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Let always $R$ be a ring with unity, but not necessarily commutative. Let $M$ be a (left-)$R$-module. If $R$ is left-noetherian, then $M$ is noetherian if and only if $M$ is finitely generated.

If $R$ is left-artinian, then so is $M$ if $M$ is finitely generated. The converse, i.e. that artinian modules over left-artinian rings are finitely generated, holds as well by the Akizuki-Hopkins-Levitzky theorem, but I was wondering if there is an elementary argument which avoids the more advanced ring theory concepts and theorems.

Bib-lost
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  • Well, the Hopkins-Levitzki theorem proves your title question, so that must not be your actual question. And you couldn't be hoping for a counterexample either (because it's true). Why not just ask something along the lines of "trying to find an elementary proof that ... " – rschwieb Oct 02 '16 at 11:42
  • Rather than "avoid" Akizuki-Hopkins-Levitzki, why not read the proof and see if you can lift a simplified approach from it, because your conditions are stronger than the hypotheses? Everyone studying noncommutative ring theory should understand the proof well at some point, so it's a good investment. – rschwieb Oct 02 '16 at 11:44
  • Thanks for your suggestions. I have reformulated the question, but am still hoping for an elementary proof. I'll see what I can do myself by simplifying a proof of A-H-L. – Bib-lost Oct 02 '16 at 12:08
  • I post an elementary proof here.https://math.stackexchange.com/a/3410402/453628 – Jian Oct 27 '19 at 01:49

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