Lemma 10.147.2. Let $R \to S$ be an étale ring map. Let $R \to B$ be any ring map. Let $A \subset B$ be the integral closure of $R$ in $B$. Let $A' \subset S \otimes _ R B$ be the integral closure of $S$ in $S \otimes _ R B$. Then the canonical map $S \otimes _ R A \to A'$ is an isomorphism.

Proof. The map $S \otimes _ R A \to A'$ is injective because $A \subset B$ and $R \to S$ is flat. We are going to use repeatedly that taking integral closure commutes with localization, see Lemma 10.36.11. Hence we may localize on $S$, by Lemma 10.23.2 (the criterion for checking whether an $S$-module map is an isomorphism). Thus we may assume that $S = R[x]_ g/(f) = (R[x]/(f))_ g$ is standard étale over $R$, see Proposition 10.144.4. Applying localization one more time we see that $A'$ is $(A'')_ g$ where $A''$ is the integral closure of $R[x]/(f)$ in $B[x]/(f)$. Suppose that $a \in A''$. It suffices to show that $a$ is in $S \otimes _ R A$. By Lemma 10.147.1 we see that $f' a = \sum a_ i x^ i$ with $a_ i \in A$. Since $f'$ is invertible in $S$ (by definition of a standard étale ring map) we conclude that $a \in S \otimes _ R A$ as desired. $\square$

Comment #7689 by nkym on

In the last sentence of the proof, I was wondering if you meant $S$ instead of $B[x]_g/(f)$.

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