Let $M$ be an $A$-module for $A$ a finite dimensional algebra. Let $\operatorname{rad}(M)=\bigcap\{N\subsetneq M\ \text{maximal}\}$. Clearly, $M/N$ is simple for any maximal submodule $N$. It seems to be standard that $M/\operatorname{rad}(M)$ is semisimple; however, I struggle to see this.
My attempt: In case of finitely many maximal submodules, this works out easily. Indeed, let $N, N'\subseteq M$ be submodules such that $M/N, M/N'$ is semisimple. Then $N/N\cap N'\cong (N+N')/N'\subseteq M/N'$, so as a submodule of a semisimple module, $N/N\cap N'$ must be semisimple itself, and the inclusion has a retraction $M/N'\to N/N\cap N'$.
The module $M/N\cap N'$ in question fits into an extension $0\to N/N\cap N'\to M/N\cap N'\to M/N\to 0$ of two semisimple modules. Additionally, there is a retraction $M/N\cap N'\to M/N\to N/N\cap N'$. Hence this extension splits, and $M/N\cap N'$ is a direct sum of semisimple modules.
Question: How to prove this for an arbitrary number of maximal submodules?
Best Answer
The accepted answer is not correct, as I explained in the comments.
The condition that $A$ is a finite-dimensional algebra can be weakened a bit: we only need that $A$ is (left or right) Artinian.
The critical part is that $\mathrm{rad}(A)$ annihilates $M/\mathrm{rad}(M)$ which is equivalent to saying that $\mathrm{rad}(A) M \subset \mathrm{rad}(M)$. (This is true without the Artinian condition)
Let $N \subsetneq M$ be a maximal submodule. Then $M/N$ is a simple module, so it is annihilated by $\mathrm{rad}(A)$, which means that $\mathrm{rad}(A)M \subset N$. Since this holds for any $N$, we get $\mathrm{rad}(A)M \subset \mathrm{rad}(M)$, so $\mathrm{rad}(A)$ annihilates $M/\mathrm{rad}(M)$ which means that $M/\mathrm{rad}(M)$ carries a canonical $A/\mathrm{rad}(A)$-module structure.
Now it is well-known that if $A$ is left or right Artinian, then $A/\mathrm{rad}(A)$ is a semisimple ring, so $M/\mathrm{rad}(M)$ is a semisimple $A/\mathrm{rad}(A)$-module and hence also a semisimple $A$-module.