Lemma 29.30.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

1. The morphism $f$ is syntomic.

2. For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is syntomic.

3. There exists an open covering $S = \bigcup _{j \in J} V_ j$ and open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that each of the morphisms $U_ i \to V_ j$, $j\in J, i\in I_ j$ is syntomic.

4. There exists an affine open covering $S = \bigcup _{j \in J} V_ j$ and affine open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that the ring map $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is syntomic, for all $j\in J, i\in I_ j$.

Moreover, if $f$ is syntomic then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_ U : U \to V$ is syntomic.

Proof. This follows from Lemma 29.14.3 if we show that the property “$R \to A$ is syntomic” is local. We check conditions (a), (b) and (c) of Definition 29.14.1. By Algebra, Lemma 10.136.3 being syntomic is stable under base change and hence we conclude (a) holds. By Algebra, Lemma 10.136.17 being syntomic is stable under composition and trivially for any ring $R$ the ring map $R \to R_ f$ is syntomic. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma 10.136.4. $\square$

Comment #1868 by Kestutis Cesnavicius on

Typo in the discussion of this section preceding this lemma: "One we have ..." --> "Once we have ..."

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