Question

Let M be a left R-module and E = End(RM), if E is a semisimple E-module, is M necessarily a semisimple R-module?

Answer 1

The answer is "no" in general. To construct an example, let $R$ be the ring of $2 \times 2$ upper triangular matrices over a field $k$, and let $M$ be the left $R$-module $k^2$ with the $R$-action given by matrix multiplication from the left. An easy computation shows that $E = \operatorname{End}(RM) = k$, in particular, $M$ is a semisimple $E$-module. We claim that $M$ is not a semisimple $R$-module. In fact, consider the $R$-submodule

$$N = \left\{\left(a, 0\right) ^ {1} \mid a \in k \right\} \subseteq M.$$

If $(\mathrm{b},\mathrm{c})^1$ is not in $N$, then $c$ is nonzero, so $R(b,c)'$ contains $\left( \begin{array}{cc}0 & ac^{-1}\\ 0 & 0 \end{array} \right)\left( \begin{array}{c}b\\ c \end{array} \right) = \left( \begin{array}{c}a\\ 0 \end{array} \right)$ for all $a \in k$. This shows that $N$ is not an $R$-direct summand of $M$, so $M$ is not semisimple.