# comparison between valuation ring and proper morphism

I came up with one question while trying to solve an exercise in algebraic geometry.

Let $A$ be a local ring with maximal ideal $m$, a prime ideal $p$, and the fraction field $K$. There is a valuation ring $R$ which dominates $A$ (and maximal with domination). Since $A_p$ is also a local ring, there is a valuation ring $R'$ which dominates $A_p$. Now my question is, can we show that the ring R is contained in R' as the ring $A$ is contained in $A_p$? In the original context, $A$ is a finitely generated algebra over a field $k$.

Let me explain a little bit why I'm concerning this problem. It's from Ex. 4.5 (c) of algebraic geometry by Hartshorne.

Let $X$ be an integral scheme of finite type over a field $k$ with function field $K$. Suppose, for any valuation of $K/k$, there exists a unique center $x$ of $X$, i.e. there exists a unique point $x$ of $X$ with the domination between local ring $\mathcal{O}_{x,X}$. Then $X$ is proper over $k$.

The given hyphothesis says that the valuative criterion for properness hold for a function field $K$ and its valuation ring $R$. However, to show the properness, we should consider other field $L$ and its valuation ring $S$.

A map from $\mathrm{Spec}L$ to $X$ factor through a closed subscheme $\{x_0\}^-$ with a reduced induced structure where $x_0$ is the image of the $\xi_0 \in \mathrm{Spec}L$. Then we should find a map from $\mathrm{Spec}S$ for a valuation ring $S$ in $L$. The residue field $k(x_0)$ is contained in $L$, and $\mathcal{O}_{x_0 ,X} \twoheadrightarrow k(x_0)$ gives a pull back of valuation on $\mathcal{O}_{x_0, X}$. Then we can find a local subring $T\subset\mathcal{O}_{x_0 ,X}$ with this valuation. Since $T \subset K$, there is a valuation ring $R$ dominates $T$. Then by the assumption, $R$ has a unique center $x_1$ which must be a specialization of $x_0$, but I can't prove this yet. However, if $R$ is also a subring of the valuation ring $R'$ dominating $\mathcal{O}_{x_0, X}$, then we can find a prime ideal in $\mathcal{O}_{x_1, X}$ by taking inverse image of the maximal ideal of $R'$, then we will get an inclusion from $\mathcal{O}_{x_1, X}$ to $\mathcal{O}_{x_0, X}$. Consider $x_1$ as a point in $\{x_0\}^-$, then we can see that $S$ dominates $\mathcal{O}_{x_1, \{x_0\}^-}$.

• In your very first question, could you specify where $R$ lives in? May 23, 2018 at 13:41
• @User0829 It is a subring of $K$.
– HLEE
May 23, 2018 at 14:38

Let $A = k[x,y]$ and $n = (x,y)$. Take the local ring $R$ as $R = A_n, m = nR$. Then $V = k[x,y/x]_{(x)}$ is a valuation ring dominating $R$. Since $R$ is normal, $R_{(x)}$ is a valuation ring. However, $V \not\subset R_{(x)}$ as $y/x \not\in R_{(x)}$.
• $k[x,y]$ is not a local ring. May 23, 2018 at 13:40
• @User0829 He is saying that $A_n$ is a local ring.