## 10.147 Integral closure and smooth base change

Lemma 10.147.1. Let $R$ be a ring. Let $f \in R[x]$ be a monic polynomial. Let $R \to B$ be a ring map. If $h \in B[x]/(f)$ is integral over $R$, then the element $f' h$ can be written as $f'h = \sum _ i b_ i x^ i$ with $b_ i \in B$ integral over $R$.

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 10.136.9. 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

$f' h \equiv \sum \nolimits _{i = 1, \ldots , d} h(\alpha _ i) (x - \alpha _1) \ldots \widehat{(x - \alpha _ i)} \ldots (x - \alpha _ d)$

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

$h(\alpha _ i)^ e + r_1 h(\alpha _ i)^{e - 1} + \ldots + r_ e = 0$

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. $\square$

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 $B[x]_ g/(f)$ (by definition of a standard étale ring map) we conclude that $a \in S \otimes _ R A$ as desired. $\square$

Example 10.147.3. Let $p$ be a prime number. The ring extension

$R = \mathbf{Z}[1/p] \subset R' = \mathbf{Z}[1/p][x]/(x^{p - 1} + \ldots + x + 1)$

has the following property: For $d < p$ there exist elements $\alpha _0, \ldots , \alpha _{d - 1} \in R'$ such that

$\prod \nolimits _{0 \leq i < j < d} (\alpha _ i - \alpha _ j)$

is a unit in $R'$. Namely, take $\alpha _ i$ equal to the class of $x^ i$ in $R'$ for $i = 0, \ldots , p - 1$. Then we have

$T^ p - 1 = \prod \nolimits _{i = 0, \ldots , p - 1} (T - \alpha _ i)$

in $R'[T]$. Namely, the ring $\mathbf{Q}[x]/(x^{p - 1} + \ldots + x + 1)$ is a field because the cyclotomic polynomial $x^{p - 1} + \ldots + x + 1$ is irreducible over $\mathbf{Q}$ and the $\alpha _ i$ are pairwise distinct roots of $T^ p - 1$, whence the equality. Taking derivatives on both sides and substituting $T = \alpha _ i$ we obtain

$p \alpha _ i^{p - 1} = (\alpha _ i - \alpha _1) \ldots \widehat{(\alpha _ i - \alpha _ i)} \ldots (\alpha _ i - \alpha _1)$

and we see this is invertible in $R'$.

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. $\square$

Lemma 10.147.5. Let $R \to S$ and $R \to B$ be ring maps. 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 $S \otimes _ R B$. If $S$ is a filtered colimit of smooth $R$-algebras, then the canonical map $S \otimes _ R A \to A'$ is an isomorphism.

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

Comment #6376 by Ola Sande on

In lemma 03GD:

What is f´?

Comment #6446 by on

It is the derivative of $f$ with respect to $x$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).