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 10.96.7. Let $R \to S$ be a local homomorphism of local rings $(R, \mathfrak m)$ and $(S, \mathfrak n)$. Let $R^\wedge $, resp. $S^\wedge $ be the completion of $R$, resp. $S$ with respect to $\mathfrak m$, resp. $\mathfrak n$. If $\mathfrak m$ and $\mathfrak n$ are finitely generated and $\dim _{\kappa (\mathfrak m)} S/\mathfrak mS < \infty $, then

  1. $S^\wedge $ is equal to the $\mathfrak m$-adic completion of $S$, and

  2. $S^\wedge $ is a finite $R^\wedge $-module.

Proof. We have $\mathfrak mS \subset \mathfrak n$ because $R \to S$ is a local ring map. The assumption $\dim _{\kappa (\mathfrak m)} S/\mathfrak mS < \infty $ implies that $S/\mathfrak mS$ is an Artinian ring, see Lemma 10.52.2. Hence has dimension $0$, see Lemma 10.59.4, hence $\mathfrak n = \sqrt{\mathfrak mS}$. This and the fact that $\mathfrak n$ is finitely generated implies that $\mathfrak n^ t \subset \mathfrak mS$ for some $t \geq 1$. By Lemma 10.95.9 we see that $S^\wedge $ can be identified with the $\mathfrak m$-adic completion of $S$. As $\mathfrak m$ is finitely generated we see from Lemma 10.95.3 that $S^\wedge $ and $R^\wedge $ are $\mathfrak m$-adically complete. At this point we may apply Lemma 10.95.12 to $S^\wedge $ as an $R^\wedge $-module to conclude. $\square$


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