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 10.40.9. Let $k$ be a field, and let $R$, $S$ be $k$-algebras. Let $S' \subset S$ be a sub $k$-algebra, and let $f \in S' \otimes _ k R$. In the commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}((S \otimes _ k R)_ f) \ar[rd] \ar[rr] & & \mathop{\mathrm{Spec}}((S' \otimes _ k R)_ f) \ar[ld] \\ & \mathop{\mathrm{Spec}}(R) & } \]

the images of the diagonal arrows are the same.

Proof. Let $\mathfrak p \subset R$ be in the image of the south-west arrow. This means (Lemma 10.16.9) that

\[ (S' \otimes _ k R)_ f \otimes _ R \kappa (\mathfrak p) = (S' \otimes _ k \kappa (\mathfrak p))_ f \]

is not the zero ring, i.e., $S' \otimes _ k \kappa (\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. The ring map $S' \otimes _ k \kappa (\mathfrak p) \to S \otimes _ k \kappa (\mathfrak p)$ is injective. Hence also $S \otimes _ k \kappa (\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. Hence $(S \otimes _ k R)_ f \otimes _ R \kappa (\mathfrak p)$ is not the zero ring. Thus (Lemma 10.16.9) we see that $\mathfrak p$ is in the image of the south-east arrow as desired. $\square$

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