Lemma 10.68.6. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $\mathfrak p$ be a prime. Let $x_1, \ldots , x_ r$ be a sequence in $R$ whose image in $R_{\mathfrak p}$ forms an $M_{\mathfrak p}$-regular sequence. Then there exists a $g \in R$, $g \not\in \mathfrak p$ such that the image of $x_1, \ldots , x_ r$ in $R_ g$ forms an $M_ g$-regular sequence.

Proof. Set

$K_ i = \mathop{\mathrm{Ker}}\left(x_ i : M/(x_1, \ldots , x_{i - 1})M \to M/(x_1, \ldots , x_{i - 1})M\right).$

This is a finite $R$-module whose localization at $\mathfrak p$ is zero by assumption. Hence there exists a $g \in R$, $g \not\in \mathfrak p$ such that $(K_ i)_ g = 0$ for all $i = 1, \ldots , r$. This $g$ works. $\square$

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