Lemma 57.16.2. Let $R \to S$ be a finite type flat ring map of Noetherian rings. Let $\mathfrak q \subset S$ be a prime ideal lying over $\mathfrak p \subset R$. Let $K \in D(S)$ be perfect. Let $f_1, \ldots , f_ r \in \mathfrak q S_\mathfrak q$ be a regular sequence such that $S_\mathfrak q/(f_1, \ldots , f_ r)$ is flat over $R$ and such that $K \otimes _ S^\mathbf {L} S_\mathfrak q$ is isomorphic to the Koszul complex on $f_1, \ldots , f_ r$. Then there exists a $g \in S$, $g \not\in \mathfrak q$ such that

1. $f_1, \ldots , f_ r$ are the images of $f'_1, \ldots , f'_ r \in S_ g$,

2. $f'_1, \ldots , f'_ r$ form a regular sequence in $S_ g$,

3. $S_ g/(f'_1, \ldots , f'_ r)$ is flat over $R$,

4. $K \otimes _ S^\mathbf {L} S_ g$ is isomorphic to the Koszul complex on $f_1, \ldots , f_ r$.

Proof. We can find $g \in S$, $g \not\in \mathfrak q$ with property (1) by the definition of localizations. After replacing $g$ by $gg'$ for some $g' \in S$, $g' \not\in \mathfrak q$ we may assume (2) holds, see Algebra, Lemma 10.68.6. By Algebra, Theorem 10.129.4 we find that $S_ g/(f'_1, \ldots , f'_ r)$ is flat over $R$ in an open neighbourhood of $\mathfrak q$. Hence after once more replacing $g$ by $gg'$ for some $g' \in S$, $g' \not\in \mathfrak q$ we may assume (3) holds as well. Finally, we get (4) for a further replacement by More on Algebra, Lemma 15.74.17. $\square$

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