A fourfold whose cylinder is affine space

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Zariski's cancellation problem asks whether an affine variety $X$ with $X\times\mathbb A^1\cong\mathbb A^{n+1}$ must itself be $\mathbb A^n$. In algebraic terms: if adjoining one variable to a finitely generated algebra $A$ gives a polynomial ring in $n+1$ variables, is $A$ a polynomial ring in $n$ variables? This lesson and the next answer the question negatively in dimension four, in characteristic zero. This lesson constructs the fourfold, a hypersurface in affine five-space given by one explicit equation, and writes down an isomorphism of its cylinder with affine five-space. The next lesson, Cancellation fails in dimension four, proves that the fourfold itself is not affine four-space.

We use Locally nilpotent derivations for exponentials of derivations, and unique factorization in polynomial rings from Integral extensions: lying over, going up and going down. Dimension is computed with Krull dimension and Noether normalization, Theorem 4.2.

Basic references are [OpenAI-cancellation] and [Gupta-survey].

1. The cancellation problem

Let $k$ be a field and write $k^{[n]}$ for a polynomial ring in $n$ variables over $k$. The cancellation problem asks: if $A$ is a finitely generated $k$-algebra and $A[w]\cong k^{[n+1]}$ for a variable $w$, must $A\cong k^{[n]}$? An isomorphism of the cylinder need not respect the projection to the line $w$, so it gives no direct way to find coordinates on $A$.

For $n=1$ the answer is yes, by Abhyankar, Eakin and Heinzer [AEH]. For $n=2$ and $k$ of characteristic zero it is yes by the work of Fujita, Miyanishi and Sugie [Fujita], [Miyanishi–Sugie]. In positive characteristic, Gupta proved that the answer is no in every dimension $n\ge3$ [Gupta]. The example of this lesson, found in 2026 [OpenAI-cancellation], answers the question negatively in characteristic zero for $n=4$. We do not use any of these earlier results.

From now on $k$ is a field of characteristic zero.

2. The fourfold

Let $P=k[p,s,u,F,J]$ be a polynomial ring in five variables, and define

\begin{gathered} x=s^2+u^3+p^2F,\\ y=s+x(x-u^3),\\ z=sx+p^2J, \end{gathered}\tag{2.1}

and make the definitions (2.2): $\begin{aligned} H&=x^2F-(1+2sx)J-p^2J^2\\ &\quad-pu,\\ A&=P/(H). \end{aligned}$

So $H$ is one explicit polynomial in the five variables $p,s,u,F,J$, and $A$ is the coordinate ring of the hypersurface $H=0$. We use the same letters for elements of $P$ and their images in $A$.

Lemma 2.1. In $P$,

\begin{aligned} &xy-z(z+1)\\ &\quad =p^2\,(H+pu). \end{aligned}\tag{2.3}

Proof. Since $xy=sx+x^2(x-u^3)$ and $z(z+1)$ equals $s^2x^2+2sxp^2J+p^4J^2+sx+p^2J$, the terms $sx$ cancel and

\begin{aligned} &xy-z(z+1)=x^2(x-u^3-s^2)\\ &\quad -p^2(1+2sx)J-p^4J^2 . \end{aligned}

Now $x-u^3-s^2=p^2F$, so the right side is $p^2\bigl(x^2F-(1+2sx)J-p^2J^2\bigr)$, which is $p^2(H+pu)$. $\square$

Over the Laurent polynomial ring $k[p,p^{-1}]$, the elements $x,y,z,u$ are coordinates on $P[p^{-1}]$. Indeed (2.1) can be solved:

\begin{gathered} s=y-x(x-u^3),\\ F=p^{-2}\bigl(x-s^2-u^3\bigr),\\ J=p^{-2}\bigl(z-sx\bigr), \end{gathered}\tag{2.4}

so $P[p^{-1}]=k[p,p^{-1}][x,y,z,u]$, with $x,y,z,u$ algebraically independent over $k[p,p^{-1}]$. In these coordinates (2.3) reads

\begin{aligned} H&=p^{-2}\bigl(xy-z(z+1)\bigr)\\ &\quad -pu . \end{aligned}\tag{2.5}

Proposition 2.2. The ring $A$ is a domain of dimension four, the image of $p$ in $A$ is nonzero, and

A[p^{-1}]=k[p,p^{-1},x,y,z]

is a Laurent polynomial ring over a polynomial ring in $x,y,z$.

