Cancellation fails in dimension four

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

The previous lesson, A fourfold whose cylinder is affine space, constructed an algebra $A=P/(H)$ with $A[w]\cong k^{[5]}$. This lesson proves that $A$ is not a polynomial ring in four variables. So Zariski's cancellation problem has a negative answer in dimension four over every field of characteristic zero.

The proof argues by contradiction. If $A$ were a polynomial ring, differentiation with respect to one of its coordinates would be a locally nilpotent derivation of $A$ that kills another coordinate. We push this derivation through three changes of ring, each keeping enough information to reach a contradiction at the end:

to the associated graded ring $G$ of a valuation along the divisor $p=0$, where it becomes a homogeneous derivation with an invariant of positive degree;
to a line bundle over $\operatorname{Spec}G$ on which an ideal with two generators becomes principal, generated by one element $v$;
to a derivation $E$ of an explicit five-variable ring $\widetilde R$ of nonpositive fibre weight with $E^2(v)=0$.

The last step is excluded by a rigidity argument using the polynomial $abc$ theorem of Mason and Stothers.

We use Locally nilpotent derivations and Additive actions on line bundles throughout. From commutative algebra we use the Jacobian criterion and the regularity of smooth algebras from Smooth algebras over a field and the Jacobian criterion, Krull's intersection theorem from Noetherian and Artinian rings, and unique factorization in polynomial rings. The field $k$ has characteristic zero, and we keep the notation of the previous lesson: $P=k[p,s,u,F,J]$, $x=s^2+u^3+p^2F$, $y=s+x(x-u^3)$, $z=sx+p^2J$, $H=x^2F-(1+2sx)J-p^2J^2-pu$, $A=P/(H)$.

Basic references are [OpenAI-cancellation] and [Kaliman–Makar-Limanov].

1. The valuation along $p=0$

Consider the algebras (1.1):

\begin{aligned} S&=k[p,x,y,z,u]\\ &\quad\big/\bigl(xy-z(z+1)-p^3u\bigr),\\ R&=S/pS\\ &=k[x,y,z,u]\big/\bigl(xy-z(z+1)\bigr). \end{aligned}

By Lemma 2.1 of the previous lesson, $xy-z(z+1)=p^2(H+pu)\equiv p^3u$ modulo $H$, so sending $p,x,y,z,u$ to the elements of the same names defines a homomorphism $S\to A$.

Lemma 1.1.

$S$ is a domain, $S\to A$ is injective, and $S[p^{-1}]=A[p^{-1}]$.
$R$ is a domain, smooth over $k$.
The ideal $pS$ is prime, $S_{pS}$ is a discrete valuation ring with uniformizer $p$, and for $b\in S$ and $i\ge0$, the order $\operatorname{ord}_p(b)$ of $b$ in this ring satisfies $\operatorname{ord}_p(b)\ge i$ exactly when $b\in p^iS$.

Proof. (1) Write $\Phi=xy-z(z+1)-p^3u$. After inverting $p$, $\Phi$ is linear in $u$ with unit coefficient $-p^3$. In a factorization $\Phi=\Phi'\Phi''$ in $k[p^{\pm1},x,y,z,u]$, one factor has degree $0$ in $u$ and divides the coefficient $-p^3$ of $u$, so it is a unit. Hence $\Phi$ is irreducible in $k[p^{\pm1},x,y,z,u]$, and solving $\Phi=0$ for $u$ gives $S[p^{-1}]=k[p^{\pm1},x,y,z]$. Now factor $\Phi=\Phi_1\cdots\Phi_r$ into irreducibles in the factorial ring $k[p,x,y,z,u]$. After inverting $p$, exactly one factor, say $\Phi_1$, remains a non-unit. Every other factor becomes a unit of $k[p^{\pm1},x,y,z,u]$, that is, $c\,p^n$ with $c\in k^\times$; so it divides a power of $p$ in $k[p,x,y,z,u]$ and is associated to the prime $p$. But $p$ does not divide $\Phi$, because $\Phi\equiv xy-z(z+1)\not\equiv0$ modulo $p$. Hence $r=1$, $\Phi$ is irreducible, and $S$ is a domain. By Proposition 2.2 of the previous lesson, $A[p^{-1}]=k[p^{\pm1},x,y,z]$ as well, and $S\to A$ becomes an isomorphism after inverting $p$. Its kernel consists of elements killed by a power of $p$, so it is zero.

