I'm trying to understand a proof of this from my textbook. It starts by saying that we can assume V is finitely generated, but I'm not sure why we can make that assumption. Is there something that goes wrong in the infinitely generated case?
Other than that the proof is straightforward. I appreciate your insight!
I think I found an answer. It occurred to me that I can choose a finite number of vectors in my module and generate a submodule out of them. The proof then reduces to the finitely generated case, which I can understand no problem.
The key fact is that any submodule of a completely reducible module is completely reducible. So if $M$ is a nonzero completely reducible module, you can pick any nonzero element $m\in M$ and consider the submodule $N$ generated by $m$. It then suffices to find an irreducible module of this nonzero module $N$, which is finitely generated. In fact, this shows you can assume your module is not just finitely generated but cyclic.
