Zariski Cancellation Problem (characteristic zero, dimension ≥ 3)

Zariski Cancellation Problem (characteristic zero, dimension $\geq$ 3)

OpenAI's claim

An explicit complex affine fourfold is claimed to become affine five-space after one stabilization while itself failing to be affine four-space, with stable-coordinate and nontrivial affine-fibration consequences. The cited manuscript argues for that result; its relation to this problem is specified below.

Mathematical statement of the claimed result

Mathematical claim 1.

In $P=\mathbb C[p,s,u,F,J]$, define \[ x=s^2+u^3+p^2F,\qquad H=x^2F-(1+2sx)J-p^2J^2-pu,\qquad A=P/(H). \] OpenAI claims that $A$ is integral of dimension four and for an independent variable $w$, \[ A[w]\simeq\mathbb C^{[5]},\qquad A\not\simeq\mathbb C^{[4]}. \] The notation $\mathbb C^{[j]}$ denotes a polynomial algebra in $j$ variables. It also claims that $H$ is a coordinate of $P[w]$ but not of $P$, where coordinate means an element belonging to a polynomial coordinate system. These assertions give one-variable cancellation and stable-coordinate counterexamples in the indicated dimensions. Writing $X=\operatorname{Spec}A$, the affine-fibration corollary claims that \[ \pi:X\to\mathbb A^1,\quad x\mapsto p(x),\qquad \rho:\mathbb A^5\to\mathbb A^2,\quad z\mapsto(p(z),H(z)) \] are smooth and surjective with relative dimension three; every fiber over a base point $y$ is $\mathbb A^3_{\kappa(y)}$, yet neither map is locally a product with $\mathbb A^3$ in the Zariski topology. Thus the examples satisfy faithful flatness and geometric integrality of the fibers.

Outline of the argument

Manuscript 1 (family 047, 2026-09-23): A locally nilpotent derivation and an explicit inverse polynomial change give the cylinder isomorphism. To obstruct cancellation, a valuation along the exceptional parameter computes the associated graded affine modification; the resulting rigidity distinguishes the original coordinate ring from a polynomial algebra.

Formal verification

For family 047, the scope page documents the following selections. The selected construction is a four-dimensional complex domain whose one-variable polynomial extension is a five-variable polynomial ring, while the original ring is not polynomial. The other affine-fibration and stable-coordinate consequences are separate. The documented statement covers the main claim of manuscript 1. Its selected declaration also covers the clause used for this problem. Comparator files: ComplexCancellation.lean. Declarations: OAI.ComplexCancellation.main.

This site has not executed the Lean files or checked their dependencies. These descriptions report the scope of the linked declarations, rather than a new verification of the proofs.

Status caveat

OpenAI attributes these results to an internal model. Its release notice cautions that results lacking full formalization can contain mistakes. This site has not independently assessed the proofs; a solved label records the scope of the reported claim.

Sources

• OpenAI, An explicit failure of complex affine-space cancellation, 2026-09-23; citation key OAI:An-explicit-failure-of-complex-affine-space-cancellation-September-23-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: thm:main; Statement source: cor:stable-coordinate; Statement source: cor:affine-fibrations.
• Family 047: Lean scope.
• Comparator ComplexCancellation.json.
• Comparator ComplexCancellation.lean.
• OpenAI catalogue entry.

Zariski Cancellation Problem (characteristic zero, dimension ≥ 3)

Definitions and mathematical statement

Let $k$ be a characteristic-zero field and let $B$ be a finitely generated commutative $k$-algebra. The cancellation problem asks whether, for every $n\ge3$, \[ B[t]\cong_k k[x_1,\ldots,x_{n+1}] \quad\Longrightarrow\quad B\cong_k k[x_1,\ldots,x_n]. \] The isomorphisms are $k$-algebra isomorphisms; $t$ is an added independent indeterminate. Geometrically, if an affine variety becomes affine $(n+1)$-space after taking its product with an affine line, must it already be affine $n$-space? The premise forces $B$ to be a domain, so no extra domain hypothesis is needed. Positive-characteristic examples, stable cancellation after several factors, and birational equivalence do not decide this exact assertion.