(2) The polynomial $xy-z(z+1)$ is linear in $y$, with coefficients $x$ and $-z(z+1)$ that have no common factor, so it is irreducible and $R$ is a domain. Its partial derivatives $y$, $x$, $-(2z+1)$ (and $0$ for $u$) vanish together only where $x=y=0$ and $z=-\tfrac12$, which is not on the hypersurface. By the Jacobian criterion Smooth algebras over a field and the Jacobian criterion, Corollary 2.2, $R$ is smooth.

(3) $S/pS=R$ is a domain, so $pS$ is prime. The local ring $S_{pS}$ is a Noetherian local domain whose maximal ideal is generated by $p$. By Krull's intersection theorem Noetherian and Artinian rings, Theorem 6.1, $\bigcap_np^nS_{pS}=0$; so every nonzero $b$ is $p^n$ times an element outside the maximal ideal, a unit. Hence $S_{pS}$ is a discrete valuation ring with uniformizer $p$. If $b\in S$ has positive order, then $tb\in pS$ for some $t\in S\setminus pS$, and primality of $pS$ gives $b\in pS$; dividing by $p$ and repeating proves the last assertion. $\square$

So every nonzero $h$ in the common fraction field of $S$ and $A$ has an order $\operatorname{ord}_p(h)\in\mathbb Z$, and every nonzero $h\in A\subset S[p^{-1}]$ can be written uniquely as $h=p^mb$ with $m=\operatorname{ord}_p(h)$ and $b\in S\setminus pS$. Define the valuation degree

\deg h=-\operatorname{ord}_p(h)\qquad(0\ne h\in A).

It is a degree function in the sense of Section 6 of Locally nilpotent derivations. Let $\mathcal F_dA$ be the elements of degree at most $d$, and $\operatorname{gr}A$ the associated graded ring.

The map

\begin{gathered} \operatorname{gr}A\longrightarrow R[\tau,\tau^{-1}],\\ [h]_d\longmapsto\overline{p^dh}\;\tau^{-d}, \end{gathered}\tag{1.2}

where the bar is reduction modulo $pS$, is a well-defined injective homomorphism of graded rings, if $R[\tau,\tau^{-1}]$ is graded by $\deg R=0$ and $\deg\tau=-1$. Indeed, for $h\in\mathcal F_dA$ the element $p^dh$ has nonnegative order, so lies in $S$ by Lemma 1.1(3), and its residue vanishes exactly when $p^dh\in pS$, that is, when $h\in\mathcal F_{d-1}A$. Sums and products are respected.

2. The associated graded ring

In $R$ put

\begin{gathered} s_R=y-x(x-u^3),\\ f=x-s_R^2-u^3,\\ g=z-s_Rx,\\ I=(f,g)\subset R . \end{gathered}\tag{2.1}

The elements $s$, $u$, $p^2F=x-s^2-u^3$ and $p^2J=z-sx$ of $A$ lie in $S$, with residues $s_R$, $u$, $f$, $g$. These residues are nonzero: at the points $x=z=0$ of $\operatorname{Spec}R$ we get $s_R=y$ and $f=-y^2-u^3$, and at $x=0$, $z=-1$ we get $g=-1$. So the degrees and the images of the initial forms are given by the table (2.2):

$h$	$\deg h$	image of $\operatorname{in}(h)$ in (1.2)
$p$	$-1$	$\tau$
$s$	$0$	$s_R$
$u$	$0$	$u$
$F$	$2$	$f/\tau^2$
$J$	$2$	$g/\tau^2$

Give $P$ the weights $\omega(p,s,u,F,J)=(-1,0,0,2,2)$, and make the definition (2.3):

\begin{aligned} H_{\mathrm{top}}&=x^2F-(1+2sx)J\\ &\quad-p^2J^2, \end{aligned}

so that $H=H_{\mathrm{top}}-pu$ and $xy-z(z+1)=p^2H_{\mathrm{top}}$.

