$R = \mathbb{Z}/m\mathbb{Z}$, and consider $R^n$ for any $n,m$. For what values of $m$ is $R$ a semisimple $R$-module?

Question

Let $R = \mathbb{Z}/m\mathbb{Z}$, and consider $R^n$ for any $n,m$. For what values of $m$ is $R$ a semisimple $R$-module?

$\textbf{Context:}$ I am interested for what $m$ would the image of an endomorphism of $R^n$ be a direct summand.

Answer 1

$R=\mathbb Z/m\mathbb Z$ is a semisimple $R$ module iff $m$ is a (nonzero) square-free integer. This is well-known because a commutative semisimple ring cannot have nonzero nilpotents, and this is exactly the condition to ensure that.

$R^n$ is a semisimple $R$ module iff $R$ is.

So the answer to your 'context' question is that if $m$ is squarefree, the image will always be a summand, but if it is not squarefree, it will only sometimes be a summand.