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.

Stalks of the scheme-theoretic fibre

Statement

For f:X→S and x∈X with s=f(x), use the corresponding point of Xs. There are canonical local-ring isomorphisms OXs,x≅OX,x/msOX,x≅OX,x⊗OS,sκ(s).

Facts & Assumptions

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

[F1]

For f:X→S and s∈S, the projection Xs→X is a homeomorphism onto f−1(s) with the subspace topology and preserves the residue field at every point. In compatible affine charts A→B, s=p, its points correspond exactly to primes q⊂B contracting to p; no extra embedding choice occurs. Also X×SSpec⁡OS,s→X is a homeomorphism onto the inverse image of the set of generalizations of s. (Points and topology of a fibre)

[F2]

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)

[F3]

For p∈Spec⁡A, there is a canonical isomorphism OSpec⁡A,p≅Ap. (The stalk of the affine structure sheaf at a prime is A_p)

Proof

1.1givenF1F2

Choose compatible affine neighbourhoods A→B, with primes p,q representing s,x. F1 identifies the relevant fibre point, and F2 gives the ring (A∖p)−1B/p(A∖p)−1B.

2.1F3step 1.1

Localize at that point, using F3. All elements of A∖p are already units in Bq, so the result is Bq/pBq. Under the stalk identifications, the extended ideal is exactly msOX,x.

3.1step 2.1algebra∎

For any ring map R→H and ideal J, the maps h⊗r‾↦hr mod JH and h mod JH↦h⊗1 are well-defined inverse ring maps H⊗RR/J≅H/JH. Apply this with R=OS,s, H=OX,x and J=ms. The ideal is contained in mx, so this stalk is nonzero; if there is no point over s, the assertion has no instance. Nilpotents are not removed, and J=0 gives the unchanged stalk.

Depends on

Used by

Dependency tree · two levels

14 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