# Noetherian prime ideals and Noetherian ring

If $R$ is a commutative ring, and for every $P$ a prime ideal of $R$, $P$ is a Noetherian $R$-module, show that $R$ is Noetherian.

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Is this homework? –  jspecter Nov 26 '12 at 17:01
In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Nov 26 '12 at 17:03
As stated, this is not true. Just take any non-noetherian domain, and consider $P=0$. Maybe you mean the following: if $R$ is a commutative ring, and every prime ideal in $R$ if finitely generated, then $R$ is noetherian. –  the L Nov 26 '12 at 17:05
yes, is homwork –  mshj Nov 26 '12 at 17:19
Nice point, @rschwieb. That comma makes world of difference. –  DonAntonio Nov 26 '12 at 18:45

I recommend you look at some proofs of

Cohen's theorem: A commutative ring $R$ is Noetherian iff all of its prime ideals are finitely generated.

If the prime ideals are Noetherian, then they are obviously finitely generated.

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As mentioned in the comments, the statement should be as follows: "Let $R$ be a commutative ring. If every prime ideal of $R$ is noetherian as an $R$-module, then $R$ itself is noetherian."

To prove this, consider an ascending chain of proper ideals $I_0\subseteq I_1\subseteq\cdots\subseteq R.$ Consider the sum $\sum_{i\ge0}I_i,$ which is an ideal $J\subseteq R.$ There are two possibilities:

1. $J=R$
2. $J\subsetneq R$

It should be relatively straightforward to proceed from here, using Krull's theorem in the second case...

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Maybe simpler: if $I$ is an ideal of $R$, then show that $I$ is finitely generated. If $I=R$ is ok. If not, take $P\supset I$ a prime ideal. Since $P$ is a noetherian $R$-module it follows that $I$ is finitely generated. –  user26857 Nov 26 '12 at 18:16
@YACP, yes, the choice of formulation will of course depend on the definitions the OP has at hand. It seems to me that either way, we invoke Krull's theorem. –  Andrew Nov 26 '12 at 18:29