The Stacks project

Lemma 10.136.17. Let $R \to S$, $S \to S'$ be ring maps.

  1. If $R \to S$ and $S \to S'$ are syntomic, then $R \to S'$ is syntomic.

  2. If $R \to S$ and $S \to S'$ are relative global complete intersections, then $R \to S'$ is a relative global complete intersection.

Proof. Proof of (2). Say $R \to S$ and $S \to S'$ are relative global complete intersections and we have presentations $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ and $S' = S[y_1, \ldots , y_ m]/(h_1, \ldots , h_ d)$ as in Definition 10.136.5. Then

\[ S' \cong R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]/(f_1, \ldots , f_ c, h'_1, \ldots , h'_ d) \]

for some lifts $h_ j' \in R[x_1, \ldots , x_ n, y_1, \ldots , y_ m]$ of the $h_ j$. Hence it suffices to bound the dimensions of the fibre rings. Thus we may assume $R = k$ is a field. In this case we see that we have a ring, namely $S$, which is of finite type over $k$ and equidimensional of dimension $n - c$, and a finite type ring map $S \to S'$ all of whose nonempty fibre rings are equidimensional of dimension $m - d$. Then, by Lemma 10.112.6 for example applied to localizations at maximal ideals of $S'$, we see that $\dim (S') \leq n - c + m - d$ as desired.

We will reduce part (1) to part (2). Assume $R \to S$ and $S \to S'$ are syntomic. Let $\mathfrak q' \subset S$ be a prime ideal lying over $\mathfrak q \subset S$. By Lemma 10.136.15 there exists a $g' \in S'$, $g' \not\in \mathfrak q'$ such that $S \to S'_{g'}$ is a relative global complete intersection. Similarly, we find $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is a relative global complete intersection. By Lemma 10.136.9 the ring map $S_ g \to S_{gg'}$ is a relative global complete intersection. By part (2) we see that $R \to S_{gg'}$ is a relative global complete intersection and $gg' \not\in \mathfrak q'$. Since $\mathfrak q'$ was arbitrary combining Lemmas 10.136.15 and 10.136.4 we see that $R \to S'$ is syntomic (this also uses that the spectrum of $S'$ is quasi-compact, see Lemma 10.17.8). $\square$


Comments (3)

Comment #6686 by WhatJiaranEatsTonight on

I think the reduction to (2) could be more simple if we use 136.15.

Suppose (2) holds. For any , we can find such that is a relative global complete intersection by 136.15. Let be the prime corresponding to . Then we can find such that is a relative global complete intersection. (I think is still a relative global complete intersection if is a relative complete global complete intersection. Is it right?) We know is a relative global complete intersection. Hence is a relative global complete intersection and obviously is a prime in . Thus is syntomic.

Comment #6687 by WhatJiaranEatsTonight on

The undefined control sequence is .

There are also:

  • 2 comment(s) on Section 10.136: Syntomic morphisms

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 00SZ. Beware of the difference between the letter 'O' and the digit '0'.