Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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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.
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A reduced field spectrum becomes nonreduced

Statement refuted

False claim: reducedness survives algebraic extension of the ground field. Let p be prime, k=Fp(v) with v transcendental, and L=k[u]/(up−v). Then L is a field, but L⊗kL≅L[ϵ]/(ϵp), so the base change of the reduced k-scheme Spec⁡L along the finite algebraic extension L/k is nonreduced.

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=Spec⁡A in XK is Spec⁡(A⊗kK). These affine charts cover XK and are compatible on overlaps and with coefficient localizations. (Affine charts after extension of the ground field)

[F2]

Let A→C be a unital ring map. For any set of variables (ti) and any ideal I⊆A[ti], (A[ti]/I)⊗AC≅C[ti]/IC[ti]. Here the extended ideal is generated by the coefficient images of all elements of I. For a multiplicative subset M⊆A, (M−1A)⊗AC≅M‾−1C. These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

[F3]

Let F have characteristic p>0, let a∈F not be a pth power in F, and let n≥1. Then xpn−a is irreducible in F[x]. (If a is not a pth power in a characteristic-p field, then xpn−a is irreducible for every n≥1)

[F4]

For every field F, the polynomial ring F[x] is a unique factorisation domain. (For every field F, F[x] is a unique factorisation domain)

Counterexample

1.1givenF3F4

By F4, Fp[v] is a UFD. If v=(a/b)p for nonzero polynomials a,b, comparison of the exponent of the prime polynomial v in vbp=ap gives 1+pord⁡v(b)=pord⁡v(a), impossible modulo p. Thus v is not a pth power in k, and F3 with exponent parameter 1 proves up−v irreducible. Its quotient L is a field of degree p over k.

2.1F1F2step 1.1algebra

By F1 and F2 the base-change ring is L[z]/(zp−v). In L one has v=up, and the characteristic-p binomial formula gives zp−up=(z−u)p. Substitution ϵ=z−u supplies the claimed ring isomorphism, with inverse z=u+ϵ.

3.1step 1.1step 2.1algebra∎

The classes 1,ϵ,…,ϵp−1 form a basis by division by the monic polynomial ϵp. Since p≥2, ϵ is nonzero but nilpotent. Thus this ring, whose unique prime is (ϵ), is nonreduced whereas L is reduced. The argument includes p=2; no integer-only Eisenstein criterion is used.

Depends on

Used by

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Dependency tree · two levels

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Sources