Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

Intersections of affine opens admit principal affine covers

Statement

If U,V are affine open subschemes of a scheme X, then U∩V is covered by open subschemes which are principal opens in U and are also principal opens in affine open charts of V.

Facts & Assumptions

Given: Affine open subschemes U,V⊆X.

[L1]

Distinguished opens form a basis in an affine scheme, and sections on D(f)⊆Spec⁡A are Af (A principal localization identifies its spectrum with a distinguished open, Sections and restrictions on distinguished opens of an affine scheme).

Proof

technique · direct
1.1L1givenchoose

Fix x∈U∩V, and write U=Spec⁡A and V=Spec⁡B. The basis property in [L1] gives f∈A and g∈B such that x∈DU(f)⊆V and x∈DV(g)⊆U. Put W=DU(f)∩DV(g); it is an open neighbourhood of x.

2.1L1step 1.1

The function g on V restricts to a section on DU(f). By [L1], write that restriction as a/fn for some a∈A and n≥0. On DU(f), its nonvanishing locus is DU(a)∩DU(f)=DU(fa). Hence W=DU(fa) is principal in U.

2.2L1step 1.1

Symmetrically, the restriction of f to DV(g) is a section b/gm for some b∈B and m≥0. Its nonvanishing locus there is DV(gb), so the same open W is principal in V.

3.1step 2.1step 2.2∎

Every x∈U∩V has such a common principal neighbourhood W. These neighbourhoods cover the intersection and prove the claim.

Depends on

Used by

Dependency tree · two levels

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Sources