I want to prove this :

Over an artinian ring every nonzero module has a simple submodule.

But the same statement for Noetherian rings is not true.

Is there any hint how to show that?

Thank you very much.


Without loss of generality, we can assume that the module is finitely generated (just pick a finitely generated submodule).

A finitely generated module over an Artinian ring is itself Artinian. In particular, every descending chain of submodules stabilises.

So if you didn't have a simple submodule, every submodule has a proper non-trival submodule, leading to an infinite strictly descending chain. This would contradict the module being Artinian.

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  • $\begingroup$ Alternatively to the contradiction approach in the last paragraph, you could say "the poset of nonzero submodules is nonempty and artinian, so it contains a minimal elements. $\endgroup$ – rschwieb Nov 25 '16 at 16:12
  • $\begingroup$ @Josh Hunt thanks,do u have any counterexample for Noetherian rings? $\endgroup$ – user115608 Nov 25 '16 at 16:13
  • $\begingroup$ A good first step would be to find a ring that is noetherian but not artinian! The standard example provides a counterexample (as a module over itself) to your question $\endgroup$ – Josh Hunt Nov 25 '16 at 16:21
  • $\begingroup$ And does $\mathbb Z$ have a simple submodule? $\endgroup$ – Josh Hunt Nov 25 '16 at 20:22
  • $\begingroup$ @JoshHunt sorry!it doesn't! $\endgroup$ – user115608 Nov 25 '16 at 20:24

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