Lemma 10.143.1. 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 $R$.
10.143 Integral closure and smooth base change
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 10.134.9. 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 10.143.2. 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 10.35.11. Hence we may localize on $S$, by Lemma 10.22.2 (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 10.141.16. 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 10.143.1 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 10.143.3. Let $p$ be a prime number. The ring extension has the following property: For $d < p$ there exist elements $\alpha _0, \ldots , \alpha _{d - 1} \in R'$ such that 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 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 and we see this is invertible in $R'$.
\[ R = \mathbf{Z}[1/p] \subset R' = \mathbf{Z}[1/p][x]/(x^{p - 1} + \ldots + x + 1) \]
\[ \prod \nolimits _{0 \leq i < j < d} (\alpha _ i - \alpha _ j) \]
\[ T^ p - 1 = \prod \nolimits _{i = 0, \ldots , p - 1} (T - \alpha _ i) \]
\[ p \alpha _ i^{p - 1} = (\alpha _ i - \alpha _1) \ldots \widehat{(\alpha _ i - \alpha _ i)} \ldots (\alpha _ i - \alpha _1) \]
Lemma 10.143.4. 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 10.143.2 we may localize on $S$. Hence we may assume that $R \to S$ is a standard smooth ring map, see Lemma 10.135.10. 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 10.143.2) 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 10.35.7) 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 10.143.3. 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 10.143.5. 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 10.143.4. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)