8.4.8. Set A D
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8.4.7. Retain the notation of the previous exercise. By Proposition there exist invertible matrices P 2 Matm.R/ and Q 2 Matn.R/ such that A0 D PAQ is diagonal, A0 D PAQ D diag.d1; d2; : : : ; ds; 0; : : : ; 0/; where s minfm; ng. Show that there is a basis fw0 ; : : : ; w0
1
mg of W such that fd1w0 ; : : : ; d
1
s w0sg is a basis of range.'/.
2 2
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1 23 8.4.8. Set A D
2
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4 4
4
5. Left multiplication by A defines a
2
2
0
6
homomorphism ' of abelian groups from Z4 to Z3. Use the diagonalization of A to find a basis fw1; w2; w3g of Z3 and integers fd1; : : : dsg (s 3), such that fd1w1; : : : ; dswsg is a basis of range.'/. (Hint: Compute invertible matrices P 2 Mat3.Z/ and Q 2 Mat4.Z/ such that A0 D PAQ is diagonal. Rewrite this as P 1A0 D AQ.)
8.4.9. Adopt the notation of Exercise 8.4.6. Observe that the kernel of ' is the set of Pj xj vj such that 2x 3
1
:
A 6 :: 7 D 0:
4
5
xn
That is the kernel of ' can be computed by finding the kernel of A (in Zn).
Use the diagonalization A0 D PAQ to find a description of ker.A/. Show, in fact, that the kernel of A is the span of the last n s columns of Q, where A0 D diag.d1; d2; : : : ; ds; 0; : : : ; 0/.
2
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1 2 8.4.10. Set A D . Find a basis fv
2
16
4 4 1; : : : ; v4g of Z4 and integers fa1; : : : ; ar g such that fa1v1; : : : ; ar vr g is a basis of ker.A/.
(Hint: If s is the rank of the range of A, then r D 4 s. Moreover, if A0 D PAQ is the Smith normal form of A, then ker.A/ is the span of the last r columns of Q, that is the range of the matrix Q0 consisting of the last r columns of Q. Now we have a new problem of the same sort as in Exercise 8.4.8.)
8.5. Finitely generated Modules over a PID, part II.
The Invariant Factor Decomposition
Consider a finitely generated module M over a principal ideal domain R. Let x1; : : : ; xn be a set of generators of minimal cardinality. Then M is the homomorphic image of a free R–module of rank n. Namely consider a free R module F with basis ff1; : : : ; fng. Define an R–module i
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8. MODULES homomorphism from F onto M by '.Pi ri fi / D Pi ri xi . Let N denote the kernel of '. According to Theorem 8.4.12, N is free of rank s n, and there exists a basis fv1; : : : ; vng of F and nonzero elements d1; : : : ; ds of R such that fd1v1; : : : ; dsvsg is a basis of N and di divides dj for i j .
Therefore M Š F=N D .Rv1 ˚ ˚ Rvn/=.Rd1v1 ˚ ˚ Rdsvs/ Lemma 8.5.1. Let A1; : : : ; An be R–modules and Bi Ai submodules.
Then .A1 ˚ ˚ An/=.B1 ˚ ˚ Bn/ Š A1=B1 ˚ ˚ An=Bn: Proof. Consider the homomorphism of A1 ˚ ˚ An onto A1=B1 ˚ ˚ An=Bn defined by .a1; : : : ; an/ 7! .a1 C B1; ; an C Bn/. The kernel of this map is B1 ˚ ˚ Bn A1 ˚ ˚ An, so by the isomorphism theorem for modules, .A1 ˚ ˚ An/=.B1 ˚ ˚ Bn/ Š A1=B1 ˚ ˚ An=Bn:
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Observe also that Rvi =Rdi vi Š R=.di /, since r 7! rvi C Rdi vi is a surjective R–module homomorphism with kernel .di /. Applying Lemma 8.5.1 and this observation to the situation described above gives M ŠRv1=Rd1v1 ˚ ˚ Rvs=Rdsvs ˚ RvsC1 ˚ Rvn Š R=.d1/ ˚ ˚ R=.ds/ ˚ Rn s:
If some di were invertible, then R=.di / would be the zero module, so could be dropped from the direct sum. But this would display M as generated by fewer than n elements, contradicting the minimality of n.
We have proved the existence part of the following fundamental theo rem:
Theorem 8.5.2. (Structure Theorem for Finitely Generated Modules over a PID: Invariant Factor Form) Let R be a principal ideal domain, and let M be a (nonzero) finitely generated module over R.
(a) M is a direct sum of cyclic modules, M Š R=.a1/ ˚ R=.a2/ ˚ ˚ R=.as/ ˚ Rk; where the ai are nonzero, nonunit elements of R, and ai divides aj for i j .
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(b) The decomposition in part (a) is unique, in the following sense:
Suppose M Š R=.b1/ ˚ R=.b2/ ˚ ˚ R=.bt / ˚ R`; where the bi are nonzero, nonunit elements of R, and bi divides bj for i j . Then s D t, ` D k and .ai / D .bi / for all i.
