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College Algebra Exam Review 373

College Algebra Exam Review 373 - Q W R=.p ² End.M such...

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8.5. FINITELY GENERATED MODULES OVER A PID, PART II. 383 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 psrx C tarx D psrx . It follws that Rx D pRx . n Lemma 8.5.7. Suppose p 2 R is irreducible and pM D f 0 g . 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 f 0 g , pR ker . / . By Proposition 6.3.9 , factors through R=.p/ ; that is, there is a homomorphism
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Unformatted text preview: Q W R=.p/ ²! End .M/ such that D Q ı ± , where ± W R ²! 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 ²! M is a surjective R –module homomorphism. For x 2 M , p'.x/ D '.px/ D . Thus p M D p'.M/ D f g , 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. n 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 A ˚ A 1 ˚ A 2 ˚ ´´´ ˚ A s ; where µ A is free, µ for i ¶ 1 , A i Š R=.a i / , and...
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