Formulation and concept references: [S1].

Short English statement

Over every characteristic-zero field, can an affine-line factor be cancelled from an isomorphism with affine space in dimension at least four?

Research attempt

GPT-6 Astra Ultra; 27 September 2026. Status: UNRESOLVED.

I translate the cancellation premise into a locally nilpotent derivation with a slice and prove that translation explicitly. This exposes a concrete kernel problem, gives a solvable triangular subfamily, and identifies why the existence of a slice alone does not supply polynomial coordinates. All arguments below use characteristic zero, as required by the selected target.

Elementary consequences of the premise.

Suppose $B[t]\cong_k A=k[x_1,\ldots,x_{n+1}]$. Since $B$ embeds in $B[t]$, it is a domain. Its units are exactly $k^*$, because its units embed into those of $A$ and the isomorphism fixes $k$. Moreover $B$ is a unique factorization domain. Indeed factor a nonzero nonunit $b\in B$ into irreducibles in the UFD $B[t]$. Degrees in $t$ add in this domain, and $b$ has degree zero, so every factor lies in $B$. Those factors are prime in $B$ because they are prime in $B[t]$: divisibility of a constant by a constant in $B[t]$ already gives a constant quotient. Uniqueness descends as well. Finally $\dim B=n$, using the polynomial-ring dimension formula for a finitely generated domain over a field.

These necessary conditions rule out many candidate counterexamples, but they do not exhibit $n$ polynomial generators of $B$. In particular, factoriality is not a coordinate construction. The task is to recover the whole algebra, not just its fraction field or its dimension.

From a cylinder to a derivation with a slice.

On $B[t]$, the derivation $\partial/\partial t$ is locally nilpotent: each element is killed by a sufficiently high derivative. Transport it across the given isomorphism to a $k$-derivation $D$ of $A$, and write $s$ for the image of $t$. Then \[ D(s)=1,\qquad D\text{ is locally nilpotent},\qquad \ker D\cong_k B. \tag{1} \] The last equality uses characteristic zero: differentiating a polynomial in $t$ kills precisely its constant coefficient part. An element $s$ with $D(s)=1$ is called a slice.

Conversely, suppose $D$ is a locally nilpotent derivation of the polynomial algebra $A$ with a slice $s$. Put $C=\ker D$. I prove \[ A=C[s]\cong C[T]. \tag{2} \] For $a\in A$ define the finite sum \[ \pi(a)=\sum_{j\ge0}\frac{(-s)^j}{j!}D^j(a). \tag{3} \] Differentiating, the terms from $D((-s)^j)=-j(-s)^{j-1}$ cancel the terms from $D^{j+1}(a)$, so $D\pi(a)=0$. The Leibniz formula also shows that $\pi$ is an algebra homomorphism: its finite double sum for $\pi(a)\pi(b)$ groups into the binomial expansion for $\pi(ab)$. It fixes $C$ and sends $s$ to zero.

An explicit Taylor identity gives \[ a=\sum_{j\ge0}\frac{s^j}{j!}\pi(D^j a). \tag{4} \] After inserting $3$, the coefficient of $s^mD^m(a)$ for $m>0$ is $\sum_{j=0}^m(-1)^{m-j}/(j!(m-j)!)=0$, whereas it is one for $m=0$. This proves generation in $2$. If $\sum_{j=0}^r c_js^j=0$ with $c_j\in C$ and $c_r\ne0$, applying $D^r$ gives $r!c_r=0$, impossible in characteristic zero. This proves algebraic independence and $2$.

The kernel is finite type here: $2$ and the evaluation map give $C\cong A/(s)$. Thus the correspondence applies exactly to the finite-type algebras in the question; no unproved general finite generation theorem for derivation kernels is needed. Taken together, $1$--$2$ show that cancellation is equivalent to saying that every kernel of a slice-bearing locally nilpotent derivation of $k[x_1,\ldots,x_{n+1}]$ is a polynomial algebra in $n$ variables.

A triangular class where the kernel is explicit.

Use coordinates $t,x_1,\ldots,x_n$ and let \[ D=\frac{\partial}{\partial t} +\sum_{i=1}^n f_i(t)\frac{\partial}{\partial x_i}, \qquad f_i\in k[t]. \] Characteristic zero permits polynomial antiderivatives $F_i$ with $F_i'=f_i$. Set $y_i=x_i-F_i(t)$. The substitution $(t,x)\leftrightarrow(t,y)$ is a polynomial automorphism, since its inverse is $x_i=y_i+F_i(t)$. In these coordinates $D(t)=1$ and $D(y_i)=0$, so $D=\partial/\partial t$ and \[ \ker D=k[y_1,\ldots,y_n]. \tag{5} \] For instance $D=\partial_t+2t\partial_x+3t^2\partial_y$, with additional variables fixed, has kernel generated by $x-t^2$, $y-t^3$, and those fixed variables. The inverse substitution checks generation as well as invariance; merely listing some invariants would not prove they generate the kernel.

The exact point at which a general proof would be circular.

If a slice $s$ were already a polynomial coordinate of $A$, then $A/(s)$ would be a polynomial algebra, and $2$ would solve that input. But $D(s)=1$ only proves that $s$ is a coordinate relative to the unknown algebra $C$ in $A=C[s]$. Showing it is part of a coordinate system over $k$ is precisely the additional claim needed. The Taylor projection $3$ produces finitely many generators $\pi(x_i)$ for $C$, but does not prove there are exactly $n$ independent ones with no relations. Finite generation and a splitting retraction are weaker than the required polynomial presentation.

For example, differentiating shows $ds$ has no common zero in the cotangent sense: if $D=\sum g_i\partial_{x_i}$, then $\sum g_i(\partial s/\partial x_i)=1$. This is a useful necessary identity, but turning it into a polynomial automorphism is not justified by the identity itself. A claim that every such derivative identity produces coordinates would introduce another major unproved step.

Characteristic and verification checks.

In characteristic $p$, even the basic derivation $\partial_t$ kills $t^p$, and the factorial denominators in $3$--$4$ are unavailable. Thus neither that kernel calculation nor the Taylor proof carries unchanged to positive characteristic. The counterexamples and related questions surveyed in [S1] have to be compared with their exact field and cylinder hypotheses. A positive-characteristic example does not settle this characteristic-zero assertion.

The displayed triangular invariants and the finite Taylor cancellation were checked by exact symbolic identities. They provide no test of all locally nilpotent derivations. The first remaining gap is to prove that the general slice kernel in $2$ is polynomial, beyond classes admitting an explicit coordinate substitution such as $5$. The attempt obtains an equivalent kernel formulation and a concrete affirmative family, but no general cancellation theorem or counterexample. The selected problem remains UNRESOLVED.

Sources

• S1 Neena Gupta, "The Zariski Cancellation Problem and related problems in Affine Algebraic Geometry," Proceedings of the International Congress of Mathematicians 2022, Vol. 4, pp. 1578-1598, EMS Press. DOI:10.4171/ICM2022/151; also arXiv:2208.14736. Independent status corroboration: A. Asok, P. A. Ostvaer, "A1-homotopy Theory and Contractible Varieties: A Survey," Lecture Notes in Mathematics vol. 2292 (Springer, 2022), pp. 145-212, Remark 5.5.1.5. https://arxiv.org/pdf/2208.14736. Formulation source.
• E1 Neena Gupta, The Zariski Cancellation Problem and related problems in Affine Algebraic Geometry, ICM 2022, arXiv:2208.14736. https://arxiv.org/pdf/2208.14736. Primary author survey corresponding to [S1], inspected for characteristic and cancellation scope. The slice lemma and the triangular subcase are proved explicitly here. Consulted 27 September 2026.