.

(1.7) Example. Let F = Q, the rational numbers, and let p be a prime. Consider the augmentation ideal Ro C R = Q(Zp). If g is a generator of Zp and T = a(g) : Ro --+ Ro, then (XP - 1) mT(X) = CT(X) = (X

which is irreducible by Corollary 2.6.12. Hence, by Remark 7.1.40 1Z0 is an irreducible Q-representation of Zp. (Strictly speaking, to apply Remark 7.1.40 we must consider not a, but a' (a) where it : Z -+ Zp is the canonical projection, and apply Exercise 2.) Note that Ro ®Q C is the augmentation ideal in C(G), and Ro ®Q C is not irreducible and, in fact, is isomorphic to

g1 ®... ® 9p-1

(In fact, this statement is true without the requirement that p be prime.) (1.8) Example. Let F be good for D2,,,. (1)

(2)

Each of the F(D2,,,)-modules Ok of Example 1.4 (7) is irreducible, and they are distinct. If, for m even, we define 40m/2 by Om/2(x) = Am/2 and Om/2(y) = B, then

(3) Let P be the set of vertices of a regular m-gon, and let D2m act on P

in the usual manner (Section 1.5). Then, as F (P)

')+ ®01 ®02 ®... ®0(.n-1)/2 ®0jp-1 lV,++®V,-+®01®02ED

if m is odd,

ifmiseven.


445

(4) For the regular representation, we have R(D2m)

ifmisodd, ® 20;j

20, S .

1 10++ ®+G+- ®+6-+

if m is even.

0

Proof. We leave this as an exercise for the reader.

(1.9) Definition. Let M and N be F-representations of G. The multiplicity m of M in N is the largest nonnegative integer with the property that mM

is isomorphic to a submodule of N. If no such m exists, i.e., if mM is isomorphic to a submodule of N for every nonnegative integer m, we say that the multiplicity of M in N is infinite.

If the multiplicity of M in N is m, we shall often say that N contains M m times (or that N contains m copies of M). If m = 0, we shall often say that N does not contain M. (1.10) Lemma. Let G be a group. Then the regular representation R contains

the trivial representation r once if G is finite, but it does not contain r if G is infinite. Proof. Let M be the submodule of R consisting of all elements on which G acts trivially. Let m E M. Then we may write m =

where

agg

ag E F.

9EG

By assumption, gom = m for every go E G. But

gom = E ag(9o9) 9EC

and

m = > ago9(909) gEG

Therefore, agog = ag for every g E G and every go E G. In particular, ago = ae

(1.2)

for every go E G. Now suppose that G is finite. Then by Equation (1.2)

aog = a., (1:

m= 9EC

g)

9EG

where

a,, E F,

446


i.e., M = (EgEG g) as an F-module. Therefore, M is one-dimensional over

F, so that M = r. On the other hand, if G is infinite, Equation (1.2) implies that a9 0 0 for all g E G whenever a, 0. But the group ring F(G) consists of finite sums E9EG a9g, so this is impossible. Hence, we must have ae = 0 and M = (0). (1.11) Corollary. If G is an infinite group, then R is not semisimple. Proof. Suppose that 1Z were semisimple. Then by Theorem 7.1.30, R would contain the simple R-module r, but by Lemma 1.10, it does not.

(1.12) Example. Consider Example 1.4 (10). This is an F-representation of Z, which is indecomposable but not irreducible: it has r as a subrepresentation (consisting of vectors in F x {0}). Indeed, if T is a linear transformation on M, Remark 7.1.40 gives a criterion for M to be a semisimple F(X)-module.

The following lemma, incorporating the technique of averaging over the group, is crucial:

(1.13) Lemma. Let G be a finite group and F a field with char(F) = 0 or prime to the order of G. Let V1 and V2 be F-representations of G defined Aut(V2) for i = 1,2. and let n = JGI. Let f E Hom(Vi, V2) and by a, : C set (1.3)

Av(f) = n Ea2(9-1)f(Q1(9)) gEG

Then Av(f) E Homc(Vi, V2) Furthermore, if f E HomG(Vl, V2), then

Av(f) = f.

(1.4)

Proof. We need to show that for every go E G and every v1 E

V1,

0`2(9o) Av(f)(vi) = Av(f)(o1(9o)(vl)) But n E a2(9-1)f(a1(9)) gEG

Av(f)(ai(9o)(vl)) =

(ai(9o)(vl))

= 1 n j a2(9-1)f(at(9)) (a1(9o)(v1)) 9EG

= 1 E a2(9-1)f(Q1(99o))(v1) n gEC = 1

E a2(9o)(12(9019-1)f(a1(99o))(VI)

n gEG


=

447

1 E C2(90)C2((99o)-1)f(Cl(99o))(ul)

n gEG

Let g' = gg0. As g runs through the elements of G, so does g'. Thus,

Av(f)(o1(9o)(vl)) = n E a2(9o)Q2(9 -1)f(a1(9))(v1) g'EG = a2(go)

1 E a2(9 _1)f(or1(9 ))

n

(vl)

g'EG

= 02 (go) Av(f)(vl)

as required, so Equation (1.3) is satisfied. Also, if f E Homc(V1, V2)1 then for every g E G,

C2(9)f = fol(9) Hence, in this case

Av(f) = ,11 EC2(9-1)fc1(9) gEC

= 1

f°1(9-1)01(9) gEC

= 1 F, fcl(e) gEC

f gEG

Hence, Equation (1.4) is satisfied.

Now we come to one of the cornerstones of the representation theory of finite groups.

(1.14) Theorem. (Maschke) Let G be a finite group. Then F(G) is a semisimple ring if and only if char(F) = 0 or char(F) is relatively prime to the order of G.

Proof. First, consider the case where char(F) = 0 or is relatively prime to the order of G. By Theorems 7.1.28 and 7.1.23, it suffices to show that every submodule M1 of an arbitrary F(G)-module M is complemented. Let t : M1 -. M be the inclusion. We will construct -ir : M --+ M1 with in = 1M,. Assuming that, we have a split exact sequence

448


0 --+ M1 --"-+ M -p M/Ml -' 0, so Theorem 3.3.9 applies to show that M 5 M1 ® Ker(rr), and hence, M1 is complemented. Now, M1 is a subvector space of M, so as an F-module M1 is comple-

mented and there is certainly a linear map p : M -. M1 with pt = 1M,. This is an F-module homomorphism, but there is no reason to expect it to be an F(G)-module homomorphism. We obtain one by averaging it. (Since we are dealing with a single representation here, for simplicity we will write g(v) instead of a(g)(v).) Let it = Av(p). By Lemma 1.13, it E HomG(M, M1). We have pt = 1M,; we need at = 1M,. Let v E M1. Then, since M1 is an F(G)-submodule, g(v) E M1 for all g E G, so p(t(g(v))) = g(v) for all g E G. Then

= 1 F, 9-1P(9(t(v))) n 9EG

= 1n F, 9-1P(t(9(v))) 9EG

= 1 >2 9-1(9(v)) n 9EG

nv

n

=v

as required. Now suppose that char(F) divides the order of G. Recall that we have

the augmentation map e as defined in Example 1.4 (11), giving a short exact sequence of F(G)-modules (1.5)

0 --+ R.o -- R. --"-+ r -- 0.

Suppose that R were semisimple. Then a would have a splitting a, so b Theorem 3.3.9, R Ro ®a(r). Since this is a direct sum, Ro fl a(r) = (0). On the other hand, a(r) is a trivial subrepresentation in R, so by the proof of Lemma 1.10,

a(r)={a1: g:aEF}. 9EG

However,

E(a1: g) =(Je(Eg) =an=OE F gEG

9EG

since char(F) I n. Thus, a(r) C Ro, contradicting Ro fl a(r) = (0).

0


449

(1.15) Example. Consider Example 1.4 (10) again, but this time with

F = Fy, the field of p elements. Then Tp = [o i] = [o i], so we may

regard T as giving an F-representation of Z.. As in Example 1.12, this is indecomposable but not irreducible. Having proven that F(G) is semisimple in favorable cases, we collect the relevant results of Section 7.1 into an omnibus theorem.

(1.16) Theorem. Let G be a finite group and F a field with char(F) = 0 or prime to the order of G. Then the following are valid.

An F-representation of G is indecomposable if and only if it is irreducible.

Every irreducible F-representation of G is isomorphic to a subrepresentation of F(G). Every F-representation of G is a projective F(G)-module. Every F-representation of G is semisimple, and if M is written as

M °_° ®siMi

for distinct irreducibles

{Mi}IEl,

iE1

(5)

then si E {0, 1, 2, ...} U {oo} is well determined. There are only finitely many distinct irreducible F-representations of G (up to isomorphism), and each has finite multiplicity in F(G).

0

Proof.

(1.17) Corollary. Let G be a finite group and F a field with char(F) = 0 or prime to the order of G. Then every irreducible F-representation of G has degree at most n and there are at most n distinct irreducible Frepresentations of G, up to isomorphism. Proof. If Mi is irreducible, then Theorem 1.16 (2) implies that Mi is a subrepresentation of F(G), so

deg(Mi) < deg(F(G)) = n. If {Mi}iEl are all of the distinct irreducible representations, then by Theorem 1.16 (2), M = ®iE1Mi is a subrepresentation of F(G). Hence, deg(M) < deg(F(G)), so that

III = E 1 < E deg(Mi) < deg(F(G)) = n, iEl

as claimed.

iE1

0

The second basic result we have is Schur's lemma, which we amplify a bit in our situation.


450

(1.18) Lemma. (Schur) Let R be an F-algebra. (1)

Let M be a simple R-module. Then EndR(M) is a division ring containing F in its center. If F is algebraically closed, then every 0 E EndR(M) is a homothety (i.e., is multiplication by an element of F) and EndR(M) = F.

(2) Let Ml and M2 be two distinct (i.e., nonisomorphic) simple R-modules.

Then HomR(Mj, M2) = (0). Proof. Much of this is a restatement of Schur's lemma (Proposition 7.1.5).

If D = EndR(M), then clearly D D F, where we regard a E F as the endomorphism given by (left) multiplication by a. If 0 E D, then 0(am) _ a4(m) for all m E M, so F is contained in the center of D. Now suppose that F is algebraically closed and 0 E EndR(M). Then

the characteristic polynomial of 0 splits into linear factors in F, so, in particular, 0 has an eigenvalue a E F. Then Ker(4 - a) is a nontrivial submodule of M. Since M is simple, this implies Ker(o - a) = M, i.e.,

0=a.

0

(1.19) Example. Here is an example to see that if M is a simple R-algebra and F is not algebraically closed, then EndR(M) need not consist solely

of homotheties. Let R = R(Z4) and M = R2 with the action of Z4 = {1, g, g2, g3} given by the matrix 0 a(g)

=

1

0

1

It is easy to check that

EndR(M)_{[b

b :a,bER}=C

under the isomorphism [ b a ] '--» a + bi. We close this section with the following lemma, which will be important later.

(1.20) Lemma. Let G be a finite group and F a good field. Let or

:

Aut(M) be an F-representation of G of finite degree. Then for each g E G, the linear transformation o(g) : M -4 M is diagonalizable. G

Proof. Since g E G, we have that gk = 1 for some k dividing m = exponent (G). Since F is good, the lemma follows immediately from Theorem 4.4.34. 0

8.2 Representations of Abelian Groups

451

8.2 Representations of Abelian Groups Before developing the general theory, we will first directly develop the representation theory of finite abelian groups over good fields. We will then see the similarities and differences between the situation for abelian groups and that for groups in general.

(2.1) Theorem. Let G be a finite group and F a good field for G. Then G is abelian if and only if every irreducible F-representation of G is onedimensional.

Proof. First assume that G is abelian and let M be an F-representation of G. By Corollary 1.17, if deg(M) is infinite, M cannot be irreducible. Thus we may assume that deg(M) < oo. Now the representation is given by a homomorphism a : G -+ Aut(M), so

a(g)a(h) = a(gh) = a(hg) = a(h)a(g)

for all

g, h E G.

By Lemma 1.20, each a(g) is diagonalizable, so S = {a(g) : g E G} is a set of mutually commuting diagonalizable transformations. By Theorem 4.3.36, S is simultaneously diagonalizable. If B = {v1, ... ,vk} is a basis of M in which they are all diagonal, then

M is semisimple). Hence M, is simple if and only if k = 1, i.e., if and only if deg(M) = 1. Conversely, let M be a one-dimensional representation of G. Then the representation is defined by a homomorphism a : G - Aut(M) 9 F', and F' is, of course, abelian, so that (2.1)

a(g)a(h) = a(h)a(g)

for all

g, h E G.

By assumption, every irreducible representation of G is one-dimensional, so Equation (2.1) is valid for every irreducible representation of G. Since every representation of G is a direct sum of irreducible representations, Equation (2.1) is valid for every representation of G. In particular, it is valid for R(G), the regular representation of G. If ao is the homomorphism defining the regular representation, and we consider 1 E 7Z(G), then for any g, h E G, (ao(g)ao(h))(1) = (ao(h)ao(g))(1)

ao(g)(h) = ao(h)(g)

gh = hg and G is abelian.

452


(2.2) Theorem. Let G be a finite group and F a good field for G. Then G is abelian if and only if G has n distinct irreducible F-representations. Proof. Let G be abelian. We shall construct n distinct F-representations of G. By Theorem 3.7.22, we know that we may write G as a direct sum of cyclic groups

G = Zn, ® Zn2 ® ... ®Zn.

(2.2)

with n = n1

n3. Since F is good for G, it is also good for each of

the cyclic groups Zni, and by Example 1.4 (6), the cyclic group Zn, has

the ni distinct F-representations 0k for 0 < k < ni - 1; to distinguish these representations for different i, we shall denote them ok' . Thus, On'

Zn, - F. If rri : G -+ Zn; denotes the projection, then on,7ri : G --+ F' = Aut(F)

(2.3)

defines a one-dimensional representation (and, hence, an irreducible representation) of G. Thus,

{ok'7ri: 1 1 for some i. Then, as in the proof of Corollary 1.17,

III = F, 1 < F, deg(Mi) < deg(F(G)) = n, iE1

iE1

so III < n, as claimed.

D

(2.3) Corollary. Let G be a finite abelian group and F a good field for G. If M1, ... , Mn denote the distinct irreducible F-representations of G, then n

F(G)

Mi. i-1

(In other words, every irreducible representation of G appears in the regular representation with multiplicity one.)

Proof. If M = ®

1

Mi, then by Theorem 1.16 (2), M is a subrepresentation

of R. But deg(M) = n = deg(R), so M = R.

D

(2.4) Corollary. Let G be a finite abelian group and F a good field for G. If M is an irreducible representation of G, then

8.3 Decomposition of the Regular Representation

453

EndG(M) = F.

Proof. Clearly, F C Endc(M) C End(M) = F. (Every F(G)-homomorphism is an F-homomorphism and M is one-dimensional.)

0

(2.5) Proposition. Let G be a group and suppose that Q = G/(G, G] is finite. Let F be a good field for Q. Then G has JQI distinct one-dimensional F-representations.

Proof. If it is the canonical projection ar : G -' Q, then for any onedimensional F-representation v : Q -. Aut(F), its pullback ar'(a) = oar is a one-dimensional F-representation of G, and by Theorem 2.2, we obtain JQI distinct representations in this way. On the other hand, if at : G -+ Aut(F) = F* is any one-dimensional representation of G, then is trivial (as F' is abelian). Thus, a' factors as air, so it is one of the representations constructed above.

0

8.3 Decomposition of the Regular Representation (3.1) Definition. Let G be a finite group. A field F is called excellent for G or simply excellent if it is algebraically closed of characteristic zero or prime to the order of G. Observe that an excellent field is good and that the field C is excellent for every G. Our objective in this section is to count the number of irreducible representations of a finite group G over an excellent field F, and to determine their multiplicities in the regular representation F(G). The answers turn out to be both simple and extremely useful.

(3.2) Definition. A representation P of G is called isotypic (of type M) if for some positive integer m and some irreducible representation M of G, P is isomorphic to mM. (3.3) Lemma. Let F be an excellent field for G, and let P and Q be isotypic representations of G of the same type. If P 5 m1M and Q m2M, then as F-algebras (3.1)

Homc(P, Q) 25 Mm2,mi (F)

Proof. Let 0 : P -+ m1M and r/i : Q - m2M be the isomorphisms (of F(G)-modules). Let a; : M -+ m1M be the inclusion of M as the i1h summand, and let 3j : m2M -' M be the projection onto the jth summand.

454


If f E HomG(P, Q), consider the following composition of F(G)-module homomorphisms (which we will call fji): Q'+ M.

Then f7i E EndG(M), and since F is excellent for G, Lemma 1.18 (Schur's lemma) implies that fji is given by multiplication by some element a3i E F. Then the isomorphism of the lemma is given by

f

A=

[ai,).

0 Remark. Note that this isomorphism depends on the choice of isomorphisms

0 and ii. This dependence is nothing more than the familiar fact that the matrix of a linear transformation depends on a choice of bases. The following theorem, proved by Frobenius in 1896, is fundamental:

(3.4) Theorem. (Frobenius) Let F be an excellent field for the finite group G. (1) The number of distinct irreducible F-representations of G is equal to

the number t of distinct conjugacy classes of elements of G. (2) If {Mi}i=1 are the distinct irreducible F-representations of G, then the

multiplicity of Mi in the regular representation R of G is equal to its degree di = deg(Mi) for l < i < t. (3) Et=1 d? = n = JGJ.

-

Convention. We adopt the notational convention henceforth that t will always denote the number of conjugacy classes of the group G.

Proof. If {Mi},° .1 are the distinct irreducible representations of G (the number of these being finite by Theorem 1.16 (5)), then by Theorem 1.16 (2) we have a

®miMi i=1

for some positive integers m1, ... , mq. (1) We shall prove this by calculating dimF C in two ways, where C is the center of R, i.e.,

C={rER:rr'=r'r forallr'ER}. C is clearly an F-algebra. First we compute dimF (C) directly. Let {Ci };_ 1 be the sets of mutually

conjugate elements of G, i.e., for each i, and each 91 and g2 E Ci, there


455

is a g E G with gl = 9929-1. Since conjugacy is an equivalence relation, {C1}i=1 is a partition of G. For 1 < i < t, let c; E R be defined by

c;= 1: g. 9EC,

For each element g of G, we have the following equality in R:

E 9i

9

g,ECi

E gig 9cEC+

1: 9(9-19:9) g EC,

99, 9; EC;

where the fourth equality holds because C1 is a conjugacy class. This immediately implies that C _D (cl, ... , ct),

(3.4)

the F-vector space spanned by these elements. On the other hand, suppose that we have an element x =

a9g E C.

gEG

Then for any go E G, we have goxg 1 = x9090 1 = x. But 9oxg61 =

a99o990 1 = g'EG

gEG

ago'9'go9'

which implies that a9 = ago ',.90

for all

g, go E G.

