Alphabeta Math
LemmaStatement: 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.

Fibres after base change

Statement

Let S′→S send s′ to s. For any X→S there is a canonical isomorphism of κ(s′)-schemes (XS′)s′≅Xs×Spec⁡κ(s)Spec⁡κ(s′).

Facts & Assumptions

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

[F1]

For a morphism f:X→S and any point s∈S, its scheme-theoretic fibre is Xs=X×SSpec⁡κ(s), viewed as a κ(s)-scheme. The map Spec⁡κ(s)→S is the canonical residue-field point from lem-field-valued-points-of-schemes, and the product is base change as in def-base-change-morphism-schemes. The point s need not be closed. A fibre over a generic point is called a generic fibre. Empty fibres are allowed. (Scheme-theoretic fibre)

[F2]

For S′′→kS′→hS and an S-scheme X, there is a canonical isomorphism (X×SS′)×S′S′′≅X×SS′′. It is functorial in X and compatible with the induced maps of S-schemes. (Iterated base change)

[F3]

For every field K and scheme X, morphisms Spec⁡K→X correspond bijectively to pairs (x,ι) with x∈X and a field embedding ι:κ(x)→K. The identity embedding gives a canonical morphism Spec⁡κ(x)→X, compatible with all scheme morphisms. More generally, for a nonzero local ring (R,m), morphisms Spec⁡R→X correspond to pairs (x,φ) with a local homomorphism φ:OX,x→R. Assuming Choice, two field-valued points have the same image in X if and only if they are dominated by a common field-valued point, by compatible embeddings of their fields into a third field. (Field-valued points and local-ring points)

Proof

1.1givenF1F2F3

By F3 the canonical point Spec⁡κ(s′)→S factors through Spec⁡κ(s) via the residue-field embedding. F1 and F2 identify the left side with X×SSpec⁡κ(s′).

2.1F2step 1.1∎

Apply F2 once more to the factorization through Spec⁡κ(s). It gives exactly the displayed right side, with the same projection to Spec⁡κ(s′). Both operations are canonical on test morphisms and are inverse regroupings; no closure, finite extension or flatness assumption is needed. If Xs is empty both sides represent only empty test schemes; identity residue-field extension gives Xs.

Depends on

Used by

Dependency tree · two levels

10 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