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R-module homomorphisms isomorphic to codomain

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data
    Let R be a commutative ring, and M be an R-module. Show that
    [tex] \text{Hom}_{\text{R-mod}}(R,M) \cong M [/tex]
    as R-modules, where the homomorphisms are R-module homomorphisms.

    3. The attempt at a solution

    This should hopefully be quick and easy. The most natural mapping to consider is
    [itex] \phi: \text{Hom}_{\text{R-mod}}(R,M) \to M [/itex] sending [itex] f \to f(1_R) [/itex]. It is simple to show that this is a R-mod homomorphism, and that it is injective. Where I am stuck is surjectivity.

    The first thing that comes to mind is that I want to use constant maps; however, these are not R-mod homs. Secondly, I've realized that I have not yet used the fact that the ring is commutative. I'm wondering if somehow f(rs) = rf(s) = sf(r) comes into play.
     
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  3. Jun 20, 2011 #2

    micromass

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    Hi Kreizhn! :smile:

    To prove surjectivity, you must prove that every m in M is in the image of the isomorphism. So you need to find a function f such that [itex]f(1_R)=m[/itex]. Can you extend this equation to fully define f? (hint: [itex]f(r)=f(r.1_R)[/itex].
     
  4. Jun 20, 2011 #3
    Ah, I think I see what you're saying. We can just define the function in such a way that it forces it to be an R-module homomorphism by demanding that
    [tex] f(r) = rf(1) = rm[/tex]
    Then the function is a homomorphism since M is an R-module.
     
  5. Jun 20, 2011 #4

    micromass

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    Yes, that's it!
     
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