Alphabeta Math
LemmaStatement: AI-adaptedProof: 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.

Independence of the chosen algebraic closure

Statement

Assume the Axiom of Choice. For two algebraic closures Ω1,Ω2 of κ(s), a chosen κ(s)-isomorphism σ:Ω1→Ω2 identifies the two geometric fibres after transport of scalars. In particular their isomorphism-invariant properties agree. No canonical choice of this identification is asserted.

Facts & Assumptions

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

[F1]

A geometric point of a scheme S is a morphism Spec⁡Ω→S with Ω algebraically closed. For a specified point s∈S, choose an algebraic closure Ω/κ(s), in the sense of def-algebraic-closure. In this page the geometric fibre at s means Xsˉ=Xs×Spec⁡κ(s)Spec⁡Ω. Here Xs is def-scheme-theoretic-fibre, and its affine charts extend as in lem-base-extension-field-coordinate-ring. The choice includes the embedding of κ(s); no preferred algebraic closure or preferred isomorphism between choices is implied. (Geometric fibres and geometric points)

[F2]

Assuming the Axiom of Choice, any two algebraic closures of a field F are F-isomorphic. No uniqueness of the isomorphism is asserted. (Assuming Choice, any two algebraic closures are base-isomorphic)

[F3]

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)

Proof

1.1givenF1F2

F2 supplies a base-field isomorphism σ:Ω1→Ω2 under Choice. By F1 the two fibres are Xs×κ(s)Spec⁡Ωi.

2.1F3step 1.1∎

Base change the first fibre along σ and apply F3 to identify it with the second fibre. On affine charts the ring isomorphism is a⊗λ↦a⊗σ(λ), whose inverse uses σ−1. Empty charts and the identity choice satisfy the same formulas. Thus the schemes are isomorphic after scalar transport, and all isomorphism-invariant properties agree, independently of the chosen isomorphism.

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