Before addressing the uniqueness statement in the theorem, we intro duce the idea of torsion.
Suppose that R is an integral domain (not necessarily a PID) and M is an R–module.
For x 2 M , recall that the annihilator of x in R is ann.x/ D fr 2 R W rx D 0g, and that ann.x/ is an ideal in R. Since 1x D x, ann.x/ R. Recall from Example 8.2.6 that Rx Š R=ann.x/
¤
as R–modules. An element x 2 M is called a torsion element if ann.x/ ¤ f0g, that is, there exists a nonzero r 2 R such that rx D 0.
If x; y 2 M are two torsion elements then sx C ty is also a torsion element for any s; t 2 R. In fact, if r1 is a nonzero element of R such that r1x D 0 and r2 is a nonzero element of R such that r2y D 0, then r1r2 ¤ 0 and r1r2.sx Cty/ D 0. It follows that the set of torsion elements of M is a submodule, called the torsion submodule, and denoted by Mtor.
We say that M is a torsion module if M D Mtor. We say that M is torsion free if Mtor D f0g. One can check that M=Mtor is torsion free. See Exercise 8.5.2
Example 8.5.3. Let G be a finite abelian group. Then G is a finitely generated torsion module over Z. In fact, every abelian group is a Z–module by Example 8.1.8. G is finitely generated since it is finite. Moreover, G is a torsion module, since every element is of finite order; that is, for every a 2 G, there is an n 2 Z such that na D 0.
Example 8.5.4. Let V be a finite dimensional vector space over a field K.
Let T 2 EndK.V /. Recall from Example 8.1.10 that V becomes a KŒx– module with .Pi ˛i xi /v D Pi ˛i T i .v/ for each polynomial Pi ˛i xi 2 KŒx and each v 2 V . V is finitely generated over KŒx because a basis over K is a finite generating set over KŒx. Moreover, V is a torsion module over KŒx for the following reason: Let n denote the dimension of V .
Given v 2 V , the set of nC1 elements fv; T .v/; T 2.v/; T 3.v/; : : : ; T n.v/g is not linearly independent over K, so there exist elements ˛0; ˛1; : : : ; ˛n of K, not all zero, such that ˛0v C ˛1T .v/ C C ˛nT n.v/ D 0. Thus
P i ˛i xi ¤ 0 and .Pi ˛i xi /v D Pi ˛i T i .v/ D 0. This means that v is a torsion element. We have shown that V is a finitely generated torsion module over KŒx.
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8. MODULES If S M is any subset, we define the annihilator of S to be ann.S/ D
\
fr 2 R W rx D 0 for all x 2 Sg D ann.x/: Note that ann.S / is an x2S
ideal, and ann.S / D ann.RS/, the annihilator of the submodule generated by S . See Exercise 8.5.1
Consider a torsion module M over R. If S is a finite subset of M then ann.S / D ann.RS/ is a nonzero ideal of R; in fact, if S D fx1; : : : ; xng and for each i , ri is a nonzero element of R such that ri xi D 0, then
Q i ri is a nonzero element of ann.S /. If M is a finitely generated torsion module, it follows that ann.M / is a nonzero ideal of R.
For the remainder of this section, R again denotes a principal idea domain and M denotes a (nonzero) finitely generated module over R.
For x 2 Mtor, any generator of the ideal ann.x/ is called a period of x. If a 2 R is a period of x 2 M , then Rx Š R=ann.x/ D R=.a/.
According to Lemma 8.4.5, any submodule of M is finitely generated.
If A is a torsion submodule of M , any generator of ann.A/ is a called a period of A.
The period of an element x, or of a submodule A, is not unique, but any two periods of x (or of A) are associates.
Lemma 8.5.5. Let M be a finitely generated module over a principal ideal domain R.
(a) If M D A ˚ B, where A is a torsion submodule, and B is free, then A D Mtor.
(b) M has a direct sum decomposition M D Mtor ˚ B, where B is free. The rank of B in any such decomposition is uniquely determined.
(c) M is a free module if, and only if, M is torsion free.
Proof. We leave part (a) as an exercise. See Exercise 8.5.3. According to the existence part of Theorem 8.5.2, M has a direct sum decomposition M D A ˚ B, where A is a torsion submodule, and B is free. By part (a), A D Mtor. Consequently, B Š M=Mtor, so the rank of B is determined.
This proves part (b).
For part (c), note that any free module is torsion free. On the other hand, if M is torsion free, then by the decomposition of part (b), M is free.
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Lemma 8.5.6. Let x 2 M , let ann.x/ D .a/, and let p 2 R be irreducible.
(a) If p divides a, then Rx=pRx Š R=.p/.
(b) If p does not divide a, then pRx D Rx.
