CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge 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.

Tensor products of domains need not be domains over a nonclosed field

Statement refuted

The tensor product of two domains over every field is a domain.

Counterexample

Given: The domains $C$ over the non-algebraically-closed field $R$ .

1.1

As an $R$ -algebra, $C ≅ R [t] / (t^{2} + 1)$ , so $C \otimes_{R} C ≅ C [t] / (t^{2} + 1)$ .

givenalgebra

2.1

In $C [t]$ , $t^{2} + 1 = (t - i) (t + i)$ with distinct factors, so the Chinese remainder calculation gives $C [t] / (t^{2} + 1) ≅ C \times C$ .

step 1.1algebra

3.1

The two nonzero coordinate idempotents have zero product, so this tensor product is not a domain. This refutes the statement and explains the base-field hypothesis in the affine product theorem.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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

J. S. Milne, Algebraic Geometry, Remark 5.18 (standard reference, not scraped)