That is, any two mutually conjugate elements have the same coefficient, and hence, C C (c1, ... , ct). Together with Equation (3.4), this implies that C = (cl, ... , ct), and since c1, ... , ct are obviously F-linearly independent elements of R, it follows that


456

dimF(C) = t.

(3.5)

Now for our second calculation of dimF(C). We know in general that (3.6)

R a, HomR(R, R) = EndR(R).

We will calculate dixF(C') where C' is the center of EndR(R). Of course, dimF(C') = duuF(C) by Equation (3.6). But

EndR(R) = HomR(R, R) 9

HomR(®m$Mi, $m, M,) i=1 9

,j=1

9

®®HomR(m;M;, m,M,). i=1 j=1

By Schur's lemma, HomR(M;, Mj) = (0) for i

j, and by Lemma 3.3,

HomR(miM;, miMi) L, M., (F)

so that q

(3.7)

R EndR(R) °-` (D Mm, (F) =1

as F-algebras. It is easy to see that (3.8)

C'

Cq

where Ci is the center of the matrix algebra Mn, (F); but by Lemma 4.1.3, Ci = FIm, C Mm, (F) so that dimF(Ci) = 1. By Equation (3.8), it follows

that dunF (C') = q.

Hence, q = t, as claimed. (2) We shall prove this by calculating dimF(M;) in two ways. First, by definition, (3.9)

dimF(Mi) = d:.

Second, we have HomR (R, Mi) dimension of this. Now

M1 as R-modules, and we calculate the 9

HomR(R, M1)

HomR((DmjM,, M1) 9=1

HomR(m:M:, M;) M1 m. (F)


457

by Schur's lemma and Lemma 3.3 again. This matrix space has dimension m; over F, so m; = d;, as claimed. (3) By part (2) and Equation (3.7),

n = dimF(R) a

= dimF (®M.,

(F))

i=1 9

_

dimp(M., (F)) :=1 a

= E m? :=1 c

=

d 2

as claimed.

D

(3.5) Remark. Note that this theorem generalizes the results of Section 8.2. For a group G is abelian if and only if every conjugacy class of G consists of a single element, in which case there are n = IGI conjugacy classes. Then G has n distinct irreducible F-representations, each of degree 1 and appearing with multiplicity 1 in the regular representation (and n = E "j 12). (3.6) Warning. Although the number of conjugacy classes of elements of G is equal to the number of distinct irreducible F-representations of G (if F is an excellent field for G), there is no natural one-to-one correspondence between the set of conjugacy classes and the set of irreducible F-representations.

(3.7) Example. Consider the dihedral group G=D2m=(x,y:xm=1,y2=1,xyyx-1).

For m odd, G has the following conjugacy classes:

l), f1

{x, xm-1}, {x2, xm-2} ,.

{X(m-1)/2 x(m+1)/2}

{y, xy, x2y,

.,

... ,x"'-1y}.

There are (m + 3)/2 conjugacy classes and in Example 1.4 (7), we constructed (m + 3)/2 distinct irreducible representations over a good field F, so by Theorem 3.4 we have found all of them. For m even G has the following conjugacy classes:

{xI-1, x"}, {x1}, {1}, {x, X'- 1), {x2, xm-2}, ... , {x;y : i is even}, {x'y : i is odd}.


458

There are i + 3 of these and in Example 1.4 (7) we also constructed z + 3 irreducible representations over a good field F, so again we have found all of them. Note also that the decomposition in Example 1.8 (4) is as predicted by Theorem 3.4.

(3.8) Example. Let us construct all irreducible C-representations of the quaternion group G = Q8. Recall that Q8 = {±1, ±1, ±j, ±k}, and it is straightforward to compute that it has conjugacy classes {1},

{-1}.

{±i},

{fj},

{±k}.

(See Example 2.1.10 (10), where we denote Q8 by Q.) Thus, we have five irreducible representations whose degrees satisfy j= I d? = 8, which forces dl = d2 = d3 = d4 = 1 and d5 = 2. (Actually, it is unnecessary here to find the number of conjugacy classes, for the Equation

;-i d? = 8 only has the solutions (2, 2),

(1, 1, 1, 1, 1, 1, 1, 1),

and

(1, 1, 1, 1, 2).

The first of these is impossible since we must have some di = 1 = deg(r), and the second cannot be right because G is nonabelian, so the third must apply.)

Note that C = {f 11 is the center of G, and so C a G, and we have an exact sequence

1-C,G'ff V-1

where V = Z2 ® Z2. Since V is abelian of order 4, it has four 1-dimensional representations and their pullbacks give four 1-dimensional (and hence certainly irreducible) representations of G. To be precise, let

V={1,I,J,K:12=J2=K2=1, IJ=K}. V is defined by 7r(±1) = 1, 7r(±i) = I, 7r(±j) = J, and Then it : G 7r(±k) = K. The representations of V are the trivial representation ao = r (and 1r' (r) = r) and the representations ai for i = 1, 2, 3 given by al

al(l) =

02

a2(I) = -1, a3(I) = -1,

C3

1,

al(J) = -1,

al(K) = -1

a2(J) =

a2(K) = -1 a3(K) = 1.

1,

a3(J) = -1,

We also need to find a two-dimensional representation p of Q8. Here it is.


P(±1) =

P(fi) = P(±j)

[d1

459

0

0

[±i 0

0

:i

0

= I±i 0

p(±k) = 0 ±1 1

0

01

Note that in the matrices, i is the complex number i. We must check that p is irreducible. This can be done directly, but it is easier to make the following

observation. If p were not irreducible it would have to be isomorphic to a'(Qi) ®7r`(aj) for some i, j E {0, 1, 2, 3}, but it cannot be, for p(-1) is nontrivial, but (7r' (vi) ®7r' (off)) (-1) is trivial for any choice of i and j. (3.9) Example. Let us construct all irreducible C-representations of the alternating group A4 of order 12. Recall that A4 is the subgroup of the symmetric group S4 consisting of the even permutations, and it is a semidirect product 1 V A4 -+ S --+ 1 where

VZ2ED Z2 _ {1, (12)(34), (13)(24), (14)(23)}

={1,I,J,K} and

S ° c Z3 = {1, (1 2 3), (13 2)} = {1, T, T2}.

We compute that A4 has 4 conjugacy classes {1},

{I, J, K},

{T, TI, TJ, TK},

{T2, T2I, ,T2J, T2K},

so we expect 4 irreducible representations whose degrees satisfy ri 1 d = 12, giving dl = d2 = d3 = 1 and d4 = 3. (Alternatively, we find that V is the commutator subgroup of G, and so we have exactly 3 one-dimensional representations of G by Proposition 2.5. Then the equation Ei=1 d? = 12 and dl = d2 = d3 = 1 with di > 1 for i > 3 forces t = 4 and d4 = 3.) The three one-dimensional representations of G are 7r* (O..) for 0 < i < 2, where Bi are the representations of the cyclic group S constructed in Example 1.4 (6). Now we need to find a three-dimensional representation. Let

M=C4={(zl,z2,z3,z4):ziEC, 1 <4}. Then S4, and hence, A4, acts on C4 by permuting the coordinates, i.e.,


460

for

9(zl, z2, z3, z4) = (z9(1), z9(2), zg(3), Z9(4))

g E S4.

Consider

M0 = {(z1, z2, z3, z4) : z1 + 22 + Z3 + z4 = 0}.

This subspace of M is invariant under S4, so it gives a three-dimensional representation a of S4, and we consider its restriction to A4, which we still denote by a. We claim that this representation is irreducible, and the argument is the same as the final observation in Example 3.8: The subgroup V acts trivially in each of the representations 7r* (0j) for i = 0, 1, 2, but nontrivially in the representation a.

(3.10) Example. Let us construct all irreducible C-representations of the symmetric group S4. We have a semidirect product

V

is the same as in Example 3.9 and

W =(T,U:T3=U2=1, TU=UT 2)?-1D6L' S3 where T = (1 2 3) as before and U = (1 2). By Corollary 1.5.10, S4 has 5 conjugacy classes, so we look for a solution of FI5=1 d? = 24 = 1541. This has the unique solution

di=d2=1,

d3 = 2,

d4 = d5 = 3.

As before, we obtain an irreducible representation of S4 from every irreducible representation of W, and by Example 1.4 (7), W has the irreducible representations tp+ = r, V'_, and 01, so we have irreducible representations ir' (r) = r, a' (ty+), and a' (01) of degrees 1, 1, and 2, respectively. We need to find two three-dimensional irreducible representations. For the first we simply take the representation a constructed in Example 3.9. Since the restriction of a to A4 is irreducible, a itself is certainly irreducible.

For the second, we take a' = lr'(r/i_) ® a. This is also a threedimensional representation of S4 (which we may now view as acting on Mo by

a'(g)(v) = 1/i_ir(g)a(v)

for

vE

1V10

since V)_7r(g) = ±1). Since a' restricted to A4 agrees with a restricted to A4 and the latter is irreducible, so is the former. To complete our construction, we need only show that a and a' are inequivalent. We see this from the fact

that a(U) has characteristic polynomial (X - 1)2(X + 1), while a'(U) has characteristic polynomial (X - 1)(X + 1)2. (The reader may wonder about the representation 7r* (-O-) ® zr' (0i ), but this is isomorphic to A' (q1), so it gives nothing new.)

We conclude this section with the following result:


461

(3.11) Theorem. (Burnside) Let F be an excellent field for the finite group G and let p : G -+ Aut(V) be an irreducible F- representation of G. Then

{p(g):gEG} spans EndF(V).

Proof. We first claim that for any field F, if po : G -+ Aut(F(G)) is the regular representation of G, then {po(g)} is a linearly independent set in Aut(F(G)). For suppose a9po(g) = 0.

a = 9EG

Then 0 = a(1) = E agg 9EG

so a9 =0 for each gE C. Now let F be an excellent field for G. Then by Theorem 3.4 we have an isomorphism 0 : Vo -+ F(G) with Vo = ®2=id2V2, po : G -+ Aut(Vo), where pt : G -+ Aut(VV) are the distinct irreducible F-representations of G. Choose a basis B, for each V and let 13 be the basis of Vo that is the union of these bases. If MM(g) = [pt(g)]B,, then for each g E G, [po(g)]B is the block diagonal matrix

diag(Mj(g), M2(g), ... M2(g),... , Mt(g), ... , Mt(9)) ,

where M2(g) is repeated d, = dim(U) times. (Of course, M1(g) = [1] appears once.) By the first paragraph of the proof, we have that the dimension of {po(g) : g E G} is equal to n, so we see that t

n<E42 where q2 is the dimension of the span of {pi(g) : g E G}. But this span is a subspace of End(V2), a space of dimension d?. Thus, we have t

t

i=1

i=1

n
Chapter8. Group Representations

462

8.4 Characters In this section we develop the theory of characters. In practice, characters are a tool whose usefulness, especially in characteristic zero, can hardly be overemphasized. We will begin without restricting the characteristic.

(4.1) Definition. Let a : G - Aut(M) be an F-representation of G of finite degree, and let B be a basis of M. The character of the representation a is the function X, : G F defined by (4.1)

Xa(9) = TrQa(9)(B)

Recall that Tr denotes the trace of a matrix or a linear transformation. This is independent of the choice of the basis B by Proposition 4.3.27. By the same logic, it is the case that if two representations are equivalent, then

their characters are equal. It is one of the great uses of characters that, under the proper circumstances, the converse of this is true as well. (4.2) Examples. (1) If a = drr, then X, (g) = d for every g E G. (2) If a is any representation of degree d, then X,(1) = d. (3) If a is the regular representation of C, then X,(1)

= n = IGI and 1. (To see this, consider [a(g)] in the basis

X,(9) = 0 for all g {g : g E G} of F(G).)

(4.3) Lemma. If g1 and 92 are conjugate elements of G, then for any representation a, (4.2)

Xa(91) = Xa(92)

Proof. If 92 = 9919-1, then Xa(92) = Xa(9919-1)

= Tr ((a(9919-1)]B)

='n ((a(9)]B(a(91)]B(a(9)]s1) = Tr ((a(9i )]B) = X" (91) -

0

(4.4) Proposition. Let a1 and a2 be two representations of the group G. Then (1) Xoi®o2 = Xa, + Xo2, and

8.4 Characters

463

(2) Xo, ®02 = Xoi Xo2 .

Proof. (1) is obvious, and (2) follows immediately from Proposition 7.2.35 0 and Lemma 4.1.20.

Our next goal is to derive the basic orthogonality results for characters. Along the way, we will derive a bit more: orthogonality for matrix coefficients.

(4.5) Proposition. Let F be an excellent field for the group G, let Vi be irreducible representations of G, given by ai : G Aut(t i), and let 8i be

a basis of Vi, fori=1,2. For g E G, let P(g) = [Pi, j, (9)] = [ai (9)]s, and

Q(g) = [gi2j2(9)( = [0`2(9)Js2. (1)

Suppose that V1 and V2 are distinct. Then for any i1, ji, i2, j2,

1n

Pi, jl (9)gi2 j2 (9-1) = 0. 9EG

(2) Suppose that V1 = V2 (so that a1 = a2) and Bl = Cit. Let d = deg(Vi). Then 1

1

n E Pil j, (9)gi2 j2 (9 ) = 9EG

1

1/d if it = j2 and ji = i2, 0

otherwise.

(Note that in this case pij(g) = gij(g), of course.) Proof. Let Qi be the projection of V1 onto its ith summand F (as determined

by the basis B1) and let aj be the inclusion of F onto the jth summand of V2 (as determined by the basis 82). Then f = aj$i E Hom(Vj, V2). Note that [aj(33JBs = Eji where Eji is the matrix with 1 in the jith position and 0 elsewhere (see Section 5.1). Let us compute Av(f). By definition

Av(f) = n F, a2(9-1)(ai#i)a1(9)gEG

Direct matrix calculation shows

pij(9)gii(9-1) Pij(9)gi2(9-1) (4.5)

(a2(9-1)(aj,3i)al(9)JBs =

P2j(9)gii(9-1)

P2j(9)gi2(9-1)


464

so the sums in question are just the entries of [Av(f ))81 (as we vary i, j and the entry of the matrix.) Consider case (1). Then Av(f) E HomG (Vi , V2) = (0) by Schur's lemma, so f is the zero map and all matrix entries are 0, as claimed. Now for case (2). Then Av(f) E HomG(V, V) = F, by Schur's lemma, with every element a homothety, represented by a scalar matrix. Thus all the off-diagonal entries of [Av(f )Jg, are 0, showing that the sum in Equation (4.4) is zero if i1 0 j2. Since a1 = 0`2i we may rewrite the sum (replacing g

by g-1) as 1

E Pi2is

(9

1

)

9EG

showing that it is zero if j1 # i2.

Consider the remaining case, where i1 = j2 = i and jl = i2 = j. As Av(f) is a homothety, all of the diagonal entries of its matrix are equal, so we obtain a common value, say x, for 1

n

F pii(9)4ii(9-1)

gEC

for any i, j (by varying the choice of f and the diagonal element in question). Now consider fo=a1f1+a2l32+....+ad$d.

Since there are d summands, we see that the diagonal entries of Av(fo) are all equal to dx. But fo is the identity! Hence, Av(fo) = fo has its diagonal entries equal to one, so dx = 1 and x = 11d, as claimed. (4.6) Corollary. Let F be an excellent field for G with char(F) = p 0, and let V be an irreducible F-representation of G. Then deg(V) is not divisible by p.

Proof. If d = deg(V), then the above proof shows that dx = 1 E F, so that

d96 0EF. (4.7) Corollary. (Orthogonality of characters) Let F be an excellent field for G, and let V1 and V2 be irreducible representations of G defined by

of:G-.Aut(V) fori=1,2. Then 1

n

1

Xo, ( 9) X1,2 ( 9

_ f 0 if V1 and V2 are distinct, ) = l 1 if V1 and V2 are isomorphic.

g

Proof. If Vi has degree di, this sum is equal to

8.4 Characters

1 1: n

d3

d2

dl

465

dz

EEPii(9)gjj(9-1) =EE n P;;(9)9jj(9-1)

gEG

i=1 j=1

i=1 j=1

gEG

which is 0 if V1 and V2 are distinct. If V1 and V2 are isomorphic of degree d, then, since isomorphic representations have the same character, we may assume that V1 = V2. The with i ,,E j are all zero, so the sum is d

EPjj(9)4jj(9-1) = d(1/d) = I. j=1

0 Proposition 4.5 and Corollary 4.7 have a generalization, as follows:

(4.8) Proposition. Given the same hypothesis and notation as Proposition .¢.5, let h be a fixed element of G. (1) Suppose that V1 and V2 are distinct. Then for any i1i j1i i2, and j2i

we have 1)

1

n E Piij, (h9)4i2j2 (9

= 0.

9EG

(2) Suppose that V1 = V2. Then if i1 # j2i 1)

1

1

piij, (h9)4isjs (9

= 0.

9EG

(3) For any two representations V1 and V2,

nI 9 Xo, (hg) X'12 (9

1

!0 (1/d)Xol (h)

if V1 and V2 are distinct,

if Vi = V2 is of degree d.

Proof. The proof follows that of Proposition 4.5, with f = aj oiol (h). Then the matrix corresponding to that in Equation (4.5) is

P1j(h9)4i1(9-1) P1j(h9)gi2(9-1) P2j(h9)gi1(9-1) P2j(h9)4i2(9-1)

..

.

First suppose that V1 and V2 are distinct. Then, again, Av(f) = 0, so all the matrix entries are zero, yielding (1), and then the first assertion of (3) follows as in the proof of Corollary 4.7.


466

Now suppose that V1 = V2. Then, again, Av(f) is a homothety, so the off-diagonal matrix entries are zero, yielding (2), while all the diagonal entries are equal. Set i = j and call the common value in this case xj, i.e., 1

xj = n F, Pij(h9)9ji(9 9EG

1),

this sum being independent of i. As in the proof of Corollary 4.7, the sum we are interested in is d

d

EE 1=1 j=1

- >P:t (h9)9jj (9 n

d 1)

_

1

1

j=1

gEG

> Pjj (h9)9jj (9

I)

gEG

x1+...+xd

where the first equality follows from part (2). Now consider fo = (0101 + ... + adQd)a1(h)

Then Av(fo) is a homothety, and all of its diagonal entries are equal to

x1+...+xd+SO

Tr(Av(fo)) = d(x1 + ... + xd).

But fo = a, (h), so Av(fo) =

n

1: ol(9-Ih9) gEG

Then

d(xi + .

