College Algebra Exam Review 367

# College Algebra Exam Review 367 - A D PAQ D diag.d 1;d 2;d...

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8.4. FINITELY GENERATED MODULES OVER A PID, PART I 377 By the linear independence of the v k we have C 2 6 4 ˛ 1 : : : ˛ n 3 7 5 D 0 . But then 2 6 4 ˛ 1 : : : ˛ n 3 7 5 D C 1 C 2 6 4 ˛ 1 : : : ˛ n 3 7 5 D 0: n We can now combine Proposition 8.4.6 and Lemma 8.4.11 to obtain our main result about bases of free R –modules and their submodules: Theorem 8.4.12. If F is a free R –module of rank n and N is a submod- ule of rank s , then there exists a basis f v 1 ; : : : ; v n g of F , and there ex- ist elements d 1 ; d 2 ; : : : ; d s of R , such that d i divides d j if i j and f d 1 v 1 ; : : : ; d s v s g is a basis of N . Proof. We already know that N is a free module. Let f f 1 ; : : : ; f n g be a basis of F and f e 1 ; : : : ; e s g a basis of N . Expand each e j in terms of the f i ’s, e j D X i a i;j f i : We can rewrite this as OEe 1 ; : : : ; e s Ł D OEf 1 ; : : : ; f n ŁA; (8.4.5) where A denotes the n –by– s matrix A D .a i;j / . According to Proposition 8.4.6 , there exist invertible matrices P 2 Mat n .R/ and Q 2 Mat s .R/ such that A 0 D PAQ is diagonal, A 0 D PAQ
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Unformatted text preview: A D PAQ D diag .d 1 ;d 2 ;:::;d s /: We will see below that all the d j are necessarily nonzero. Again, accord-ing to Proposition 8.4.6 , P and Q can be chosen so that d i divides d j whenever i ± j . We rewrite ( 8.4.5 ) as Œe 1 ;:::;e s ŁQ D Œf 1 ;:::;f n ŁP ± 1 A : (8.4.6) According to Lemma 8.4.11 , if we deﬁne f v 1 ;:::;v n g by Œv 1 ;:::;v n Ł D Œf 1 ;:::;f n ŁP ± 1 and f w 1 ;:::;w s g by Œw 1 ;:::;w s Ł D Œe 1 ;:::;e s ŁQ; then f v 1 ;:::;v n g is a basis of F and f w 1 ;:::;w s g is a basis of N . By Equation ( 8.4.6 ), we have Œw 1 ;:::;w s Ł D Œv 1 ;:::;v n ŁA D Œd 1 v 1 ;:::;d s v s Ł:...
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