Alphabeta Math
CounterexampleConstruction: 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.

An integral real scheme splits over the complex numbers

Statement refuted

False claims: an integral real scheme must stay irreducible or connected after extension to C; an injective morphism of schemes must remain injective after arbitrary base change. The morphism SpecCSpecR refutes all these assertions.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

For a field extension K/k and a k-scheme X, the inverse image of every affine open U=SpecA in XK is Spec(AkK). These affine charts cover XK and are compatible on overlaps and with coefficient localizations. (Affine charts after extension of the ground field)

[F2]

Let AC be a unital ring map. For any set of variables (ti) and any ideal IA[ti], (A[ti]/I)ACC[ti]/IC[ti]. Here the extended ideal is generated by the coefficient images of all elements of I. For a multiplicative subset MA, (M1A)ACM1C. 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 I1,,Ir be pairwise comaximal ideals, where r1. Then the canonical map Ri=1rR/Ii,x(x+I1,,x+Ir) is surjective, its kernel is i=1rIi, and i=1rIi=i=1rIi. Equivalently, R/i=1rIii=1rR/Ii. (Chinese remainder theorem for pairwise comaximal ideals)

Counterexample

1.1

The source has one point with field local ring C, so it is integral and connected, and the map to the one-point spectrum of R is injective. By F1 and F2 its complex base change has ring CRCC[z]/(z2+1).

givenF1F2
2.1

The factors zi,z+i are comaximal because 2i is a unit. F3 gives C[z]/(z2+1)C×C. The two prime ideals are C×0 and 0×C; the complementary idempotents exhibit them as two disjoint nonempty clopen points. Thus the base change is reduced but disconnected and reducible, and its map to SpecC is not injective.

F3step 1.1algebra

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