LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-09

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Affine-source morphisms agreeing on a dense open agree everywhere

Statement

Assume the Axiom of Choice, inherited from the Nullstellensatz route. If $X, Y$ are affine varieties and two morphisms $ϕ, ψ : X \to Y$ agree on a dense open subset of $X$ , they agree everywhere.

Facts & Assumptions

Given: AC, affine varieties $X, Y$ over algebraically closed $k$ , and morphisms $ϕ, ψ : X \to Y$ agreeing on a dense open of $X$ .

[F1]

Global regular functions on X are elements of its coordinate ring (Global regular functions on a classical affine variety are its coordinate ring).

[F2]

Coordinate functions on Y pull back to regular functions (A morphism from an open subset of a classical affine variety to an affine variety).

[F3]

Zero sets of regular functions on an open source are relatively closed (A classical morphism pulls Zariski closed sets back to closed sets).

Proof

technique · direct

1.1

Embed $Y \subseteq k^{m}$ and write $y_{j}$ for its coordinates. The functions $r_{j} = y_{j} \circ ϕ - y_{j} \circ ψ$ are global regular functions by F1 and F2. Their zero sets are closed by F3. The common dense open is contained in every such zero set, hence each zero set is all of $X$ .

F1F2F3given

2.1

Thus for every $x \in X$ and every $j$ , $y_{j} (ϕ (x)) = y_{j} (ψ (x))$ . Equality of all coordinates is equality of the tuples, so $ϕ (x) = ψ (x)$ . For $m = 0$ the target has just the empty tuple and this conclusion is immediate as well.

step 1.1∎

Sources

Source comparison: Milne, Algebraic Geometry, v6.10, Proposition 3.26 and the separatedness calculation of §5c, pp. 67, 102. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.

Depends on

Used by

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Dependency tree · two levels

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Sources

J. S. Milne, Algebraic Geometry v6.10, Proposition 3.26 and the separatedness calculation of §5c, pp. 67, 102 (standard reference, not scraped)