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Free modules

  1. Dec 31, 2008 #1
    View attachment Exampleoflatex.pdf

    Let R be a commutative ring with 1. If F is a free module of rank n < 1, then show that
    HomR(F;M) is isomorphic to M^n, for each R-module M.

    I was thinking about defining a map
    Psi : HomR(F;M)--> M^n by psi(f) = (f(e1); f(e2); ... ; f(en))
    where F is free on (e1; ... ; en) and
    show Psi is an isomorphism. But I am having difficulties showing it is onto.
    Last edited: Dec 31, 2008
  2. jcsd
  3. Dec 31, 2008 #2
    Oh, I think I can use the Universal property of free modules to get the onto part.
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