Is $\mathrm{Hom}(eR, M) \cong Me$ as right $eRe$-modules?

Question

Let $R$ be a ring with identity and $e \in R$ is an idempotent. It is well-known that $\mathrm{Hom}_R(R, M)$ is isomorphic to $M$ as right $R$-modules, for any right $R$-module $M$. Is it true that $\mathrm{Hom}(eR, M)$ is isomorphic to $Me$ as right $eRe$-modules? If not, is a condition such as “$eR$ being a generator for the category of right $R$-modules” necessary? Thanks go to anybody answering my question!

Answer 1

Yes, that's true: You have the decomposition $R = eR \oplus (1-e) R$ as right $R$-modules, hence $$\text{Hom}_R(eR,M)\cong\text{ker}\left(\text{Hom}_R(R,M)\xrightarrow{\text{res}}\text{Hom}_R((1-e)R,M)\right)\stackrel{(\dagger)}{\cong}\text{ker}(M\xrightarrow{\cdot (1-e)} M) = Me,$$ Explicitly, this map is given by $\varphi\mapsto\varphi(e)$ and hence right $eRe$-linear.

Addendum The isomorphism $(\dagger)$ results from the isomorphism $\tau:\text{Hom}_R(R,M)\cong M, \varphi\mapsto\varphi(1)$ you already know together with the fact that the restriction $\text{res}(\varphi)$ of a morphism $\varphi: R\to M$ to $(1-e)R$ vanishes if and only if $0=\varphi(1-e) = \varphi(1)\cdot (1-e)$ - hence, $\tau$ restricts to $\text{ker}(\text{res})\cong \text{ker}(M\xrightarrow{\cdot (1-e)} M)$.