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 10.96.8. Let $R$ be a Noetherian ring. Let $R \to S$ be a finite ring map. Let $\mathfrak p \subset R$ be a prime and let $\mathfrak q_1, \ldots , \mathfrak q_ m$ be the primes of $S$ lying over $\mathfrak p$ (Lemma 10.35.21). Then

\[ R_\mathfrak p^\wedge \otimes _ R S = (S_\mathfrak p)^\wedge = S_{\mathfrak q_1}^\wedge \times \ldots \times S_{\mathfrak q_ m}^\wedge \]

where the $(S_\mathfrak p)^\wedge $ is the completion with respect to $\mathfrak p$ and the local rings $R_\mathfrak p$ and $S_{\mathfrak q_ i}$ are completed with respect to their maximal ideals.

Proof. The first equality follows from Lemma 10.96.1. We may replace $R$ by the localization $R_\mathfrak p$ and $S$ by $S_\mathfrak p = S \otimes _ R R_\mathfrak p$. Hence we may assume that $R$ is a local Noetherian ring and that $\mathfrak p = \mathfrak m$ is its maximal ideal. The $\mathfrak q_ iS_{\mathfrak q_ i}$-adic completion $S_{\mathfrak q_ i}^\wedge $ is equal to the $\mathfrak m$-adic completion by Lemma 10.96.7. For every $n \geq 1$ prime ideals of $S/\mathfrak m^ nS$ are in 1-to-1 correspondence with the maximal ideals $\mathfrak q_1, \ldots , \mathfrak q_ m$ of $S$ (by going up for $S$ over $R$, see Lemma 10.35.22). Hence $S/\mathfrak m^ nS = \prod S_{\mathfrak q_ i}/\mathfrak m^ nS_{\mathfrak q_ i}$ by Lemma 10.52.6 (using for example Proposition 10.59.6 to see that $S/\mathfrak m^ nS$ is Artinian). Hence the $\mathfrak m$-adic completion $S^\wedge $ of $S$ is equal to $\prod S_{\mathfrak q_ i}^\wedge $. Finally, we have $R^\wedge \otimes _ R S = S^\wedge $ by Lemma 10.96.1. $\square$


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