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If a commutative ring with unity has finite Krull dimension, then it satisfies a.c.c. and d.c.c. on prime ideals. The converse is not true in general, as can be seen from here An infinite dimensional local domain whose chains of primes are finite

But, if $R$ is a Valuation ring (https://en.wikipedia.org/wiki/Valuation_ring) and satisfies both a.c.c. and d.c.c. on prime ideals, then $R$ has only finitely many prime ideals due to the comparability of any two ideals in a Valuation ring. So then it is natural to ask, what happens if we assume only of the chain conditions on prime ideals for a Valuation ring ...

Let $R$ be a Valuation ring .

1) If $R$ satisfies a.c.c. on prime ideals, then does $R$ have finite Krull dimension ?

2) If $R$ satisfies d.c.c. on prime ideals, then does $R$ have finite Krull dimension ?

By the remarks preceding the question it is enough to show whether any of the a.c.c. or d.c.c. on prime ideals imply the other ... but I don't see whether this is true or not ...

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    $\begingroup$ Since there are valuation rings having any given ordered abelian group as value group, it suffices to construct such a group whose set of convex subgroups is infinite but satisfies a.c.c. (resp. d.c.c.). That shouldn't be too hard. $\endgroup$ – Laurent Moret-Bailly Oct 5 '18 at 15:12
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For any totally ordered set $W, R=k[x_i/x_j^n: i,j \in W, i \lt j, n \ge 0]$ is a valuation ring with value group $\oplus_{i \in W} \mathbb{Z}$ and Spec($R$) = {initial segments of $W$}. Thus $W$=negative integers gives a counterexample to (1) and $W$=positive integers gives a counterexample to (2).;

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