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*}

[Theorem 15, Matlis]. The slick proof given here is from an email of Bjorn Poonen dated Nov 5, 2016.

Lemma 10.95.3. Let $R$ be a ring. Let $I$ be a finitely generated ideal of $R$. Let $M$ be an $R$-module. Then

  1. the completion $M^\wedge $ is $I$-adically complete, and

  2. $I^ nM^\wedge = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ nM) = (I^ nM)^\wedge $ for all $n \geq 1$.

In particular $R^\wedge $ is $I$-adically complete, $I^ nR^\wedge = (I^ n)^\wedge $, and $R^\wedge /I^ nR^\wedge = R/I^ n$.

Proof. Since $I$ is finitely generated, $I^ n$ is finitely generated, say by $f_1, \ldots , f_ r$. Applying Lemma 10.95.1 part (2) to the surjection $(f_1, \ldots , f_ r) : M^{\oplus r} \to I^ n M$ yields a surjection

\[ (M^\wedge )^{\oplus r} \xrightarrow {(f_1, \ldots , f_ r)} (I^ n M)^\wedge = \mathop{\mathrm{lim}}\nolimits _{m \geq n} I^ n M/I^ m M = \mathop{\mathrm{Ker}}(M^\wedge \to M/I^ n M). \]

On the other hand, the image of $(f_1, \ldots , f_ r) : (M^\wedge )^{\oplus r} \to M^\wedge $ is $I^ n M^\wedge $. Thus $M^\wedge / I^ n M^\wedge \simeq M/I^ n M$. Taking inverse limits yields $(M^\wedge )^\wedge \simeq M^\wedge $; that is, $M^\wedge $ is $I$-adically complete. $\square$


Comments (2)

Comment #3282 by Nicolas on

The proof doesn't seem to use that is finitely generated.

Comment #3283 by Dario on

Rather subtle: The terms 'finite direct sum' and 'finite product' coincide. As limits commute we have .


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