Noetherian prime ideals and Noetherian ring


Solution 1

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...

Solution 2

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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Updated on June 20, 2022


  • mshj
    mshj about 2 hours

    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.

    • Julian Kuelshammer
      Julian Kuelshammer over 9 years
      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. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post.
    • Rankeya
      Rankeya over 9 years
      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.
  • Admin
    Admin over 9 years
    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.
  • Andrew
    Andrew over 9 years
    @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.