Lemma 15.31.5. 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

1. $A' \to B'$ is flat and of finite presentation,

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 flat over $A$.

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

Proof. Set $C' = B'/(f'_1, \ldots , f'_ r)$. We have to show $A' \to C'$ is flat. Let $\mathfrak r' \subset C'$ be a prime ideal lying over $\mathfrak p' \subset A'$. We let $\mathfrak q' \subset B'$ be the inverse image of $\mathfrak r'$. By Algebra, Lemma 10.39.18 it suffices to show that $A'_{\mathfrak p'} \to C'_{\mathfrak q'}$ is flat. Algebra, Lemma 10.128.6 tells us it suffices to show that $f'_1, \ldots , f'_ r$ map to a regular sequence in

$B'_{\mathfrak q'}/\mathfrak p'B'_{\mathfrak q'} = B_\mathfrak q/\mathfrak p B_\mathfrak q = (B \otimes _ A \kappa (\mathfrak p))_\mathfrak q$

with obvious notation. What we know is that $f_1, \ldots , f_ r$ is a quasi-regular sequence in $B$ and that $B/(f_1, \ldots , f_ r)$ is flat over $A$. By Lemma 15.31.4 the images $\overline{f}_1, \ldots , \overline{f}_ r$ of $f'_1, \ldots , f'_ r$ in $B \otimes _ A \kappa (\mathfrak p)$ form a quasi-regular sequence. Since $(B \otimes _ A \kappa (\mathfrak p))_\mathfrak q$ is a Noetherian local ring, we conclude by Lemma 15.30.7. $\square$

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).