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.
Noetherian prime ideals and Noetherian ring
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abstract-algebra
commutative-algebra
ring-theory
modules
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0A slightly more challenging exercise would be to show that if every prime ideal of $R$ is finitely generated, then $R$ is a Noetherian ring. – 2012-11-26
2 Answers
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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:
- $J=R$
- $J\subsetneq R$
It should be relatively straightforward to proceed from here, using Krull's theorem in the second case...
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0@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. – 2012-11-26
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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.