If $N$ is semisimple and and $M/N$ semisimple then $M$ semisimple? Let $R$ be finite dimensional $k$ algebra, $k$ is a field possibly $\mathbb{C}$. Let $_RN\leq{}_RM$ (submodule as  a left module). Is the following statement true (if yes could you give me a hint) , if $_RN$ is semisimple and $_RM/N$ semisimple then $_RM$ semisimple?
 A: No, for example let $k$ be any field you like and consider the ring $R = k[x]/x^2$.
The ideal $(x) \leq R$ is one dimensional and the quotient $R/(x)$ is one dimensional so both of these are simple (hence semisimple) $R$-modules.  But the ring $R$ is not semisimple as a module over itself because the ideal $(x)$ is the only simple submodule of $R$.
If the ring $R$ that you're working with is a semisimple ring (i.e. semisimple as a module over itself) then the statement becomes true.
A: Hint:
The idea is that finite dimensional ( over $k$) $R$ modules $M$ have finite length, that is,
there exists a sequence of submodules 
$$0=M_0 \subset M_1 \subset \ldots \subset M_n = M$$
that can not be refined any further, that is, $M_{l-1}$ is maximal inside $M_l$, in other words, $M_{l}/M_{l-1}$ is simple. So any module is a composition of simple modules (and so, of semisimple modules). But if $R$ itself is not semisimple, there are some non-semisimple modules $M$ out there ( for instance, $R$ itself)
No.
