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The normalization is an isomorphism over the normal locus

Statement

Assume the Axiom of Choice. Let $X$ be an irreducible affine variety with normalization $ν ⁣ : X^{ν} \to X$ . The normal locus of $X$ is a dense open subset, and at every normal point $x$ there is a principal open $D (f) ∋ x$ with $A_{f}$ integrally closed; on $ν^{- 1} (D (f))$ the map $ν$ restricts to an isomorphism onto $D (f)$ . Consequently $ν$ is an isomorphism over the normal locus.

Facts & Assumptions

Given: AC, the algebraically closed field $k$ , the irreducible affine variety $X$ with coordinate ring $A = k [X]$ and function field $k (X)$ , the integral closure $B$ of $A$ in $k (X)$ , the normalization $ν ⁣ : X^{ν} \to X$ with pullback $A ↪ B$ , and a point $x \in X$ with maximal ideal $m = m_{x} \subseteq A$ .

[F1]

$B$ is a finite $A$ -module, $A \subseteq B \subseteq k (X)$ , $B$ is an integrally closed domain, and $k [X^{ν}] = B$ (The normalization of an irreducible affine variety).

[F2]

The point $x$ is normal exactly when $O_{X, x} ≅ A_{m}$ is integrally closed, and $A$ is integrally closed exactly when all its maximal localisations are (Normality is checked on affine open charts, The local ring at a point of an affine variety is the localization at its maximal ideal, A domain is integrally closed if and only if its prime localisations are, equivalently if and only if its maximal localisations are).

[F3]

For a finite $A$ -module $M$ , $M_{m} = 0$ if and only if there is $f \notin m$ with $f M = 0$ ; equivalently $Supp (M)$ is closed and $M_{p} \neq 0$ exactly for primes containing the annihilator (For a finite module, support is the set of primes containing the annihilator).

[F4]

For $0 \neq f \in A$ , the principal open $D (f)$ is affine with coordinate ring $A_{f}$ , and the integral closure of $A_{f}$ in $k (X)$ is $B_{f}$ , a finite $A_{f}$ -module; principal opens form a basis of the topology, and $ν^{- 1} (D (f))$ is an affine open with coordinate ring $B_{f}$ under the pullback (Regular functions on a principal open are the principal localization of the coordinate ring, Principal opens form a basis for the Zariski topology on an affine variety, Finite normalization commutes with principal localization, The normalization of an irreducible affine variety).

Proof

1.1F1F2F3F4given

Let $x$ be a normal point. Then $A_{m}$ is an integrally closed domain by [F2]. Since $B$ is a finite, hence integral, $A$ -module, $B_{m}$ is integral over $A_{m}$ and lies in the common fraction field $k (X)$ ; an integrally closed domain contains every element of its fraction field integral over it, so $B_{m} \subseteq A_{m}$ and therefore $(B / A)_{m} = 0$ . By [F3] applied to the finite module $B / A$ there is $f \notin m$ with $f B \subseteq A$ . Then $B_{f} = A_{f}$ : the inclusion $A_{f} \subseteq B_{f}$ is clear, and every $b / f^{n} \in B_{f}$ equals $f^{k} b / f^{n + k}$ with $f^{k} b \in A$ . In particular $A_{f} = B_{f}$ is integrally closed (it is the integral closure of $A_{f}$ by [F4], and that closure is $B_{f}$ ), and by [F4] the map $ν$ restricts over $D (f)$ to the morphism of affine varieties with pullback $A_{f} \tilde{\to} B_{f}$ ; by the anti-equivalence that restriction is an isomorphism $ν^{- 1} (D (f)) \tilde{\to} D (f)$ .

1.2F1F2F3given

The normal locus of $X$ is the set of points $x$ with $A_{m_{x}} = B_{m_{x}}$ , equivalently the complement of $Supp (B / A)$ [F2, F3]. Since $B / A$ is a finite $A$ -module, its support is closed [F3], so the normal locus is open. It is nonempty: write finitely many $A$ -module generators of $B$ as fractions in $Frac (A)$ and multiply their nonzero denominators to obtain $0 \neq d \in A$ with $d B \subseteq A$ . Then $B_{d} = A_{d}$ , so every point of the nonempty principal open $D (d)$ is normal. Since $X$ is irreducible, a proper closed subset has empty interior, so this nonempty open set is dense. Hence the normal locus is a dense open subset of $X$ .

2.1F4step 1.1step 1.2∎

By step 1.1, at each normal point $x$ there is a principal open $D (f) ∋ x$ with $A_{f}$ integrally closed on which $ν$ restricts to an isomorphism; by step 1.2 the normal locus is dense open and is covered by those principal opens. Hence the normalization restricts to an isomorphism over the normal locus, and in particular over some principal open neighbourhood of each normal point.

Depends on

Used by

The conductor of a normalization Definition
Unibranch points of a classical variety Definition

Dependency tree · two levels

52 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Geometry (2025 version), Ch. 8 §a-b: the normalization is an isomorphism over the normal locus (standard reference, not scraped)
Ravi Vakil, The Rising Sea: Foundations of Algebraic Geometry (November 18, 2017 public draft), §9.7 (standard reference, not scraped)