The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.49.11. Let $A$ be a normal domain with fraction field $K$.

  1. For every $x \in K$, $x \not\in A$ there exists a valuation ring $A \subset V \subset K$ with fraction field $K$ such that $x \not\in V$.

  2. If $A$ is local, we can moreover choose $V$ which dominates $A$.

In other words, $A$ is the intersection of all valuation rings in $K$ containing $A$ and if $A$ is local, then $A$ is the intersection of all valuation rings in $K$ dominating $A$.

Proof. Suppose $x \in K$, $x \not\in A$. Consider $B = A[x^{-1}]$. Then $x \not\in B$. Namely, if $x = a_0 + a_1x^{-1} + \ldots + a_ d x^{-d}$ then $x^{d + 1} - a_0x^ d - \ldots - a_ d = 0$ and $x$ is integral over $A$ in contradiction with the fact that $A$ is normal. Thus $x^{-1}$ is not a unit in $B$. Thus $V(x^{-1}) \subset \mathop{\mathrm{Spec}}(B)$ is not empty (Lemma 10.16.2), and we can choose a prime $\mathfrak p \subset B$ with $x^{-1} \in \mathfrak p$. Choose a valuation ring $V \subset K$ dominating $B_\mathfrak p$ (Lemma 10.49.2). Then $x \not\in V$ as $x^{-1} \in \mathfrak m_ V$.

If $A$ is local, then we claim that $x^{-1} B + \mathfrak m_ A B \not= B$. Namely, if $1 = (a_0 + a_1x^{-1} + \ldots + a_ d x^{-d})x^{-1} + a'_0 + \ldots + a'_ d x^{-d}$ with $a_ i \in A$ and $a'_ i \in \mathfrak m_ A$, then we'd get

\[ (1 - a'_0) x^{d + 1} - (a_0 + a'_1) x^ d - \ldots - a_ d = 0 \]

Since $a'_0 \in \mathfrak m_ A$ we see that $1 - a'_0$ is a unit in $A$ and we conclude that $x$ would be integral over $A$, a contradiction as before. Then choose the prime $\mathfrak p \supset x^{-1} B + \mathfrak m_ A B$ we find $V$ dominating $A$. $\square$


Comments (1)

Comment #3227 by Giulio on

With little effort, if is local this can be strengthened: is the intersection of all the valuation rings dominating . Instead of proving that is not trivial, the same argument actually shows that is not trivial.


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