Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A dominant affine map factors finitely over relative affine space after shrinking the base

Statement

Let f:X→Y be dominant between irreducible affine varieties, put A=k[Y]⊆B=k[X], and let r=trdeg⁡k(Y)k(X). There are 0≠a∈A and elements t1,…,tr∈Ba, algebraically independent over Aa, such that Ba is module-finite over Aa[t1,…,tr].

Work over a fixed algebraically closed field k, with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[F1]

For irreducible classical X, the fraction fields of all nonempty affine charts identify canonically; denote the resulting field by k(X). A dominant morphism f:X→Y between irreducible classical varieties induces an injection f∗:k(Y)↪k(X). Dominant means that the image is dense. Work over a fixed algebraically closed field k, with the Axiom of Choice. Classical varieties are separated and admit finite affine covers; they may be reducible or empty unless irreducibility is specified. Irreducible means nonempty. All fibres and points below are classical closed-point fibres and points. (Function fields and dominant pullbacks on general varieties).

[F2]

Let k be a field and let A be a nonzero finite-type k-algebra. Then there exist algebraically independent elements z1,…,zd∈A such that A is a module-finite algebra over the polynomial ring k[z1,…,zd]. (Noether normalisation yields module finiteness over a polynomial subring).

[F3]

Assume the Axiom of Choice. Let X be a classical affine variety over an algebraically closed field k, and let f‾∈k[X]. Put U=DX(f‾). A function φ:U→k is called regular on U if there exist finitely many pairs (g‾i,h‾i) in k[X]×k[X] such that U=⋃i=1rDX(h‾i) and φ(x)=g‾i(x)h‾i(x)for every x∈DX(h‾i). Write OX(U) for the ring of regular functions on U. Then evaluation induces a ring isomorphism k[X]f‾→∼OX(U). If U=∅, both sides are the zero ring. (Regular functions on a principal open are the principal localization of the coordinate ring).

Proof

1.1F1

Dominance makes A→B injective and identifies their fraction fields with k(Y)⊆k(X). Put K=Frac⁡A. The localization BK=B⊗AK is a nonzero finite-type K-domain inside Frac⁡B, with that same fraction field.

2.1F2step 1.1

Apply normalization over the field K to obtain algebraically independent t1,…,tr∈BK over which BK is module-finite. Their number is r because the fraction field is algebraic over the fraction field of the normalization polynomial ring. Choose finite A-algebra generators b1,…,bs of B. Each satisfies a monic equation over K[t1,…,tr].

3.1F3step 2.1∎

Every ti is a fraction with numerator in B and nonzero denominator in A. Invert the product a of these denominators and all denominators in the finitely many monic-equation coefficients. The product is nonzero because A is a domain, and an empty product is 1. Now ti∈Ba, and the same equations are monic over Aa[t1,…,tr]. Independence descends from K. If the equation degrees are dj, the finitely many monomials ∏jbjej with 0≤ej<dj span Ba over this polynomial subring by repeated monic reduction. The principal-open supplier identifies the localized rings with the corresponding open-chart rings. This works also for r=0.

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