Let R be a ring with identity. An R-module M is Artinian if it satisfies the descending chain condition on submodules. What is an example of an Artinian module with a proper submodule that is not finitely generated?


Take a look at http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition.

There you can find a concrete example of an Artinian module over the integers which is non-Noetherian and hence has a non-finitely generated submodule.

In the example at hand, the module itself is not finitely generated, and there are also lots of non-finitely generated proper submodules.

  • $\begingroup$ @Jon, of course not: every module has finitely generated submodules, independently of how monstrous is the ring. $\endgroup$ – Mariano Suárez-Álvarez Oct 20 '10 at 18:37
  • $\begingroup$ I wonder if there is an example of a finitely generated Artinian ring which is not Noetherian. $\endgroup$ – Rasmus Oct 20 '10 at 18:38
  • $\begingroup$ @Rasmus: Hmmm... A left Artinian ring is necessarily left Noetherian; the example in Wikipedia is of an Artinian module that is not Noetherian. So when you say "a concrete example of an Artinian ring which is non-Noetherian", did you mean "Artinian module"? $\endgroup$ – Arturo Magidin Oct 20 '10 at 18:47
  • $\begingroup$ @Arturo Magidin: Yes, thank you! $\endgroup$ – Rasmus Oct 20 '10 at 18:49
  • $\begingroup$ Thanks for the quick answer, guys! @Mariano: Yep, should have typed "there exists". $\endgroup$ – Jon Bannon Oct 20 '10 at 19:03

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