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We know that Finitely generated modules over a Noetherian ring are Noetherian and finitely generated modules over a Artinian ring are artinian.

Now I want to know whether the converse is also right, that is: Are noetherian modules over a noetherian ring finitely generated? Are artinian modules over a artinian ring finitely generated?

To noetherian case, I think it is right: Let $M$ be a left noetherian module over a noetherian ring $R$. Suppose $M$ is not finitely generated, take $m_1 \in M$, then $M_1=Rm_1 \subset M$. Since $M$ is not finitely generated, there is $m_2 \in M$ \ $M_1$,$M_2 :=M_1 +Rm_2$, then $M_1 \subset M_2 $. Take $m_3 \in M$\ $M_2$, $M_3 := M_2 +Rm_3$, then $M_1 \subset M_2 \subset M_3$. Continue this process, we get an infinite increasing chain of submodules of $M$. This contradicts with $M$ noetherian. So $M$ is finitely generated.

I can't make sure the above is right, thank you for checking. Also I don't know whether the artinian case is right or wrong. Thank you for any help.

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    $\begingroup$ Noetherian and Artinian $\endgroup$ – Tachyon Jun 22 '17 at 3:06
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    $\begingroup$ Are noetherian modules over a noetherian ring finitely generated? Noetherian modules over any ring are finitely generated. You may have noticed this since your proof is basically the right idea and does not reference anything about $R$ being noetherian. The Artinian case does use different mechanisms. At the linked half-duplicate, you can see two explanations for that. $\endgroup$ – rschwieb Jun 22 '17 at 10:49
  • $\begingroup$ Just a quick question, the answer in the link explicitly uses commutativity, is this necessary? $\endgroup$ – SEWillB Feb 3 '18 at 22:19
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for $1:$ a Noetherian $R$-module $M$ is finitely generated irrespective of whether or not $R$ is Noetherian. This is for the reason you give in the body of the question. Nothing about $R$ is required, as f.g follows from just the ascending chain condition.

The answer for question $2$, is yes, and in fact $M$ is Noetherian, as shown by hopkin's theorem.

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