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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?

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I'm assuming a semisimple module is one which is a direct sum of (possibly infinitely many) simple modules.

Let $P:=\bigoplus M/N$ where the sum is taken over all (nontrivial) simple quotients $M/N$, and consider the natural map $M\rightarrow P$. The kernel of this map is exactly $\mathrm{rad}(M)$, so there is an embedding $M/\mathrm{rad}(M)\hookrightarrow P$.

Clearly $P$ is semisimple. Can you show that semisimplicity is preserved for submodules?

(There are several equivalent characterizations of semisimplicity: $P$ is a sum of simple submodules, $P$ is a direct sum of simple modules, and every submodules of $P$ is a direct summand.

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  • $\begingroup$ Ah, thanks. Sure, if $M/\operatorname{rad}(M)\subseteq P$ for $P$ semisimple, then any submodule $N$ of $M/\operatorname{rad}(M)$ is also a submodule of $P$ and thus has a retraction. The composition with $M/\operatorname{rad}(M)\to P$ then gives a retraction of $N\to M/\operatorname{rad}(M)$. $\endgroup$ – Bubaya May 8 at 16:31

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