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If $M$ is a left $R$-module that is Noetherian then I know $R/\text{ann}_R(M)$ is Noetherian. I believe that if instead $M$ is Artinian then it isn’t necessarily the case that $R/\text{ann}_R(M) $ is Artinian but I can’t think of any examples or how to show this? Any pointers?

Jean-Claude Arbaut
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Anonmath101
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  • You must mean a commutative ring, because it doesn't hold either of Noetherian or Artinian without that. – rschwieb Nov 23 '22 at 22:23
  • Lukas: i misread the question initially as containing "finitely generated". A prufer $p$-group itself is already Artinian and faithful as a $\mathbb Z$ module, and $\mathbb Z$ isn't Artinian, so that idea you had works. – rschwieb Nov 23 '22 at 22:26

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Consider the Prüfer p-group $\Bbb Z [1/p]/\Bbb Z$ as a $\Bbb Z$-module.

The main difference here is that an Artinian module is not necessarily finitely generated and this is crucial for the proof in the Noetherian case, as you can also see in the other answer.

Lukas Heger
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I assume we are talking about commutative rings, because neither statement (with Noetherian or Artinian) is correct otherwise.

What's true is that a commutative ring $R$

  1. has a faithful, finitely generated Noetherian module iff $R$ is Noetherian
  2. has a faithful, finitely generated Artinian module iff $R$ is Artinian

In the first one, you can omit "finitely generated" because it is implied by "Noetherian." Not so for the second one: you can have an infinitely generated Artinian module. A good example is the Prüfer $p$-group (considered as a module over $\mathbb Z$), which is Artinian, faithful, and not finitely generated. Of course $\mathbb Z$ is not Artinian.

All this applies, of course, beause $M_R$ becomes a faithful $R/Ann(M)$ module.

Lukas Heger was in the middle of posting something like this first but I think I disrupted things by misreading the question and closing as a dupe. I reopened shortly after I realized my inaccuracy. Sorry about that. (The answer had been self-deleted for a time but now is thankfully back.)

rschwieb
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