The element $x$ has weight zero and $H_{\mathrm{top}}$ is homogeneous of weight two.

Proposition 2.1. Identify $\operatorname{gr}A$ with its image under (1.2). Then

\begin{aligned} G:=\operatorname{gr}A&=R[\tau,f/\tau^2,g/\tau^2]\\ &\subset R[\tau,\tau^{-1}], \end{aligned}

the generators $p,s,u,F,J$ of $A$ are exact for the valuation degree, and $G\cong P/(H_{\mathrm{top}})$ as graded rings, with $p,s,u,F,J$ going to $\tau,s_R,u,f/\tau^2,g/\tau^2$. The ring $G$ is a smooth domain, finitely generated over $k$.

Proof. Let $\theta:P\to R[\tau^{\pm1}]$ substitute the five classes in (2.2). Then $\theta(x)=s_R^2+u^3+f=x$, $\theta(y)=s_R+x(x-u^3)=y$ and $\theta(z)=s_Rx+g=z$. So $\theta(P)$ contains $R$, and equals $R[\tau,f/\tau^2,g/\tau^2]$. After inverting $p$, Section 2 of the previous lesson gives $P[p^{-1}]=k[p^{\pm1},x,y,z,u]$, and $\theta$ becomes the quotient map onto $R[\tau^{\pm1}]$ sending $p$ to $\tau$, with kernel generated by $xy-z(z+1)=p^2H_{\mathrm{top}}$. The polynomial $H_{\mathrm{top}}$ is prime in $P$: it is irreducible after inverting $p$, and $p$ does not divide it, because $H_{\mathrm{top}}\equiv x_0^2F-(1+2sx_0)J$ modulo $p$ with $x_0=s^2+u^3$; the factorization argument of Lemma 1.1 applies. So $(H_{\mathrm{top}})$ is saturated with respect to $p$, and $\ker\theta=(H_{\mathrm{top}})$.

Exactness: let $Q\in P$ represent $0\ne h\in A$, with maximal monomial weight $M$. Each monomial has valuation degree at most its weight, so $\deg h\le M$. Suppose $\deg h<M$. The class of $h$ in degree $M$ is zero, and it equals $\theta(Q_M)$ where $Q_M$ is the weight-$M$ part of $Q$, since the other monomials lie in $\mathcal F_{M-1}A$. So $Q_M\in(H_{\mathrm{top}})$, say $Q_M=H_{\mathrm{top}}C$ with $C$ homogeneous of weight $M-2$; homogeneity of $C$ can be arranged because $P$ is a graded domain. Then $Q-HC=Q-Q_M+puC$ represents $h$ and has maximal weight less than $M$, since $puC$ has weight $M-3$. The maximal weight decreases at each step and is bounded below by $\deg h$, so we reach a representative of maximal weight exactly $\deg h$. Its top part maps under $\theta$ to the initial form of $h$. So $G=\theta(P)\cong P/(H_{\mathrm{top}})$.

Smoothness: $G$ is a domain inside $R[\tau^{\pm1}]$. After inverting $\tau$ it is $R[\tau^{\pm1}]$, smooth by Lemma 1.1(2). At a prime containing $\tau$, the partial derivatives $\partial H_{\mathrm{top}}/\partial F\equiv x_0^2$ and $\partial H_{\mathrm{top}}/\partial J\equiv-(1+2sx_0)$ modulo $p$ do not both vanish, because

(1-2sx_0)(1+2sx_0)+4s^2x_0^2=1 .

The Jacobian criterion applies at every prime. $\square$

The ring $G$ is an affine modification of $R[\tau]$ with centre $(\tau^2,I)$, in the sense of Kaliman and Zaidenberg [Kaliman–Zaidenberg]. By Theorem 2.1 of Smooth algebras over a field and the Jacobian criterion, $G$ is a regular Noetherian domain.

3. What a polynomial structure would force

Proposition 3.1. If $A\cong k^{[4]}$, then $G$ has a nonzero homogeneous locally nilpotent derivation $D_0$ and a nonzero homogeneous element $h_0\in\ker D_0$ of positive degree.

