Lemma 15.30.5. 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 $A'$.
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 15.30.4 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 10.135.16.
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