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