Proof. Let $q_1,\ldots,q_4$ be polynomial coordinates of $A$. The elements of degree at most zero form a $k$-subalgebra $\mathcal F_0A$, which does not contain $F$, of degree two. So some coordinate $q_r$ has positive degree, since otherwise the coordinates would generate $A$ inside $\mathcal F_0A$. Choose $i\ne r$ and let $D=\partial/\partial q_i$, a nonzero locally nilpotent derivation of $A$ with $D(q_r)=0$. The generators $p,s,u,F,J$ are exact by Proposition 2.1, so Proposition 6.2 of Locally nilpotent derivations gives a nonzero homogeneous locally nilpotent derivation $D_0$ of $G$, and Proposition 6.1 there gives $D_0(\operatorname{in}q_r)=0$. Put $h_0=\operatorname{in}q_r$, of degree $\deg q_r>0$. $\square$

The derivation $D_0$ need not preserve the subring $R$ of $G$. For this reason the next step lifts the action on $\operatorname{Spec}G$ itself, not an action on $\operatorname{Spec}R$.

4. The line bundle that makes $I$ principal

Let $\widetilde R=k[a,d,b,c,u]/(ac-bd-1)$, and consider the assignments (4.1):

\begin{gathered} x\mapsto ab,\quad y\mapsto dc,\\ z\mapsto db,\quad u\mapsto u . \end{gathered}

This defines $R\to\widetilde R$, since $xy=(ab)(dc)=(db)(ac)=z(z+1)$. Grade $\widetilde R$ by the fibre weights $\operatorname{wt}(a)=\operatorname{wt}(d)=1$, $\operatorname{wt}(b)=\operatorname{wt}(c)=-1$, $\operatorname{wt}(u)=0$, and put

\begin{aligned} v&=a^3b-a^2u^3-d^2\\ &=a^2(x-u^3)-d^2, \end{aligned}\tag{4.2}

which has weight two.

Proposition 4.1.

$\widetilde R$ is a domain, and a Laurent bundle algebra over $R$ in the sense of Additive actions on line bundles, with the elements $z+1$ and $-z$ of $R$ and fibre coordinates $a$ and $d$. In particular $R\to\widetilde R$ is injective with image $\widetilde R_0$.
In $\widetilde R$, with $C_1=c^2-b^2(x-u^3)$, $\alpha=a^2(1-2bd)$ and $\beta=d^2(3+2bd)+\alpha(x-u^3)$,

\begin{gathered} f=C_1v,\\ g=b^2v,\qquad v=\alpha f+\beta g . \end{gathered}\tag{4.3}

So $I\widetilde R=v\widetilde R$.

Proof. (1) The relation $ac-(bd+1)$ is linear in $c$ with coprime coefficients $a$ and $-(bd+1)$, so $\widetilde R$ is a domain. After inverting $z+1=ac$, the element $a$ is a unit, and $b=x/a$, $c=(z+1)/a$, $d=ay/(z+1)$. These formulas define mutually inverse homomorphisms between $\widetilde R_{z+1}$ and $R_{z+1}[a,a^{-1}]$; the relation is respected because $(z+1)-(x/a)(ay/(z+1))-1$ equals $(z(z+1)-xy)/(z+1)=0$ in $R$. After inverting $z=db$, the element $d$ is a unit, and $b=z/d$, $c=y/d$, $a=dx/z$ identify $\widetilde R_z$ with $R_z[d,d^{-1}]$. Since $z+1$ and $-z$ sum to one, $\widetilde R$ is a Laurent bundle algebra over its weight-zero part; and $R\subset\widetilde R_0\subset R_z\cap R_{z+1}=R$, because an element of $R_z\cap R_{z+1}$ is multiplied into $R$ by powers of $z$ and of $z+1$, which generate the unit ideal.

