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Let $R$ be a ring and $$0\rightarrow L\rightarrow M\rightarrow N\rightarrow 0$$ a short exact sequence. If $M$ is semisimple what can we say about $L$ and $N$?

I guess they are also semisimple. Is this true? What is the idea of the proof?

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If $M$ is semisimple, then any submodule $L$ and the corresponding quotient module $N=M/L$ are semisimple. The proof is based on the fact that a semisimple module is the direct sum of simple submodules - see here.

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    You are right but since I want to get your remark as an corollary out of my statement I guess I should not use this :( – 2017-01-05
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    You can use it, because a short exact sequence for $M$ means in particular, that $L$ is a submodule with quotient $M/L$. – 2017-01-05
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    I mean: I want to prove the question in my post. Later I want to prove (with that statement of exact sequences) that every submodule and quotient module of a semisimple module is semisimple. – 2017-01-05
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    Again, this proves already the question of your post. You ask" If $M$ is semisimple what can we say about $L$ and $N$"? Answer: since the map $L\rightarrow M$ is injective, $L$ is a submodule of $M$, hence semisimple. So this *is* already your "later" proof. – 2017-01-05
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    Yes, but how can I answer my question "What can we say about $L$ and $N$? – 2017-01-05
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    As I said: Since the sequence is exact, $L$ is a submodule of $M=\oplus_iM_i$, so that $L$ itself is a direct sum of simple modules, hence semisimple - done. Similarly for $N$. – 2017-01-06