LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 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.

Base change and composition of affine morphisms

Statement

Arbitrary base change preserves affine morphisms. Composites of affine morphisms are affine, and every closed immersion is affine.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

A morphism of schemes $f : X \to S$ is affine when $f^{- 1} (U)$ is affine for every affine open subscheme $U \subseteq S$ . Here the inverse image carries the restricted structure sheaf, as in def-affine-open-subscheme, and $f$ is a morphism of locally ringed spaces as in def-morphism-of-schemes. The empty scheme is affine, being $Spec 0$ . Affineness of a morphism does not require its total source or target to be affine. (Affine morphisms)

[F2]

A morphism $f : X \to S$ is affine if and only if there exists an affine open cover $S = ⋃_{i} U_{i}$ for which every $f^{- 1} (U_{i})$ is affine. (Affineness is local on the target)

[F3]

Let $A \to B$ and $A \to C$ be maps of commutative unital rings, allowing the zero ring. In the category of all schemes, $Spec B \times_{Spec A} Spec C ≅ Spec (B \otimes_{A} C) .$ The projections correspond to $b \mapsto b \otimes 1$ and $c \mapsto 1 \otimes c$ . (Affine fibre products are spectra of tensor products)

[F4]

Suppose $P = X \times_{S} Y$ exists, with projections $p, q$ . If opens $V \subseteq X$ , $W \subseteq Y$ map into an open $U \subseteq S$ , then the open subscheme $Q = p^{- 1} (V) \cap q^{- 1} (W)$ represents $V \times_{U} W$ , and also $V \times_{S} W$ . Independently, for $f : X \to S$ and an open $U \subseteq S$ , the open subscheme $f^{- 1} (U)$ represents $X \times_{S} U$ . (Restricting fibre products to open subschemes)

[F5]

For a ring $A$ , closed immersions $Z \to Spec A$ are, up to unique isomorphism over $Spec A$ , precisely the morphisms $Spec (A / I) \to Spec A$ for ideals $I \subseteq A$ . (Closed immersions into affine schemes are quotient spectra)

[F6]

For $f : X \to S$ , the following are equivalent: $f$ is quasi-compact; the inverse image of every affine open in $S$ is quasi-compact; some affine open cover of $S$ has quasi-compact inverse images. Moreover any arbitrary base change of a quasi-compact morphism is quasi-compact. (Quasi-compactness is local on the target and survives base change)

Proof

1.1

Let $f : X \to S$ be affine and $S^{'} \to S$ arbitrary. Around each point of $S^{'}$ choose an affine $V^{'}$ mapping into an affine $U \subset S$ . By F1, $f^{- 1} (U)$ is affine. F4 identifies the inverse image of $V^{'}$ with $f^{- 1} (U) \times_{U} V^{'}$ , which is affine by F3.

givenF1F3F4

2.1

These $V^{'}$ cover the new base. Apply F2 to conclude that $f_{S^{'}}$ is affine. The proof includes empty charts and zero tensor rings.

F2step 1.1

3.1

For affine $X \to Y \to Z$ and affine open $U \subset Z$ , the successive inverse images are affine by F1, hence the composite is affine. For a closed immersion, its restriction over any affine open $Spec A$ is $Spec (A / I)$ by F5, again affine by F1. The cases $I = 0, (1)$ are the identity and empty closed immersion. Affine inverse images are quasi-compact, so the target-local criterion F6 also proves every affine morphism, and in particular every closed immersion, quasi-compact.

F1F5F6∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

Stacks 29.11.8–10 (standard reference, not scraped)