+ xd) = Tr(Av(fo)) = 1 n > Tr(aj(9-Ih9)) gEG

= 1 E Tr(a1(h)) n gEG _ Tr(o, I(h))

= xo, (h),

yielding the desired equality. (4.9) Remarks.

(1) We should caution the reader that there are in fact three alternatives in Proposition 4.5: that VI and V2 are distinct, that they are equal, or that they are isomorphic but unequal. In the latter case the sum in Equation (4.4) and the corresponding sum in the statement of Proposition 4.8 may vary (see Exercise 14). Note that we legitimately reduced the third

8.4 Characters

467

case to the second in the proof of Corollary 4.7; the point there is that while the individual matrix entries of isomorphic representations will differ, their traces will be the same. (2) We should caution the reader that the sum in Proposition 4.8 (2) with

i1 = j2 but jl 36 i2 may well be nonzero and that the quantities XI, ... , xd in the proof of the proposition may well be unequal (see Exercise 15).

Given a finite group G and an excellent field F, let G have conjugacy classes

C1i...,Ct (in some order) and irreducible representations

01i...,Ot (in some order). (4.10) Definition. The character table of G is the matrix A E Mt (F) defined by A = [aij] = [Xi(cj)] where Xi = X ,j and cj E Cj.

Let ICiI be the number of elements of Ci. Then Corollary 4.7 gives a sort of orthogonality relation on the rows of A; namely, (4.7)

n E Xi(9)Xj(9-1) = n E ICkIXi(ck)Xj(ck)-1 = 6ij k=1

9EG

(Recall that 8ij = 1 if i = j and 0 if not.) From this, we can immediately write down B = A'1; namely, (4.11) Lemma. B = [bij] =

[i4XiC')]

Proof. The equation AB = I is the equation >k=1 aikbkj = bij, which is immediate from Equation (4.7).

Now let us interpret B.

(4.12) Definition. Let f : G -+ F be a function with the property that whenever gl and g2 E G are conjugate, then f(g1) = f(92). Then f is called a class function on G.

Clearly, f : G - F is a class function if and only if it is constant on each conjugacy class Ci (1 < i < t). Thus the space of class functions is clearly an F-vector space of dimension t, with basis 5 = if,, ... , ft} where


468

fi (9)

1

0

if g E C;, otherwise.

(4.13) Lemma. Let F be an excellent field for G. Then

A={Xi,...,Xt} is a basis for the space of class functions on G.

Proof. By Lemma 4.3, the characters are class functions. There are t of them, so to show that they are a basis it suffices to show that they are linearly independent. Suppose aix, + a2X2 + - + atXt = 0

where

a, E F

where, as in Definition 4.10, we have written X; for X,,. Note that X,!9 defined by X; (9) = Xi(9-1), is also a class function. Then for each i, (a1X1 + a2X2 + ... + atXt)X* = 0 and averaging over the group

1

a1X1(9)X,(9) + a2X2(9)X.(9) +

+ atXt(9)Xc(9) = 0.

n 9EG

But, by the orthogonality relations, this sum is just a;/n, so that ai/n = 0 and a; = 0 for each i, as required. 0

(4.14) Remark. If x is the character of the representation V defined by o : G - Aut(V), then X*(9) = X(9-1) is indeed the character of a representation; namely, with the given action of G, V is a left F(G)-module. The action of G given by g '- o(g-1) gives V the structure of a right F(G)module. Then V' = HomG(V, F(G)) is a left F(G)-module with character X'. (In of matrices, if B is a basis for V, B' is the dual basis of V', and X' is defined by a' : G -+ Aut(V*), then (a*(9)le- =

(4.15) Proposition. Let A and B be the above matrices, and let Xi and f; be defined as above. Then for every 1 .

..1 bijxj.

Proof. (1) is easy. We need only show that both sides agree on ck for 1 k < t. Since fj(ck) = 6jk, t

F, a11fj(Ck) = aik = Xi(Ck) j=1

as claimed.

8.4 Characters

469

Now (1) says that the change of basis matrix PB is just A. Then

PA = (Ps')' = A-' = B, 0

giving (2).

As a corollary, we may derive another set of orthogonality relations.

'(4.16) Corollary. Let 9i, 92 E G. Then 0

Xj(g1)Xj(gV)=fT7,vT

if g, and g2 are not conjugate, if gl, 92 E Ci-

j=1

Prof. We may assume g1 = ci and g2 = ck for some i, k. Then by the definition of fk and Proposition 4.15 (2), bki = !k(C-i)

ft

bkjXj

(ci)

j=1

(

t

E-1

1)

l Xj(ci)

n EXj(ci)Xj(ck1) u

is

.

j-1

p

yielding the corollary.

We now define an important quantity.

(4.17) Definition. Let G and F be arbitrary and let V and W be two Frepresentations of G. The intertwining number of V and W is i(V, W) = (V, W) =dimFHomc(V, W).

(4.18) Lemma. Let G be finite, F a field of characteristic 0 or prime to the order of G, and let V and W be two f -representation of G. Then (1)

i(V, W) = i(W, V); ®iEl piMi and W 25 ®iE! 4iMi are the decomposition of V and W into irreducibles, then

(2) if V

i(V, W) = 1: pigidimF(End(Mi)); iE!

470


(3) if F is excellent for G and V is irreducible, then i(V, W) is the multi-

plicity of V in W; (4) if F is excellent for G, then V is irreducible if and only if i(V, V) = 1; (5) if F is excellent for G and char(F) = 0, then

i(V, W) = (V, W) = (Xv, Xw) =

1

n

1: Xv(g)Xw(g-') gEG

Proof. Note that the hypotheses imply that R is semisimple, and recall Schur's lemma. The proof then becomes straightforward, and we leave it for the reader. 0 Remark. If G is finite, F is of characteristic zero or prime to the order of G, and V is of finite degree, then we have i(V, W) = dimF V. ®F(G) W.

This follows directly from Theorem 7.1.28 and Proposition 7.2.32.

Here we see the great utility of characters-if F is excellent for G and of characteristic zero, we may use them to compute intertwining numbers. In particular, we have:

(4.19) Corollary. If F is excellent for G and is of characteristic zero and Wl and W2 are two F-representations of G of finite degree, then Wl and W2 are isomorphic if and only if Xw, = Xw2 Proof. Clearly, Wl and W2 are isomorphic if and only if i(V, W1) = i(V, W2) for every irreducible V. But, by Lemma 4.18 (3), this is true if Xw, = Xw2. The converse is trivial. 0

We could continue to work over a suitable excellent field of characteristic 0, but instead, for the sake of simplicity, we will take F = C (the field of complex numbers). We note that C has the field automorphism z -, z of complex conjugation, with zl + z2 = zl + 12i ziz2 = TI-Z2, and z = z.

If a : G -+ Aut(V) is any complex representation, its conjugate o : G --- Aut(V) is another representation, and X- = Xo By Lemma 4.18 (3), if G is finite, then v is isomorphic too if and only if

Xo=Xo=Xv+ i.e., if and only if Xo is real valued. We have the following important result:

(4.20) Lemma. Let a : G , Aut(V) be a complex representation of the finite group G. Then for every g E G,

8.4 Characters

471

Xo(9-1) = Xo(9)-

Proof. Let g E G. Then g has finite order k, say. By Lemma 1.20, V has a basis B with [o(g)[B diagonal, and in fact, [0'(g)]B = diag(S°', C°', ... , (ad)

where C = exp(27ri/k), d = deg(V), and a1, ... , ad E Z. Then

[°(9-1))B = ([o(9))B)-1 = diag(S 61, S-a', ... ,S-ad).

But (-° =l;° for any a, so d

d

i=1

i=1

X"(9-1) = Tr ([0(9-1)]B)

([o(9)]B) = X.,(9) 0

(4.21) Consequence. If F = C, then X, (g-1) may be replaced by X, (g) in results 4.7, 4.8, 4.11, 4.16, and 4.18. A further consequence is that complex characters suffice to distinguish conjugacy classes. To be precise: (4.22) Proposition. Let g1 and g2 E G. Then g1 and g2 are conjugate if and only if for every complex character X of G, X(91) = X(92)

Proof. The only if is trivial. As for the if part, suppose that X(gl) = X(92) for every complex character. Then by Lemma 4.20, X(92

1)

= X(91),

, Xt are the irreducible characters,

so if XI, t

t

j=1

j=1

t E IXi(91)12 > 0 EXj(91)Xj(92 1) = EXj(91)Xj(91) =

j=1

since each term is nonnegative and X1(91) = 1 (X1 being the character of the trivial representation r). Then by Corollary 4.16, gl and g2 are conjugate.

0 (4.23) Proposition. Let G be a finite group. Then every complex character of G is real valued if and only if every element in G is conjugate to its oum inverse.

472


Proof. Suppose that every g is conjugate to g-1. Then for every complex character X, X(g) = X(g-1) = (g), so X(g) is real. Conversely, suppose that Xi(g) is real for every irreducible complex representation ai and every g E G. Since a1 = r, X1(g) = 1, so e

0<

e

XI(g)2

=

j=1

Xi(g)Xi((g_1)-1) j=1

so g and g-1 are conjugate by Corollary 4.16 again. Lemma 4.20 also gives a handy way of encoding the orthogonality relations (Corollaries 4.7 and 4.16) for complex characters. Recall the character table A = [a11] = [X,(c1)J of Definition 4.10. Let

C = [ciiJ = [4/TTa]id EMt(C).

(4.24) Proposition. The above matrix C is unitary, i.e.,

C-1

= Ct . (See

Definition 4.6.16.)

Proof. We leave this for the reader. Using this result, we may generalize Proposition 4.23. Let us call a conjugacy class C, self-inversive if for some (and hence for every) ci E Ci, we also have c,- 1 E Ci.

(4.25) Proposition. The number of self-inversive conjugacy classes of G is equal to the number of irreducible complex characters of G that are real valued.

Proof. We outline the proof and leave the details to the reader. We make use of the above matrix C. Consider the diagonal elements of CCt. From the fact that Xi = Xi if Xi is real valued and is the character of a distinct :reducible if not, we see that the ith diagonal entry of CCt is 1 if Xi is real valued and zero if not. Thus Tr(CCt) is equal to the number of real-valued characters. Now consider the diagonal elements of C1C. From the fact that ci 1 E Ci if Ci is self-inversive, and is in a different conjugacy class if not, we see that the ith diagonal entry of C1C is 1 if Ci is self-inversive and zero if not. Thus 'Ir(CLC) is equal to the number of self-inversive conjugacy classes, and since Tr(C1C) = Tr(CCt), the proposition follows. (4.26) Corollary. If G has odd order, no nontrivial irreducible character of G is real valued.

8.4 Characters

473

Proof. Suppose that ghg-1 = h-1. Then ging-` = h if i is even or h'1 if i is odd. In particular, g"hg-' = h'1 (recall that n = IGI). i ut g" = 1, so this gives h = h-1, i.e., h2 = 1. Since n is odd, this give h = 1, and the

0

corollary follows from Proposition 4.25.

We now determine the idempotents in C(G). In our approach the irreducible representations and their characters have been central, and the idempotents are of peripheral importance, but there is an alternative approach in which they play a leading role. (4.27) Proposition. For i = 1, ... , t let e; E C(G) be defined by (4.8)

ei =

dt

X1(9-1)9 9EC

where Xi are the characters of the irreducible representations of of G and di = deg(oi) = Xi(1). Then (1) e, = e1 and ejei = 0 for i O j, and

(2) el+...+eg=1.

Proof. (1) We compute (4.9)

ej ei = (

l / n l1 E f

nJ

I

J\

X.1(92 1)Xi(9i 1)9291

/ 9i.92EC

Setting gl = g, g2 = hg-1, and recalling that X1(gh-1) = Xi(h-1g), Equation (4.9) becomes

ejet =

(L) \n

)hEG (9F- Xi(h-19)X1(9-1) I h.

The interior sum, and hence the double sum, is zero if i $ j by Proposition 4.8. If i = j the interior sum is nXi(h-1)/di (again by Proposition 4.8), and so in this case the double sum is d'2

n hEC

(nXi(h-1)/di) h = ei.

(2) Let X be the character of the regular representation. If el+ +et _ EhEG ahh, then by Example 4.2 (3) ah =

Now

1

n

X(h_1(el + ... + et)).

474

Chapter 8. Group Representations t

X(h-1(el + ... + et)) =

L

n, Xj (g-1)X(h-lg)

j=1 gEG

t

1: jXj

- j=1gEG n t

t

_ E j=1 i=1

n

_

diXi(h-1g)

(g-1) i=1

didj

n

Xi(h-1g)Xj(g-1)

gEG

2

(d=)

(nXi(h-1)/di)

n

n

_ dixi(h-1) i=1 n

_

Xi(1)Xi(h-1)

i=1

=

f0 ifh54 1

In ifh=1

where the third equality is by Proposition 4.8 and the last is by Corollary 0 + et = 1. 4.16. Thus, al = 1 and ah = 0 for h # 1, giving el + We will use idempotents to prove the following result:

(4.28) Theorem. Let di = deg(ai) for i = 1, ... , t be the degrees of the irreducible complex representations of G. Then for each i, di divides n, the order of G.

Proof. We begin with the following observation: Let M be a finitely generated torsion-free (and, hence, free) Z-module, and set N = Q ® M. If r E Q has the property that rM C M, then r E Z. To see this, let b be a primitive element of M, which exists by Lemma 3.6.14. If r = p/q with p and q relatively prime, then rb E M implies that b = qb' for some b' E M, which by primitivity implies that q = ±1. Let ( = exp(27ri/m), with m the exponent of G. For fixed i, let Mi be the Z-submodule of C(G) spanned by

{(kgei:k=0,...,m-1, gEG). (Here we consider C(G) as an additive abelian group.) Since e? = ei, we have

n

n -e2 = (Xi(g-1)g) ei E Mi.

di ei = di

This immediately implies (n/di)(Mi) C Mi, so n/di E Z and di divides n, 0 as claimed.

8.4 Characters

475

We now present some typical examples of the use of characters and their properties. In the first two examples we show how to use the basic Theorem 3.4 (and the orthogonality relations (Corollary 4.7)) to find all complex characters of a group (without first finding all the irreducible representations), and in the third we show how to use Lemma 4.18 to find the decomposition of a complex representation into irreducibles. (4.29) Example. Let us determine all the irreducible complex characters of the alternating group As, of order 60. First we determine that As has 5 conjugacy classes: 111, { all products of two dist 2-cycles }, { all 3-cycles } each constitute a conjugacy class, while { all 5-cycles } splits into two conjugacy classes (with the property that if g is in one of them, g2 is in the other). Thus, as representatives for

the conjugacy classes we may take cl = 1, c2 = (1 2)(3 4), c3 = (1 2 3), c4 = (1 2 3 4 5), and c5 = (13 5 2 4). It may be checked that the conjugacy classes have sizes 1, 15, 20, 12, and 12 respectively. Thus there are five irreducible representations whose degrees d, satisfy E.1 d? = 60, which has the unique solution dl = 1, d2 = 3, d3 = 3, d4 = 4, and d5 = 5. Thus, we have determined the degrees of the irreducible representations. Denote the irreducible representations by al, a3, 03, a4i and as with

characters Xi, X3, X3, X4, and Xs We of course have al = r, the trivial representation, with Xi (g) = 1 for all g E G. Consider the representation $ on C5 given by letting As act by permuting the coordinates, i.e.,

9(zi, ... z5) = (z9(1), ... , zg(5)) ,

This is a permutation representation and the trace of any element is easy to compute. For g E As, X,6(g) = I{i : g(i) = ill. Of course Xp(1) = 5, and we observe that Xp(c2) = 1, x#(c3) = 2, and Xp(c4) = X,9(CS) = 0- We

compute (Xp, x) (noting that Xp = Xp) and obtain

1.52+15.12+20.22 1 (X0, X0) = 60( +12.02+12.02) =2 so ,6 has two irreducible components. We compute (XI, X'6) = 60(1 1 .5+ 1 5 . 1 . 1 +20. 1 .2

+12.1.0+12.1.0) =1

so r = al of degree one is one of them. The complem irreducible representation of degree 4, so is a4. Furthermore, since 3 =

476


r ® a4i we have x = X1 + X4, i.e., X4 = X0 - X1 We thus compute X4(1) = 4, X4 (C2) = 0, X4(c3) = 1, and X4 (C4) = X4 (CO = -1.

Next we consider the representation y on C10 given by the action of A5 permuting coordinates, where now C10 is coordinatized by the 10 sets {i, j} of unordered pairs of distinct elements of {1, ... , 5}. Again,

X7(9) = I{{i, j} : g({i, j}) = {i, j) N. Then X7(1) = 10, X, (C2) = 2, X-, (q) = 1, and X7(c4) = x, (c5) = 0. Again, and Xy =

_

(X7, X7

(1 . 102+15.22+20. 12 +12.02+12.02)

= 3,

so y has three irreducible components. We compute

(r,X7)=(1.1.10+15.1.2+20.1.1 +12.1.0+12.1.0) = 1, and since we have already computed X4, we may compute (X4, X7)

60

+20.1. 1

(1 .4.10 + 15.0

1,

so the complement of r ® 04 in y is an irreducible representation of degree 5, so it is 05. Also, Xs = X4 - X1, and we compute X5(1) = 5,

X5(c2) = 1,

X5(c4) = X5(c5) = 0.

X5 (CO = -1,

We are left with the task of finding X3 and X'3 Let us at this point write down what we know of the character table of A5: Cl

C2

C3

C4

Oil

1

1

1

1

C5 1

a3 a3 a4

3

x2

x3

x4

x4

3 4

Y2

113

3/4

Y5

0

1

-1

-1

-1 0 0 Our task is to determine the unknown entries. Set z; = x, + y; for (k5

5

1

2 < i < 5. First we note that if 1Z is the regular representation 7Z =a1 ®3a38)3a3ED 404 ED 5a5,

8.4 Characters

477

so

X?t = Xi + 3X3 + 3X'3 + 4X4 + 5X5.

Evaluating this on c2 gives (using Example 4.2 (3))

0=1+3z2+4.0+5.1 so Z2 = -2, and by evaluation on c3, c4, and c5 we obtain z3 = 0, z4 = z5 = 1. Now let us use orthogonality.

