The Stacks project

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

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

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. $\square$