It's true that a valuation ring $R$ in the quotient field of a normal ring $A$ contain $A$?

Question

Let $A$ be a finitely generated $k$-algebra ($k$ algebraically closed) of dimension one, integrally closed in its quotient field $K$. Let $R\subseteq K$ be a valuation ring. It's true that $A\subseteq R$ ?

Thank you!

Answer 1

No. Let $A=k[x]$ be the ring of univariate polynomials with coefficients in $k$. Then $K=k(x)$ is the field of rational functions, and $A$ is integrally closed in $K$. The valuation ring known as the "place at infinity" $$ R=\left\{\frac{p(x)}{q(x)}\in K\,\bigg\vert\,\deg p(x)\le \deg q(x)\right\} $$ is a DVR in $K$. Clearly it does not contain $A$.

Answer 2

By Krull-Akizuki theorem every valuation ring containing a Dedekind domain is a DVR. If take $R$ a valuation ring which is not a DVR you can conclude.

An explicit example: let $L$ be a field, $k=L(Y)$, and $A=k[X]$. Then $K=k(X)=L(X,Y)$ and now consider for $K$ a non-noetherian valuation ring.