1

(-1) so we obtain a linear system 15x2

+ 20x3 20x3

15x2

-

+ -

12(x4 + x5)

12(x4 + x5)

20x3

= -3 = -12 = -15

with solution x2 = -1, x3 = 0, x4 + xs = 1. Then also y2 = -1, y3 = 0,

y4+ys=1, and so, since z4 =1,x4=x5=x, say, y4=x5=y, say, with x + y = 1. Thus, it remains to determine x and y. We observe that A5 is a group in which each element is conjugate to its inverse, so by Proposition 4.23 all of its characters are real valued, and so X3 = X3,

X3 = (4),

We use orthogonality once more to get

x,

so xy = -1, which together with x + y = 1 gives x = (1 + vr5)/2

and

y = (1 - f)/2

(or vice-versa, but there is no order on X3 and X3, so we make this choice). Hence, we find the complete "expanded" character table of A5:

478

Chapter 8. Group Representations C,

C2

C3

C4

Cs

1

15

20

12

12

a,

1

1

1

1

a3

3

0

(1+v"5-)/2

3

0

(1-v')/2

(1-f)/2

C"3

-1 -1

4

0

1

-1

-1

5

1

-1

0

0

a4 as

1

(1+f)/2

(We call this the "expanded" character table of As because we have included on the second line, as is often but not always done, the number of elements in each conjugacy class. Note that there is no canonical order for either conjugacy classes or representations, so different character tables for the same group may "look" different.) (4.30) Example. Let us show how to determine all the irreducible complex characters of the symmetric group Ss of order 120. First we see from Corollary 1.5.10 that S5 has seven conjugacy classes, with representatives 1, (12), (123), (12)(34), (1234), (123)(45), and (12345), so we expect seven irreducible representations. One of them is r, of course,

and another, also of degree 1, is e given by e(g) = sgn(g) = ±1 E Aut(C), the sign of the permutation g. Observe that the representations 3 and ry of As constructed in Example 4.29 are actually restrictions of representations of Ss, so the representations a4 and as of As constructed there are restrictions of representations a4 and as of S5. Since a4 and as are irreducible, so are a4 and as. Furthermore, a's are irreducible, so are e ® 64 and e ® as (by Exercise 13), since a4 and and we may compute that the characters of a4 and e ® 54 (respectively, as and e (& &5) are unequal, so these representations are distinct. Hence, we have found six irreducible representations, of degrees 1, 1, 4, 4, 5, 5, so we expect one more of degree d, with

12+12+42+42+52+52+d2 = 120 so d = 6. If we call this a6i then we have (using X(a) for )(o, for convenience), X(R) = X(T) + X(e) + 4X(&4) + 4X(e ® & 4) + 5X(as) + WE ® &5) + 6X(&6)

enabling us to determine X(a6). We leave the details for the reader.

(4.31) Example. It is easy to check from Example 3.9 that A4 has the following expanded character table (where we have listed the conjugacy classes in the same order as there):

8.5 Induced Representations

1

C2 3

C3 4

r = ir'(8o)

1

1

1

7T (91)

1

1

(2

it (92 )

1

1

( S2

3

-1

0

0

Cl

a

479

C4 4 1

with C = exp(27ri/3). We wish to find the decomposition of the tensor product of any two irreducible representations. The only nontrivial case is a ® a. Recall that Xa®a = (Xa)(Xo), and thus has values 9, 1, 0, 0 on the four conjugacy classes. We compute

and for i = 0, 1, or 2

(a®a,7r'(9;))=12(1.1.9+3.1.1)=1, so

a®a= 2a ED 7r* (0o) ED ir* (01) ED 7r* (02).

(4.32) Remark. It is perhaps natural to conjecture that two groups with identical character tables must be isomorphic. This is, in fact, false! The groups D8 and Q8 (the dihedral and quaternion groups of order 8) are two distinct groups with the same character table. We leave the verification to the reader. (4.33) Remark. It is clearly a necessary condition for a representation to be defined over a field F (cf. Example 1.4 (12)) that its character takes its values in F. However, this condition is not sufficient. For example, if a is the representation p of Q8 defined in Example 3.8, then X, is real valued but a cannot be defined over R. (It is known in this situation that some finite multiple of the given representation can be defined over F. For example, here 2a can be defined over R.) Again we leave the verification to the reader.

8.5 Induced Representations In this section we develop an important and powerful method of constructing representations, the technique of induction. First, however, we consider restriction.

480


(5.1) Definition. Let G be a group, H a subgroup of G, and V an F(G)module. The restriction of V to H, ResH(V), is V regarded as an F(H)module. (In other words, we have the same underlying vector space, but we

restrict our attention to the action of the subring F(H) of F(G).)

If G is understood, we will often write ResH(V); if both G and H are understood, we will often write Res(V). As well, we may often write VH = ResN(Vc), using subscripts to denote the group that is operating. Finally, we may sometimes simply write V for Rest' (V) when it is clear from the context what is meant. Let Vg be a representation of G and H C G a subgroup. Clearly, if VH is irreducible, then so is VG, but the converse need not hold. For example, if H = (1), ResH(V) = deg(V)r. As a nontrivial example of restriction, let G = D2,.a and H = Z,,,. Then, in the notation of Examples 1.4 (6) and (7), Res(4i) = 0, ®8,,,_i. (5.2) Example. Let H be a subgroup of G. Then

ResH(F(G)) = [G: H]F(H), since F(G) is a free left F(H)-module of rank [G : HI (with a basis given by a set of right coset representatives). Now we come to induction.

(5.3) Definition. Let G be a group, H a subgroup of G, and W an F(H)module. Then the induction of W to G, denoted IndH(W), is given by Indo(W) = F(G) ®F(H) W. Note that this makes sense as we may regard F(G) as an (F(G), F(H))-

bimodule, and then the result of induction is an F(G)-module. If V = IndH (W ), we say that V is induced from W and call V an induced representation. (5.4) Lemma. deg(IndH(W)) = (G : HI deg(W).

Proof. As a right F(H)-module, F(G) is free of rank [G : H] (with basis given by a set of left coset representatives).

0

(5.5) Example. Let G be an arbitrary group and H a subgroup of G. Then for the regular representation F(H) of H, we have

IndH(F(H)) = F(G), the regular representation of G. This is immediate, for it is just the equality


481

F(G) ®F(H) F(H) = F(G).

In particular, setting H = (1), we have IndN(r) = F(G). Let us now give a criterion that will enable us to recognize induced representations.

(5.6) Theorem. Let V = Ind°(W) and identify W with W1 = F(H) ®F(H) W C F(G) ®F(H) W = V.

Let {gi}iEl be a complete set of left coset representatives of H in G, with gl = 1, and let Wi = gi(W1). Then

(1) V = ®iEIWi; (2) the action of G on V permutes the Wi, i.e., for every g E G and for every i E I, g(WW) = Wj for some j E I; (3) this permutation is transitive, i.e., for every i, j, there is a g E G with

g(Wi) = W3; and

(4) H={gEG:g(Wi)=W1}. Conversely, let V be an F(G)-module and suppose there are subspaces

{Wi}iEl of V such that (1), (2), and (3) hold. Define H by (4), and set W = W1. Then V = IndN(W). Proof. The first statement of the proposition is clear from the isomorphism

F(G) = ®9iF(H) iEI

of right F(H)-modules. As for the converse, note that there is a one-to-one correspondence {Wi} «--..

given by Wi

«-'

{ left cosets of H }

{g E G : g(Wi) = Wi }.

Pick coset representatives {9i}iEl with gi = 1. Define a function a G x W V by a(g, w) = g(w). This clearly extends to an F-linear transformation a : F(G) x W

V, and since

a(9h, w) = 9h(w) = g(hw) = a(9, hw)

it readily follows that a is F(H)-middle linear and so defines

a:F(G)®F(H)W V. Now we define Q : V -4 F(G) ®F(H) W. Since V = ®Wi, it suffices to define /3 : Wi - F(G) ®F(H) W for each i E I. We let /3(wi) = gi ® g,-1(wi),


482

for wi E Wi. Let us check that & and 6 are inverses of each other, establishing the claimed isomorphism. First,

ai/(wi) = &(gi (9.9i'(WO) = 9i(9i 1(wi)) = wi.

Now each g E G can be written as g = gih for a unique i E I and h E H. Thus, for w E W, )3a(9 ® w) = 3(g(w)) = Q(9ih(w)) = Q(9i(hw))

=gi®hw =gih®w =g®w, as required. We make two observations. First, if V itself is irreducible, the condition of transitivity is automatic (for >9EG g(W1) is a subrepresentation of V, which must then be V itself). Second, our choice of W1 was arbitrary; we could equally well have chosen some other W. This second observation yields the following useful result:

(5.7) Corollary. Let H be a subgroup of G. Let go E G be fixed and set K = go 1Hgo. Let au : H -+ Aut(W) be a representation of H, and let aK : Aut(W) be the representation of K defined by aK (9) = aH (90990 1) K for g E K. Then IndH(aH) = IndK(aK). Proof. Let V = IndH (W). Then we may identify W with W, in the statement of Theorem 5.6. Then K = {g E G : g(WW) = Wi} acting on W by the above formula, so V = IndK(W) as well. As a corollary of Theorem 5.6, we may identify two types of representations as induced representations. (5.8) Corollary.

(1) Let V be a transitive permutation representation on the set P = {pi }iEI

and letH={gEG:g(p,)=p,}. Then V=IndH(r). (2) Let V be a transitive monomial representation With respect to the basis

B = {bi}, and let H = {g E G : g9Fbi) = Fb,}. Let a = Fb1 as a representation of H. Then V = Indi(a). Proof.


483

Observe from Theorem 5.6 that W is certainly a subrepresentation of ResH IndH(W). We will consider this point in more detail later. However, this observation is enough to enable us to identify some induced representations already. For example, if G = D2m and 0, is one of the 2-dimensional representations of G defined in Example 1.4 (7), then each O,, which is a monomial representation, is induced from H = Z,,,, a subgroup of G of index two. If

9, is the representation of that name of H in Example 1.4 (6), i < m/2, This also illusthen IndH(91) _ 0,, and if i > m/2, IndN(91) trates Corollary 5.7, for if g = y (in the notation of Example 1.4 (7)) and (Of all = 8;, then K = H and aK = 9,,,_, and IndX(91) = course, Res$(0,) = 9, 9 9. Let us now concentrate on the case of a normal subgroup H of G and

a representation a : H - Aut(W). If a' is defined by a'(h) = a(ghg-1) for some fixed g E G, we call a' a conjugate of a, or more precisely, the conjugate of a by g. Let {aJ } be a complete set of conjugates of a = al . Note that if we let

N(a) _ {g E G : a' : H --+ Aut(W) defined by a'(h) = a(ghg-') is a representation of H isomorphic to a} then N(a) is a subgroup of G containing H and [G: N(a)] is the number of conjugates of a. (Note also that all the subgroups N(a') are mutually conjugate.) The subgroup N(a) is known as the inertia group of a. (5.9) Corollary. Let H be a normal subgroup of G and a : H -+ Aut(W) a representation of H. Then if a' is any conjugate of a,

Indo(a') = Indy(a). Proof. This is a special case of Corollary 5.7.

0

(5.10) Corollary. Let H be a normal subgroup of G and a : H -+ Aut(W) a representation of H. Let jail be a complete set of conjugates of a = al. Then

ResH IndH(a) = [N(a) : HI ®,, ai.

Proof. Let {g,} be a set of right coset representatives of H, with gl = 1, and let a, : H - Aut(W) be defined by a,(h) = a(g,-'hg,). Then

ResH IndN(a) = ®a, by Theorem 5.6. But in the statement of the corollary we have just grouped

the a, into isomorphism classes, there being [N(a) : H] of these in each class.

0

484


(5.11) Theorem. (Clifford) Let H be a normal subgroup of G. Let p G - Aut(V) be an irreducible representation of G, and let or : H Aut(W) be any irreducible component of Res°(p). Let Jai) be a complete set of conjugates of or = al, and set Vj = E9ECg(W1), where the sum is taken over the left coset representatives of H such that the conjugate of or

by g is isomorphic to of. Set K = {g E G : g(V,) = Vi}. Then V, is an irreducible representation of K and V = IndK(Vi). Furthermore, K = N(a).

Proof. First consider V' = E9EG g(Wi ). This is an F(G)-submodule of V, but V was assumed irreducible, so V' = V. Next observe that instead of summing over g E G, we may instead sum over left coset representatives of H. Further note that the representation of H on g(Wj) is the conjugate of or by g, so we may certainly group the together to get V = F, Vi. Now each Vj is a sum of subspaces W;, which are isomorphic to o& as an F(H)-module, so by Lemma 7.1.20, each Vj is in fact isomorphic to a direct sum of these. We claim that V = ®j V? . Consider U = V, fl F , V.. Then U is an F(H)-submodule of V,, and so is isomorphic to for some ml. Also, U is an F(H)-submodule of E»i Vj, and so is isomorphic to e3>, m3W, for some {mj I. By Corollary 7.1.19, ml = m2 = = 0, so U = (0), as required. If a- is the conjugate of a by gj E G, then gj(Vl) = V,, so G permutes the subspaces Vj transitively. Then by Theorem 5.6, we obtain V = IndG (Vi).

Also, V, is irreducible, for if it contained a nonzero proper subrepresentation Vi, then IndG(V,') would be a nonzero proper subrepresentation of V, contradicting the irreducibility of V. Finally, g E K if and only if the conjugate by g of W, is isomorphic to W1. But this is exactly the condition for g E N(a). (5.12) Corollary. Let H be a normal subgroup of G. Let p : G -+ Aut(V) be

an irreducible representation of G, and let or = a' : H - Aut(W,) be any irreducible component of Re$(p). Then (1) (2)

Res$(p) is a semisimple representation of H. In the decomposition of Rest' (p) into a direct sum of irreducible rep-

resentations ®mjUj, all of the U, are conjugate to of (and so are

(3)

mutually conjugate), all conjugates of al appear among the U;, and all the multiplicities m3 are equal. Res4(p) = m ®j of for some m E {1, 2, 3, ...} U {oo}, where {a'} is a complete set of conjugates of a1.

8.5 Induced Representations (4)

485

Let H' = {g E G : g(W,) = W,}. Then H C Hl C N(a), and m < [N(a) : H1].

Proof. (1) In the notation of the proof of Theorem 5.11, we have that, as an F(H)-module, V = Eg(W1), so V is a sum of simple F(H)-modules and hence is semisimple by Lemma 4.3.20. (2) Clearly, all the Uj are conjugate to a1 and all conjugates appear.

Let V, be as in Theorem 5.11, with V, = ®;_i k, (W1) for some group elements ki E G. Then if gj is as in the proof of Theorem 5.11, m1 Wj

Vj

= gi (Vi ) M1

= ®gi(ki(W,)) i=1

m, Wi

so mi = m1 by Corollary 7.1.18. (3) This is merely a restatement of (2). (4) Since g(W1) = W, for g E H', V1 = >J g(W,) where the summa0 tion is over left coset representatives of H1 in N(a).

Remark. It is not true that m = [N(a)

: H1], or even that m divides [N(a) : H'], in general. Let G = Z3 have the representation p = r 9 01

on V=C2 and let H=(1).LetW={(z,z)EC2}. Then N(a)=G, H1=H,so [N(a):H1]=3,but m=2. As a consequence of Clifford's theorem we have the following result:

(5.13) Corollary. Let H be a normal subgroup of G, and let p : G - Aut(V) be an irreducible representation of C. Then if ResH(V) is not isotypic, there is a proper subgroup K of G containing H and an irreducible representation V, of K such that V = IndK(VI).

Proof. Suppose ResH(V) is not isotypic. Let W1 be an irreducible F(H)submodule of ResH (V), and set V, = >g g(W,) where the sum is over all

g E G with g(W,) isomorphic to W1 as an F(H)-module, and let K = {g E G : g(V,) = V,}. Then, as above, V1 is an irreducible F(K)-module. Note K C H and K 54 G as ResH (V) is not isotypic. Then IndX (V1) is an F(G)-submodule of V, but as V is irreducible, it is equal to V. 0 We may use Corollary 5.13 to sharpen Theorem 4.28. (5.14) Theorem. Let G be a finite group and A an abelian normal subgroup of G. If V is an irreducible complex representation of G, then d = deg(V) divides [G : A].

486


Proof. We prove this by induction on n = IGI. Let V be defined by a : G -' Aut(VG ). Let W = ResA (V ). By Corollary 5.13, if W is not isotypic, V = IndK (V1) for some proper subgroup K of G containing H and some irreducible representation V1 of K. Then deg(V1) divides 1K : H), so deg(V) = [G : K] deg(V1) divides [G : K] [K : H) = [G : H]. If W is isotypic, then W = dW1 for some one-dimensional representation W1 of A, or, in other words, a : A Aut(W) = Aut(V) is given by a(g) = multiplication by some complex number X(g) for each g E A. (Also, for any h E G, \(hgh-1) = A(g), g E A.) Now for each m consider aem on V ® .. ® V. By Exercise 20, this is an irreducible representation of G"' = G x x G. Let

H is a subgroup of A' and hence a normal subgroup of Gm, and it acts trivially on V®"', so we obtain an irreducible representation of G'IH, which is irreducible as o®m is. Hence, by Theorem 4.28, the degree dm of this representation divides the order of Gl"IH. Note, however, that H has a subgroup

{(gi, ... ,gm-i,g1 1 "'9'nl l) : gi E A}

isomorphic to A'"-1, so if a = JAI, then d'^ divides nm/a'-1 = a(n/a)"' for every m, which implies that d divides n/a = [G: A], as claimed. We now return to the general study of induction. (5.15) Lemma. Let K C H C G be subgroups. (1) (Transitivity of restriction) For any representation V of G,

ResK ResN(V) = ResK(V). (2) (Transitivity of induction) For any representation W of K,

IndN IndK (W) = IndK (W).

Proof. (1) is trivial. As for (2), IndN IndK(W) = F(G) ®F(H) (F(H) ®F(K) W) = (F(G) ®F(H) F(H)) ®F(K) W = F(G) ®F(K) W

= IndK(W)

where the second equality is just the associativity of the tensor product (Theorem 7.2.17).


487

The next formula turns out to be tremendously useful, and we will see many examples of its use. Recall that we defined the intertwining number

i(V,W)=(V,W) of two representations in Definition 4.17.

(5.16) Theorem. (Flobenius reciprocity) Let F be an arbitrary field, G an arbitrary group, and H a subgroup of G. Let W be an F-representation of H and V an F-representation of G. Then (IndH(W), V) = (W, ResH(V)). Proof. By definition,

(IndH(W), V) = dimF Homp(G)(IndH(W), V).

But HomF(G)(IndH(W), V) = HomF(G)(F(G) ®F(H) W, V) = HomF(H)(W,HomF(G)(F(G), V)) = HomF(H)(W, V)

where the second equality is the adt associativity of Hom and tensor product (Theorem 7.2.20). Again, by definition, dimFHomF(H)(W, V) = (W, ResH(V)) 0

One direct consequence of Fobenius reciprocity occurs so often that it is worth stating explicitly. (5.17) Corollary. Let F be an excellent field for G. Let W be an irreducible F-representation of H and V an irreducible F-representation of G. Then the multiplicity of V in IndH (W) is equal to the multiplicity of W in ResH (V). Proof. Immediate from Theorem 5.16 and Lemma 4.18 (3).

