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 10.95.8. Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. If $M$ is $J$-adically complete and $I$ is finitely generated, then $M$ is $I$-adically complete.

Proof. Assume $M$ is $J$-adically complete and $I$ is finitely generated. We have $\bigcap I^ nM = 0$ because $\bigcap J^ nM = 0$. By Lemma 10.95.7 it suffices to prove the surjectivity of $M \to \mathop{\mathrm{lim}}\nolimits M/I^ nM$ in case $I$ is generated by a single element. Say $I = (f)$. Let $x_ n \in M$ with $x_{n + 1} - x_ n \in f^ nM$. We have to show there exists an $x \in M$ such that $x_ n - x \in f^ nM$ for all $n$. As $x_{n + 1} - x_ n \in J^ nM$ and as $M$ is $J$-adically complete, there exists an element $x \in M$ such that $x_ n - x \in J^ nM$. Replacing $x_ n$ by $x_ n - x$ we may assume that $x_ n \in J^ nM$. To finish the proof we will show that this implies $x_ n \in I^ nM$. Namely, write $x_ n - x_{n + 1} = f^ nz_ n$. Then

\[ x_ n = f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots ) \]

The sum $z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots $ converges in $M$ as $f^ c \in J^ c$. The sum $f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots )$ converges in $M$ to $x_ n$ because the partial sums equal $x_ n - x_{n + c}$ and $x_{n + c} \in J^{n + c}M$. $\square$


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