Lemma 10.136.16 (07BT)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.136: Syntomic morphisms
Lemma 10.136.16 (cite)

Lemma 10.136.16. Let $R$ be a ring. Let $S = R[x_1, \ldots , x_ n]/I$ for some finitely generated ideal $I$. If $g \in S$ is such that $S_ g$ is syntomic over $R$, then $(I/I^2)_ g$ is a finite projective $S_ g$-module.

Proof. By Lemma 10.136.15 there exist finitely many elements $g_1, \ldots , g_ m \in S$ which generate the unit ideal in $S_ g$ such that each $S_{gg_ j}$ is a relative global complete intersection over $R$. Since it suffices to prove that $(I/I^2)_{gg_ j}$ is finite projective, see Lemma 10.78.2, we may assume that $S_ g$ is a relative global complete intersection. In this case the result follows from Lemmas 10.134.16 and 10.136.12. $\square$

Comments (2)

Comment #6487 by Ishan Levy on August 07, 2021 at 13:09

Can I suggest this Lemma be written more concisely? I would prefer "If $R \to S = R[x_1,\dots,x_n]/I$ is syntomic, then $I/I^2$ is a finite projective $S$ -module." Unless I'm mistaken, no more generality is added from $g$ .

Comment #6559 by Johan on September 13, 2021 at 13:30

@#6487. Certainly the lemma implies what you say, but I think it is strictly stronger. Namely, the problem is exactly what is solved in Lemma 10.134.16: we need to compare the conormal module for the algebra $S$ and its presentation with a conormal module of some presentation of $S_g$ . To explain this better we could have a remark somewhere discussing more clearly how well defined this "conormal" module really is.

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