0

As an example of the use of Frobenius reciprocity let us use it to provide an alternate proof of part (2) of the fundamental Theorem 3.4. Let F be an excellent field for G and M an irreducible F-representation of G of degree d. Let m be the multiplicity of M in F(G). We wish to show m = d. Let H = (1) and recall from Example 5.5 that F(G) = IndH(r). Then

488


m = (M, F(G)) _ (F(G), M) _ (IndH(T), M) _ (T, ROH(M)) _ (T, dr)

=d as claimed.

(5.18) Example. Consider A4 from Example 3.9, and let F = C. We have a split extension

with V Z2 ® Z2. V has 4 irreducible representations, all of degree 1: T, and three others, which we shall simply denote by A1i A2, and A3. A4 has 4 irreducible representations: 7r'(0i) for i = 0, 1, 2 of degree 1 and a of degree 3. Then (IndV`(T), i'(e:)) _ (T, R.esyA '(ir`(9t))) _ (T, T) = 1,

so 2

Indv°(T) = ®Tr'(9,) i=O

(since

deg(Indv' (T)) = [A4 : V] deg(T) = 3. 1 = 3 and the right-hand side is a representation of degree 3). Also, for i = 0, 1, 2

andj=1,2,3

0 = (A,, T) = (Ai, Reso(x'(91))) = (Indy(A,),fr'(9 )), so we must have IndX(A,) = a (as both are of degree 3). Continuing with this example, since we have a split extension, we have a subgroup S of A4 isomorphic to Z3, and we identify S with Z3 via this isomorphism. (We have given S and this isomorphism explicitly in Example 3.9). Now S has three irreducible representations 90 = r, 91 i and 92. Because we have a splitting, R.ess' (7r' (6;)) = 9;, or, more generally,

a,; = (ReSA'(ir'(9,)), 0,) = (ir'(ei), Inds'(91)); since IndA'(91) has degree 4, this gives

Inds'(0,)=7r'(0j)®a

for

i=0, 1,2.


489

Here is another useful consequence of Frobenius reciprocity:

(5.19) Proposition. Let G be a group and H an abelian subgroup of G. Let F be an excellent field for G. Then every irreducible F-representation of G has degree at most [G : H).

Proof. Let V be an irreducible representation of G. Then ResX (V) is a representation of H and so contains an irreducible representation W, which is one-dimensional, since H is abelian. Then

1 < (Res$(V), W) = (V, IndH(W))

the equality being Frobenius reciprocity, so V is a subrepresentation of 0 IndX(W). But deg (IndH(W)) = [G: H], so deg(V) < [G: H]. Example. Note that D2m has an abelian subgroup of index 2 and its irreducible complex representations all have dimension at most 2. The same is true for Q8. Also, A4 has an abelian subgroup of index 3 and its irreducible complex representations all have dimension at most 3. Now we determine the character of an induced representation. In order to state this most simply, we adopt in this theorem the following nonstan-

dard notation: If H is a subgroup of G and W an F-representation of H with character Xw, we let X'w be the function on G defined by Xw(g) =

f ow(g) if g V H.

(5.20) Theorem. Let H be a subgroup of G of finite index k, and let W be an F-representation of H of finite degree. Set V = Indy(W). If {gi}k 1 is a complete set of left coset representatives of H, then for any g E G, k

XV(g)

i.l

Xw(gi 'ggi)

Proof. We know that we may write k

k

V =®gi(W)=®Wi i=1

icl

and that every element of G acts by permuting { Wi }. To be precise, if we let Hi = g1Hgi 1, then

Hi= {gEG:g(W1)=Wi}, Hi, then g(W1) = Wj for some j 0 i. On the other hand, if the representation of Hi on Wi is given by ai : H; -' Aut(Wi), with a, = a, so if g


490

then for g E H,, a,(9) = a(9; 199i) = a(g'). Now fi : W1 - W, by fi(w) = gi(w) is an isomorphism with

(fia(9')f, 1)(wi) = ai(9)(wi)

for allg H, and wiE W,, so ` r(ai(9)) = Tr(a(9 )) Now Xv(g) is the trace of a matrix representing the operation of g. Choose a basis for V that is a union of bases for the Wi. Then if g E G with g 0 Hi, the action of g on W; contributes nothing to Xv(g); while if g E Hi, it contributes Tr(a(9')) = Xw(9') to Xv(g), yielding the theorem. 0

(5.21) Example. In Example 4.29, we found the character table of A5. Let us here adopt an alternate approach, finding the irreducible characters via induced representations. We still know, of course, that they must have degrees 1, 3, 3, 4, 5 and we still denote them as in Example 4.29. Of course, al = r. The construction of a4 was so straightforward that an alternative is hardly necessary, but we shall give one anyway (as we shall need most of the work in any case). Let G = As and let H = A4 included in the obvious way (as permutations of {1, 2, 3, 4) C- (1, 2, 3, 4, 5). Then a system of left coset representatives for H is

{1,(12)(45), (12)(35), (13)(25), (23)(15)} = {gi, ... ,g5}. We choose as representatives for the conjugacy classes of C

{1, (14)(23), (123), (12345), (13524)} = {cl, ... ,c5}.

Of course, g' (c1)gi = g, for every i. Otherwise, one can check that 9i 1(cj)gi 0 H except in the following cases: 91 1(c2)91 = c2,

91 1(c3)91 = c3,

and

91 1((123))91 = (132).

(Note that (12 3) = C3 and that while (13 2) is conjugate to it in As, it is not conjugate to it in A4, so we have written the permutation explicitly.) Let W = r and consider IndN (W) = V. Then by Theorem 5.20, it is easy to compute Xv: Xv(c1) = 5,

Xv(c2) = 1,

Xv(C3) = 2,

XV(C4) = XV(C5) = 0

Then 1

(XV,XV)=10(52+15.1+20.22)=2 and r appears in V with multiplicity 1 by Frobenius reciprocity (or by calculating (Xv, X1) = 1), so V contains one other irreducible representation, V = r ® a4, and X4 = Xv - X1. giving the character of a4.


491

Now, following Example 3.9, let W = ir'(01) (or 7r'(02)) and consider Indo (W) = V. Again, it is easy to compute Xv: Xv(c1) = 5,

Xv(c2) = 1,

Xv(c3) = exp(21ri/3) +exp(4iri/3) _ -1, XV (C4) = XV (C5) = 0-

Now r does not appear in W by Frobenius reciprocity, so that implies here (by considering degrees) that V is irreducible (or alternatively one may calculate that (Xv, XV) = 1) so V = a5 and its character is given above. Now we are left with determining the characters of the two irreducible

representations of degree 3. To find these, let H = Z5 be the subgroup generated by the 5-cycle (1 2 3 4 5). Then H has a system of left coset representatives

{1, (14)(23), (243), (142), (234), (143), (12)(34), (13)(24), (123), (134), (124), (132)}

= (91,...,912) Again, g, 1(c1)gj = c1 for every i. Otherwise, one can check that gi 1(cj)gi

H except in the following cases: g11(c4 )91 = c4, 91 1(c5)91 = cs, and

g21((12345))g2 = (15432) _ (12345)-1 and gz1((13524))g2 = (14253) = (13524)-1. Now let W = 9j and let V = IndH (W). Again by Theorem 5.20 we compute Xv(c1) = 12, Xv(c2) = Xv(c3) = 0, and Xv(c4) = exp(2iri/5) + exp(8ai/5) Xv(c5) = exp(47ri/5) +exp(6ai/5)

if

Xv(c4) = exp(47ri/5) + exp(6ai/5) Xv(c5) = exp(2iri/5) + exp(8iri/5)

if j - 2 or 3.

1 or 4,

In any case, one has that r does not appear in V (by either Frobenius reciprocity or calculating (Xv,X1) = 0) and a4 and as each appear in V with multiplicity 1 (by calculating (Xv, X4) = (Xv, Xs) = 1), so their complement is an irreducible representation of degree 3 (which checks with (Xv, Xv) = 3), whose character is Xv - X4 - X5Choose j = 1 (or 4) and denote this representation by a3, and choose

j = 2 (or 3) and denote this representation by a3 (and note that they are distinct as their characters are unequal). Then we may calculate that X3(c1) = X3(c,) = 3

X3(2) = X3(2) = -1 X3(C3) = X3(03) = 0

X3(c4) = 1 + exp(2iri/5) + exp(87ri/5) = (1 + f)/2 %3(cs) = 1 + exp(4iri/5) + exp(6ai/5) = (1 - f)12 and vice-versa for Xs, agreeing with Example 4.29.


492

Our last main result in this section is Mackey's theorem, which will generalize Corollary 5.10, but, more importantly, give a criterion for an induced representation to be irreducible. We begin with a pair of subgroups K, H of G. A K-H double coset is

KgH={kgh:kEK, hEH}. It is easy to check that the K-H double cosets partition G (though, unlike for ordinary cosets, they need not have the same cardinality). We shall also refine our previous notation slightly. Let a : H -+ Aut(W) be a representation of H. For g E C, we shall set H9 = g-'Hg and we will let a9 be the representation a9 : H9 -. Aut(W) by a9(h) = a(ghg-1), for h E H. Finally, let us set H. = H9 fl K. We regard any representation of H, given by a : H - Aut(W), as a representation of H9 by o9 and, hence, as a representation of H. by the restriction of a9 to H9. (In particular, this applies to the regular representation F(H) of H.) (5.22) Theorem. (Mackey) As (F(K), F(H))-bimodules,

F(G) ® F(K) ®F(H,) F(H) 9

where the sum is taken over a complete set of K-H double coset representatives.

Proof. For simplicity, let us write the right-hand side as ®9(F(K)(&F(H))9. Define maps a and f3 as follows:

For g' E G, write g' as g' = kgh for k E K, h E H, and g one of the given double coset representatives, and let a(g') _ (k ® h),. We must

check that a is well defined. Suppose that g' = kgh with k E K and h E H. We need to show (k ®h)9 = (k (& h)9. Now kgh = g' = kgh gives

9_1k-'!g = hh 1, and then (k ®h)g = (k(k-11) ®(hh-1)h)9 ®g_1(k-lk)9(hh-1)h)9 = (k _ (k ®h)9, as required. Then a extends to a map on F(G) by linearity. Conversely, define 3, on K x H by 3.9 (k, h) = kgh E G and extend Q9 to a map

fag : F(K) x F(H) -. F(G) by linearity. Then for any x E H9, we have 39(k,9-lxgh),

Q9(kx,h) = kx9h = k9(9-1x9)h =


493

so i3g is F(H9)-middle linear and so defines

fag : (F(K) (9 F(H))9 -' F(G).

Set Q = rjQg. Then it is easy to check that a and p are inverses of each 0 other, yielding the theorem. Note that the subgroup H. depends not only on the double coset KgH, but on the choice of representative g. However, the modules involved in the statement of Mackey's theorem are independent of this choice. We continue to use the notation of the preceding proof.

(5.23) Proposition. Let g and g be in the same K-H double coset. Then

(F(K) ® F(H))g is isomorphic to (F(K) ® F(H))y as (F(K), F(H))bimodules.

Proof. Let g= kgh with k E K, h E H, and define a: K x H K x H by a(k, h) = (kk, hh). Extend to

a : F(K) x F(H) -' F(K) x F(H) by linearity, thus giving

a : F(K) x F(H) -+ (F(K) 0 F(H))g. We show that a is middle linear: Let x E Hg be arbitrary. Then

a(kx, h) = kxk ®hh = k(xk) ®hh = k ®g-'(xk)ghh and

a(k, g-'xgh) = kk ®hg-'xgh = k ® (9-lkg)hg-'xgh. But g = kgh, so g-lkgh(h-1g_lk-1)x(kgh)h

(9 'kg)hg-'xgh = = g-lxkghh,

and these are equal. Hence, we obtain a map

&: (F(K) (9 F(H))g - (F(K) (& F(H))y. Its inverse is constructed similarly, so i7 is an isomorphism.

0

Remark. Note that a depends on the choice of k and h, which may not be unique, but we have not claimed that there is a unique isomorphism, only that the two bimodules are isomorphic.

494


(5.24) Corollary. For any F-representation W of H, ItesKlndN(W) = ®(F(K) (&F(11,) W) _ ®IndHB(W9). 9

9

Proof. By Theorem 5.22 and associativity of the tensor product (Proposition 7.2.17), we have the following equalities among F(K)-modules. IndH(W) = F(G) ®F(H) W

= (®F(K) ®F(H9) F(H)) ®F(H) W 9

_ ® F(K) ®F(H9) (F(H) ®F(H) W) 9

F(K) ®F(H9) W. (Note that although H. is a subgroup of H, its action on F(H) is not the usual action, so one needs to check that F(H) is indeed an (F(H9), F(H))bimodule. but this is immediate.) 0 In the final term in the statement of the corollary, we have denoted the representation by W. to remind the reader of the action of H9. (5.25) Lemma. Let G be a finite group, H a subgroup of G, and F a field of characteristic zero or relatively prime to the order of H. Let W be an F-representation of H defined by a : H --. Aut(W) and set V = Indy(W). Then

EndG(V) =

HomH9(W9, Re4Hi(W)) 9

where the direct sum is over a complete set of H-H double cosets, H9 = H9 n H, and W9 is the representation of H9 on W defined by a9. Proof. The required isomorphism is a consequence of the following chain of equalities and isomorphisms.

Endc(V) = HomG(V, V) = HomH(W1 V)

as in the proof of Frobenius reciprocity L' HomG(V, W) as by our assumption on F, F(H) is semisimple


495

= HomF(H) (®F(H) (&F(H9) W, W) 9

= ®HomF(H) (F(H) ®F(H9) W, W) 9

Now note that in the first W above, H9 is operating by o'9; while in the second, it is operating by a _ ® HomF(H9) (W9, ResH9 (W)) 9

by adt associativity of Hom and tensor product (Theorem 7.2.20).

11

(5.26) Corollary. (1) Let G be a finite group, H a subgroup of G, and F a field of character-

istic zero or relatively prime to the order of H. Let W be a represen-

tation of H and set V = IndH(W). Then EndG(V) = F if and only if EndH (W) = F and HomH9 (W9, ResH9 (W)) = 0 for every g E G,

gVH. (2) Let F be an algebraically closed field of characteristic zero or relatively

prime to the order of G. Then V is irreducible if and only if W is irreducible and, for each g E G, g V H, the H9-representations W9 and ResH9 (W) are dist (i.e., have no mutually isomorphic irreducible components). In particular, if H is a normal subgroup of G, V is irreducible if and only if W is irreducible and distinct from all its conjugates.

Proof. Note Endc(V) contains a subspace isomorphic to EndH(W) (given by the double coset representative g = 1). Also, under our assumptions V 11 is irreducible if and only if EndG(V) = F, and similarly for W.

(5.27) Example. Let H = Z,,, and G = D2,,,,. Then the representations O and 0,,,-i of H are conjugate, so for i # m/2, IndH(0i) is irreducible. Of course, this representation is just Oi. (5.28) Example. Let H = A5 and G = S5. As a system of coset representatives, we choose {1, (12 5 4)} = {gl, g2}. Note that

92(12345)g21 = (13524) = (12345)2.

Thus, if we let a1 be the representation a3 of A5 (in the notation of Example 4.29) we see that its conjugate a2 = a3, a distinct irreducible representation. Hence by Corollary 5.26, IndH(a3) = IndH(a') is an irreducible representation of degree 6, and by Theorem 5.20 we may compute its character, giving an alternative to the method of Example 4.30.

496


(5.29) Example. Let H = V and G = A4 in the notation of Example 5.18. Then the representations A1i A2, and A3 of H are mutually conjugate, so Indy(Ai) is irreducible and is equal to a for i = 1, 2, 3, ing the result of Example 5.18. (5.30) Example. Let m be such that p = 2' - 1 is prime. Then S = Zy acts on V = (Z/2)' by cyclically permuting the elements of V other than the identity. (This may most easily be seen by observing that GL(m, F2) has an element of order p.) Thus, we may form the semidirect product

1--iV--iG--*-+S-+1 with G a group of order n = 2"' (27` -1). G has the one-dimensional complex

representations n' (01) for i = 0, ... , p - 1. Also, if a is any nontrivial complex representation of V, G has the representation Indy(a) of degree [G : V) = 21-1. Now a is dist from all its conjugates (as Ker(a) may be considered to be an F2-vector space of dimension m-1, and GL(m-1, F2) does not have an element of order p), so by Lemma 5.25, a is irreducible. As (2" - 1)2 + 2m-1(1)2 = n, these 2' complex representations are all of the irreducible complex representations of G. (Note that if m = 2, then G = A4, so this is a generalization of Example 5.29.)

8.6 Permutation Representations We have already encountered permutation representations, but because of their particular importance, we wish to discuss them here further. We shall restrict our attention to complex representations, so that we may use the full power of character theory. We begin with a bit of recapitulation.

(6.1) Definition. Let P = {pifiEl be a set and a : G

Aut(P) a homo-

morphism. Then a defines a representation on (=complex vector space with basis P) by

F, a99 9EG

(aiPi) 1: sEl

_

agaia(9)(Pi) g, i

We will often call this representation or as well.

For simplicity, we shall assume throughout that G and P are both finite, though some of our results hold more generally.

8.6 Permutation Representations

497

(6.2) Definition. Let p E P. Then Op = Orbit(p) = {p' E P : v(g)(p) = p' for some g E G}

Gp = Stab(p) = {h E G : a(h)(p) = p}.

Note that Orbit(p) is a subset of P and Stab(p) is a subgroup of G. (Orbit(p) and Stab(p) have been previously defined in Definition 1.4.9.)

(6.3) Definition. A nonempty subset Q of P is a domain of transitivity for

o, if a(g)(q) E Q for every g E G and q E Q. If P is the only domain of transitivity for a (which holds if and only if Op = P for every p E P) then or is called transitive, otherwise intransitive. The following is obvious:

(6.4) Lemma. Let Ql,

... , Qk partition P into domains of transitivity. Then

=CQi®...®CQk, Proof.

Recall the following basic result of Corollary 5.8:

(6.5) Theorem. Let a be a transitive permutation representation of G on P. Let p E P and set H = Gp. Then or = IndH(r).

0

Proof.

(Recall that in this situation all of the subgroups Gp, for P E P, are conjugate and we may choose H to be any one of them.) Because of Lemma 6.4, we shall almost always restrict our attention to transitive representations, though we state the next two results more generally.

(6.6) Proposition. Let P be partitioned into k domains of transitivity under the representation a of G. Then the multiplicity of r in is equal to k.

