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The socle of a module $M$ is semisimple.

I know that the socle is $soc(M)=\sum\{N\leq M| \text{$N$ is a simple submodule of $M$}\}$. So $soc(M)=N_1+\cdots+N_k$, where $N_j$ are simple submodules, for $j=1,\ldots, k$. I Have to prove that $soc(M)=N_1\oplus\cdots\oplus N_k$ for this I'm trying to prove that $N_j\cap (N_1+\cdots+N_{j-1})=0$. For this I suppose that $N_j\cap (N_1+\cdots+N_{j-1})\neq0$ so $N_j\cap (N_1+\cdots+N_{j-1})\subseteq N_j$ then $N_j\cap (N_1+\cdots+N_{j-1})= N_j$ and I don't know how to continue.

Stefan4024
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user292200
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  • See this question and the ensuing discussion: https://math.stackexchange.com/questions/179177/submodules-of-direct-sums-of-simple-modules – Tomo Jul 10 '20 at 22:32

1 Answers1

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Well two things went wrong already. For one thing, you don't know that the socle is a finite sum of simple submodules. For another, you have no reason to believe the sum of all simple submodules is a direct sum (it usually is not.).

You have to deduce why there is a subset of the simple submodules which does do the job though.

Consider the collection of sets of simple modules which form direct sums. Use inclusion to partially order these sets. Argue by Zorn's lemma this set has maximal elements, and choose one.

The sum of submodules in our special set is a direct sum of simples, call it $M$. If $M$ contains all simple submodules we are done. Any simple submodule $S$ not contained in $M$ intersects $M$ trivially so that $M\oplus S$ is a larger direct sum. This contradicts the maximality of our choice, so it is an absurdity.

Thus the direct sum of elements in our maximal set of simple submodules equals the socle, showing the socle is a direct sum of simple submodules.

rschwieb
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