VibeMathedMath problems solved with AI

The Zariski cancellation problem for complex affine space in dimension four

Zariski's cancellation problem asks whether affine space is determined by its cylinder: if XX is an affine variety with X×A1≅An+1X\times\mathbb A^1\cong\mathbb A^{n+1}, must X≅AnX\cong\mathbb A^n? Algebraically, over C\mathbb C: if AA is a finitely generated complex algebra with A[w]≅C[n+1]A[w]\cong\mathbb C^{[n+1]}, is A≅C[n]A\cong\mathbb C^{[n]}? It holds for n=1n=1 (Abhyankar-Eakin-Heinzer) and n=2n=2 (Fujita, Miyanishi-Sugie). Gupta (2014) showed it fails for every n≥3n\ge3 in positive characteristic, and the characteristic-zero case in dimensions n≥3n\ge3 is listed as open in Gaifullin-Petrov (July 2026) and Gupta's ICM 2022 survey. Does cancellation hold for complex affine space in dimension n≥3n\ge3, in particular for n=4n=4?

Result
Disproved(see note)
Status
Partial result
AI contribution
AI-discovered
Method
Construction
Field
Affine algebraic geometry
Posed by
Oscar Zariski; the manuscript cites Abhyankar-Eakin-Heinzer (1972), Fujita (1979), Gupta's ICM 2022 survey and Gaifullin-Petrov (2026) for the problem and its status
Year posed
—
Years open
—
Solved
2026-09-23
Model
Unreleased internal OpenAI model
Vendor
OpenAI
Collaborators
—
Verification
Lean-checked, statement unaudited
Publication
Announced
Collection
OpenAI math release (October 2026), version adc7f12
Significance
60 / 100
Disclosed cost
—
Wikipedia
No dedicated article

What was actually shown

Theorem 1.1: with P=C[p,s,u,F,J]P=\mathbb C[p,s,u,F,J], x=s2+u3+p2Fx=s^2+u^3+p^2F and H=x2F−(1+2sx)J−p2J2−puH=x^2F-(1+2sx)J-p^2J^2-pu, the algebra A=P/(H)A=P/(H) is an integral complex domain of dimension four with A[w]≅C[5]A[w]\cong\mathbb C^{[5]} but A≇C[4]A\not\cong\mathbb C^{[4]}. The stabilization uses that HH is a residual coordinate (El Kahoui-Ouali, Dutta-Lahiri); non-polynomiality is proved by degenerating to a graded algebra, lifting a putative additive action through a torus bundle over a quadric, and excluding it by a Mason-Stothers abc argument. It answers only n=4n=4: dimension three is not treated, and the paper notes that adjoining variables to a counterexample does not by itself give counterexamples in higher dimensions.

What the AI did

Produced by an unreleased internal OpenAI model as part of an OpenAI evaluation on open research problems. The release README says the vast majority of results used one fixed procedure, averaging about three hours of ChatGPT Pro thinking compute per result; this result is not among the README's stated exceptions (the Riemann zeta zero-free region work and the Hodge conjecture for CM abelian varieties). The manuscript is authored as OpenAI with no human author named. The README also cautions that unformalized results could have issues.

Verification

No independent mathematician has checked this yet. Checked here: the abstract, introduction and Theorem 1.1, read against the cancellation problem as the manuscript states it. The degeneration and rigidity arguments were not refereed. Lean-checked on the release's Comparator challenge ComplexCancellation (declaration OAI.ComplexCancellation.main, file OAI/Algebra/AffineCancellation/Main.lean, listed in lean/formalization.yaml). Its statement was read here: it defines the explicit H in MvPolynomial (Fin 5) over C and asserts that A = P/(H) is of finite type, a domain, of Krull dimension 4, that Polynomial A is C-algebra isomorphic to MvPolynomial (Fin 5) C, and that A is not isomorphic to MvPolynomial (Fin 4) C. That is the headline claim exactly. Not rebuilt here. The release's Lean scope note says the stable-coordinate and line-bundle consequences are not formalized.

Sources

Changelog1 change

Discussion