College Algebra Exam Review 374

College Algebra Exam Review 374 - s D t by uniqueness of...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
384 8. MODULES ± the ring elements a i are nonzero and noninvertible, and a i di- vides a j for i ² j ; and also M D B 0 ˚ B 1 ˚ B 2 ˚ ³³³ ˚ B t ; where ± B 0 is free, ± for i ² 1 , B i Š R=.b i / , and ± the ring elements b i are nonzero and noninvertible, and b i di- vides b j for i ´ j ; We have to show that rank .A 0 / D rank .B 0 / , s D t , and .a i / D .b i / for all i ´ 1 . By Lemma 8.5.5 , we have M tor D A 1 ˚ ³³³ ˚ A s D B 1 ˚ B 2 ˚ ³³³ ˚ B t : Hence A 0 Š M=M tor Š B 0 . By uniqueness of rank (Lemma 8.4.1 ), rank .A 0 / D rank .B 0 / . It now suffices to prove that the two decompositions of M tor are es- sentially the same, so we may assume that M D M tor for the rest of the proof. Note that a s and b t are periods of M . So we can assume a s D b t 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 b i and a j are associates of m . In this case, we have only to show that s D t . Since mM D f 0 g , 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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 A i into itself. According to Lemma 8.5.6 , if p divides a i then A i =pA i Š R=.p/ , but if p is relatively prime to a i , then A i =pA i D f g . We have M=pM Š .A 1 ˚ A 2 ˚ ³³³ ˚ A s /=.pA 1 ˚ pA 2 ˚ ³³³ ˚ pA s / Š A 1 =pA 1 ˚ A 2 =pA 2 ˚ ³³³ ˚ A s =pA s Š .R=.p// k ; where k is the number of a i such that p divides a i . Since p.M=pM/ D f g , 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...
View Full Document

This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

Ask a homework question - tutors are online