Does totally ordered prime ideals in a domain imply valuation ring?

A domain is a valuation ring if and only if the partially ordered set of ideals in this ring is totally ordered.

Does this equivalence hold when we restrict to prime ideals only? I.e. is a domain $R$ a valuation ring, if the partially ordered set $\operatorname{Spec}(R)$ is totally ordered?

Let $$K$$ be an algebraically closed field with characteristic $$0$$, $$B=K[X,Y]$$. The polynomial $$P(X,Y)=X(X^2+Y^2)+X^2-Y^2$$ is irreducible, hence generates a prime ideal in $$B$$. $$B/PB$$ is a noetherian domain with Krull dimension $$1$$, so its localisation $$A=\bigl(B/PB)_{(X,Y)}\;$$ at the maximal ideal generated by the canonical images of $$X$$ and $$Y$$ is a local noetherian domain of dimension $$1$$ and henceforth $$\operatorname{Spec}A$$ is totally ordered by inclusion.
• Dear Bernard, to find a local noetherian domain of dimension one which is not a DVR is the easiest task in the world: just take the local ring of a singularity of a curve. The ring $A$ of this answer is an example of this procedure. Of course Bourbaki cannot say so because He hasn't published His Magnum Opus on Algebraic Geometry... yet :-). Anyway, +1. – Georges Elencwajg Oct 19 '18 at 21:03
• Nitpick: all fields are integrally closed, so this hypothesis is superfluous! You probably meant to write that $K$ is algebraically closed, but that is irrelevant for the answer. We don't need $\operatorname {char}(K)= 0$ either: the only needed condition on the field $K$ is $\operatorname {char}(K)\neq 2$, which ensures the irreducibility of $P$. – Georges Elencwajg Oct 19 '18 at 21:17
Another example of a $1$-dimensional Noetherian local domain which isn’t a valuation ring is $k[[x^2,x^3]]$, whose integral closure is $k[[x]]$.