TITLE: Is a normal domain whose prime ideals are totally ordeded a valuation ring?
QUESTION [7 upvotes]: Recall one of the definition of a valuation ring is a domain whose ideals are totally ordered. (Then it will be a normal domain.) But if we restrict to all prime ideals the reverse is not true. The stalk at a non-normal closed point of a one-dimensional scheme is a counter-example. For example, $k[[x^2, x^3]]$  or $\mathbb{Z}_2[\sqrt{-3}]$. I just found that this has been already discussed in Does totally ordered prime ideals in a domain imply valuation ring?
All counter-examples so far are non-normal rings, so the question in the title naturally appears.

REPLY [4 votes]: Let $L\subset K$ be a field extension such that $L$ is algebraically closed in $K$, and $R=L+XK[[X]]$ a subring of power series ring $K[[X]]$. Then $R$ is integrally closed, and it has only two prime ideals: $(0)$ and $XK[[X]]$, so its spectrum is totally ordered. It's not hard to show that $R$ is not a valuation ring.