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Tuesday, June 4, 2013

Quotients and Finite Generation of Modules (10.3.7)

Dummit and Foote Abstract Algebra, section 10.3, exercise 7:

MathJax TeX Test Page If N is a submodule of the R-module M, prove that if M/N and N are finitely generated, then so is M.

Proof: Let A={a1,...,an} generate N, and let B={b1,...,bm} be such that ¯B generates M/N. We claim AB generates M. Note that for an arbitrary element mM we have ¯m=¯r1b1+...+rmbm for some riR so that m=r1b1+...+rmbm+n for some nN, so that m=r1b1+...+rmbm+s1a1+...+snan for some siR, and now M is generated by AB. 

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