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](https://mathoverflow.net/questions/311907/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 ... 



Valuation ring satisfying either a.c.c. or d.c.c. on prime ideals
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).;


