Alphabeta Math
ExampleConstruction: AI-generatedVerification: 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.

The fibres of xy=t

Example

Over any field k, consider X=Spec⁡k[x,y,t]/(xy−t)→Spec⁡k[t]. At t=a∈k the fibre is Spec⁡k[x,y]/(xy−a). At zero it is the union of two reduced axes; at a≠0 it is Gm,k=Spec⁡k[x,x−1]. The generic fibre is Spec⁡k(t)[x,x−1].

Facts & Assumptions

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

[F1]

Let A→B be a ring map, p∈Spec⁡A, and M=A∖p, acting on B through the ring map. The fibre over p is canonically Spec⁡(B⊗Aκ(p))≅Spec⁡(M−1B/pM−1B). The residue field is κ(p)=Ap/pAp. No reduction of the tensor ring is taken. (Coordinate ring of an affine fibre)

[F2]

Let A→C be a unital ring map. For any set of variables (ti) and any ideal I⊆A[ti], (A[ti]/I)⊗AC≅C[ti]/IC[ti]. Here the extended ideal is generated by the coefficient images of all elements of I. For a multiplicative subset M⊆A, (M−1A)⊗AC≅M‾−1C. These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1givenF1F2

F1 computes a fibre by tensoring with its residue field. F2 substitutes t=a to give k[x,y]/(xy−a), and at the generic point it extends coefficients to k(t) with relation xy=t.

2.1step 1.1algebra

At a=0, (xy)=(x)∩(y): a polynomial divisible by both x and y has every monomial divisible by xy. The ideals (x),(y) are prime, so their intersection is radical and the quotient is reduced. Every prime containing xy contains x or y; these two incomparable minimal primes give exactly the two axes, meeting at (x,y).

3.1step 1.1algebra∎

At a≠0, the inverse maps send x↦x, y↦ax−1 in one direction and x−1↦a−1y in the other. They identify the ring with k[x,x−1]. The same formulas with the nonzero unit t∈k(t) give the generic fibre. In particular a=1 has the same form; all three rings are nonzero, so none of these fibres is empty.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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