R-module homomorphisms isomorphic to codomain

[Kreizhn]

Homework Statement

Let R be a commutative ring, and M be an R-module. Show that
$$\text{Hom}_{\text{R-mod}}(R,M) \cong M$$
as R-modules, where the homomorphisms are R-module homomorphisms.

The Attempt at a Solution

This should hopefully be quick and easy. The most natural mapping to consider is
$\phi: \text{Hom}_{\text{R-mod}}(R,M) \to M$ sending $f \to f(1_R)$. 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.

 

[micromass]

Hi Kreizhn!

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 $f(1_R)=m$. Can you extend this equation to fully define f? (hint: $f(r)=f(r.1_R)$.

 

[Kreizhn]

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
$$f(r) = rf(1) = rm$$
Then the function is a homomorphism since M is an R-module.

 

[micromass]

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
$$f(r) = rf(1) = rm$$
Then the function is a homomorphism since M is an R-module.

Yes, that's it!