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 10.51.4. Let $R$ be a local ring with maximal ideal $\mathfrak m$. Let $M$ be an $R$-module.

  1. If $M$ is a finite module and $\mathfrak m^ n M \not= 0$ for all $n\geq 0$, then $\text{length}_ R(M) = \infty $.

  2. If $M$ has finite length then $\mathfrak m^ nM = 0$ for some $n$.

Proof. Assume $\mathfrak m^ n M \not= 0$ for all $n\geq 0$. Choose $x \in M$ and $f_1, \ldots , f_ n \in \mathfrak m$ such that $f_1f_2 \ldots f_ n x \not= 0$. By Nakayama's Lemma 10.19.1 the first $n$ steps in the filtration

\[ 0 \subset R f_1 \ldots f_ n x \subset R f_1 \ldots f_{n - 1} x \subset \ldots \subset R x \subset M \]

are distinct. This can also be seen directly. For example, if $R f_1 x = R f_1 f_2 x$ , then $f_1 x = g f_1 f_2 x$ for some $g$, hence $(1 - gf_2) f_1 x = 0$ hence $f_1 x = 0$ as $1 - gf_2$ is a unit which is a contradiction with the choice of $x$ and $f_1, \ldots , f_ n$. Hence the length is infinite, i.e., (1) holds. Combine (1) and Lemma 10.51.2 to see (2). $\square$


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