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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.

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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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