We know the following proposition.

Proposition. Let $R$ be a Noetherian/Artinian ring and $M$ an $R$-module. If $M$ is finitely generated, then the $R$-module $M$ is Noetherian module.

I was wondering if the condition of "finitely generation" of $M$ is necessary. And for this purpose if tried to find an example.


  1. We consider the $\Bbb Z$-module $\Bbb Q$, which is the abelian group $(\Bbb Q,+).$ The ring $\Bbb Z$ is a Noetherian ring, the $\Bbb Z$-module $\Bbb Q$ is not finitely generated (since the abelian group $(\Bbb Q,+)$ is not finitely generated) and of course $\Bbb Z$-module $\Bbb Q$ is not a Noetherian module (since itself is again a submodule, not f.g.). Is this idea in the right way?

  2. But $\Bbb Z$ is not Artinian, to consider an example for this case. So, what if we take a field $K$ and construct the $K\!$-module $K[X]$? What can we conclude for this case? Is $K[X]$ Noetherian/Artinian as a $K$-module and why?

  3. Any other ideas?

Thank you.

  • $\begingroup$ So, to be clear, you are asking "If $M$ is Noetherian, is it finitely generated?" and not "Can I drop the condition of 'finitely generated' from the proposition?" At first glance, it looks like the latter, but after further reading the former one (which makes more sense) is the likely one. $\endgroup$ – rschwieb Mar 6 at 14:51

Any Noetherian module is finitely generated, so the condition is clearly necessary for a module being Noetherian.

Artinian modules can be infinitely generated. The simplest example is the Prüfer group.

Your $K[X]$ cannot be Artinian, because vector spaces over a field are Artinian if and only if they're Noetherian, that is, finite dimensional.

  • $\begingroup$ Nice example, thank you, I 'll check this group! $\endgroup$ – Chris Mar 5 at 22:05

For a commutative ring, we have that an artinian ring is noetherian, this by the Hopkins-Levitzki theorem.


If $A$ is a noetherian ring then $A[x]$ is a noetherian ring, but $A[x]$ is not noetherian as an $A$-module.


In general for a non zero commutative ring $A$, $A[x]$ is not a noetherian $A$-module. Consider the ideals $I_n=\sum_{k=0}^nAx^k$. So we have a strict ascending chain of $A$-submodules in $A[x]$. So you have the ascending chain $I_0\subset I_1\subset i_2\subset\dots$, so have found a non statioanary ascending chain.Showing that $A[x]$ is not noetherian as $A$-module.

To see that is no artinian as an a $A$-module, we build a descending chain given by $J_n=\sum_{k=n}^\infty Ax^k$. So you have a strict descending chain of $A$-modules. Showing that $A[x]$ is not artinian as $A$-module.

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    $\begingroup$ Thank you for your answer. Do you mean the descending chain $\langle 1_K \rangle \supseteq \langle X \rangle \supseteq \langle X^2 \rangle \supseteq \dots$? So $K-$module $K[X]$ is not Artinian? $\endgroup$ – Chris Mar 5 at 21:57
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    $\begingroup$ The chain you mention is only a chain if you consider what you generate as $K[x]$-modules, that's it, as an ideal of $K[x]$. But as you want it as $K$-module, they are no in general $K[x]$-modules. $\endgroup$ – Murphy Mar 5 at 22:27
  • $\begingroup$ Do you mean that my idea is valid only for the $K-$module $K[X]$ and not for $K[X]-$module $K[X]$? $\endgroup$ – Chris Mar 5 at 22:31
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    $\begingroup$ Viceversa, you idea is valid for $K[x]$ as $K[x]$-module. But is no valid for $K[x]$ as $K$-module. $\endgroup$ – Murphy Mar 5 at 22:34
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    $\begingroup$ Exactly, you got it right. $\endgroup$ – Murphy Mar 7 at 19:11

No, $K[X]$ is not artinian, because a commutative artinian ring has Krull dimension $0$.

Actually, for commutative rings, artinian $\iff$ noetherian and of Krull dimension $0$.

However, you may consider its field of fractions $K(X)$, which is artinian, being a field.

  • $\begingroup$ Thank you for your answer. Why $K[X]$ is not a Noetherian $K-$module? $\endgroup$ – Chris Mar 5 at 21:17
  • $\begingroup$ Because it is not finitely generated. $\endgroup$ – Bernard Mar 5 at 21:24
  • $\begingroup$ We know that $K[X]$ is a $K-$vector space $\iff K[X]$ is a $K-$module as we know. And $K[X]=\langle 1_K,X,X^2,\dots,X^n,\dots \rangle$ is not finitely generated. It's ok for Noetherian. Can we prove $K[X]$ is not Artinian without use of Hopkins' Theorems (maybe only with definition)? $\endgroup$ – Chris Mar 5 at 21:31
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    $\begingroup$ It is enough to mention the descending chain: $\;(X)\supsetneq (X^2)\supsetneq (X^3)\supsetneq \dots$. $\endgroup$ – Bernard Mar 5 at 21:56
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    $\begingroup$ Oh! yes. I saw it, but I was doing several things simultaneously, and I forgot. $\endgroup$ – Bernard Mar 5 at 22:08

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