Unformatted text preview: dent over R and Œv 1 ;:::;v n ŁC D , then C is the zero matrix. See Exercise 8.4.1 Let us show next that any two bases of a ﬁnitely generated free R – module have the same cardinality. Lemma 8.4.1. Let R be a commutative ring with identity element. Any two bases of a ﬁnitely generated free R –module have the same cardinality. Proof. We already know that any basis of a ﬁnitely generated R module is ﬁnite. Suppose that an R module M has a basis f v 1 ;:::;v n g and a spanning set f w 1 ;:::;w m g . We will show that m ± n . Each w j has a unique expression as an R –linear combination of the basis elements v j , w j D a 1;j v 1 C a 2;j v 2 C ²²² C a n;j v n : Let A denote the n –by– m matrix A D .a i;j / . The m relations above can be written as a single matrix equation: Œv 1 ;:::;v n ŁA D Œw 1 ;:::;w m Ł: (8.4.1)...
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 Fall '08
 EVERAGE
 Algebra, Vector Space, Ring, FINITELY GENERATED MODULES

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