local (not Noetherian) domain satisfying a.c.c. on radical ideals and satisfying d.c.c on prime ideals

Question

Let $R$ be a local (not Noetherian) domain satisfying a.c.c. on radical ideals and satisfying d.c.c on prime ideals.

Then is the Krull dimension of $R$ finite ?

EDIT: Note that if our ring is a Valuation ring, then due to the comparability of any two ideals, the ring must have only finitely many prime ideals if it has to satisfy ac.c. on radical ideals, hence the dimension would be finite.

Answer 1

It seems that if $R$ is a valuation ring, then with the acc and dcc on prime ideals, it is of finite dimension. I tried this :

Because of the dcc and the fact that the set $\mathcal{P}$ of prime ideals of $R$ is totally ordered, there exists a $p_0$ such that $\forall p \in \mathcal{P}, p_0 \subset p$. If we consider $\mathcal{P} \setminus \{ p_0 \}$, it has a minimal element too, say $p_1$. Thus we construct a chain $p_0 \subset p_1 \subset \cdots $ which must be finite due to the acc. Then there is a finite number of prime ideals, so the dimension is finite.