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*}

Being syntomic is fpqc local on the base.

Lemma 10.134.2. Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = R'\otimes _ R S$. Then $R \to S$ is syntomic if and only if $R' \to S'$ is syntomic.

Proof. By Lemma 10.125.2 and Lemma 10.38.8 this holds for the property of being flat and for the property of being of finite presentation. The map $\mathop{\mathrm{Spec}}(R') \to \mathop{\mathrm{Spec}}(R)$ is surjective, see Lemma 10.38.16. Thus it suffices to show given primes $\mathfrak p' \subset R'$ lying over $\mathfrak p \subset R$ that $S \otimes _ R \kappa (\mathfrak p)$ is a local complete intersection if and only if $S' \otimes _{R'} \kappa (\mathfrak p')$ is a local complete intersection. Note that $S' \otimes _{R'} \kappa (\mathfrak p') = S \otimes _ R \kappa (\mathfrak p) \otimes _{\kappa (\mathfrak p)} \kappa (\mathfrak p')$. Thus Lemma 10.133.11 applies. $\square$


Comments (2)

Comment #980 by on

Suggested slogan: Syntomic ring maps descend along faithfully flat ring maps.

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  • 2 comment(s) on Section 10.134: Syntomic morphisms

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