Proof. By Lemma 6.4, we may assume k = 1. But then, by Theorem 6.5 and Frobenius recipocity,

(r, v) = (r, Indo(r)) = (Resy(7), r) = 1. 0


498

We record the following simple but useful result:

(6.7) Lemma. Let ai be a permutation representation of o, on a set Pi for i = 1, 2. Then al and a2 are equivalent if and only if for each g E G, I {pi E Pi : 0`1 (9)(pi)}I = I{p2 E P2 :0.2(9)(P2) =p2}I

Proof. We know that al and 0`2 are equivalent if and only if their characters Xi and X2 are equal. But if X is the character of a permutation representation or of G on a set P, then its character is given by X(9) = I{p E P : a(9)(p) = p}I 0

(6.8) Definition. A partition of P into subsets {Qi}iEr is called a partition into domains of imprimitivity for a if for every g E G and every i E I there

exists a j E I with a(g)(Qi) = Qj. If the only partitions into domains of imprimitivity are either the partition consisting of the single set Q = P, or the partition into subsets of P consisting of single elements, then a is called primitive, otherwise imprimitive. (Note that an intransitive representation of G is certainly imprimitive.) Let G be transitive, Q a subset of P with the property that for every g E G, either a(g)(Q) = Q or a(g)(Q) fl Q = 0. Set

H = {gEG:a(g)(Q)=Q}. Then H is a subgroup of G, and if {gi} are a set of left coset representatives of H, then Qi = gi(Q) partitions P into domains of imprimitivity. Furthermore, all partitions into domains of imprimitivity arise in this way. Note that H acts as a group of permutations on Q. We have the following result, generalizing Theorem 6.5, which also comes from Corollary 5.8.

(6.9) Theorem. Let a be a transitive permutation representation of G on P, Q a domain of imprimitivity for a, and H = {g E G : a(g)(Q) = Q}. If p denotes the permutation representation of H on Q given by p(h)(q) = a(h)(q) for h E H, q E Q, then p is a transitive permutation representation

ofH onQ and a = IndH(p). Proof.

We have the following useful proposition:

(6.10) Proposition. Let a be a transitive permutation representation of G on a set P, and let p E P.


499

(1) Let P be partitioned into a set of domains of imprimitivity {Qi} with

p E Qi, and let H = {g E G : a(g)(Qi) = Qi}. Then there are [G: HI sets in the partition, each of which has [H : Gp) elements. (2) Let H be a subgroup of G containing Gp, and let

Q = {a(g)(p) : g E H}.

Then Q is one element in a partition of G into [G : H) domains of imprimitivity, each of which contains (H : G,J elements.

Proof. Clear from the remark preceding Theorem 6.9, identifying domains 0 of imprimitivity with left cosets.

(6.11) Corollary. A transitive permutation representation a of G on a set P is primitive if and only if some (and hence every) Gp is a maximal subgroup of G. Proof.

0

(6.12) Corollary. Let a be a transitive permutation representation on a set P with IPI prime. Then or is primitive. Proof. By Proposition 6.10, the cardinality of a domain of imprimitivity for G must divide [G : Gp] = JP1, and so consists of either a single element or 0 all of P.

We now introduce another sort of property of a representation.

(6.13) Definition. A permutation representation a of a group C on a set P is k-fold transitive if P has at least k elements and for any pair (pl, ... , pk) and (ql, ... , qk) of k-tuples of distinct elements of P, there is a g E G with a(g)(pi) = qi for i = 1, ... , k. 2-fold transitive is called doubly transitive. (6.14) Examples.

(1) Note that 1-fold transitive is just transitive. (2) The permutation representation of D2n on the vertices of an n-gon is doubly transitive if n = 3, but only singly (= 1-fold) transitive if n > 3. (3) The natural permutation representation of S on {1, ... ,n} is n-fold transitive, and of An on 11, ... , n} is (n - 2)-fold (but not (n- I)-fold) transitive.

(4) The natural permutation representation of Sn on (ordered or unordered) pairs of elements of {1, ... ,n} is transitive but not doubly


500

transitive for n > 3, for there is no g E S,, taking {(1, 2), (2,3)1 to {(1,2), (3,4)}. Doubly transitive permutation representations have two useful properties.

(6.15) Proposition. Let a be a doubly transitive permutation representation of G on a set P. Then a is primitive.

Proof. Suppose a is not primitive, and let Q be a domain of imprimitivity, with pi, p2 E Q and p3 0 Q. Then there is no g E G with a(g)(pi) = Pl, and a(g)(p2) = p3, so a is not doubly transitive. The proposition follows by contraposition.

(6.16) Theorem. Let a be a transitive permutation representation of G on

a set P. Then a is doubly transitive if and only if a = r ® a' for some irreducible representation a' of G. Proof. Since a is a permutation representation, its character X is real valued. Note that X2 is the character of the permutation representation a ® a

on P x P. Let k be the number of orbits of a ® a on P x P. Note that k = 2 if a is doubly transitive (the orbits being {(p,p) : p E P} and {(p, q) : p, q E P, p # q}) and k > 2 otherwise. Then, by Lemma 6.4,

k=(a®a,r) 1

= n F, X 2 (9) 9EG

=1

X(g)X(g) gEG

=1

X(g)X(g) gEG

= (X, X).

Note that r is a subrepresentation of a by Lemma 6.4; also note that (X, X) = 2 if and only if in the decomposition of a into irreducibles there are exactly two distinct summands, yielding the theorem. (6.17) Example. The representation 0 of Example 4.29 was doubly transitive and decomposed as r® a4, a4 irreducible. The representation y of Example

4.29 was not doubly transitive; -y (D -y had three orbits on {1, ... ,5} x 11, ... , 5}, and y decomposed into a sum r ®a4 ED a5 of three distinct irreducibles.

(6.18) Remark. The converse of Proposition 6.15 is false. For example, let p be a prime and consider the permutation representation of D2p on the


501

vertices of a p-gon. By Corollary 6.12, this representation is primitive, but for p > 3 it is not doubly transitive.

(6.19) Example. The action of a group on itself by left multiplication is a permutation representation. Of course, this is nothing other than the regular representation, which we have already extensively studied. Instead, we consider the action of a group on itself by conjugation, i.e., y(g)(h) = ghg-1 for all g, h E G. Let us determine some examples of this representation. Of course, if G is abelian, this representation is just nr. In any case, the orbits of this representation are just the conjugacy classes. In general, if C1, ... , Ct are the conjugacy classes of G (in some order), we will let -ti be

'yonC1,so

7=11®...®7t (1) Consider D2m for m odd. Then (cf. Example 3.7) we have C1 = {1},

C2 = {x, xm-1},

... , C(m+l)/2 = {x(m-1)/2,x(m+l)/2},

C(m+3)/2 = {y) xy, ... ,

xm-1y}.

Then 'y1 = r, and for i = 2, ... , (m + 1)/2, -ti is a nontrivial representation containing r as a subrepresentation. Since D2,,, only has one nontrivial one-dimensional representation, namely, 0-, we see that

y1=rD

...

for

2<(m+1)/2.

We are left with the action on C(,,,+3)/2, and we claim this is

FP=r®01 ®...®0(m_1)/2 (cf Example 1.4 (7)), where P = { vertices of a regular m-gon }. This may be seen as follows: C(m+3)/2 consists of the elements of order exactly two in D2m, and each such fixes exactly one vertex of P. Thus we have a one-to-one correspondence C(m+3)/2 41 P by

x'y

vertex pi fixed by x'y

and further, for any g E D2m, g(x'y)g-1 fixes the vertex g(p;), so the two actions of G are isomorphic. (2) Consider D2m for m even. Then (cf. Example 3.7) we have {x?-1, x+l}, ... , Cm/2 = C;J+1 = {X?), CJ+2 = {x'y : i is even}, C;1+3 = WY: i is odd}.

C1 = {1}, C2 = {x,

xm-1},

r and ry; = r®t/i+- for i = 2, ... , m/2 (as conjugation by x is trivial on C; but conjugation by y is not). We will determine the Again, ry1 =


502

last two representations by computing their characters. Of course, 1 acts trivially, as does xm/2, being central, so

X2+21=X+3(1)=XZ+2(xm/2)=X;+3(xm/2)= Also,

xi(x'y)x-j

M 2

= xi+23y, so for i 0 m/2, xi fixes no element and

Xz+2(x3)=XZ+3(x3)=0

for j# 2.

Now y(xiy)y-1 = yxi = x-ly, so y fixes x:y when 2i = 0, i.e., i = 0 or m/2. Hence, if m/2 is odd, X -+2(y) = XZ+3(y) = 1,

while if m/2 is even X 22 +2(y) = 2

and

Xzi+3(y) = 0-

and

XIY+3(xy) = 1,

and

X2+3(xy) = 2.

Similarly, if m/2 is odd,

Xz+2(xy) = 0 while if m/2 is even,

XZ+2(xy) = 0

Now by computation with the irreducible characters of D2,,,, (which may be read off from Example 1.4 (7)) we see the following: For m/2 odd, 7Y

;

+2 = 7n+3 +3 = Y,++ ® 7 .

For m/2 even,

'Y2+2='+b++®-+®y and 'Y

2

+3 =

'Y

where _ 02 ®04 ®06

®...(D Ok,

with k = (m - 1)/2 for m odd and k = z - 1 for m even. (3) Consider A4. We adopt the notation of Example 3.9. One may then check that the following table is correct, where an entry is the number of elements of the given conjugacy class fixed by the given element (and hence the trace of that element in the representation on that conjugacy class):

8.7 Concluding Remarks 1

1

T

T2

{1}

1

1

1

1

{I, J, K} {T, TI, TJ, TK}

3 4

3

0

0

0 0

1

1

1

1

{T2, T2I, T2J, T2K}

503

Then computation with the characters of A4 (cf. Example 4.31) gives 'Vi = T

12 = r®rr'(01) ®rr'(02)

rya=1'4=T®a.

8.7 Concluding Remarks In this section we make three remarks. The first is only a slight variation of what we have already done, while the last two are results that fit in quite naturally with our development, but whose proofs are beyond the scope of this book. (7.1) Definition. Let F be a field of characteristic zero or prime to the order of the finite group G. The F-representation ring RF(G) of the group G is the free Z-module with basis {Qi}i_1, the irreducible F-representations of G.

The elements a = Li-1 miai of RF(G) are called virtual representations, and those with mi > 0 for all i are called proper representations. The formation of RF (G) is a special case of a more general construction.

(7.2) Definition. Let R be a ring. Let F be the free abelian group with basis

(P: P is a projective R-module}, and let N be the subgroup spanned by

{M - M1 - M2 : there is a short exact sequence of 7Z-modules:

-+M--iM2--i0}. Then let K(R) = F/N. Thus RF(G) = K(F(G)). This equality is as Z-modules. But RF(G), in fact, has more structure. Recall that we have defined the intertwining number of two representations V and W of G, which we shall here denote

504


by iG. Then iG extends by linearity to RF(G). It is easy to check that iG is then a symmetric bilinear form on RF(G). (If F is algebraically closed, this form is isometric to t[1], and the irreducible F-representations of G form an orthonormal basis of RF(G) with respect to iG.) Also, let H be a subgroup of G, W a virtual F-representation of H, and V a virtual F-representation of G. Then Frobenius reciprocity is the equality

ic(IndH(W), V) = iH(W, ResH(V)), which says that in a certain sense induction and restriction are adts of each other. This can be made precise: induction and restriction are an example of what is known as a pair of "adt functors." Finally, note that the tensor product of representations extends by linearity to RF (G), so RF (G) has the structure of a commutative ring (or Z-algebra) as well. Note too that we may form the character of a virtual representation-the character of F_1=1 mini is Ei=1 miXai. In the special case where F = C, if {ai}i=1 are the irreducible complex representations of G with characters {Xi}i=1, then the map a --* Xa gives a ring isomorphism between Rc(G) (under direct sum and tensor product) and Z[X1, ... Xt] (under sum and product).

(7.3) Remark. The reader will recall that we defined a good field F for G in Definition 1.3 and an excellent one in Definition 3.1. We often used the hypothesis of excellence, but in our examples goodness sufficed. This is no accident. If F is a good field and F' an excellent field containing F, then all F'-representations of G are in fact defined over F, i.e., every F'representation is of the form V = F' ®F W, where W is an F(G)-module. (In other words, if V is defined by a : G Aut(V), then V has a basis 13 such that for every g E G, [a(g)]g is a matrix with coefficients in F.) This was conjectured by Schur and proven by Brauer. We remark that there is no proof of this on "general principles"; Brauer actually proved more, showing how to write all representations as linear combinations of particular kinds of induced representations. (Of course, in the easy case where G is abelian, we showed this result in Corollary 2.3.)

(7.4) Remark. In Theorem 4.28, we proved that the degrees di of complex irreducible representations of C are divisors of n, the order of G. If F is an algebraically closed field of characteristic p prime to the order of G, all we showed was that the degrees of the irreducible F-representations are

prime to p (Corollary 4.6). In fact, it is the case that the degrees di are independent of the characteristic (assuming, of course, that (p, n) = 1), and, indeed, one can obtain the characteristic p representations from the complex ones.

8.8 Exercises

505

8.8 Exercises 1.

the assertions of Example 1.8 directly (i.e., not as consequences of our general theory or by character computations). In particular, for (3) and (4), find explicit bases exhibiting the isomorphisms.

2.

Let 7r : C -i H be an epimorphism. Show that a representation a of H is

irreducible if and only if 7r' (a) is an irreducible representation of G. Let H be a subgroup of G. Show that if Resy(a) is irreducible, then so is a, but not necessarily conversely. 4. the last assertion of Example 1.7. 5. Show that r and Ra are the only irreducible Q-representations of Zp. 6. Find all irreducible and all indecomposable Fy-representations of Zp. (Fp denotes the unique field with p elements.) 7. In Example 3.8, compute 7r'(8;) ®p in two ways: by finding an explicit basis and by using characters. 8. Do the same for the representations a' (9) ® a of Example 3.9. 9. In Example 3.8 (resp., 3.9) prove directly that p (reap., a) is irreducible. 3.

10.

11. 12.

In Example 3.10, that the characteristic polynomials of a(U) and a'(U) are as claimed. Also, that 7r' (7/'_) ® w'(01) = 7r' (&1) both directly and by using characters. that Ds and Qs have the same character table (cf. Remark 4.32). Show that the representation p of Q8 constructed in Example 3.8 cannot be defined over R, although its character is real valued, but that 2p can (cf. Remark 4.33).

13.

Let F and G be arbitrary. If a is an irreducible F-representation of G and

(3 is a 1-dimensional F-representation of G, show that a ®Q is irreducible. 14. Find an example to illustrate Remark 4.9 (1). (Hint: Let G = D6.) 15. (a) Find an example to illustrate both phenomena remarked on in Remark 4.9 (2). (Hint: Let G = D6.) (b) Show that neither phenomenon remarked on in Remark 4.9 (2) can occur if h is in the center of the group G. 16. Prove Lemma 4.18. 17. Prove Proposition 4.24. 18. Show that the following is the expanded character table of S4: C1

C2

C3

C4

C5

1

6

3

8

6

r

1

1

1

1

a 7r'(Ol) 7r'(7(i_)®a

3

1

-1

1

2

0

2

19.

1

1

-1

-1

-1 -1

3

7r'(7/i_)

0

0

0

1

-1

1

1

Show that the following is the expanded character table of S5 in two waysby the method of Example 4.30 and the method of Example 5.28.

r 54 as as

Ci

C2

C3

C4

C5

C6

C7

1

10

15

20

20

30

24

1

1

1

1

1

1

1

4 5

2

0

1

1

1

-1

1

-1

-1

0

1

1

0

-1

1

1

-1

-1

1

0

E®a4

6 5 4

-1

C

1

-1

E®&5

-2

-2

0

-1

0

1

1

-1

0

0

0

1

1

0

506


20. Let F, G1, and G2 be arbitrary. If o, is an irreducible F-representation of G;

for i = 1, 2, show that of ®02 is an irreducible F-representation of Gl x G2.

21.

22.

23.

If Gl and G2 are finite and F is excellent for Gl x C2, show that all irreducible F-representations of Gl and G2 are as in the last problem. (Hint: Use Theorem 3.4.) Also, show this without any restrictions. Compute the "multiplication table" for irreducible complex representations of Se, i.e., for any two irreducible representations of and a2, decompose of ® u2 into a direct sum of irreducibles. (Hint: Use characters.) Do the same for A5.

24. Do the same for S5. 25. Let P1 and P2 be dist sets and let a, be a permutation representation of G on P1 for i = 1, 2. Show that of ® a2 is a permutation representation of G on P1 U P2 and that O1 ® a2 is a permutation representation of G on P1 X P2.

all the group and character computations of Example 5.21. Find all systems of domains of imprimitivity for the permutation action of D2. on the vertices of a regular m-gon. 28. Determine the conjugation representation of S4 on itself (cf. Example 6.19). 29. (a) Let G be the nonabelian group of order 21. Find all irreducible complex representations of C. (b) More generally, let p and q be primes with p dividing q - 1, and let G be the nonabelian group of order pq. Find all irreducible complex representations of G. (Hint: They are all either one-dimensional or induced from one-dimensional representations.) 30. Let a : C -. Aut(V) be an irreducible representation of G. If 9 E Z(G), the center of G, show that a(g) is a homothety. 26. 27.

Appendix

A.1 Equivalence Relations and Zorn's Lemma If X is a set, a binary relation on X is a subset R of X x X. If (x, y) E X x X

it is traditional to write xRy or x R y instead of (x, y) E R. The second notation is frequently shortened to x - y and the relation is denoted by rather than R. (A.1) Definition. A binary relation - on X is an equivalence relation if the following conditions are satisfied.

(1) x -r x for all x E X. (reflexivity) (2) x - y implies y - x. (symmetry) (3) x - y and y - z implies x - z. (transitivity) If x E X then we let [x] = {y E X : x - y}. [x] is called the equivalence class of the element x E X.

Let X be a nonempty set. A partition of X is a family of nonempty subsets {At}iEI of X such that A; n Aj = 0 if i # j and X = UIEI A;. Thus a partition of X is a collection of subsets of X such that each element of X is in exactly one of the subsets. Partitions and equivalence relations are essentially the same concept as we see from the following two propositions. (A.2) Proposition. If - is an equivalence relation on X, then the family of all equivalence classes is a partition of X. Proof. By Definition A.1 (1), if x E X, then x E [x]. Thus the equivalence classes are nonempty subsets of X such that X = U [x]. zEX

508

Appendix

Now suppose that [x] n [y] ; 0. Then there is an element z E [x] n [y].

Then x - z and y - z. Therefore, by symmetry, z - y and then, by transitivity, we conclude that x - y. Thus y E [xJ and another application of transitivity shows that [y] C [x] and by symmetry we conclude that

0

[xi C [y] and hence [x] = [y].