(2) Write $h=x-u^3$. Since $s_R=dc-abh$ and $ac-bd=1$, we get $as_R-d=a\,dc-a^2bh-d$, which is $d(ac-1)-a^2bh=b(d^2-a^2h)=-bv$. Hence $g=db-s_Rab=b(d-as_R)=b^2v$. Next, after expanding, $\begin{aligned} &f-C_1v=h-(dc-abh)^2\\ &\quad-(c^2-b^2h)(a^2h-d^2)\\ &\quad=h\bigl(1-(ac-bd)^2\bigr)=0. \end{aligned}$ Finally, with $q=bd$ and $ac=1+q$, $\begin{aligned} &\alpha C_1+\beta b^2\\ &\quad=(ac)^2(1-2bd)+b^2d^2(3+2bd)\\ &\quad=(1+q)^2(1-2q)+3q^2+2q^3=1, \end{aligned}$ and so $\alpha f+\beta g=(\alpha C_1+\beta b^2)v=v$. $\square$

Now base change the line bundle to $\operatorname{Spec}G$: put

\widetilde G=G\otimes_R\widetilde R .

Proposition 4.2. $\widetilde G$ is a Laurent bundle algebra over $G$, and a domain. It is the subring $\widetilde R[\tau,v/\tau^2]$ of $\widetilde R[\tau,\tau^{-1}]$, and writing $V=v/\tau^2$,

\widetilde G\cong\widetilde R[\tau,V]/(\tau^2V-v).

It carries two commuting gradings: the valuation degree, with $\widetilde R$ in degree $0$, $\deg\tau=-1$ and $\deg V=2$; and the fibre weight, with $\operatorname{wt}(\tau)=0$ and $\operatorname{wt}(V)=2$.

Proof. A base change of a Laurent bundle algebra is one. The $R$-module $\widetilde R$ is a direct sum of invertible modules, hence flat, and it contains $R=\widetilde R_0$ as a direct summand, so $R\to\widetilde R$ is faithfully flat. Tensoring the injection $G\subset R[\tau^{\pm1}]$ with $\widetilde R$ gives an injection $\widetilde G\to\widetilde R[\tau^{\pm1}]$, whose image is generated by $\widetilde R$, $\tau$, $f/\tau^2$, $g/\tau^2$; by (4.3) it is $\widetilde R[\tau,v/\tau^2]$, a domain. For the presentation, let $Q=\widetilde R[\tau,W]/(\tau^2W-v)$. If $\tau\,\Psi=(\tau^2W-v)\Theta_1$ in $\widetilde R[\tau,W]$, then setting $\tau=0$ gives $v\,\Theta_1(0,W)=0$. Here $v\ne0$: by (4.3) $f=C_1v$, the element $f$ of $R$ is nonzero (Section 2), and $R\to\widetilde R$ is injective (Proposition 4.1). As $\widetilde R[W]$ is a domain, so $\Theta_1(0,W)=0$, $\Theta_1=\tau\Theta_2$ and $\Psi=(\tau^2W-v)\Theta_2$. So $\tau$ is not a zero divisor in $Q$, and $Q$ embeds in $Q[\tau^{-1}]=\widetilde R[\tau^{\pm1}]$ with image $\widetilde R[\tau,v/\tau^2]$. The relation $\tau^2V=v$ is homogeneous for both gradings. $\square$

Every element of $\widetilde G$ of valuation degree $e$ is a combination of monomials $\tau^iV^h$ with $2h-i=e$, and $\tau^2V=v$ rewrites them as follows.

Lemma 4.3. The valuation-degree components of $\widetilde G$ are

\widetilde G_e=\begin{cases}\tau^{-e}\widetilde R,&e<0,\\ V^\ell\widetilde R,&e=2\ell\ge0,\\ \tau V^{\ell+1}\widetilde R,&e=2\ell+1\ge1,\end{cases}

and in each case the coefficient in $\widetilde R$ is unique. In particular every component of negative degree and every component of positive odd degree lies in $\tau\widetilde G$.

Proof. The monomials $\tau^iV^h$ of degree $e$ become $\tau^{-e}v^h$ when $e<0$, $V^\ell v^{h-\ell}$ when $e=2\ell$, and $\tau V^{\ell+1}v^{h-\ell-1}$ when $e=2\ell+1$, with nonnegative exponents of $v$. Conversely the displayed multipliers lie in $\widetilde G$ and have degree $e$. Uniqueness holds because the multipliers are nonzero in the domain $\widetilde G$. $\square$

5. Lifting the derivation

Proposition 5.1. Let $D_0$ be a nonzero locally nilpotent derivation of $G$, homogeneous of valuation-degree shift $j$. There is a nonzero locally nilpotent derivation $\widetilde D$ of $\widetilde G$ extending $D_0$, preserving the fibre weights, and homogeneous of valuation-degree shift $j$.

