Let R be a ring with identity; then there is a ring isomorphism $\operatorname{Hom}_R(R,R) \cong R^{\mkern 1mu\text{op}}$.

Question

In particular, if R is commutative, then show there exists a ring isomorphism $\operatorname{Hom}_R(R,R) \cong R$.

Answer 1

Given a ring $R$, for any $r \in R$, the map $x \mapsto xr$ is a left $R$-module endomorphism of $R$. This gives a map $f:R \to \operatorname{Hom}_{R}(R, R)$, which is easily seen to preserve addition and the multiplicative identity. However, it reverses multiplication. Indeed, $\forall r,s,x \in R \, [f(rs)](x)=x(rs)=(xr)s=[f(r)](x)s=[f(s)]([f(r)](x))=[f(s) \circ f(r)](x)$, so $\forall r,s \in R \, f(rs)=f(s) \circ f(r)$. Hence, $f$ is at least a ring homomorphism from $R^{op}$ to $\operatorname{Hom}_{R}(R, R)$.

To show that $f$ is actually an isomorphism, we have to show that it has an inverse. In this case, the inverse sends a left $R$-module endomorphism $\alpha$ of $R$ to $\alpha(1)$. Indeed, for any $r \in R$, $[f(r)](1)=1r=r$; and for any $\alpha \in \operatorname{Hom}_{R}(R, R)$ and any $x \in R$, $[f(\alpha(1))](x)=x\alpha(1)=\alpha(x1)=\alpha(x)$, showing that $f(\alpha(1))=\alpha$.

Note that the "op" is not needed if one instead considers endomorphisms of $R$ as a right $R$-module.