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 $X = Spec k [x, y, t] / (x y - t) \to Spec k [t]$ . At $t = a \in k$ the fibre is $Spec k [x, y] / (x y - a)$ . At zero it is the union of two reduced axes; at $a \neq 0$ it is $G_{m, k} = Spec k [x, x^{- 1}]$ . The generic fibre is $Spec k (t) [x, x^{- 1}]$ .

Facts & Assumptions

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

[F1]

Let $A \to B$ be a ring map, $p \in Spec A$ , and $M = A ∖ p$ , acting on $B$ through the ring map. The fibre over $p$ is canonically $Spec (B \otimes_{A} κ (p)) ≅ Spec (M^{- 1} B / p M^{- 1} B) .$ The residue field is $κ (p) = A_{p} / p A_{p}$ . No reduction of the tensor ring is taken. (Coordinate ring of an affine fibre)

[F2]

Let $A \to C$ be a unital ring map. For any set of variables $(t_{i})$ and any ideal $I \subseteq A [t_{i}]$ , $(A [t_{i}] / I) \otimes_{A} C ≅ C [t_{i}] / I C [t_{i}] .$ Here the extended ideal is generated by the coefficient images of all elements of $I$ . For a multiplicative subset $M \subseteq A$ , $(M^{- 1} A) \otimes_{A} C ≅ {\overline{M}}^{- 1} C .$ These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1

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

givenF1F2

2.1

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

step 1.1algebra

3.1

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

step 1.1algebra∎

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

Specialization of the polynomial fibre computation in Vakil 10.3.1–2 (standard reference, not scraped)