Let $M$ be a semsimple module and let $\{V_i\}_{i \in I}$ be a complete set of representatives of the isomorphism classes of simple modules. Let $M_i$ be the sum of all simple submodules isomorphic to $V_i$.
Now, if all the $M_i$ are finitely generated we have
$$M = \oplus m_i V_i$$
for suitable integers $m_i$.
Now, if $N$ is any submodule of $M$, then it too is semisimple ( I wish to show that $N$ is of the form
$$N = \oplus n_iV_i$$
for suitable integers $n_i \leq m_i$. Either $N \cap V_i = 0$ or $N \cap V_i = V_i$ since the $V_i$ are simple. Now, it is clear that $N$ can be projected onto the decomposition of $M$ which feels insufficient to state the result. Further, I believe if I assume $N$ is not in fact a sum of copies of the simple modules it contains, then there would have to be some other $W$ (at least one) such that $W$ is simple and $W \not \cong V_i$ for all $i$. I feel like this would force a composition series which is too long for $M$ hence $W$ must be 0 but have been unable to confirm my hunch. Can someone offer some hints?
Best Answer
Let $M$ be a semsimple module and let $\{V_i\}_{i \in I}$ be a complete set of representatives of the isomorphism classes of simple modules. Let $M_i$ be the sum of all simple submodules isomorphic to $V_i$.
Now, if all the $M_i$ are finitely generated we have
$$M = \oplus m_i V_i$$
for suitable integers $m_i$.
Now, if $N$ is any submodule of $M$, then it too is semisimple. Notice that since the $M_i$ are finitely generated they are noetherian (the $M_i$ have a finite composition series) and so every submodule of the $M_i$ finitely generated. Define the $N_i$ similarly to the $M_i$ except for simple submodules of $N$ not $M$ (notice that all simple modules of $M$ give all the simple modules of $N$) and so similar to $M$ we have
$$N= \oplus N_i$$
Then, since the $N_i$ are finitely generated we have
$$N = \oplus n_i V_i$$
for appropriate $n_i$ and $V_i$. We must have $n_i \leq m_i$ since
$$M \cong \oplus n_iV_i \oplus (m_i-n_i)V_i=N\oplus(m_i-n_i)V_i$$
and so if $n_i > m_i$ we would violate the length of $M$'s composition series. Finally, notice that this implies that $M/N \cong (m_i-n_i)V_i$.