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.

An empty fibre from a zero tensor ring

Example

For the open immersion $Spec k [t, t^{- 1}] \to Spec k [t]$ , the fibre at $t = 0$ is empty: its tensor coordinate ring is the zero ring.

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 its ring as $k [t, t^{- 1}] \otimes_{k [t]} k$ , with $t$ acting as zero on $k$ . By F2 this is the localization of $k$ at the image of the powers of $t$ .

givenF1F2

2.1

In this ring $t$ is simultaneously zero and invertible, forcing $1 = t t^{- 1} = 0$ . Thus the ring is zero and has no prime ideals, so the fibre is empty. This agrees with the absence of $(t)$ from $D (t)$ . At $t = 1$ the same formula instead gives $k$ , since 1 is already invertible; the emptiness depends on the specified point.

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

Direct localization special case of Vakil 10.2.F and 10.3.2 (standard reference, not scraped)