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 10.101.7. In Situation 10.101.1. Suppose $R$ is a local ring with maximal ideal $\mathfrak m$. Suppose that $0 \to R^{n_ e} \to R^{n_{e-1}} \to \ldots \to R^{n_0}$ is exact at $R^{n_ e}, \ldots , R^{n_1}$. Let $x \in \mathfrak m$ be a nonzerodivisor. The complex $0 \to (R/xR)^{n_ e} \to \ldots \to (R/xR)^{n_1}$ is exact at $(R/xR)^{n_ e}, \ldots , (R/xR)^{n_2}$.

Proof. Denote $F_\bullet $ the complex with terms $F_ i = R^{n_ i}$ and differential given by $\varphi _ i$. Then we have a short exact sequence of complexes

\[ 0 \to F_\bullet \xrightarrow {x} F_\bullet \to F_\bullet /xF_\bullet \to 0 \]

Applying the snake lemma we get a long exact sequence

\[ H_ i(F_\bullet ) \xrightarrow {x} H_ i(F_\bullet ) \to H_ i(F_\bullet /xF_\bullet ) \to H_{i - 1}(F_\bullet ) \xrightarrow {x} H_{i - 1}(F_\bullet ) \]

The lemma follows. $\square$

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