Lemma 10.97.8. 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 10.36.21). 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 10.97.1. 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 10.97.7. 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 10.36.22). Hence $S/\mathfrak m^ nS = \prod S_{\mathfrak q_ i}/\mathfrak m^ nS_{\mathfrak q_ i}$ by Lemma 10.53.6 (using for example Proposition 10.60.6 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 10.97.1.
$\square$

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