The Stacks project

Lemma 35.14.7. Let

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[dl]^ q \\ & S } \]

be a commutative diagram of morphisms of schemes. Assume that

  1. $f$ is surjective, flat, and locally of finite presentation,

  2. $p$ is syntomic.

Then both $q$ and $f$ are syntomic.

Proof. By Lemma 35.14.3 we see that $q$ is of finite presentation. By Morphisms, Lemma 29.25.13 we see that $q$ is flat. By Morphisms, Lemma 29.30.10 it now suffices to show that the local rings of the fibres of $Y \to S$ and the fibres of $X \to Y$ are local complete intersection rings. To do this we may take the fibre of $X \to Y \to S$ at a point $s \in S$, i.e., we may assume $S$ is the spectrum of a field. Pick a point $x \in X$ with image $y \in Y$ and consider the ring map

\[ \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x} \]

This is a flat local homomorphism of local Noetherian rings. The local ring $\mathcal{O}_{X, x}$ is a complete intersection. Thus may use Avramov's result, see Divided Power Algebra, Lemma 23.8.9, to conclude that both $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ are complete intersection rings. $\square$

Comments (0)

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Your email address will not be published.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 05B7. Beware of the difference between the letter 'O' and the digit '0'.