Suppose $N_1,..,N_r$ are submodules such that $\cap_{i=0}^r N_i=\{0\}$ and $M/N_i$ are semisimple for all $i$. Then M is semisimple.

Question

Suppose $N_1,..,N_r$ are submodules such that $\cap_{i=0}^r N_i=\{0\}$ and $M/N_i$ are semisimple for all $i$. Then M is semisimple.

I am stuck with the above problem. All I can show is that if $N$ is a submodule of M then there exits $M_i$ such that $N_i\subseteq M_i$ and $\phi_i(N)+M_i/N_i=M/N_i$ where $\phi_i$ is the usual map from $M\rightarrow M/N_i$. This implies that $N+M_i=M$ for all $i$. But that is where I'm stuck.

Any ideas?

Answer 1

HINT:

Use that $$M/(\cap_{i=1}^r N_i) \to \oplus_{i=1}^r M/N_i$$ is an injection, and the fact that a submodule of a semisimple module is semisimple.