The Stacks project

Being syntomic is fpqc local on the base.

Lemma 10.136.2. 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 10.126.2 and Lemma 10.39.8 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 10.39.16. 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 10.135.11 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 10.136: Syntomic morphisms

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