Proof. The ring $G$ is a regular Noetherian domain containing $k$ (Proposition 2.1), and $\widetilde G$ is a Laurent bundle algebra over it. Theorem 3.2 of Additive actions on line bundles gives a unique weight-preserving locally nilpotent derivation $\widetilde D$ extending $D_0$; it is nonzero because $D_0$ is.

For $\lambda\in k^*$, let $\sigma_\lambda$ be the automorphism of $G$ multiplying elements of valuation degree $e$ by $\lambda^e$. It fixes $R$, so $\widetilde\sigma_\lambda=\sigma_\lambda\otimes\mathrm{id}_{\widetilde R}$ is an automorphism of $\widetilde G$ preserving fibre weights, with $\widetilde\sigma_\lambda(G)=G$; it multiplies valuation degree $e$ by $\lambda^e$ on all of $\widetilde G$. Homogeneity of $D_0$ says $\sigma_\lambda D_0\sigma_\lambda^{-1}=\lambda^jD_0$. By Corollary 3.3 of that lesson, $\widetilde\sigma_\lambda\widetilde D\widetilde\sigma_\lambda^{-1}=\lambda^j\widetilde D$. So if $r$ has valuation degree $e$, then $\widetilde\sigma_\lambda(\widetilde D(r))=\lambda^{e+j}\widetilde D(r)$ for all $\lambda\in k^*$. Writing $\widetilde D(r)=\sum_mc_m$ in valuation-degree components, $\sum_m\lambda^mc_m=\lambda^{e+j}\sum_mc_m$ for infinitely many $\lambda$, so $c_m=0$ for $m\ne e+j$. $\square$

6. A derivation of $\widetilde R$ of nonpositive weight

Proposition 6.1. If $A\cong k^{[4]}$, there is a nonzero locally nilpotent derivation $E$ of $\widetilde R$, homogeneous of some fibre-weight shift $q\le0$, with $E^2(v)=0$.

Proof. Take $D_0$ and $h_0$ from Proposition 3.1 and $\widetilde D$ from Proposition 5.1, with valuation-degree shift $j$. The image of $h_0$ in $\widetilde G$ is a nonzero invariant of positive degree, so by Lemma 4.3 it is divisible by $V$. The kernel of $\widetilde D$ is factorially closed (Proposition 3.2 of Locally nilpotent derivations), so

\widetilde D(V)=0 . \tag{6.1}

Case 1: $\widetilde D(\tau)=0$. Then $\widetilde D$ extends to a locally nilpotent derivation of $\widetilde G[\tau^{-1}]=\widetilde R[\tau^{\pm1}]$ (Corollary 3.3 of that lesson). Its degree-$j$ component is $\tau^{-j}\widetilde R$, so for $r\in\widetilde R$ there is a unique $E(r)\in\widetilde R$ with $\widetilde D(r)=\tau^{-j}E(r)$. Cancelling $\tau^{-j}$ in the linearity and product rules shows that $E$ is a derivation of $\widetilde R$, of fibre-weight shift zero. Since $\widetilde D(\tau)=0$, induction gives $\widetilde D^n(r)=\tau^{-nj}E^n(r)$, so $E$ is locally nilpotent. By (6.1), $0=\widetilde D(\tau^2V)=\widetilde D(v)=\tau^{-j}E(v)$, so $E(v)=0$. If $E=0$, then $\widetilde D$ would kill $\widetilde R$, $\tau$ and $V$, which generate $\widetilde G$. So $E\ne0$, and $q=0$.

