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This question was asked in my assignments of commutative algebra and I am not able to make much progress on it.

Question: (a) Show that A is noetherian iff every prime ideal is of finite type.

(b) Let A be a commutative ring and I , J are ideals. Assume that J is of the finite type and A/I , A/J are noetherian. Show that A/IJ is still noetherian.

Attempt:(a) If A is noetherian,then I have shown that every prime ideal is of the finite type. But I am not able to proceed in the opposite direction. Please help me!

(b) assuming that J is of finite type, I have shown that J is noetherian and I have been given that A/I , A/J are noetherian . I showed that $A/IJ \subseteq A+ I$ and as any descending chain in A+I stabilizes then so does in A/IJ and hence A/IJ is noetherian. Is my proof fine?

Thanks!

Apass.Jack
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1 Answers1

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(a) You have shown the "$\implies$" implication. Let us show the "$\impliedby$" implication.

We are given a ring $A$ such that every prime ideal is of finite type. We are asked to prove $A$ is noetherian, i.e. every ideal is of finite type.

Towards a contradiction, suppose there is an ideal that is not of finite type, i.e., the set of ideals that are not of finite type is not empty. Using Zorn’s lemma we can see that there exists a maximal element $m$ in this set.

Since $m$ is not of finite type, $m$ is not prime. Hence $m$ is properly contained in two ideals $a,b$ such that $ab\subseteq m$. By the maximal property of $m$, we know $a$ and $b$ are finite type and $R/a$ is Noetherian.

Since $b/ab$ as $R/a$-module is finitely generated and $R/a$ is Noetherian, every submodule is finitely generated over $R/a$, hence likewise over $R$.

In particular, $m/ab$ is a finitely generated $R$-module and since $ab$ is finite type, so is $m$. We have thus arrived at a contradiction.

This proof was given in commutative rings with restricted minimum condition by I. S. Cohen, 1950.

(b) "... as any descending chain in $A+I$ stabilizes then so does in $A/IJ$ and hence $A/IJ$ is noetherian." How is "descending chain" relevant to "noetherian"?

Apass.Jack
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  • @ApassJack I don't think your answer proves what I asked in (a) . I was asking for a proof that A is noetherian if every ideal is given to be of finite type. But I think your hypothesis in proof of (a) is different. –  Dec 13 '22 at 17:50
  • @3ibfwcbi what is your definition of noethrian ring? My hypotheses for (a) are 1) a ring is noetherian ring iff all ideals are finite type and 2) a ring $R$ is notherian ring iff all submodule of any $R$-module of finite type are finite type. Please tell me the first place in my answer that is not consistent with your expectations/hypotheses. Thank you! – Apass.Jack Dec 13 '22 at 23:06
  • @ApassJack In (a) you took the hypothesis that m is not prime. But my hypothesis would be to assume that let every prime ideal is of finite type and using it trying to prove that A is noetherian. –  Feb 10 '23 at 19:39
  • @3ibfwcbi I just added more explanation to my answer. Is it fine now? – Apass.Jack Feb 10 '23 at 20:12