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 10.121.6. Let

\[ \xymatrix{ S \ar[r] & S' & & \mathfrak q \ar@{-}[r] & \mathfrak q' \\ R \ar[u] \ar[r] & R' \ar[u] & & \mathfrak p \ar@{-}[r] \ar@{-}[u] & \mathfrak p' \ar@{-}[u] } \]

be a commutative diagram of rings with primes as indicated. Assume $R \to S$ of finite type, and $S \otimes _ R R' \to S'$ surjective. If $R \to S$ is quasi-finite at $\mathfrak q$, then $R' \to S'$ is quasi-finite at $\mathfrak q'$.

Proof. Write $S \otimes _ R \kappa (\mathfrak p) = S_1 \times S_2$ with $S_1$ finite over $\kappa (\mathfrak p)$ and such that $\mathfrak q$ corresponds to a point of $S_1$ as in Lemma 10.121.1. Because $S \otimes _ R R' \to S'$ surjective the canonical map $(S \otimes _ R \kappa (\mathfrak p)) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p') \to S' \otimes _{R'} \kappa (\mathfrak p')$ is surjective. Let $S_ i'$ be the image of $S_ i \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p')$ in $S' \otimes _{R'} \kappa (\mathfrak p')$. Then $S' \otimes _{R'} \kappa (\mathfrak p') =S'_1 \times S'_2$ and $S'_1$ is finite over $\kappa (\mathfrak p')$. The map $S' \otimes _{R'} \kappa (\mathfrak p') \to \kappa (\mathfrak q')$ factors through $S_1'$ (i.e. it annihilates the factor $S_2'$) because the map $S \otimes _ R \kappa (\mathfrak p) \to \kappa (\mathfrak q)$ factors through $S_1$ (i.e. it annihilates the factor $S_2$). Thus $\mathfrak q'$ corresponds to a point of $\mathop{\mathrm{Spec}}(S_1')$ in the disjoint union decomposition of the fibre: $\mathop{\mathrm{Spec}}(S' \otimes _{R'} \kappa (\mathfrak p')) = \mathop{\mathrm{Spec}}(S_1') \amalg \mathop{\mathrm{Spec}}(S_1')$. (See Lemma 10.20.2.) Since $S_1'$ is finite over a field, it is Artinian ring, and hence $\mathop{\mathrm{Spec}}(S_1')$ is a finite discrete set. (See Proposition 10.59.6.) We conclude $\mathfrak q'$ is isolated in its fibre as desired. $\square$

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