Lemma 10.147.4. Let $R \to S$ be a smooth 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. Arguing as in the proof of Lemma 10.147.2 we may localize on $S$. Hence we may assume that $R \to S$ is a standard smooth ring map, see Lemma 10.137.10. By definition of a standard smooth ring map we see that $S$ is étale over a polynomial ring $R[x_1, \ldots , x_ n]$. Since we have seen the result in the case of an étale ring extension (Lemma 10.147.2) this reduces us to the case where $S = R[x]$. Thus we have to show

$f = \sum b_ i x^ i \text{ integral over }R[x] \Leftrightarrow \text{each }b_ i\text{ integral over }R.$

The implication from right to left holds because the set of elements in $B[x]$ integral over $R[x]$ is a ring (Lemma 10.36.7) and contains $x$.

Suppose that $f \in B[x]$ is integral over $R[x]$, and assume that $f = \sum _{i < d} b_ i x^ i$ has degree $< d$. Since integral closure and localization commute, it suffices to show there exist distinct primes $p, q$ such that each $b_ i$ is integral both over $R[1/p]$ and over $R[1/q]$. Hence, we can find a finite free ring extension $R \subset R'$ such that $R'$ contains $\alpha _1, \ldots , \alpha _ d$ with the property that $\prod _{i < j} (\alpha _ i - \alpha _ j)$ is a unit in $R'$, see Example 10.147.3. In this case we have the universal equality

$f = \sum _ i f(\alpha _ i) \frac{(x - \alpha _1) \ldots \widehat{(x - \alpha _ i)} \ldots (x - \alpha _ d)}{(\alpha _ i - \alpha _1) \ldots \widehat{(\alpha _ i - \alpha _ i)} \ldots (\alpha _ i - \alpha _ d)}.$

OK, and the elements $f(\alpha _ i)$ are integral over $R'$ since $(R' \otimes _ R B)[x] \to R' \otimes _ R B$, $h \mapsto h(\alpha _ i)$ is a ring map. Hence we see that the coefficients of $f$ in $(R' \otimes _ R B)[x]$ are integral over $R'$. Since $R'$ is finite over $R$ (hence integral over $R$) we see that they are integral over $R$ also, as desired. $\square$

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