Proof. Consider the module homomoprhism of R onto Rx, r 7! r x, which has kernel .a/. If p divides a, then .p/ .a/, and the image of .p/ in Rx is pRx. Hence by Proposition 8.2.8, R=.p/ Š Rx=pRx. If p does not divide a, then p and a are relatively prime. Hence there exist s; t 2 R such that sp C ta D 1. Therefore, for all r 2 R, r x D 1rx D ps rx C tarx D psrx. It follws that Rx D pRx.
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Lemma 8.5.7. Suppose p 2 R is irreducible and pM D f0g. Then M is a vector space over R=.p/. Moreover, if ' W M ! M is a surjective R–module homomorphism, then M is an R=.p/–vector space as well, and ' is R=.p/–linear.
Proof. Let W R ! End.M / denote the homomorphism corresponding to the R–module structure of M , .r /.m/ D rm. Since pM D f0g, pR ker. /. By Proposition 6.3.9, factors through R=.p/; that is, there is a homomorphism Q W R=.p/ ! End.M / such that D Q ı , where W R ! R=.p/ is the quotient map. Hence M is a vector space over the field R=.p/. The action of R=.p/ on M is given by .r C .p//x D Q .r C .p//.x/ D .r/.x/ D rx: Suppose that ' W M ! M is a surjective R–module homomorphism.
For x 2 M , p'.x/ D '.px/ D 0. Thus pM D p'.M / D f0g, and M is a also an R=.p/–vector space. Moreover, '..r C .p//x/ D '.rx/ D r'.x/ D .r C .p//'.x/; so ' is R=.p/–linear.
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We are now ready for the proof of uniqueness in Theorem 8.5.2.
Proof of Uniqueness in Theorem 8.5.2 Suppose that M has two direct sum decompositions: M D A0 ˚ A1 ˚ A2 ˚ ˚ As; where A0 is free, for i 1, Ai Š R=.ai /, and
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8. MODULES the ring elements ai are nonzero and noninvertible, and ai di vides aj for i j ; and also M D B0 ˚ B1 ˚ B2 ˚ ˚ Bt ;
where B0 is free, for i 1, Bi Š R=.bi /, and the ring elements bi are nonzero and noninvertible, and bi di vides bj for i j ;
We have to show that rank.A0/ D rank.B0/, s D t, and .ai / D .bi / for all i 1.
By Lemma 8.5.5 , we have Mtor D A1 ˚ ˚ As D B1 ˚ B2 ˚ ˚ Bt : Hence A0 Š M=Mtor Š B0. By uniqueness of rank (Lemma rank.A0/ D rank.B0/.
It now suffices to prove that the two decompositions of Mtor are es sentially the same, so we may assume that M D Mtor for the rest of the proof.
Note that as and bt are periods of M . So we can assume as D bt D m.
We proceed by induction on the length of m, that is, the number of irreducibles (with multiplicity) occuring in an irreducible factorization of m. If this number is one, then m is irreducible, and all of the bi and aj are associates of m. In this case, we have only to show that s D t. Since mM D f0g, by Lemma 8.5.7, M is an R=.m/–vector space; moreover, the first direct sum decomposition gives M Š .R=.m//s and the second gives M Š .R=.m//t as R=.m/–vector spaces. It follows that s D t by uniqueness of dimension.
We assume now that the length of m is greater than one and that the uniqueness assertion holds for all finitely generated torsion modules with a period of smaller length.
Let p be an irreducible in R. Then x 7! px is a module endomor phism of M that maps each Ai into itself. According to Lemma 8.5.6, if p divides ai then Ai =pAi Š R=.p/, but if p is relatively prime to ai , then Ai =pAi D f0g.
We have M=pM Š .A1 ˚ A2 ˚ ˚ As/=.pA1 ˚ pA2 ˚ ˚ pAs/ Š A1=pA1 ˚ A2=pA2 ˚ ˚ As=pAs Š .R=.p//k; where k is the number of ai such that p divides ai .
Since p.M=pM / D f0g, according to Lemma 8.5.7, all the R–modules in view here are actually R=.p/–vector spaces and the isomorphisms are R=.p/–linear. It follows that the number k is the dimension of M=pM
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as an R=.p/–vector space. Applying the same considerations to the other direct sum decomposition, we obtain that the number of bi divisible by p is also equal to dimR=.p/.M=pM /.
If p is an irreducible dividing a1, then p divides all of the ai and exactly s of the bi . Hence s t. Reversing the role of the two decompositions, we get t s. Thus the number of direct summands in the two decompositions is the same.
Fix an irreducible p dividing a1. Then p divides aj and bj for 1 j s. Let k0 be the last index such that ak0=p is a unit. Then pAj is cyclic of period aj =p for j > k0, while pAj D f0g for j k0, and pM D pAk0C1 ˚ ˚ pAs. Likewise, let k00 be the last index such that bk00=p is a unit. Then pBj is cyclic of period bj =p for j > k00, while pBj D f0g for j k00, and pM D pBk00C1 ˚ ˚ pBs.
Applying the induction hypothesis to pM (which has period m=p) gives k0 D k00 and .ai =p/ D .bi =p/ for all i > k0. It follows that .ai / D .bi / for all i > k0. But for 1