CounterexampleConstruction: 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.

Equal fibre points can have different multiplicities

Statement refuted

False claim: the underlying set of a scheme fibre together with its pointwise residue fields determines that fibre up to scheme isomorphism. For any field $k$ , the zero fibres of $x^{2} = t$ and $x^{3} = t$ have the same one-point underlying space and residue field $k$ , but are not isomorphic as schemes. Their coordinate rings have $k$ -dimensions two and three.

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]

For commutative unital rings $A, B$ , the assignment $φ \mapsto Spec (φ)$ gives a natural bijection ${Hom}_{C R i n g} (A, B) ≅ {Hom}_{L R S} (Spec B, Spec A) .$ Consequently $A \mapsto Spec A$ is a contravariant equivalence from commutative rings to affine schemes, with quasi-inverse global sections. (Affine schemes are contravariantly equivalent to commutative rings)

Counterexample

1.1

F1 identifies the two fibre rings as $R_{2} = k [x] / (x^{2})$ and $R_{3} = k [x] / (x^{3})$ . Each has exactly one prime, $(x)$ , since every prime contains the nilpotent class of $x$ and the quotient is the field $k$ . Hence their underlying spaces and pointwise residue fields agree.

givenF1algebra

2.1

The monomial classes $1, x, \dots, x^{r - 1}$ form a $k$ -basis of $R_{r}$ by division by $x^{r}$ . Thus their dimensions are two and three, excluding a $k$ -algebra isomorphism. To exclude even an abstract ring isomorphism, note that the nilradical of $R_{2}$ has square zero, whereas the nilradical of $R_{3}$ has nonzero square generated by $x^{2}$ . Every ring isomorphism preserves the nilradical and its powers. F2 therefore excludes any scheme isomorphism. These rings are nonzero and the argument works in every characteristic.

F2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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

Variant of Vakil 10.3.3(ii) (standard reference, not scraped)