College Algebra Exam Review 370

College Algebra Exam Review 370 - 1 / R=.d s / R n s : If...

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380 8. MODULES homomorphism from F onto M by '. P i r i f i / D P i r i x i . Let N denote the kernel of ' . According to Theorem 8.4.12 , N is free of rank s ± n , and there exists a basis f v 1 ;:::;v n g of F and nonzero elements d 1 ;:::;d s of R such that f d 1 v 1 ;:::;d s v s g is a basis of N and d i divides d j for i ± j . Therefore M Š F=N D .Rv 1 ˚ ²²² ˚ Rv n /=.Rd 1 v 1 ˚ ²²² ˚ Rd s v s / Lemma 8.5.1. Let A 1 ;:::;A n be R –modules and B i ³ A i submodules. Then .A 1 ˚ ²²² ˚ A n /=.B 1 ˚ ²²² ˚ B n / Š A 1 =B 1 ˚ ²²² ˚ A n =B n : Proof. Consider the homomorphism of A 1 ˚²²²˚ A n onto A 1 =B 1 ˚²²²˚ A n =B n defined by .a 1 ;:::;a n / 7! .a 1 C B 1 ; ²²² ;a n C B n / . The kernel of this map is B 1 ˚ ²²² ˚ B n ³ A 1 ˚ ²²² ˚ A n , so by the isomorphism theorem for modules, .A 1 ˚ ²²² ˚ A n /=.B 1 ˚ ²²² ˚ B n / Š A 1 =B 1 ˚ ²²² ˚ A n =B n : n Observe also that Rv i =Rd i v i Š R=.d i / , since r 7! rv i C Rd i v i is a surjective R –module homomorphism with kernel .d i / . Applying Lemma 8.5.1 and this observation to the situation described above gives M Š Rv 1 =Rd 1 v 1 ˚ ²²² ˚ Rv s =Rd s v s ˚ Rv s C 1 ²²² ˚ Rv n Š R=.d
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Unformatted text preview: 1 / R=.d s / R n s : If some d i were invertible, then R=.d i / 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) nitely generated module over R . (a) M is a direct sum of cyclic modules, M R=.a 1 / R=.a 2 / R=.a s / R k ; where the a i are nonzero, nonunit elements of R , and a i divides a j for i j ....
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