LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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.

Separatedness survives base change

Statement

Let $f : X \to S$ be a separated morphism and let $g : S^{'} \to S$ be any morphism. Then the base change $f^{'} : X \times_{S} S^{'} \to S^{'}$ is separated. No quasi-compactness, Noetherian or finite-type hypothesis is used, and $S^{'}$ or $X$ may be empty.

Facts & Assumptions

Given: A separated morphism $f : X \to S$ , an arbitrary morphism $g : S^{'} \to S$ , and the base change $X^{'} = X \times_{S} S^{'}$ with structure morphism $f^{'} : X^{'} \to S^{'}$ .

[F1]

A morphism $f : X \to S$ is separated when $Δ_{X / S}$ is a closed immersion. (Separated morphism of schemes)

[F2]

For $g : S^{'} \to S$ and $X \to S$ there is a canonical isomorphism $X^{'} \times_{S^{'}} X^{'} ≅ (X \times_{S} X) \times_{S} S^{'}$ ; under it $Δ_{X^{'} / S^{'}}$ is the base change of $Δ_{X / S}$ , and the square with the two diagonals and the projections to $X$ and $X \times_{S} X$ is Cartesian. (The diagonal commutes with base change)

[F3]

Closed immersions remain closed immersions after arbitrary base change; this holds with no flatness or finiteness hypothesis and includes the empty and zero-ring cases. (Base change of immersions)

[F4]

The base change $X_{S^{'}} = X \times_{S} S^{'}$ carries the second projection as structure morphism, and the base change of an $S$ -morphism is defined by its two projections. (Base change of objects, morphisms and properties)

Proof

technique · direct

1.1

The morphism $f^{'}$ is $X^{'} \to S^{'}$ with $X^{'} = X \times_{S} S^{'}$ as in [F4], and [F2] supplies the canonical isomorphism $X^{'} \times_{S^{'}} X^{'} ≅ (X \times_{S} X) \times_{S} S^{'}$ together with the Cartesian square comparing the two diagonals over $X$ and $X \times_{S} X$ .

F2F4given

1.2

Since $f$ is separated, [F1] says that $Δ_{X / S}$ is a closed immersion.

F1given

2.1

Under the identification of step 1.1 the diagonal $Δ_{X^{'} / S^{'}} : X^{'} \to X^{'} \times_{S^{'}} X^{'}$ is the base change of $Δ_{X / S}$ along the projection $(X \times_{S} X) \times_{S} S^{'} \to X \times_{S} X$ , by the second assertion of [F2].

F2step 1.1

3.1

By [F3] the base change of the closed immersion $Δ_{X / S}$ is again a closed immersion; with step 2.1 this exhibits $Δ_{X^{'} / S^{'}}$ as a closed immersion, and the empty or zero-ring case is covered by the same statement.

F3step 2.1step 1.2

4.1

By [F1] the morphism $f^{'} : X^{'} \to S^{'}$ is separated, which is the assertion.

F1step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

The Stacks Project, Schemes, Lemma 26.21.12, printed p.41 (standard reference, not scraped)
Vakil, The Rising Sea, Section 11.3.3, printed p.308 (standard reference, not scraped)