Question

Let U, V be semisimple modules over a commutative ring. Is HomR(U, V ) also semisimple?

Answer 1

If one of $U, V$ is finitely generated, the answer is "yes". In general, however, the answer is "no", as we can show by the following example over $R = \mathbb{Z}$. Take $U = V = \oplus_{p} \mathbb{Z}_{p}$, where $p$ ranges over all primes. Let $\varepsilon \in E := \operatorname{Hom}_{\mathbb{Z}}(V, V)$ be the identity map from $V$ to $V$. We claim that $\varepsilon$ has infinite additive order in $E$ (which certainly implies that $E$ is not a semisimple $\mathbb{Z}$-module). For any natural number $n$, take a prime $p > n$. Then $(n \cdot \varepsilon)(0, \ldots, 1, 0, \ldots) = (0, \ldots, n, 0, \ldots) \neq 0$.

if the 1 appears in the coordinate corresponding to $\mathbb{Z}_p$. Therefore, $n \cdot \varepsilon \neq 0 \in E$, as claimed.