Proof. In (2.5) the coefficient of $u$ is the unit $-p$, so replacing $u$ by $H$ is another change of coordinates of $P[p^{-1}]$ over $k[p,p^{-1}]$. Hence $H$ generates a prime ideal of $P[p^{-1}]$, and $P[p^{-1}]/(H)=k[p,p^{-1},x,y,z]$.

Reducing $H$ modulo $p$ gives $x_0^2F-(1+2sx_0)J$ with $x_0=s^2+u^3$, which is not zero. So $p$ does not divide $H$ in $P$. Factor $H$ into irreducibles in the factorial ring $P$. In $P[p^{-1}]$ it becomes irreducible, so all but one of its irreducible factors become units there, that is, divide a power of $p$; such a factor is associated to $p$, which does not divide $H$. So $H$ is irreducible, hence prime, and $A$ is a domain. Since $A[p^{-1}]\ne0$, the element $p$ is not zero in $A$. The fraction field of $A$ is $k(p,x,y,z)$, of transcendence degree four, so $\dim A=4$ by Krull dimension and Noether normalization, Theorem 4.2. $\square$

3. An exponential that changes the equation

On $P[p^{-1}]=k[p,p^{-1}][x,y,z,u]$, let $\Delta=-p^2\,\partial/\partial u$, the derivation that kills $p,x,y,z$ and sends $u$ to $-p^2$. It is locally nilpotent, by Lemma 1.2 of Locally nilpotent derivations.

Lemma 3.1. The derivation $\Delta$ maps $P$ into itself, with

\begin{aligned} &\Delta(p)=0,\quad \Delta(u)\\ &\quad =-p^2,\quad \Delta(s)\\ &\quad =-3p^2xu^2,\quad \Delta(F)\\ &\quad =(6sx+3)u^2,\quad \Delta(J)\\ &\quad =3x^2u^2 . \end{aligned}\tag{3.1}

Moreover $\Delta(H)=p^3$ and $\Delta^2(H)=0$.

Proof. Apply $\Delta$ to (2.4), using $\Delta(x)=\Delta(y)=\Delta(z)=0$. From $s=y-x(x-u^3)$, $\Delta(s)=3xu^2\Delta(u)=-3p^2xu^2$. From $p^2F=x-s^2-u^3$, $p^2\Delta(F)=-2s\Delta(s)-3u^2\Delta(u)$, which is $6p^2sxu^2+3p^2u^2$. From $p^2J=z-sx$, $p^2\Delta(J)=-x\Delta(s)=3p^2x^2u^2$. These values lie in $P$, and $P$ is generated by $p,s,u,F,J$, so $\Delta(P)\subset P$. By (2.5), $\Delta(H)=-p\,\Delta(u)=p^3$, which $\Delta$ kills. $\square$

Extend $\Delta$ to $P[w]$ by $\Delta(w)=0$. By Proposition 2.1 of Locally nilpotent derivations, $\Phi=\exp(w\Delta)$ is an automorphism of $P[w]$, and by Lemma 3.1

\Phi(H)=H+w\,\Delta(H)=H+p^3w .

Hence $\Phi$ induces the isomorphism (3.2):

\begin{aligned} A[w]=P[w]/(H)&\xrightarrow{\ \sim\ }T,\\ T&:=P[w]/(H+p^3w). \end{aligned}

So it suffices to show that $T$ is a polynomial ring in five variables. Since $\Phi$ fixes $p$ and $A$ is a domain, $T$ is a domain in which $p\ne0$.