Case 2: $\widetilde D(\tau)\ne0$. Write $\nu$ for the order function of $\widetilde D$ on $\widetilde G\setminus0$, and $m=\nu(\tau)\ge1$. By Proposition 3.2(4) of that lesson, $\widetilde D(\tau)\notin\tau\widetilde G$. It is homogeneous of degree $j-1$, so by Lemma 4.3, $j-1$ is even and nonnegative: $j=2\ell+1$ with $\ell\ge0$. For $r\in\widetilde R$, $\widetilde D(r)$ has degree $j$, so Lemma 4.3 gives a unique $E(r)\in\widetilde R$ with

\widetilde D(r)=\tau V^{\ell+1}E(r). \tag{6.2}

Cancelling the nonzero factor $\tau V^{\ell+1}$, of fibre weight $2(\ell+1)$, shows that $E$ is a derivation of $\widetilde R$ homogeneous of fibre-weight shift $q=-2(\ell+1)<0$.

Whenever $E(r)\ne0$, comparing orders in (6.2) and using $\nu(V)=0$ from (6.1) gives $\nu(r)-1=m+\nu(E(r))$, that is,

\nu(E(r))=\nu(r)-1-m . \tag{6.3}

So $\nu(E^n(r))=\nu(r)-n(m+1)$ as long as $E^n(r)\ne0$, and since orders are nonnegative, $E$ is locally nilpotent. Differentiating $\tau^2V=v$ gives $\tau V^{\ell+1}E(v)=\widetilde D(v)=2\tau V\widetilde D(\tau)\ne0$, so $E(v)\ne0$ and $E\ne0$. Also $\nu(v)=\nu(\tau^2V)=2m$, so (6.3) gives $\nu(E(v))=m-1$, and if $E^2(v)\ne0$ a second application would give $\nu(E^2(v))=-2$, which is impossible. So $E^2(v)=0$. $\square$

The sign of the weight matters: the derivation $E_+$ with $E_+(b)=a$, $E_+(c)=d$, $E_+(a)=E_+(d)=E_+(u)=0$ kills $ac-bd-1$, is locally nilpotent, has $E_+(v)=a^4$ and $E_+^2(v)=0$, and has fibre-weight shift $+2$.

7. The polynomial $abc$ theorem

Theorem 7.1 (Mason, Stothers). Let $K$ be a field of characteristic zero, and let $X,Y,Z\in K[q]$ be nonzero, pairwise coprime, not all constant, with $X+Y=Z$. Then

\begin{aligned} &\max\{\deg X,\deg Y,\deg Z\}\\ &\quad \le r(XYZ)-1, \end{aligned}

where $r(XYZ)$ is the number of distinct roots of $XYZ$ in an algebraic closure of $K$.

Proof. Let $W=X'Y-XY'$, with primes denoting $d/dq$. If $W=0$, then $X$ divides $X'Y$, hence $X'$, so $X$ is constant; then $Y'=0$, and $Y$, $Z$ are constant too, which is excluded. So $W\ne0$. Since $Z=X+Y$, the same $W$ arises up to sign from any two of the three polynomials. A root of multiplicity $h$ of one of them is a root of $W$ of multiplicity at least $h-1$, computing $W$ from a pair containing that polynomial; the roots of the three polynomials are distinct by coprimality. So, with $N=\deg X+\deg Y+\deg Z$, we get $\deg W\ge N-r(XYZ)$. On the other hand, computing $W$ from the two polynomials other than one of maximal degree $M$ gives $\deg W\le N-M-1$. Together, $M\le r(XYZ)-1$. $\square$

The theorem is due to Stothers and, independently, to Mason. The proof through the Wronskian is the one written out in [OpenAI-cancellation, Section 6].

8. The rigidity argument

Proposition 8.1. There is no nonzero locally nilpotent derivation $E$ of $\widetilde R$, homogeneous of fibre-weight shift $e\le0$, with $E^2(v)=0$.

Proof. Suppose $E$ is one.

