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 10.124.9. Let $R$ be a valuation ring with residue field $k$ and field of fractions $K$. Let $S$ be a domain containing $R$ such that $S$ is of finite type over $R$. If $S \otimes _ R k$ is not the zero ring then

\[ \dim (S \otimes _ R k) = \dim (S \otimes _ R K) \]

In fact, $\mathop{\mathrm{Spec}}(S \otimes _ R k)$ is equidimensional.

Proof. It suffices to show that $\dim _{\mathfrak q}(S/k)$ is equal to $\dim (S \otimes _ R K)$ for every prime $\mathfrak q$ of $S$ containing $\mathfrak m_ RS$. Pick such a prime. By Lemma 10.124.6 the inequality $\dim _{\mathfrak q}(S/k) \geq \dim (S \otimes _ R K)$ holds. Set $n = \dim _{\mathfrak q}(S/k)$. By Lemma 10.124.2 after replacing $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ there exists a quasi-finite ring map $R[t_1, \ldots , t_ n] \to S$. If $\dim (S \otimes _ R K) < n$, then $K[t_1, \ldots , t_ n] \to S \otimes _ R K$ has a nonzero kernel. Say $f = \sum a_ I t_1^{i_1}\ldots t_ n^{i_ n}$. After dividing $f$ by a nonzero coefficient of $f$ with minimal valuation, we may assume $f\in R[t_1, \ldots , t_ n]$ and some $a_ I$ does not map to zero in $k$. Hence the ring map $k[t_1, \ldots , t_ n] \to S \otimes _ R k$ has a nonzero kernel which implies that $\dim (S \otimes _ R k) < n$. Contradiction. $\square$


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