Is $\operatorname{Hom}(E,M)$ simple as an $\operatorname{End}(M)$-module if $E$ is simple?

Question

Let $R$ be a ring, let $M$ be an $R$-module and let $R' := \operatorname{End}_R(M)$. Let $E$ be a simple $R$-module with $\operatorname{Hom}_R(E,M) \neq 0$.

Question: Is $\operatorname{Hom}_R(E,M)$ simple as an $R'$-module?

Background: While trying to understand (a version of) the double centralizer theorem I learned that the $E$-isotypical component of $M$ can be described as $M_E \cong \operatorname{Hom}_R(E,M) \otimes_{\operatorname{End}_R(E)} E$ as an $(R' \otimes_{\mathbb{Z}} R)$-module. Because $M_E$ is simple as an $(R' \otimes_{\mathbb{Z}} R)$-module I am wondering about the $R'$-simplicity of the tensor factor $\operatorname{Hom}_R(E,M)$.

I think the statement holds if $M$ is semisimple: We can then assume that $M = M_E \cong E^{\oplus I}$ for some index set $I$ (because every $R$-endomorphism of $M_E$ extends to an $R$-endomorphism of $M$). For the skew field $D := \operatorname{End}_R(E)$ we then have that $R' \cong \operatorname{M}_I(D)$ (the algebra of column-finite $(I \times I)$-matrices) and $\operatorname{Hom}_R(E,M) \cong D^{\oplus I}$, with $\operatorname{M}_I(D)$ acting transitively on $D^{\oplus I} \smallsetminus \{0\}$.

I assume that the statement does not hold for general $M$, but I haven’t found a counterexample.

Any help is welcome.

Answer 1

Let $k$ be a field, let $R=k[x,y]$, and let $I$ be the maximal ideal $(x,y)\subset R$. Let $M=R/I^2$, and let $E=R/I$. Then $R'$ is just the quotient ring $R/I^2$ and $\operatorname{Hom}_R(E,M)$ is naturally identified with $I/I^2$. This is not a simple $R'$-module, since $x$ and $y$ act trivially on it so that submodules are the same thing as $k$-subspaces, and $\dim_k I/I^2=2$ ($x$ and $y$ are a basis).

(The key point is that the submodule $N$ of $M$ spanned by copies of $E$ is a direct sum of copies of $E$, but not every $\operatorname{End}_R(E)$-endomorphism of it extends to an endomorphism of $M$. In this case, $N\cong E^2$ and $\operatorname{End}_R(E)\cong k$, but $R'$ only contains the diagonal action of $k$ on $E^2$, not the full $M_2(k)$ action, since in $M$ the two copies of $E$ in $N$ are actually both multiples of a single generator over $R$.)