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Normalization of an affine integral curve

PDF p. 13: Normalization of the support curve

Statement

Assume the Axiom of Choice. Let k be a perfect field and T=Spec⁡A an integral affine curve of finite type over k: thus A is a finitely generated k-algebra which is a domain of Krull dimension one. Let K=Frac⁡(A) and let B be the set of elements of K integral over A, meaning roots of monic polynomials with coefficients in A. The normalization of T is T~=Spec⁡B, with its morphism ν:T~→T induced by inclusion. Then T~ is affine and smooth over k, and ν is finite. If k is algebraically closed, every closed point of T and T~ has residue field k, and the set of points in the fiber over each closed point of T is nonempty and finite; each induced residue-field map in such a fiber is an isomorphism.

Facts & Assumptions

Given: AC, a perfect field k, and the domain A and normalization ring B in the statement.

Goal: Establish finiteness and smoothness of the normalization, then the asserted properties of its closed fibers when k is algebraically closed.

[F1]

The integral closure of a finite-type domain over a field in its fraction field is a finite module over that domain (A finite-type domain over a field has finite normalization). Integral elements form a subring, and integrality is transitive (Integral elements over a nonzero base ring form a subring, Integral extensions are transitive).

[F2]

Assuming AC, an injective integral extension of nonzero rings preserves Krull dimension (Injective integral extensions preserve Krull dimension). A morphism with an affine cover of its target whose inverse images are affine with module-finite coordinate rings is finite (Finite is affine and local on its target).

[F3]

Assuming AC, an integral separated finite-type one-dimensional normal scheme over a perfect field is smooth over that field. Normal means that each local ring is an integrally closed domain (A normal curve over a perfect field is nonsingular).

[F4]

Assuming AC, a prime of a ring lifts to a prime of any integral extension containing it; for an integral extension, an upstairs prime is maximal exactly when its contraction is maximal (Lying over for integral ring maps, Under an integral extension, a prime is maximal if and only if its contraction is maximal).

[F5]

Over an algebraically closed field, every maximal ideal of a polynomial ring in finitely many variables is an evaluation ideal, and its quotient is the base field (Over an algebraically closed field, every maximal ideal is an evaluation ideal). For finitely many pairwise comaximal ideals, the map to the product of their quotients is surjective (Chinese remainder theorem for pairwise comaximal ideals).

[A1]

AC is the assertion that every family of nonempty sets has a choice function; it is used through the cited dimension, finite-morphism, normal-curve and lying-over results (The Axiom of Choice).

Proof

1.1F1F2A1given

By [F1], B is a subring of K containing A and a finite A-module. It is therefore a domain, and its fraction field is K: it lies inside K and contains A. It is also a finite-type k-algebra, since a finite set of A-module generators generates it as an A-algebra and A is finite type over k. The inclusion A⊆B is integral by the definition of B. Its spectrum is affine by construction, and the one-affine-cover criterion in [F2] makes ν finite.

2.1F1step 1.1algebra

The ring B is integrally closed in K: if x∈K is integral over B, transitivity in [F1] makes x integral over A, so x∈B. For any prime q of B, the localization Bq is integrally closed in K. Indeed, if x∈K satisfies a monic equation over Bq, choose s∈B∖q clearing all coefficient denominators. Multiplying the equation of degree n by sn shows that sx satisfies a monic equation over B; hence sx∈B and x∈Bq. Thus Spec⁡B is normal.

3.1F2F3A1givenstep 1.1step 2.1

By [F2], dim⁡B=dim⁡A=1. The scheme Spec⁡B is integral, separated and finite type over k by step 1.1, and normal by step 2.1. The perfect-field theorem [F3] therefore makes it smooth over k.

4.1F5givenstep 1.1step 3.1

Suppose k is algebraically closed. For either R=A or R=B, write R as a quotient of a finite polynomial algebra over k. A maximal ideal of R pulls back to a maximal ideal of that polynomial algebra, whose quotient is k by [F5]. Consequently every closed point of Spec⁡R has residue field k.

5.1F4F5A1step 1.1step 4.1algebra∎

Let m be a maximal ideal of A. By [F4], there is a prime of B over m, and every such prime is maximal. The quotient C=B/mB is a finite-dimensional vector space over A/m=k, because B is a finite A-module. Its prime ideals are exactly the primes of B over m, and they are maximal. If n1,…,nr are distinct such ideals in C, they are pairwise comaximal; [F5] gives a surjection C→∏iC/ni. Each factor is a nonzero k-vector space, so r≤dim⁡kC. Thus there can be only finitely many points in the fiber, and it is nonempty. By step 4.1 both residue fields at each such point are k, and their map is a k-algebra map, hence the identity under these identifications. This proves all assertions.

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