Counterexample for the infinitely many primes between two primes in a Noetherian ring

Question

Consider the following Proposition:

Proposition: Let $R$ be a noetherian ring. If $p_0 \subsetneq p_1 \subsetneq p_2$ is a chain of distinct prime ideals in $R$, then there exist infinitely many distinct primes $q$ such that $p_0 \subsetneq q \subsetneq p_2$.

For a proof, see for instance this question. I would like to see a counterexample if we drop the noetherian hypothesis. Should such a ring exists I would find it rather interesting because it would be an example where a "finiteness" hypothesis implies that there are infinitely many of something!

today almost a same question was asked. you can see it here: http://math.stackexchange.com/questions/1040186/ring-with-nested-prime-ideals — Krish, Nov 26 '14 at 23:02
That does it, thanks. My apologies for missing that my question was already asked. — Daniel Bragg, Nov 27 '14 at 01:19

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

1

Consider a non-noetherian valuation ring of rank two. For such an example you can take a look at Examples of Non-Noetherian Valuation Rings.

edited Apr 13 '17 at 12:21

Community

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answered Nov 27 '14 at 08:57

user26857

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Counterexample for the infinitely many primes between two primes in a Noetherian ring

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