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 10.95.12. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

  1. $R$ is $I$-adically complete,

  2. $\bigcap _{n \geq 1} I^ nM = (0)$, and

  3. $M/IM$ is a finite $R/I$-module.

Then $M$ is a finite $R$-module.

Proof. Let $x_1, \ldots , x_ n \in M$ be elements whose images in $M/IM$ generate $M/IM$ as a $R/I$-module. Denote $M' \subset M$ the $R$-submodule generated by $x_1, \ldots , x_ n$. By Lemma 10.95.1 the map $(M')^\wedge \to M^\wedge $ is surjective. Since $\bigcap I^ nM = 0$ we see in particular that $\bigcap I^ nM' = (0)$. Hence by Lemma 10.95.11 we see that $M'$ is complete, and we conclude that $M' \to M^\wedge $ is surjective. Finally, the kernel of $M \to M^\wedge $ is zero since it is equal to $\bigcap I^ nM = (0)$. Hence we conclude that $M \cong M' \cong M^\wedge $ is finitely generated. $\square$


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