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

1. $A' \to B'$ is flat and of finite presentation (for example smooth),

2. $I$ is locally nilpotent,

3. the images $f_1, \ldots , f_ r \in B$ form a quasi-regular sequence,

4. $B/(f_1, \ldots , f_ r)$ 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. $\square$

Comment #3854 by Zhiyu Zhang on

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

Comment #4919 by Matthieu Romagny on

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

Comment #4920 by on

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

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