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.

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