For a semi-simple ring $R$ and $R$-module $M$ show TFAE: M finitely generated, artinian and noetherian.

Question

Let $R$ be a semi-simple ring, and let $M$ be an $R$-module. Show the following are equivalent:
(i) $M$ is finitely generated.
(ii) $M$ is noetherian.
(iii) $M$ is artinian.

I know the following:

Let $R$ be a semi-simple ring. Then i) $R$ is left-artinian and left-noetherian. ii) Every $R$-module is semi-simple.

Let R be a noetherian ring. Then an R-module is noetherian if and only if it is finitely generated.

So (i) $\iff$ (ii).

Let R be a artinian ring. Then an R-module is artinian if and only if it is finitely generated.

So (i) $\iff$ (iii).

Am I on the right tracks and what is left to prove?

Answer 1

Am I on the right tracks and what is left to prove?

Well, if you really assume those two lemmas you're using, no, of course not, there's nothing left to prove. But whether or not these lemmas can be used is up to your judgement.

I'm not sure why the problem was presented in this way, but really something much more general is true, and it is not really any harder to prove.

Proposition For a semisimple left module $M$ over any ring, t.f.a.e.:

1. $M$ is finitely generated
2. $M$ is a finite direct sum of simple submodules
3. $M$ has finite composition length
4. $M$ is Noetherian
5. $M$ is Artinian

If you are familiar with how having a composition series is equivalent to a module being Noetherian and Artinian, then it should be fairly easy to use that angle too.