The Stacks project

Lemma 10.136.15. Let $R \to S$ be a ring map. Let $\mathfrak q \subset S$ be a prime lying over the prime $\mathfrak p$ of $R$. The following are equivalent:

  1. There exists an element $g \in S$, $g \not\in \mathfrak q$ such that $R \to S_ g$ is syntomic.

  2. There exists an element $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is a relative global complete intersection over $R$.

  3. There exists an element $g \in S$, $g \not\in \mathfrak q$, such that $R \to S_ g$ is of finite presentation, the local ring map $R_{\mathfrak p} \to S_{\mathfrak q}$ is flat, and the local ring $S_{\mathfrak q}/\mathfrak pS_{\mathfrak q}$ is a complete intersection ring over $\kappa (\mathfrak p)$ (see Definition 10.135.5).

Proof. The implication (1) $\Rightarrow $ (3) is Lemma 10.135.8. The implication (2) $\Rightarrow $ (1) is Lemma 10.136.13. It remains to show that (3) implies (2).

Assume (3). After replacing $S$ by $S_ g$ for some $g \in S$, $g\not\in \mathfrak q$ we may assume $S$ is finitely presented over $R$. Choose a presentation $S = R[x_1, \ldots , x_ n]/I$. Let $\mathfrak q' \subset R[x_1, \ldots , x_ n]$ be the prime corresponding to $\mathfrak q$. Write $\kappa (\mathfrak p) = k$. Note that $S \otimes _ R k = k[x_1, \ldots , x_ n]/\overline{I}$ where $\overline{I} \subset k[x_1, \ldots , x_ n]$ is the ideal generated by the image of $I$. Let $\overline{\mathfrak q}' \subset k[x_1, \ldots , x_ n]$ be the prime ideal generated by the image of $\mathfrak q'$. By Lemma 10.135.8 the equivalent conditions of Lemma 10.135.4 hold for $\overline{I}$ and $\overline{\mathfrak q}'$. Say the dimension of $\overline{I}_{\overline{\mathfrak q}'}/ \overline{\mathfrak q}'\overline{I}_{\overline{\mathfrak q}'}$ over $\kappa (\overline{\mathfrak q}')$ is $c$. Pick $f_1, \ldots , f_ c \in I$ mapping to a basis of this vector space. The images $\overline{f}_ j \in \overline{I}$ generate $\overline{I}_{\overline{\mathfrak q}'}$ (by Lemma 10.135.4). Set $S' = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$. Let $J$ be the kernel of the surjection $S' \to S$. Since $S$ is of finite presentation $J$ is a finitely generated ideal (Lemma 10.6.2). Consider the short exact sequence

\[ 0 \to J \to S' \to S \to 0 \]

As $S_\mathfrak q$ is flat over $R$ we see that $J_{\mathfrak q'} \otimes _ R k \to S'_{\mathfrak q'} \otimes _ R k$ is injective (Lemma 10.39.12). However, by construction $S'_{\mathfrak q'} \otimes _ R k$ maps isomorphically to $S_\mathfrak q \otimes _ R k$. Hence we conclude that $J_{\mathfrak q'} \otimes _ R k = J_{\mathfrak q'}/\mathfrak pJ_{\mathfrak q'} = 0$. By Nakayama's lemma (Lemma 10.20.1) we conclude that there exists a $g \in R[x_1, \ldots , x_ n]$, $g \not\in \mathfrak q'$ such that $J_ g = 0$. In other words $S'_ g \cong S_ g$. After further localizing we see that $S'$ (and hence $S$) becomes a relative global complete intersection by Lemma 10.136.10 as desired. $\square$

Comments (1)

Comment #6685 by WhatJiaranEatsTonight on

Suggested slogan: Syntomic maps locally are relative global complete intersections.

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



View Lemma 10.136.15 as pdf