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 10.101.3. In Situation 10.101.1. Suppose $R$ is a local Noetherian ring with maximal ideal $\mathfrak m$. Assume $\mathfrak m \in \text{Ass}(R)$, in other words $R$ has depth $0$. Suppose that $0 \to R^{n_ e} \to R^{n_{e-1}} \to \ldots \to R^{n_0}$ is exact at $R^{n_ e}, \ldots , R^{n_1}$. Then the complex is isomorphic to a direct sum of trivial complexes.

Proof. Pick $x \in R$, $x \not= 0$, with $\mathfrak m x = 0$. Let $i$ be the biggest index such that $n_ i > 0$. If $i = 0$, then the statement is true. If $i > 0$ denote $f_1$ the first basis vector of $R^{n_ i}$. Since $xf_1$ is not mapped to zero by exactness of the complex we deduce that some matrix coefficient of the map $R^{n_ i} \to R^{n_{i - 1}}$ is not in $\mathfrak m$. Lemma 10.101.2 then allows us to decrease $n_ e + \ldots + n_1$. Induction finishes the proof. $\square$


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