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Let $N$ be a submodule of the module $M$. Suppose $M/N$ and $N$ are semi-simple. Does it follow that $M$ is semi-simple?

I think the answer is yes but I am not sure how to prove it. Any help would be appreciated.

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No. In fact this is false even if you assume that $M/N$ and $N$ are simple, and even for a ring as well-behaved as $\mathbb{Z}$. – Qiaochu Yuan May 15 '12 at 22:21

No. Consider the ring $R=\mathbb C[t]/(t^2)$. The regular module $M=R$ contains the submodule $N=(t)$ which is simple, and $M/N$ is also simple, yet $M$ is not semisimple.

In fact, under sensible hypotheses if a ring is such that your statement does hold, then the ring must be semisimple. For example, if the ring is a finite dimensional algebra over a ring (or even an artin algebra)

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Adding on Qiaochu's comment and trying to come up with an easier, perhaps, counter-example than that given by Mariano: the only simple $\mathbb{Z}-\,$modules (which are the same as abelian groups) are the cyclic groups of prime order, thus: take Klein's viergrup $$\,M:=C_2\times C_2\,\,,\,\,N:= \{1\}\times C_2\,\,,\,\,M/N\cong C_2$$ Both $M\,,\,M/N\,$ are even simple, as mentioned by Qiaochu

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But the big module in your example is semisimple :) – Mariano Suárez-Alvarez May 16 '12 at 2:12
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If you replace $M$ with $C_4$ and $N$ with the cyclic group of order $2$, though, then you'll be cooking with gas. – Arturo Magidin May 16 '12 at 2:29

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