The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.68.4. Let $R$ be a Noetherian ring. Let $M$ be a finite $R$-module. Let $\mathfrak p$ be a prime. Let $x_1, \ldots , x_ c$ be a sequence in $R$ whose image in $R_{\mathfrak p}$ forms an $M_{\mathfrak p}$-quasi-regular sequence. Then there exists a $g \in R$, $g \not\in \mathfrak p$ such that the image of $x_1, \ldots , x_ c$ in $R_ g$ forms an $M_ g$-quasi-regular sequence.

Proof. Consider the kernel $K$ of the map (10.68.0.1). As $M/JM \otimes _{R/J} R/J[X_1, \ldots , X_ c]$ is a finite $R/J[X_1, \ldots , X_ c]$-module and as $R/J[X_1, \ldots , X_ c]$ is Noetherian, we see that $K$ is also a finite $R/J[X_1, \ldots , X_ c]$-module. Pick homogeneous generators $k_1, \ldots , k_ t \in K$. By assumption for each $i = 1, \ldots , t$ there exists a $g_ i \in R$, $g_ i \not\in \mathfrak p$ such that $g_ i k_ i = 0$. Hence $g = g_1 \ldots g_ t$ works. $\square$


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