College Algebra Exam Review 377

College Algebra Exam Review 377 - p m 2 s p m t t , where...

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8.5. FINITELY GENERATED MODULES OVER A PID, PART II. 387 Let R be a principal ideal domain. Let a 2 R be a nonzero nonunit element. Let a D p m 1 1 p m 2 s ±±± p m t t , where the p i are pairwise relatively prime irreducible elements of R . Consider the cyclic R –module M D R=.a/ . Since a is a period of M , according to Theorem 8.5.9 , the primary decomposition of M is M D MŒp 1 Ł ˚ :::MŒp t Ł: For 1 ² i ² t , let r i D a=p m i i D Q k ¤ i p m k k . For x 2 R , let ŒxŁ denote the class of x in R=.a/ . We claim that MŒp i Ł is cyclic with generator Œr i Ł and period p m i i for each i , so MŒp i Ł Š R=.p m i i / . Note that Œr i Ł 2 MŒp i Ł and the period of Œr i Ł is p m i i . So it suf- fices to show that Œr i Ł generates MŒp i Ł . Since greatest common divisor of f r 1 ;:::;r s g is 1, there exist u 1 ;:::;u t in R such that u 1 r 1 C±±±C u t r t D 1 . Hence for any x 2 R , x D xu 1 r 1 C ±±± C xu t r t , and ŒxŁ D xu 1 Œr 1 Ł C :::xu t Œr t Ł: In particular, if ŒxŁ 2 MŒp i Ł , then ŒxŁ D xu 1 Œr 1 Ł , and Œr i Ł generates MŒp i Ł as claimed. We have shown: Lemma 8.5.11. Let R be a principal idea domain. Let a 2 R be a nonzero nonunit element. Let a D p m 1 1
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Unformatted text preview: p m 2 s p m t t , where the p i are pairwise relatively prime irreducible elements of R . Then R=.a/ R=.p m 1 1 / R=.p m t t /: Example 8.5.12. Consider a.x/ D .x 2 C 1/.x 1/.x 3/ 3 2 Q x . Let M D Q x=.a.x// . Let b.x/ denote the class of a polynomial b.x/ in M . We have M D Mx 2 C 1 Mx 1 Mx 3 Q x=. .x 2 C 1// Q x=. .x 1// Q x=. .x 3/ 3 /: Set 1 .x/ D x 2 C 1 , 2 .x/ D .x 1/ , and 3 .x/ D .x 3/ 3 . Put r 1 .x/ D 2 .x/ 3 .x/ , r 2 .x/ D 1 .x/ 3 .x/ , and r 3 .x/ D 1 .x/ 2 .x/ . Using the method of Example 8.4.10 , we can compute polynomials u 1 .x/;u 2 .x/;u 3 .x/ , such that u 1 .x/r 1 .x/ C u 2 .x/r 2 .x/ C u 3 .x/r 3 .x/ D 1:...
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