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I have a question about whether Noetherian-ness and Artinian-ness of modules are preserved under changes of the base ring. More precisely:

Let $R$ be a commutative ring and $S \subseteq R$ a subring. If $E$ is a Noetherian (resp. Artinian) $R$-module, under what conditions (on $R$ and $S$) is $E$ a Noetherian (resp. Artinian) $S$-module? Conversely, if $E$ is a Noetherian (resp. Artinian) $S$-module, under what conditions is $E$ a Noetherian (resp. Artinian) $R$-module?

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  • $\begingroup$ If $E$ is an $S$-module, how could you ask about its properties as an $R$-module? And if $E$ is an $R$-module which is Noetherian/Artinian as an $S$-module, then obviously it is the same as $R$-module. So, the second part of your question could be removed. $\endgroup$ – user26857 Apr 26 '13 at 4:43
  • $\begingroup$ Yes, I realized this a little while after posting the question. Also, thanks for your answer! $\endgroup$ – JHF Apr 26 '13 at 23:16
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If the extension $S\subset R$ is finite, then every Noetherian, respectively Artinian and finitely generated $R$-module is Noetherian, respectively Artinian and finitely generated as an $S$-module.

If the extension is not finite, then there is no much hope to get this. For instance, consider $R=\mathbb Z\subset \mathbb Q=S$ and $M=\mathbb Q$. Then $M$ is Noetherian and Artinian as an $S$-module, but is not Noetherian or Artinian as $R$-module.

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  • $\begingroup$ E-n-l-i-g-h-t-e-n-e-d! $\endgroup$ – Bombyx mori Apr 26 '13 at 5:01

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