4. The cylinder is affine five-space

Linear coordinates adapted to $p=0$. Put $x_0=s^2+u^3$ and

\begin{gathered} L=x_0^2F-(1+2sx_0)J,\\ M=(1-2sx_0)F+4s^2J . \end{gathered}\tag{4.1}

This is a linear change of the variables $F,J$ over $k[p,s,u]$, with determinant $4s^2x_0^2+(1+2sx_0)(1-2sx_0)=1$. Its inverse is given by (4.2):

\begin{gathered} F=4s^2L+(1+2sx_0)M,\\ J=-(1-2sx_0)L+x_0^2M . \end{gathered}

So $P=B[L]$ with $B=k[p,s,u,M]$, a polynomial ring in four variables. From now on we regard $F$, $J$ and $H$ as polynomials in $L$ over $B$ through (4.2). Expanding $x=x_0+p^2F$ in powers of $p$ gives (4.3):

\begin{gathered} H=L-pu+p^2Q+p^4F^3,\\ Q=2x_0F^2-2sFJ-J^2 . \end{gathered}

Indeed $x^2F=x_0^2F+2p^2x_0F^2+p^4F^3$ and $(1+2sx)J=(1+2sx_0)J+2p^2sFJ$, and $x_0^2F-(1+2sx_0)J=L$.

A root modulo $p^3$. Write $C=Q|_{L=0}\in B$ and $Q=C+L\,Q_1$ with $Q_1\in B[L]$. Explicitly $C=M^2\bigl(2x_0+6sx_0^2+4s^2x_0^3-x_0^4\bigr)$. Put

L_*=pu-p^2C\in B .

Then (4.3) gives $\begin{aligned} &H(L_*)=L_*-pu\\ &\quad+p^2\bigl(C+L_*Q_1(L_*)\bigr)+p^4F(L_*)^3\\ &\quad=p^3\bigl((u-pC)Q_1(L_*)+pF(L_*)^3\bigr), \end{aligned}$ so

H(L_*)\in p^3B . \tag{4.4}

The fifth coordinate. In $B[L,w]$ put

\begin{gathered} h=u-pQ-p^3F^3-p^2w,\\ e_0=-w-pF^3-Q_1h . \end{gathered}

Then $H+p^3w=L-ph$ by (4.3), and a direct expansion, using $Q=C+LQ_1$, gives the identity (4.5):

\begin{aligned} &p^3e_0-(L-L_*)\\ &\quad=\bigl(p^2Q_1-1\bigr)\bigl(H+p^3w\bigr). \end{aligned}

So the image $e$ of $e_0$ in $T$ satisfies $p^3e=L-L_*$.

Theorem 4.1. $T=B[e]=k[p,s,u,M,e]$ is a polynomial ring in five variables. Consequently $A[w]\cong k^{[5]}$.

Proof. $B$ embeds in $T$, and $e$ is transcendental over it. Inverting $p$ in $T=B[L,w]/(H+p^3w)$ lets us solve for $w=-p^{-3}H$, so $T[p^{-1}]\cong B[p^{-1}][L]$, the identity on $B$ and $L$. So $B\to T$ is injective, and in $T[p^{-1}]$ the element $e=p^{-3}(L-L_*)$ is a linear polynomial in the variable $L$ with unit leading coefficient. Hence $e$ is transcendental over $B[p^{-1}]$, and so over $B$.

$B$ and $e$ generate $T$. We have $L=L_*+p^3e\in B[e]$. In $B[Z]$, the difference $H(L_*+p^3Z)-H(L_*)$ is divisible by $p^3$, and $H(L_*)$ is divisible by $p^3$ by (4.4). So $W(Z)=-p^{-3}H(L_*+p^3Z)$ lies in $B[Z]$. The relation $H+p^3w=0$ in $T$ gives $p^3\bigl(w-W(e)\bigr)=0$, and $p$ is a nonzero element of the domain $T$, so $w=W(e)\in B[e]$. Thus $T=B[L,w]=B[e]$.

Finally $A[w]\cong T$ by (3.2). $\square$

The five coordinates of $A[w]$ itself are the images of $\Phi^{-1}(p)=p$, $\Phi^{-1}(s)$, $\Phi^{-1}(u)$, $\Phi^{-1}(M)$ and $\Phi^{-1}(e_0)$, where $\Phi^{-1}=\exp(-w\Delta)$; each is an explicit polynomial, because the exponential is a finite sum.

5. What remains

Theorem 4.1 shows that the hypersurface $H=0$ in $\mathbb A^5$ becomes affine five-space after multiplying by a line. The equation was chosen so that the hypersurface looks like affine four-space in two other ways as well: after inverting $p$, Proposition 2.2 identifies it with $\mathbb A^3\times(\mathbb A^1\setminus0)$, and modulo $p$ the change of coordinates (4.1) makes $H\equiv L$ a coordinate. The next lesson proves that nevertheless $A$ is not a polynomial ring.

6. Exercises

Exercise 6.1 (easy). Verify that the linear change (4.1) has determinant one, and that (4.2) inverts it.

Exercise 6.2 (medium). Show that $\Delta(H)=p^3$ directly from (3.1), without passing to the coordinates $x,y,z,u$.

Exercise 6.3 (medium). Show that the fibre $p=0$ of $\operatorname{Spec}A$ is affine three-space: $A/pA\cong k^{[3]}$.

7. Solutions

6.1. The coefficient matrix of $(F,J)\mapsto(L,M)$ is $\begin{pmatrix}x_0^2&-(1+2sx_0)\\1-2sx_0&4s^2\end{pmatrix}$, with determinant $4s^2x_0^2+(1+2sx_0)(1-2sx_0)$, which is $4s^2x_0^2+1-4s^2x_0^2=1$. Its inverse is $\begin{pmatrix}4s^2&1+2sx_0\\-(1-2sx_0)&x_0^2\end{pmatrix}$, which is (4.2).

6.2. Write $H=x^2F-(1+2sx)J-p^2J^2-pu$ with $x=s^2+u^3+p^2F$. Then $\Delta(x)=2s\Delta(s)+3u^2\Delta(u)+p^2\Delta(F)$, which is $-6p^2sxu^2-3p^2u^2+p^2(6sx+3)u^2=0$. So $\begin{aligned} &\Delta(H)=x^2\Delta(F)-2x\Delta(s)J\\ &\quad-(1+2sx)\Delta(J)-2p^2J\Delta(J)\\ &\quad-p\Delta(u). \end{aligned}$ Substituting (3.1), the first four terms are $\begin{aligned} &x^2(6sx+3)u^2+6p^2x^2u^2J\\ &\quad-3(1+2sx)x^2u^2-6p^2x^2u^2J=0, \end{aligned}$ and the last is $p^3$.

6.3. Modulo $p$, (4.3) gives $H\equiv L$, and $P/pP=k[s,u,M,L]$ by (4.1) and (4.2). So $A/pA=P/(p,H)$, which is $k[s,u,M,L]/(L)\cong k[s,u,M]$.

References

[OpenAI-cancellation] OpenAI, An explicit failure of complex affine-space cancellation, preprint, 23 September 2026. https://github.com/openai/math/blob/main/preprints/An-explicit-failure-of-complex-affine-space-cancellation-September-23-2026/paper.pdf
[Gupta-survey] N. Gupta, The Zariski cancellation problem and related problems in affine algebraic geometry, Proceedings of the International Congress of Mathematicians 2022, vol. 3, EMS Press, 2023, 1578–1598. https://doi.org/10.4171/ICM2022/151
[Gupta] N. Gupta, On Zariski's cancellation problem in positive characteristic, Adv. Math. 264 (2014), 296–307. https://arxiv.org/abs/1309.1368
[AEH] S. S. Abhyankar, P. Eakin, W. Heinzer, On the uniqueness of the coefficient ring in a polynomial ring, J. Algebra 23 (1972), 310–342. https://doi.org/10.1016/0021-8693(72)90134-2
[Fujita] T. Fujita, On Zariski problem, Proc. Japan Acad. Ser. A Math. Sci. 55 (1979), 106–110. https://doi.org/10.3792/pjaa.55.106
[Miyanishi–Sugie] M. Miyanishi, T. Sugie, Affine surfaces containing cylinderlike open sets, J. Math. Kyoto Univ. 20 (1980), 11–42. https://doi.org/10.1215/kjm/1250522319