(A.3) Proposition. If {Ai}iEJ is a partition of X, then there is an equivalence relation on X whose equivalence classes are precisely the subsets Ai for i E I.

Proof. Define the relation - on X by the rule x - y if and only if x and y both are in the same subset Ai. Properties (1) and (2) of the definition of an equivalence relation are clear. Now check transitivity. Suppose that x - y and y - z. Then x and y are in the same Ai, while y and z are in the same subset, which must be the same Ai that contains x since the family {Ai} is pairwise dist. Thus, - is in fact an equivalence relation. Now suppose that x E Ai. Then from the definition of - we have that x - y if and only if y E Ai. Thus [x] = Ai and the proposition is proved. 0

Remark. The partition of a set X into equivalence classes following an equivalence relation is a concept that is used repeatedly in the construction of quotient objects in algebra. Now we give a brief introduction to Zorn's lemma.

(A.4) Definition. Let X be a nonempty set. A binary relation < on X is said to be a partial order if it satisfies the following: (1) x < x for all x E X. (reflexivity) (2) x < y and y x implies x = y. z implies x < z. (transitivity) (3) x < y and y The standard example of a partially ordered set is the power set P(Y) of a nonempty set Y, where A < B means A C B.

If X is a partially ordered set, we say that X is totally ordered if whenever x, y E X then x < y or y < x. A chain in a partially ordered set X is a subset C C X such that C with the partial order inherited from X is a totally ordered set. If S C X is nonempty, then an upper bound for S is an element xo E X (not necessarily in S) such that s < x0

for all

s E S.

A maximal element of X is an element m E X such that

if m < x,

then m = x.

A.1 Equivalence Relations and Zorn's Lemma

509

Thus, m is maximal means that, whenever m and x are comparable in the partial order on X, x < m. It is possible for a partially orderei. set to have many maximal elements, and it is possible for a partially ordered set to have no maximal elements. The most important criterion for the existence of maximal elements in a partially ordered set is Zorn's lemma.

(A.5) Proposition. (Zorn's Lemma) Let X be a partially ordered set and assume that every chain in X has an upper bound. Then X has a maximal element.

It turns out that Zorn's lemma is equivalent to the axiom of choice. We refer the reader to Section 16 of Naive set theory by P.R. Halmos for a detailed discussion of the derivation of Zorn's lemma from the axiom of choice. It is worth pointing out that many of the results proved using the axiom of choice (or its equivalent, Zorn's lemma) turn out to be equivalent to the axiom of choice. We mention only that the existence of a maximal ideal in a ring with identity (Theorem 2.2.16) is one such result.

Bibliography

The following is a collection of standard works in algebra:

2.

E. Artin, Galois Theory, Second Edition, University of Notre Dame Press, South Bend, IN, 1959. N. Bourbaki, Algebra I, Chapters 1 -- 3, Springer-Verlag, New York,

3.

N. Bourbaki, Algebra II, Chapters 4 - 7, Springer-Verlag, New York,

4.

P. Cohn, Algebra, Second Edition, John Wiley, New York, Vol. I, 1982; Vol. II, 1989; Vol. III, 1991.

5.

C.W. Curtis and I. Reiner, Representation Theory of Finite Groups

6.

and Associative Algebras, Interscience Publishers, New York, 1962. L.G. Grove, Algebra, Academic Press, New York, 1983.

7.

T.W. Hungerford, Algebra, Graduate Texts in Math. 73, Springer-

1.

1989. 1990.

Verlag, New York, 1974.

N. Jacobson, Lectures in Abstract Algebra, Van Nostrand, Princeton, NJ, Vol. 1, 1951; Vol. II, 1953; Vol. III, 1964. 9. N. Jacobson, Basic Algebra I, W. H. Freeman, New York, 1985. 10. N. Jacobson, Basic Algebra 11, W. H. Freeman, New York, 1989. 11. S. Lang, Algebra, Addison-Wesley, Reading, MA, 1965. 12. J.-P. Serre, Linear Representations of Finite Groups, Graduate Texts in Math. 42, Springer-Verlag, New York, 1977. 13. B.L. van der Waerden, Modern Algebra, Vol. I, II, Frederick Unger Publishers, New York, 1955. 8.

Index of Notation

This list consists of all the symbols used in the text. Those without a page reference are standard set theoretic symbols; they are presented to establish the notation that we use for set operations and functions. The rest of the list consists of symbols defined in the text. They appear with a very brief description and a reference to the first occurrence in the text. Symbol

Description

C

set inclusion

n

set intersection set union everything in A but not in B cartesian product of A and B cardinality of A (E N U {oo}) function from X to Y

Page

A C B but A #,B U

A\B

AxB JAI

f:X-+Y a'-+ f(a)

IX :X-*X

fIZ

N

e, 1 Z Z+ Q

R C

Q' R' C.

Z Zn

a is sent to f (a) by f identity function from X to X restriction of f to the subset Z natural numbers = { 1, 2, ... } group identity 1 integers 2 nonnegative integers 54 rational numbers 2 real numbers 2 complex numbers 2 nonzero rational numbers 2 nonzero real numbers 2 nonzero complex numbers 2 integers modulo n 2 integers relatively prime to n (multiplication mod n) 2

512

Index of Notation

Sx, S. U.

GL(n, R) SL(n, R) P(X)

ADB C(R)

D(R)

al fin

:-la

am

Ker(f) Im(f) (S)

T(n, R) Z(G) C(x) [a, b]

G', [G, G] Q D2n JGJ

o(a)

aH, Ha [G : H] [a]c

P-(G) G/H

NaG Ld

Aut(G) Inn(G) la (m, n) Gx G(x) (il i2 ... ir)

sgn(a)

An

NxH N yam H R'

char(R) Z[i]

symmetric group complex nth roots of unity general linear group over R of degree n special linear group over R of degree n power set of X symmetric difference of A and B continuous real-valued functions on R differentiable real-valued functions on R inverse of a in G

product of al, ..., an in G a to the meh power in G kernel of homomorphism f image of homomorphism f subgroup generated by S upper triangular n x n matrices center of G centralizer of x E G commutator of a and b commutator subgroup of G quaternion group dihedral group order of G order of a E G left and right cosets of H index of H in G conjugacy class of a nonempty subsets of G coset space of H N is a normal subgroup of G is isomorphic to automorphism group of G inner automorphisms of G conjugation by a greatest common divisor of m and n permutation representation on G/H orbit of x stabilizer of x r-cycle sign of the permutation a alternating group external direct product of N and H semidirect product group of units of ring R characteristic of R Gaussian integers

Index of Notation Q[V "a-1

quadratic field

Mm,n(R)

m x n matrices over R

R[[X ]] R(G)

ring of n x n matrices over R ijeh entry of matrix A kronecker delta matrix units diagonal matrices upper triangular matrices lower triangular matrices definite quaternion algebra ring of quaternions endomorphism ring of abelian group A polynomial ring in indeterminate X polynomial ring in X1, ..., Xn formal power series ring in X group ring

(r)

binomial coefficient

MM(R) ent,3 (A)

bii

Efi D, ,(R)

Tn(R) TT(R)

Q(-x, -y; F)

H

End(A) R[X] R[X1, ... , Xn]

R/I

quotient ring

RXR, RX (X) PID

> I, rj 1;

[j R; R5 Q (RI

ideal generated by set X principal ideal domain sum of ideals product of ideals direct product of rings localization of R away from S quotient field of integral domain R

Z[N/d]

F(X) R[u)

deg(f (X) ) Pn(F) gcd

1cm

ACC FP

UFD cont(f (X)) R°P

Aut(R) F((R))

HomR(M, N) EndR(M) AutR(M)

rational functions in X "polynomials" in u degree of polynomial f (X) polynomials of degree < n greatest common divisor least common multiple ascending chain condition field with p elements unique factorization domain content of polynomial f (X) opposite ring of R automorphism group of ring R formal Laurent series R-module homomorphisms from M to N R-module endomorphims of M R-module automorphisms of M

Rn C(S)

center of S

513

514

Index of Notation

VT (S)

µ(M) r, N.

Ann(X) M7 ®iEIM, r1,EJ M) Coker(f ) 0.1 0* free-rankR(M) c(x) dimR(M) me(M) co(M) MP

row; (A) col; (A)

F[X]-module structure on V determined by T 113, 232 submodule generated by S 114 rank of M 115 sum of submodules NQ 116 annihilator of X 116 117 torsion submodule of M direct sum of modules M; 118 120 direct product of modules Mi 121 cokernel of homomorphism f 123 induced homomorphisms 143 free rank of M over R 150 content of x 152 dimension of a vector space 161 minimal exponent of M 161 characteristic order of M 162 p-component of M 183 ith row of matrix A 183 jth column of matrix A

I.

n x n identity matrix general linear group over R

At TI (A) T'.i (a)

GL(n, R)

P;j

transpose of A trace of A elementary transvection elementary dilation elementary permutation matrix

diag(a1, ... , an)

diagonal matrix En, a;E

LA

left multiplication by A right multiplication by A direct sum of matrices tensor product of matrices

Di(a)

RA

A®B A®B nn PW

det(A) Cofac(A) Adj(A)

permutation matrix determinant of A cofactor matrix of A adt matrix of A

Qp.m

A[aIQ] D-rank(A)

Ft (A)

submatrix of A determinantal rank of A tth Fitting ideal of A

206 208 208 209

M-rank(A)

row-rank(A) col-rank(A) rank(A)

op fv1B

184 186 186 186 187 187 187 188 189 189 192 193 197 197 199 204 204 205 205 205 205

opposite function coordinate matrix of v using basis B

214 215

Index of Notation

P' If ]C If] 13

det(f) 11'(f) TA

mT(X) CT(X)

C(f(X)) CA(X) JA,n valg(A) vgeom (A)

JL(XI,n

(:) A*

W T*

CC(A)

BJA BRA BLA

change of basis matrix matrix of f E HomR(M, N) matrix of f E EndR(M) determinant of f E EndR(M) trace of f E EndR(M) multiplication by A minimal polynomial of T characteristic polynomial of T companion matrix of f (X) characteristic polynomial of matrix A Jordan block algebraic multiplicity of eigenvalue A geometric multiplicity of eigenvalue A generalized Jordan block real Jordan block inner product Hermitian transpose norm orthogonal complement adt of T pth compound matrix of A B is left equivalent to A B is right equivalent to A B is equivalent to A

P(M)

M* 13*

M**

Hull(N) K(A) K*(B) C(M) f*

Ad(f) T

215 217 217 227 227 229 235 235

237 240 246 249 249 251 254 269 269 270 273 273 281

296 296 296 329 329

(M: N) C(T)

515

centralizer of T dual module of M dual basis double dual of M

annihilator of A C M annihilator of B C_ M* set of complemented submodules of M

adt of f adt of f conjugate of r

am, 0«

matrix of form 0

335 341 341 342 345 345 346 346 349 349 351 352

Bilin(M) Seslin(M) (±1)-symmetric e-symmetric

bilinear forms on 111

353 353

sesquilinear forms on M

353

Nl 1 N2

orthogonal direct sum

356 356 357

[0113

516

Index of Notation

M. = M° (4)

Nl

kernel of b/s-linear form orthogonal complement

no

m 1 ... 1 0 (n summands)

Pf(A)

Pfaffian of A

signature (0) fT

adt of f

41114)2

orthogonal sum of 01 and 02 e(m), o(m) Arf (4i) fM,

fv

f(M) sM, FM MI HomR(M, N)

Hom_S(M, N) Hom(R,S)(M, N)

M®sN .f ® 9 RMS RMS ®S SNT

SPIR 1R RNT DCC n

( (or (k) T

a1 ®a2 a1 ®a2

Arf invariant of 4i isometry group of 0 reflection determined by M; hyperplane reflection determined by y length of module M direct sum of s (I') copies of M

left R-module homomorphisms right S-module homomorphisms (R, S)-bimodule homomorphisms tensor product of modules tensor product of homomorphisms M is an (R, S)-bimodule bimodule type of tensor product bimodule type of Horn descending chain condition in Chapter 8, n = IGI primitive root of X k - 1 in F trivial representation direct sum of representations tensor product of representations one-dimensional representation augmentation map

pullback of a by f average of f Q8 Xo

i(V, W) In d#(W )

N(o)

number of conjugacy classes of G quaternion group character of a intertwining number restriction of V to H induction of W to G inertia group of a

H9 H9 RF(G)

K(R)

F-representation ring of G

358 359 361 364 372 375 378 380 381 384 386 386 399 401 403 414 414 414 418 421 423 423 423 434 439 440 440 441 441 441

442 443 446 454 458 462 469 480 480 483 492 492 502 503

Index of Terminology

abelian groups, 1 generators and relations, 325 representations of, 451 structure theorem for finitely generated, 168 action of a group, 25 adt associativity of Hom and tensor product, 423 of homomorphism, 349, 375 of linear transformation, 273 matrix, 204 algebra definite quaternion, 54 exterior, 437 fundamental theorem of, 79 R-, 110, 184 algebraic, 79, 91 integer, 91, 100, 284 multiplicity, 249 algebraically closed, 79 alternating, 195 group, 31, 43, 47, 459, 475 annihilator, 116, 345 antiautomorphism, 108 Arf invariant, 381 ascending chain condition (ACC),

in a group, 1, 4 in a ring, 5, 49 of matrix multiplication, 219 augmentation ideal, 65, 442 map, 65, 442 automorphism group, 21, 54, 109 inner, 21 outer, 21 ring, 101 R-module, 109 averaging over a group, 446

basis, 129 extension of, 171 symplectic, 382 of vector space, 134 bialgebra, 413 bijective function, 2 bimodule, 183, 412 binomial theorem, 57 bisubmodule, 413 b/s-linear form, 352 Burnside's theorem, 461 Byrnes, C. I., 329

82 associates, 79 associativity

generalized, 57

cancellation laws in a group, 4 Cartan-Dieudonne theorem, 388

518


Cauchy-Binet theorem, 210 Cauchy-Schwartz inequality, 270 Cauchy's theorem, 25 Cayley diagram, 4 Cayley-Hamilton theorem, 241, 284

Cayley's theorem, 23 center of a group, 8 of a ring, 111, 184 of EndR(M), 220 centralizer of element, 8 in a ring, 335 chain, 508 of invariant ideals, 160 of submodules, 397 length of, 397 refinement of, 397 change of basis, 215 for b/s-linear form, 354 change of rings, 420 character, 462 orthogonality of, 464 table, 467 characteristic of a ring, 51 polynomial, 235, 240 Chinese remainder theorem, 65, 88 Choi's pivotal condensation, 282 class

equation, 15 function, 226, 467 Clifford's theorem, 484 cofactor expansion, 203 matrix, 204 Cohn, P. M., 303 cokernel, 121, 290 column index, 183 commutator of elements, 9 matrix, 278 subgroup, 9 companion matrix, 236

complemented submodule, 121, 171

complete set of nonassociates, 301 set of representatives, 300 set of residues, 301 composition series, 397 compound matrix, 281 congruence, 355 conjugate, 355 conjugacy classes, 15, 454 self-inversive, 472 conjugate, 14 conjugation, 21, 350, 501 content of element, 150 of polynomial, 93 convolution product, 57 coprime ideals, 65 correspondence theorem for groups, 20 for modules, 114 for rings, 60 coset double, 492 left/right, 11 representative, 11

space, 16

cycle, 28 dist, 28

structure, 32 cyclic

group, 8 module, 115, 156 VT, 113, 232

Dedekind domain, 142, 178 degree of polynomial, 73 of representation, 439 derivation, 280 descending chain condition (DCC), 434

determinant, 199


cofactor expansion of, 203 function, 195 of b/s-linear form, 355 of homomorphism, 227 Laplace expansion of, 203 determinantal criteria for homomorphisms, 222 divisor, 205, 312 formula for invariant factors, 314

rank, 205 diagonalizable, 229 form, 364 linear transformation, 243 simultaneously, 229 dihedral group, 9, 33 representations of, 441, 457 dimension, 152 direct product (external) of groups, 34 (internal) of groups, 35 of modules, 120 of rings, 66 direct sum of matrices, 192 of modules, 118 of representations, 441 direct summand, 121 divided differences, 102 divides, 79 division algorithm, 74 noncommutative, 103 division ring, 50, 396 Dixon, J. D., 329, 334 domain Euclidean, 67, 77, 86 integral, 50 of imprimitivity, 498 of transitivity, 497 principal ideal, 62, 142 unique factorization, 92 dual basis, 341 double, 342, 431 module, 341, 430

space, 273

eigenmodule, 229 eigenspace, 229 generalized, 250 eigenvalue, 229, 243, 247 eigenvector, 228, 243, 247 generalized 248 Eisenstein's criterion, 97 elementary dilation, 187 divisors, 163, 316 matrix, 187 permutation matrix, 187 symmetric functions, 105 transvection, 187 endomorphism of abelian group, 54 of R-module, 109 equivalence class, 507 relation, 507 equivalent homomorphisms, 223, 290 left, 296 matrices, 223, 290, 296 right, 296 Euclidean algorithm, 86 domains 67, 77, 86 even bilinear form, 365, 381 permutation, 31 exact half, 428 left, 428 right, 428 exact sequence of groups, 37 of R-modules, 121 short, 121 split, 37, 121 exponent of a group, 13 extension, 37

519


520

split, 37 exterior algebra, 437

field, 50, 170 algebraically closed, 79 excellent for representation, 453 good for representation, 439 of constants, 281 of p-adic numbers, 106 quadratic, 52 finite free presentation, 289 Fitting ideal, 205, 340 flat, 430 form bilinear, 351 b/s-linear, 352 even, 365, 381 Hermitian, 356 negative definite, 371 non-degenerate, 356 non-singular, 356 odd, 365, 381 positive definite, 371 quadratic, 376 rank of, 359 sesquilinear, 351 skew-Hermitian, 356 skew-symmetric, 356 symmetric, 356 free on an index set, 130 module, 129, 142 presentation, 133 rank, 143 torsion-, 117 Frobenius, 336, 454 formula for minimal polynomial, 315

reciprocity, 487 fundamental theorem of algebra, 79 of arithmetic, 83, 91 Gauger, M. A., 329 Gaussian integers, 52

Gauss's lemma, 93 general linear group, 3, 53, 186 Gram-Schmidt, 272, 394 greatest common divisor (gcd), 22, 81

group, 1 abelian, 1 alternating, 31, 43, 47, 459 automorphism, 21, 54 character of, 462 cyclic, 8 dihedral, 9, 33, 441, 457 finitely generated, 7 general linear, 3, 53, 186 given by generators and relations, 8, 325 inertia, 483 isometry, 384 Klein 4-, 3 of order < 5, 14 of order < 15, 44 p-, 26 permutation, 2, 23 quaternion, 8, 46, 54, 458 quotient, 17 simple, 47 special linear, 3 symmetric, 2, 22, 460 unitary, 394

