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 10.136.10 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 10.17.10). $\square$


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

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