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. $\square$

## Comments (2)

Comment #1868 by Kestutis Cesnavicius on

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

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01UD. Beware of the difference between the letter 'O' and the digit '0'.