Lemma 15.31.6 (0CER)—The Stacks project

Table of contents
Part 1: Preliminaries
Chapter 15: More on Algebra
Section 15.31: More on Koszul regular sequences
Lemma 15.31.6 (cite)

previous lemma

Lemma 15.31.6. Let $A' \to B'$ be a ring map. Let $I \subset A'$ be an ideal. Set $A = A'/I$ and $B = B'/IB'$. Let $f'_1, \ldots , f'_ r \in B'$. Assume

$A' \to B'$ is flat and of finite presentation (for example smooth),
$I$ is locally nilpotent,
the images $f_1, \ldots , f_ r \in B$ form a quasi-regular sequence,
$B/(f_1, \ldots , f_ r)$ is smooth over $A$.

Then $B'/(f'_1, \ldots , f'_ r)$ is smooth over $A'$.

Proof. Set $C' = B'/(f'_1, \ldots , f'_ r)$ and $C = B/(f_1, \ldots , f_ r)$. Then $A' \to C'$ is of finite presentation. By Lemma 15.31.5 we see that $A' \to C'$ is flat. The fibre rings of $A' \to C'$ are equal to the fibre rings of $A \to C$ and hence smooth by assumption (4). It follows that $A' \to C'$ is smooth by Algebra, Lemma 10.137.17. $\square$

Comments (4)

Comment #3854 by Zhiyu Zhang on December 27, 2018 at 21:44

Maybe a little type: "Set $A=A/I$ " shall be "Set $A=A'/I$ ".

Comment #3939 by Johan on January 23, 2019 at 08:54

THanks and fixed here.

Comment #4919 by Matthieu Romagny on February 08, 2020 at 07:50

I still read $A=A/I$ . (Am I silly?)

Comment #4920 by Johan on February 08, 2020 at 12:46

Oops, I fixed it in another lemma and not this one. Now I did it again, see here.

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