Hamilton, 54 Hermite normal form, 301 Hermitian form, 356 matrix, 274 skew-, 356 transpose, 269 Hilbert basis theorem, 77 homomorphism anti-, 351 bimodule, 412 group, 7 ring, 58 R-module, 109


homothety, 450 hull, 345 hyperplane reflection, 386

ideal, 58 augmentation, 65, 442 Fitting, 205 generated by a set, 61 invertible, 140 left and right, 58 maximal, 62 order, 116 prime, 63 principal, 61 product of, 62 sum of, 62 idempotent, 98, 473 matrix, 279 image of group homomorphism, 7 of ring homomorphism, 58 of R-module homomorphism 109 indeterminate, 55 index of a subgroup, 12 induction of representation, 480 transitivity of, 486 injection, natural, 35 inner product, 269 space, 269, 351 integral domain, 50 intertwining number, 469, 487 invariant factor, 156, 161 determinantal formula for, 314 of linear transformation, 235 of matrix, 308 theorem for submodules, 153, 224

invariant submodule, 227 subspace, 113, 227 inverse, left/right, 221, 283 irreducible element, 79 module, 179, 395

521

representation, 443 isometric, 353 isometry, 353 group, 384 isomorphism group, 7 ring, 58 isomorphism theorems (noether) for groups, 19 for modules, 113 for rings, 60

Jordan block, 246

canonical form, 247, 264

generalized - block, 251 generalized - canonical form, 252

irreducible generalized - block, 251

real - block, 254 real - canonical form, 254 kernel of b/s-linear form, 358 of group homomorphism, 7 of ring homomorphism, 58 of R-module homomorphism, 109

Kronecker, 97 delta, 52 product, 193, 279

Lagrange interpolation, 76, 280 Lagrange's theorem, 12 Laplace expansion theorem, 203, 212

least common multiple, 81 length, 329, 399 linear

n- - transformation, 194 subspace, 112

522


transformation, 109 linearly

dependent,128 independent, 128, 209 localization, 68, 175

Mackey's theorem, 492 Maschke's theorem, 447 matrix adt, 204 block, 190 block diagonal, 192 change of basis, 215 cofactor, 204 column, 182 commutator, 278 companion, 236 compound, 281 coordinate, 215 diagonal, 53, 188 diagonalizable, 229 diagonally dominant, 393 elementary, 188 Hermitian, 274 invertible, 186 lower triangular, 53 nilpotent, 278, 284 normal, 274 of homomorphism, 217 orthogonal, 274 partitioned, 190 permutation, 197 positive definite, 394 product, 52 reduced row echelon, 302 representation of b/s-linear form, 353 ring, 52 row, 182 scalar, 185 self-adt, 274 similar, 187 skew-Hermitian, 287 skew-symmetric, 279

strictly diagonally dominant, 394

strictly lower triangular, 53 strictly upper triangular, 53 symmetric, 274, 279 transpose, 186 unimodular, 186 unitary, 274 units, 53, 184 upper triangular, 53 Vandermonde, 280 Wronskian, 282 maximal element, 82, 508 middle linear map, 418 minimal polynomial, 235 Frobenius formula for, 315 minor, 205 modular law, 175 module cyclic, 115, 156 divisible, 179 finitely generated, 115 free, 129 indecomposable, 179, 395 irreducible, 179, 395

left/right, 107 primary, 162 projective, 138, 145, 231, 345, 425

R-, 109 rank of, 115 reflexive, 343, 431 semisimple, 397 simple, 395 torsion, 117 torsion-free, 117, 146 monoid, 15, 442 multiplication left/right, 189, 332 scalar, 107, 183 multiplicatively closed, 68 multiplicity algebraic, 249 geometric, 249 of representation, 445


negative definite, 371 Newton identities, 105, 286 Newton interpolation, 76 nilpotent, 98 matrix, 279, 284 Noetherian ring, 82 non-degenerate b/s-linear form, 356 part of b/s-linear form, 359 quadratic form, 376 skew-symmetric form, 361 non-singular bilinear form, 356 quadratic form, 376 norm, 270 normal, 274 normal form generalized Jordan, 252 Hermite, 301 Jordan, 247, 264 real Jordan, 254 Smith, 308

odd bilinear form, 365, 381 permutation, 31 opposite ring, 98, 108, 214 orbit, 26, 328, 496 order of a group, 9 of an element, 9 orthogonal, 271 complement, 272, 359 direct sum, 357, 378 group, 385 submodules, 357 transformation 274 orthogonality of characters, 464 orthonormal, 271 basis, 272

P_

component, 162

523

group, 26 primary, 162 subgroup, 26 Sylow subgroup, 26 partial fraction expansion, 178 partial order, 508 partition, 507 permutation, 2 even, 31 group, 23 matrix, 197 odd, 31 sign of, 31 similar, 282 Pfafi an, 364 PID (principal ideal domain), 62, 142, 329

polarization identity, 392 polynomial constant term of, 73 content of, 93 degree of, 73 function, 72 leading coefficient of, 73 monic, 73 primitive, 93 regular, 103 substitution, 73 positive definite, 371, 394 power set, 3 primary decomposition theorem, 235

primary module, 162 prime divisors, 92

element in a ring, 63, 79 factor, 92 factorization, 92 ideal, 63 primitive element, 148 polynomial, 93 product convolution, 57 direct, 34, 35, 66, 120

524


kronecker, 193

of ideal and module, 117 of ideals, 62 of matrices, 183 of partitioned matrices, 191 scalar, 183 semidirect, 35 tensor, 193, 417, 441 projection, 287 natural, 18, 34 orthogonal, 288 projective module, 138, 145, 231, 345, 406, 425

quadratic form, 376 isometric, 379 orthogonal direct sum of, 378 quadratic refinement, 378 quaternion group, 9, 46, 54, 458 ring, 54 quotient field, 68 group, 17 module, 113 ring, 59

rank column, 208 determinantal, 205, 211 free, 143 invariance of, 152 M-, 206 of a matrix, 209 of a module, 115, 152, 172, 347 of form, 359 row, 208 rational canonical form, 238, 310 rational function field of, 72 proper, 178 rational root theorem, 104 reflection, 386

relation binary, 507 equivalence, 507 relatively prime ideals, 65 set of elements, 81 remainder theorem, 75 noncommutative, 103 representation, 438 defined over F, 442 degree of, 439 direct sum of, 441 equivalent, 439 faithful, 443 imprimitive, 498 indecomposable, 443 induced, 479 intransitive, 497 irreducible, 443 isomorphic, 439 isotypic, 453 left regular, 23, 440 monomial, 442 multiplicity of, 445 permutation, 23, 440, 442, 496 primitive, 498 proper, 503 pullback, 443 restriction of, 480 ring, 502 tensor product of, 441, 443 transitive, 497 trivial, 440 virtual, 503 restriction of representation, 480 transitivity of, 486 ring, 49 characteristic of, 51 commutative, 49 division, 50 of formal power series, 56 group, 57, 108, 438 local, 72 matrix, 52, 67 Noetherian, 82


of p-adic integers, 106 of quaternions, 54 opposite, 99, 108, 214 polynomial, 55 quotient, 59 representation, 502 semigroup, 57 semisimple, 406, 447 simple, 409 with identity, 49 row index, 183

scalar multiplication, 107 product, 183 Schur's lemma, 396, 450 self-adt, 274 semidirect product, 35 semigroup, 57 semisimple linear transformation, 435 module, 397 ring, 406, 447 sesquilinear form, 351 signature, 372 similarity of homomorphisms, 226 of matrices, 187, 310, 328 simple factorization, 401 group, 47 module, 395 ring, 409 Smith normal form, 308 solvability of linear equations, 178, 206, 208 over Z, 326 spectral theorem, 276, 287 self-adt case, 277 split, 121 stabilizer, 26, 497 subgroup, 6 commutator, 9 generated by S, 8

525

normal, 17

of F', 170 p-, 26 p-Sylow, 26 submodule, 112 complemented, 121, 171 generated by a set, 114 generators of, 114 pure, 172 sum of, 116 torsion, 117, 158 subring, 51 generated by a set, 61 subspace, 113 invariant, 113, 227 substitution homomorphism, 73 sum direct, 118, 120, 127 of submodules, 116 Sylow's theorem, 26 Sylvester's law of inertia, 371 symmetric difference, 3 ,--,356 bilinear form, 356 group, 2, 22, 460, 478 matrix, 274 skew-, 356 transformation, 274 symmetry, 33 symplectic basis, 382

tensor product of bimodules, 417 of matrices, 193, 202, 283, 333, 433

of representations, 441, 443 torsion element, 117 free, 117 module, 117 submodule, 117 totally isotropic, 357 trace, 186

526


transitive, 499 transpose, 186 Hermitian, 269 transposition, 28 triangle inequality, 270

unique factorization domain (UFD), 92 unit, 50 unitary, 274

vector space, 109

Wedderburn, 50, 407 Witt's cancellation theorem, 371 Witt's theorem, 368, 372, 387 zero divisor, 50 Zorn's lemma, 62, 135, 144, 154, 377, 509

Graduate Texts in Mathematics (continued from )mgr ii)

62

KARGAPOLOV/MERLZJAKOV. Fundamentals

93

Part 1. 2nd ed.

63 64

BOLLOBAS. Graph Theory. EDWARDS. Fourier Series. Vol. 1 2nd ed.

94

65

WELLS. Differential Analysis on Complex Manifolds. 2nd ed.

95

66 WATERHOUSE. Introduction to Affine Group Schemes. 67

SERRE. Local Fields.

68 WEIDMANN. Linear Operators in Hilbert Spaces. 69

LANG. Cyclotomic Fields 11.

70 MASSEY. Singular Homology Theory. 71 72

FARKAS/KRA. Riemann Surfaces. 2nd ed. STILLWELL. Classical Topology and

96

Theory. 2nd ed. 75

Analysis. 2nd ed. KOBLITZ. Introduction to Elliptic Curves and Modular Forms. 2nd ed.

98

BROCKERITom DIECK. Representations of

Compact Lie Groups. 99 GROVE/BENSON. Finite Reflection Groups. 2nd ed.

100 BERG/CHRISTENSEN/RESSEL. Harmonic

76

lrrAKA. Algebraic Geometry.

77

HECKE. Lectures on the Theory of Algebraic Numbers.

78

BURRIS/SANKAPPANAVAR. A Course in

Universal Algebra. 79 WALTERS. An Introduction to Ergodic Theory. 80

ROBINSON, A Course in the Theory of

Groups. 2nd ed. 81

FORSTER. Lectures on Riemann Surfaces.

82

Borr/Tu. Differential Forms in Algebraic Topology.

83

WASHINGTON. Introduction to Cyclotomic Fields. 2nd ed. IRELAND/ROSEN. A Classical Introduction

84 85 86

Analysis on Semigroups: Theory of Positive Definite and Related Functions. EDWARDS. Galois Theory. 102 VARADARAJAN. Lie Groups, Lie Algebras

101

and Their Representations.

HOCHSCHILD. Basic Theory of Algebraic

Groups and Lie Algebras.

103 LANG. Complex Analysis. 3rd ed. 104 DuBROVIN/FOMENK<)/NOVIKOV. Modern

Geometry-Methods and Applications. Part II. 105 LANG. SL2(R). 106

SILvaRMAN. The Arithmetic of Elliptic Curves.

OLVER. Applications of Lie Groups to Differential Equations. 2nd ed. 108 RANGE. Holomorphic Functions and Integral Representations in Several Complex Variables. 109 LENTO. Univalent Functions and Teichmuller Spaces. 110 LANG. Algebraic Number Theory. 107

III

HUSEMOLLER. Elliptic Curves.

to Modern Number Theory. 2nd ed.

112 LANG. Elliptic Functions.

EDWARDS. Fourier Series. Vol. 11. 2nd ed. VAN LINT. Introduction to Coding Theory.

113 KARATZAS/SHREVE. Brownian Motion and

2nd ed.

114

88

PIERCE. Associative Algebras.

89

LANG. Introduction to Algebraic and Abelian Functions. 2nd ed.

90 BRONDSTED. An Introduction to Convex Polytopes. BEARDON. On the Geometry of Discrete Groups.

92 DIESTEL. Sequences and Series in Banach Spaces.

Stochastic Calculus. 2nd ed. KOBLITZ. A Course in Number Theory and Cryptography. 2nd ed.

87 BROWN. Cohomology of Groups.

91

SHIRYAEV. Probability. 2nd ed. CONWAY. A Course in Functional

97

HUNGERF>RD. Algebra.

74 DAVENPORT. Multiplicative Number

WARNER. Foundations of Differentiable

Manifolds and Lie Groups.

Combinatorial Group Theory. 2nd ed. 73

DUBROVIN/FOMENKO/NOVIKOV. Modem

Geometry-Methods and Applications.

of the Theory of Groups.

115

BERGER/GOSTtAUx. Differential Geometry: Manifolds, Curves, and Surfaces.

116

KELLEY/SRtNIVASAN. Measure and

117

Integral. Vol. 1. SERRE. Algebraic Groups and Class Fields.

118 PEDERSEN. Analysis Now.

119 ROTMAN. An Introduction to Algebraic Topology.

120 ZIEMER. Weakly Differentiable Functions:

Sobolev Spaces and Functions of Bounded Variation. 121 LANG. Cyclotomic Fields I and Il. Combined 2nd ed. 122 REMMERT. Theory of Complex Functions. Readings in Mathematics 123 EBBINGHAUS/HERMES et al. Numbers.

Readings in Mathematics 124 DUBROVIN/FOMENKO/NoVIKOV. Modem

Geometry-Methods and Applications. Part III. 125 BERENSTEIN/GAY. Complex Variables: An Introduction. 126 BOREL. Linear Algebraic Groups. 2nd ed. 127 MASSEY. A Basic Course in Algebraic Topology. 128 RAUCH. Partial Differential Equations. 129 FuLTON/HARRIS. Representation Theory: A First Course. Readings in Mathematics

130 DODSGN/POSTGN. Tensor Geometry.

LAM. A First Course in Noncommutative Rings. 132 BEARDON. Iteration of Rational Functions. 133 HARRIS. Algebraic Geometry: A First Course. 134 ROMAN. Coding and Information Theory. 135 ROMAN. Advanced Linear Algebra. 131

136 ADKINS/WEtNTRAUB. Algebra: An Approach via Module Theory. 137 AXLER/BOURDON/RAMEY. Harmonic

Function Theory. 138 COHEN. A Course in Computational Algebraic Number Theory. 139 BREDON. Topology and Geometry. 140 AuBIN. Optima and Equilibria. An Introduction to Nonlinear Analysis.

149 RATCLIFFE. Foundations of Hyperbolic Manifolds. 150 EISENBUD. Commutative Algebra

with a View Toward Algebraic Geometry. 151 SILVERMAN. Advanced Topics in the Arithmetic of Elliptic Curves. 152 ZIEGLER. Lectures on Polytopes. 153 FULTON. Algebraic Topology: A First Course. 154 BROWN/PEARCY. An Introduction to Analysis.

155 KASSEL. Quantum Groups. 156 KECHRIS. Classical Descriptive Set Theory.

157 MALLIAVIN. Integration and Probability. 158 ROMAN. Field Theory. 159 CONWAY. Functions of One

Complex Variable ll. 160 LANG. Differential and Riemannian Manifolds. 161

BORWEIN/ERDELYI. Polynomials and

Polynomial Inequalities. 162 ALPERIN/BELL. Groups and

Representations. 163 DIXON/MORTIMER. Permutation Groups. 164 NATHANSON. Additive Number Theory:

The Classical Bases. 165 NATHANSON. Additive Number Theory:

Inverse Problems and the Geometry of Sumsets. 166 SHARPE. Differential Geometry: Cartan's Generalization of Klein's Erlangen Program. 167 MORANDI. Field and Galois Theory. 168 EWALD. Combinatorial Convexity and Algebraic Geometry.

141

BECKER/WEISPFENNING/KREDEL. GrObner

169 BHATIA. Matrix Analysis.

142

Bases. A Computational Approach to Commutative Algebra. LANG. Real and Functional Analysis. 3rd ed.

172 REMMERT. Classical Topics in Complex

143 DOOB. Measure Theory. 144 DENNIS/FARB. Noncommutative Algebra.

145 VICK. Homology Theory. An Introduction to Algebraic Topology. 2nd ed. 146 BRIDGES. Computability: A Mathematical Sketchbook. 147 ROSENBERG. Algebraic K-Theory and Its Applications. 148 ROTMAN. An Introduction to the

Theory of Groups. 4th ed.

170 BREDON. Sheaf Theory. 2nd ed. 171

PETERSEN. Riemannian Geometry. Function Theory.

173 DIESTEL. Graph Theory.

174 BRIDGES. Foundations of Real and Abstract Analysis.

175 LICKORISH. An Introduction to Knot Theory. 176 LEE. Riemannian Manifolds. 177 NEWMAN. Analytic Number Theory. 178 CLARKEILEDYAEV/STERNIWOLENSKI.

Nonsmooth Analysis and Control Theory.

179 DOUGLAS. Banach Algebra Techniques in Operator Theory. 2nd ed.

180 SRIVASTAVA. A Course on Borel Sets.

KRESS. Numerical Analysis. 182 WALTER. Ordinary Differential 181

Equations.

183 MEGGINSON. An Introduction to Banach Space Theory. 184 BOLLOBAS. Modern Graph Theory. 185 Cox/LrTTLefO'SHEA. Using Algebraic Geometry.

186 RAMAKRISHNAN/VALENZA. Fourier

Analysis on Number Fields.

187 HARRISIMORRISON. Moduli of Curves. 188 GOLDBLATT. Lectures on the Hyperreals:

An Introduction to Nonstandard Analysis. 189 LAM. Lectures on Modules and Rings. 190 ESMONDFJMURTY. Problems in Algebraic Number Theory. 191 LANG. Fundamentals of Differential Geometry. 192 HIRSCHILACOMBE. Elements of Fuctional Analysis.

This book is designed as a text for a first-year graduate algebra course. The choice of topics is guided by the underlying theme of modules as a basic unifying concept in mathematics. Beginning

with standard topics in group and ring theory, the authors then develop basic module theory, culminating in the fundamental structure theorem for finitely generated modules over a principal ideal domain. They then treat canonical form theory in linear algebra as an application of this fundamental theorem. Module theory is also used in investigating bilinear, sesquilinear, and quadratic forms. The authors develop some multilinear algebra (Hom and tensor product) and the theory of semisimple rings and modules and apply these results in the final chapter to study group representations by viewing a representation of a group G over a field F as an F (G)-module.

The book emphasizes proofs with a maximum of insight and a minimum of computation, in order to promote understanding. However, extensive material on computation (for example, computation of canonical forms) is provided.

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