ExampleConstruction: Literature-sourcedVerification: 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.

Quadratic fibres over rational points and the generic point

Example

For any field $k$ , the family $Spec k [x, t] / (x^{2} - t) \to Spec k [t]$ has fibre $Spec k [x] / (x^{2})$ at zero, nonreduced in every characteristic. Over $Q$ , the fibre rings at $t = 1, 0, - 1$ and at the generic point are respectively $Q \times Q, Q [ϵ] / (ϵ^{2}), Q (i), Q (x),$ where the generic base field $Q (t)$ embeds in $Q (x)$ via $t = x^{2}$ , with degree two.

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)

[F3]

Let $R$ be a commutative ring and let $I_{1}, \dots, I_{r}$ be pairwise comaximal ideals, where $r \geq 1$ . Then the canonical map $R ⟶ \prod_{i = 1}^{r} R / I_{i}, x ⟼ (x + I_{1}, \dots, x + I_{r})$ is surjective, its kernel is $⋂_{i = 1}^{r} I_{i}$ , and $⋂_{i = 1}^{r} I_{i} = \prod_{i = 1}^{r} I_{i} .$ Equivalently, $R / \prod_{i = 1}^{r} I_{i} ≅ \prod_{i = 1}^{r} R / I_{i} .$ (Chinese remainder theorem for pairwise comaximal ideals)

Verification

1.1

F1 and F2 give $k [x] / (x^{2} - a)$ at $t = a$ . At $a = 0$ , the classes of $1, x$ are a basis, so $x$ is nonzero with square zero, in every characteristic. The ideal $(x)$ is the only prime and the residue field is $k$ .

givenF1F2

1.2

The generic fibre is $Q (t) [x] / (x^{2} - t)$ , equivalently $Q [x]$ with all nonzero polynomials in $x^{2}$ inverted. For nonzero $g (x)$ , the product $g (x) g (- x)$ is a nonzero even polynomial, hence a polynomial in $x^{2}$ . Thus $g (x)^{- 1} = g (- x) / (g (x) g (- x))$ belongs to this ring. It is exactly $Q (x)$ . Its dimension over $Q (t)$ is two, because division by the monic polynomial $x^{2} - t$ leaves unique remainders of degree less than two.

F1F2algebra

2.1

Over $Q$ , at $a = 1$ the factors $x - 1, x + 1$ generate comaximal ideals since their difference is 2. F3 therefore gives $Q \times Q$ . At $a = - 1$ , $x^{2} + 1$ has no rational root, hence is irreducible of degree two and its quotient is the field $Q (i)$ .

F3step 1.1algebra

3.1

The displayed fibres are nonempty: each coordinate ring contains 1 nonzero. Over an algebraic closure of any residue field in this rational family, $x^{2} - a$ splits into two distinct factors when $a \neq 0$ , since the characteristic is zero and a root then has nonzero derivative $2 x$ . At $a = 0$ it remains a doubled point. CRT proves the two-point assertion just as at $a = 1$ .

F3step 1.1step 2.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

Vakil Example 10.3.3(i)–(iv) (standard reference, not scraped)