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.
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.
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)
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