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?

  • 1
    $\begingroup$ Assuming your claims are correct, your proof is complete, because you've shown that (i)$\implies$(ii), (ii)$\implies$(i), (i)$\implies$(iii) and (iii)$\implies$(i). You only need to deduce from these that (iii)$\implies$(ii) and vice versa, but this follows as (iii)$\implies$(i)$\implies$(ii) and similarly (ii)$\implies$(i)$\implies$(iii). Just make sure everything is justified properly and that you keep track of left vs. right business as necessary. $\endgroup$ – Stahl Dec 2 '16 at 18:08
  • $\begingroup$ Semisimple is a red herring: the statement holds about left modules for every left Artinian ring, because a left Artinian ring is left Noetherian. $\endgroup$ – egreg Dec 2 '16 at 18:33

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.

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  • $\begingroup$ The things I know (the lemmas) have been proved. $\endgroup$ – Bob Dec 2 '16 at 18:10
  • $\begingroup$ Where I can find the proof of this proposition? $\endgroup$ – Naheel Ghaith Nov 22 '19 at 16:14
  • $\begingroup$ @NaheelGhaith I’m not sure... lectures on rings and modules, or maybe Rings and categories of modules. Either way it is easy to prove, and you don’t really have to rely on a book... $\endgroup$ – rschwieb Nov 22 '19 at 20:01
  • $\begingroup$ @rschwieb Many thanks $\endgroup$ – Naheel Ghaith Nov 22 '19 at 20:14

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