Free modules

1. Dec 31, 2008

math8

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. Dec 31, 2008

math8

Oh, I think I can use the Universal property of free modules to get the onto part.