Hom(R,M) and M are isomorphic as R-modules

Let R be a ring with 1 and let M be a left unital R-module. Prove that \mathsf{Hom}_R(R,M) \cong_R M. (We use \cong_R to say “isomorphic as R-modules”.)

Define \Phi : \mathsf{Hom}_R(R,M) \rightarrow M by \Phi(\varphi) = \varphi(1). We claim that \Phi is an R-module isomorphism.

Let \varphi, \psi \in \mathsf{Hom}_R(R,M) and r \in R. Then \Phi(\varphi + r \cdot psi) = (\varphi + r \cdot \psi)(1) = \varphi(1) + r \cdot (\psi(1)) = \Phi(\varphi) + r \cdot \Phi(\psi). Thus \Phi is an R-module homomorphism.

Suppose now that \varphi \in \mathsf{ker}\ \Phi. Then for all r \in R, \varphi(r) = r \cdot \varphi(1) = r \cdot \Phi(\varphi) = r \cdot 0 = 0. Hence \varphi = 0, and so \Phi is injective.

Now let m \in M. Define \psi_m : R \rightarrow M by \psi_m(r) = r \cdot m. Since for all x,y \in R and r \in R we have \psi_m(x + r \cdot y) = (x + r \cdot y) \cdot m = x \cdot m + r \cdot (y \cdot m) = \psi_m(x) + r \cdot \psi_m(y), \psi_m is an R-module homomorphism, and so \psi_m \in \mathsf{Hom}_R(R,M). Moreover, \Phi(\psi_m) = \psi_m(1) = 1 \cdot m = m. Thus \Phi is surjective.

Thus we have \mathsf{Hom}_R(R,M) \cong_R M.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: