The Stacks project

Proof. Say $h^ e + r_1 h^{e - 1} + \ldots + r_ e = 0$ in the ring $B[x]/(f)$ with $r_ i \in R$. There exists a finite free ring extension $B \subset B'$ such that $f = (x - \alpha _1) \ldots (x - \alpha _ d)$ for some $\alpha _ i \in B'$, see Lemma 10.136.14. Note that each $\alpha _ i$ is integral over $R$. We may represent $h = h_0 + h_1 x + \ldots + h_{d - 1} x^{d - 1}$ with $h_ i \in B$. Then it is a universal fact that

as elements of $B'[x]/(f)$. You prove this by evaluating both sides at the points $\alpha _ i$ over the ring $B_{univ} = \mathbf{Z}[\alpha _ i, h_ j]$ (some details omitted). By our assumption that $h$ satisfies $h^ e + r_1 h^{e - 1} + \ldots + r_ e = 0$ in the ring $B[x]/(f)$ we see that

in $B'$. Hence $h(\alpha _ i)$ is integral over $R$. Using the formula above we see that $f'h \equiv \sum _{j = 0, \ldots , d - 1} b'_ j x^ j$ in $B'[x]/(f)$ with $b'_ j \in B'$ integral over $R$. However, since $f' h \in B[x]/(f)$ and since $1, x, \ldots , x^{d - 1}$ is a $B'$-basis for $B'[x]/(f)$ we see that $b'_ j \in B$ as desired. $\square$

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. $\square$

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. $\square$

Proof. This follows from the straightforward fact that taking tensor products and taking integral closures commutes with filtered colimits and Lemma 10.147.4. $\square$