The Stacks project

Proof. Pick $x_1, \ldots , x_ n \in S$ which generate $S$ over $R$. Let $K$ be the fraction field of $R$ and $S_ K = S \otimes _ R K$. By Lemma 10.115.4 we can find $y_1, \ldots , y_ d \in S$ such that $K[y_1, \ldots , y_ d] \to S_ K$ is a finite injective map. Note that $y_ i \in S$ because we may pick the $y_ j$ in the $\mathbf{Z}$-algebra generated by $x_1, \ldots , x_ n$. As a finite ring map is integral (see Lemma 10.36.3) we can find monic $P_ i \in K[y_1, \ldots , y_ d][T]$ such that $P_ i(x_ i) = 0$ in $S_ K$. Let $f \in R$ be a nonzero element such that $fP_ i \in R[y_1, \ldots , y_ d][T]$ for all $i$. Then $fP_ i(x_ i)$ maps to zero in $S_ K$. Hence after replacing $f$ by another nonzero element of $R$ we may also assume $fP_ i(x_ i)$ is zero in $S$. Set $x_ i' = fx_ i$ and let $S' \subset S$ be the $R$-subalgebra generated by $y_1, \ldots , y_ d$ and $x'_1, \ldots , x'_ n$. Note that $x'_ i$ is integral over $R[y_1, \ldots , y_ d]$ as we have $Q_ i(x_ i') = 0$ where $Q_ i = f^{\deg _ T(P_ i)}P_ i(T/f)$ which is a monic polynomial in $T$ with coefficients in $R[y_1, \ldots , y_ d]$ by our choice of $f$. Hence $R[y_1, \ldots , y_ d] \subset S'$ is finite by Lemma 10.36.5. Since $S' \subset S$ we have $S'_ f \subset S_ f$ (localization is exact). On the other hand, the elements $x_ i = x'_ i/f$ in $S'_ f$ generate $S_ f$ over $R_ f$ and hence $S'_ f \to S_ f$ is surjective. Whence $S'_ f \cong S_ f$ and we win. $\square$