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.

Morphisms defined on an open source and agreeing on a dense open agree on their common domain

Statement

Let $U, V$ be nonempty opens of an affine variety $X$ , and let $ϕ : U \to Y$ , $ψ : V \to Y$ be morphisms to an affine variety. If they agree on a nonempty open subset of $U \cap V$ , they agree on all of $U \cap V$ . More generally agreement on any subset dense in their common domain suffices.

Facts & Assumptions

Given: Affine varieties $X, Y$ over algebraically closed $k$ , nonempty opens $U, V \subseteq X$ , and morphisms $ϕ : U \to Y$ , $ψ : V \to Y$ agreeing on a dense subset of $U \cap V$ or on a nonempty common open.

[F1]

A nonempty open of an irreducible variety is dense, and finite such intersections are nonempty (Every nonempty open of a classical affine variety is dense).

[F2]

Target coordinates pull back to regular functions (A morphism from an open subset of a classical affine variety to an affine variety).

[F3]

A regular function on an open source has closed zero set (A classical morphism pulls Zariski closed sets back to closed sets).

Proof

technique · direct

1.1

Set $W = U \cap V$ . For target coordinates $y_{1}, \dots, y_{m}$ , each difference $r_{j} = y_{j} \circ ϕ - y_{j} \circ ψ$ restricted to W is regular, so $E = ⋂_{j} {r_{j} = 0}$ is closed in W. Since points of $Y \subseteq k^{m}$ are determined by their coordinates, E is exactly the equalizer.

F2F3given

2.1

If the maps agree on a subset dense in W, its containing closed set E is all W. In particular any nonempty open of W is dense there by F1 (or by intersecting nonempty opens of X), so the nonempty-open hypothesis suffices. If $m = 0$ , the intersection defining E is the whole W and the same conclusion holds.

F1step 1.1∎

Sources

Source comparison: Milne, Algebraic Geometry, v6.10, Lemma 5.6 and Proposition 5.8, pp. 102–103. Conventions here distinguish arbitrary affine algebraic sets from nonempty irreducible varieties.

Depends on

Used by

Dependency tree · two levels

7 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 v6.10, Lemma 5.6 and Proposition 5.8, pp. 102–103 (standard reference, not scraped)