CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

A dominant image contains a dense open

Statement

If $f : X \to Y$ is a dominant morphism of classical varieties and $Y$ is irreducible, then $f (X)$ contains a nonempty open subset of $Y$ . The source need not be irreducible.

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]

Every morphism $f : X \to Y$ of classical varieties sends every constructible subset of $X$ to a constructible subset of $Y$ . In particular $f (X)$ is constructible. 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. (Chevalley: images of constructible sets are constructible).

[F2]

If a constructible subset $C \subseteq X$ has nonempty irreducible closure $Z$ , then $C$ contains a nonempty open subset of $Z$ . 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. (Dense constructible subsets contain an open).

Proof

1.1F1

Chevalley makes $f (X)$ constructible, and dominance says its closure is $Y$ .

2.1F2step 1.1∎

The closure is nonempty and irreducible, so the dense-constructible lemma supplies the claimed nonempty open. If $Y$ has one point, dominance forces that point into the image and the open is $Y$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

Milne Theorem 9.1 and Proposition 9.6 (standard reference, not scraped)