## 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 $\mathsf{Hom}_R(R,M) \cong_R M$.