# 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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A slightly more challenging exercise would be to show that if every prime ideal of $R$ is finitely generated, then $R$ is a Noetherian ring. – Rankeya Nov 26 '12 at 23:24

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

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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