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 10.49.2. Let $K$ be a field. Let $A \subset K$ be a local subring. Then there exists a valuation ring with fraction field $K$ dominating $A$.

Proof. We consider the collection of local subrings of $K$ as a partially ordered set using the relation of domination. Suppose that $\{ A_ i\} _{i \in I}$ is a totally ordered collection of local subrings of $K$. Then $B = \bigcup A_ i$ is a local subring which dominates all of the $A_ i$. Hence by Zorn's Lemma, it suffices to show that if $A \subset K$ is a local ring whose fraction field is not $K$, then there exists a local ring $B \subset K$, $B \not= A$ dominating $A$.

Pick $t \in K$ which is not in the fraction field of $A$. If $t$ is transcendental over $A$, then $A[t] \subset K$ and hence $A[t]_{(t, \mathfrak m)} \subset K$ is a local ring distinct from $A$ dominating $A$. Suppose $t$ is algebraic over $A$. Then for some $a \in A$ the element $at$ is integral over $A$. In this case the subring $A' \subset K$ generated by $A$ and $ta$ is finite over $A$. By Lemma 10.35.17 there exists a prime ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates $A$. If $A = A'_{\mathfrak m'}$, then $t$ is in the fraction field of $A$ which we assumed not to be the case. Thus $A \not= A'_{\mathfrak m'}$ as desired. $\square$


Comments (2)

Comment #1238 by Jonathan Wise on

This proof does not appear to be complete. In the last paragraph, the element may be in , so and does not contradict the maximality of .

Comment #1251 by on

If then is in the fraction field of which we assumed not to be the case. I have added this explanation to the proof, see here.


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