CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-09-26 (gpt-6-sol)

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.

A scheme is not determined by its underlying topological space

Statement refuted

“The underlying topological space determines a scheme.”

Facts & Assumptions

Given: A field $k$ .

[F1]

Points of $Spec A$ are prime ideals of $A$ (The underlying space of an affine spectrum). A field has exactly one proper ideal, $(0)$ , and it is prime.

[F2]

The dual-numbers scheme is $Spec D$ for $D = k [ϵ] / (ϵ^{2})$ (The affine scheme of dual numbers, The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ). Passing to the quotient by the nilradical does not change the prime spectrum or its topology (Passing to the reduced quotient does not change the prime spectrum). An affine scheme is reduced exactly when its coordinate ring is reduced (Reduced affine schemes).

Counterexample

technique · direct

1.1F1F2algebra

Every element of $D$ has a unique form $a + b ϵ$ with $a, b \in k$ and $ϵ^{2} = 0$ . Such an element is nilpotent exactly when $a = 0$ , since its image under $D \to D / (ϵ) ≅ k$ must then be nilpotent in a field, and every $b ϵ$ is square-zero. Thus $Nil (D) = (ϵ)$ and $D / Nil (D) ≅ k$ . By [F2], $Spec D$ is homeomorphic to $Spec k$ , which has one point by [F1].

2.1F2step 1.1algebra

The first scheme is reduced because a field has no nonzero nilpotents. The second is not reduced: $ϵ$ is nonzero in $D$ because the polynomial $ϵ$ does not belong to $(ϵ^{2})$ , but its square is zero.

3.1step 1.1step 2.1∎

Scheme isomorphisms preserve affine coordinate rings up to isomorphism and hence reducedness, so these schemes are not isomorphic.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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

James S. Milne, Algebraic Geometry, 10.28 (standard reference, not scraped)