TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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Graphs of morphisms to a projective classical variety are closed

Statement

Let $f : X \to Y$ be a classical variety morphism for which a classical product $P = X \times Y$ has been constructed, and suppose $Y$ is projective. Define $Γ_{f}$ as the image of the product morphism $⟨ {id}_{X}, f ⟩ : X \to P$ . Then $Γ_{f}$ is closed in $P$ . No assertion is made for unconstructed products or arbitrary nonseparated prevarieties.

Facts & Assumptions

Given: $f : X \to Y$ and a constructed product $X \times Y$ , with $Y$ projective.

Proof

1.1

The diagonal $Δ_{Y} \subseteq Y \times Y$ is closed: embed $Y$ projectively and use the Segre/projective coordinate model, where equality of two projective points is cut out on the standard charts by coordinate differences.

given

2.1

Let $p_{X} : P \to X$ and $p_{Y} : P \to Y$ be the structure maps. The morphism $F = ⟨ f \circ p_{X}, p_{Y} ⟩ : P ⟶ Y \times Y$ is defined by the two product universal properties, without choosing a pair-set realization of $P$ . Its inverse image of $Δ_{Y}$ is exactly the image of $⟨ {id}_{X}, f ⟩$ .

step 1.1

3.1

Since inverse images of closed algebraic sets under a morphism are closed, $F^{- 1} (Δ_{Y}) = Γ_{f}$ is closed. This is precisely Milne's projective-target graph criterion under the stated construction hypothesis.

step 2.1∎

Depends on

Used by

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Dependency tree · two levels

8 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, Corollaries 5.27 and 5.28 (standard reference, not scraped)
MIT 18.725 Algebraic Geometry, Lecture 7, separatedness (standard reference, not scraped)