## Hom(R,R) (over R) and R are isomorphic as rings

Let $R$ be a commutative ring with 1. Recall that $\mathsf{Hom}_R(R,R)$ is a ring under pointwise addition and composition. Prove that as rings, $\mathsf{Hom}_R(R,R) \cong R$.

In this previous exercise, we saw that $\mathsf{Hom}_R(R,R)$ and $R$ are isomorphic as left $R$-modules via the mapping $\Phi : \varphi \mapsto \varphi(1)$. In particular, this is an isomorphism of abelian groups. Thus it suffices to show that $\Phi(\varphi \circ \psi) = \Phi(\varphi) \Phi(\psi)$. To that end, note that $\Phi(\varphi \circ \psi) = (\varphi \circ \psi)(1) = \varphi(\psi(1)) = \psi(1) \cdot \varphi(1)$ $= \varphi(1) \psi(1) = \Phi(\varphi) \Phi(\psi)$ since $R$ is commutative.

Thus $R$ and $\mathsf{Hom}_R(R,R)$ are isomorphic as rings.