An auxiliary grading. Grade $\widetilde R$ by the auxiliary degrees $(a,d,b,c,u)\mapsto(0,3,-3,0,2)$; the relation $ac-bd-1$ is homogeneous of degree zero, and this grading commutes with the fibre weights. Let $E'$ be the component of $E$ of largest auxiliary shift $\lambda$. By Proposition 5.1 of Locally nilpotent derivations, $E'$ is a nonzero locally nilpotent derivation, still of fibre-weight shift $e$. Put $N=d^2+a^2u^3$, homogeneous of auxiliary degree $6$, so that $v=a^3b-N$ with $a^3b$ of auxiliary degree $-3$. The component of $E^2(v)$ of auxiliary degree $6+2\lambda$ receives nothing from $E^2(a^3b)$, whose components have degree at most $-3+2\lambda$, and from $E^2(N)$ it receives $(E')^2(N)$. So

(E')^2(N)=0 . \tag{8.1}

The orbits of $a$, $d$, $u$. Inverting $a$ gives $\widetilde R[a^{-1}]=k[a^{\pm1},d,u][b]$ with $c=(1+bd)/a$, so $a,d,b,u$ are algebraically independent and $K=\operatorname{Frac}\widetilde R=k(a,d,b)(u)$. Let $a(q)=\exp(qE')(a)$, $d(q)=\exp(qE')(d)$, $u(q)=\exp(qE')(u)$, polynomials in $K[q]$, nonzero because their values at $q=0$ are nonzero. Since $\exp(qE')$ is multiplicative, (8.1) gives

\begin{aligned} &d(q)^2+a(q)^2u(q)^3\\ &\quad =N+q\,E'(N), \end{aligned}\tag{8.2}

a nonzero polynomial of degree at most one. The two summands on the left are coprime: a common root would be a root of multiplicity at least two of each, hence of their sum, which is impossible for a nonzero polynomial of degree at most one. So the three polynomials in (8.2) are pairwise coprime.

Write $D_a,D_d,D_u$ for the degrees of $a(q),d(q),u(q)$. Suppose a summand on the left is not constant. A nonconstant summand has degree at least two, and the right side has degree at most one, so both summands have the same degree $M=2D_d=2D_a+3D_u\ge2$. The distinct roots of the product of the three polynomials number at most $D_d+D_a+D_u+1$, so Theorem 7.1 gives $2D_d\le D_d+D_a+D_u$, that is, $D_d\le D_a+D_u$. With $2D_d=2D_a+3D_u$ this gives $3D_u\le2D_u$, so $D_u=0$ and $D_a=D_d>0$. The leading terms of degree $M$ must cancel: if $\alpha_0,\delta_0\in K^*$ are the leading coefficients of $a(q)$ and $d(q)$, then $\delta_0^2+\alpha_0^2u^3=0$, so $(\delta_0/\alpha_0)^2=-u^3$ in $K$. But in $K=K_0(u)$, with $K_0=k(a,d,b)$, the order of vanishing at $u=0$ of a square is even, while $-u^3$ has order three. This contradiction shows that both summands are constant, so $D_a=D_d=D_u=0$, and comparing coefficients of $q$,

\begin{aligned} E'(a)&=E'(d)\\ &=E'(u)=0 . \end{aligned}\tag{8.3}

The remaining values. Differentiating $ac-bd=1$ and using (8.3) gives $aE'(c)=dE'(b)$. Put $h=cE'(b)-bE'(c)$. Then $ah=(ac-bd)E'(b)=E'(b)$ and $dh=(ac-bd)E'(c)=E'(c)$. Let $C_0=k[a,d,u]$ and $Q=k(a,d,u)$. Every nonzero element of $C_0$ lies in $\ker E'$ by (8.3), so $E'$ extends to a locally nilpotent $Q$-linear derivation of the localization $(C_0\setminus0)^{-1}\widetilde R=Q[b]$ (Corollary 3.3 of the lesson on locally nilpotent derivations). By Proposition 4.1 of that lesson, $E'(b)\in Q$, so $h=E'(b)/a\in Q$. Moreover $\widetilde R\cap Q=C_0$: solving for $b$ instead gives $\widetilde R[d^{-1}]=k[a,d^{\pm1},u][c]$, so an element of $Q\cap\widetilde R$ lies in $k[a^{\pm1},d,u]\cap k[a,d^{\pm1},u]=C_0$, the last equality by unique factorization in $C_0$. Hence $h\in C_0$. It is nonzero, since otherwise $E'$ would kill $a,d,u,b,c$. Its fibre weight is $\operatorname{wt}(c)+\operatorname{wt}(b)+e=e-2<0$. But every nonzero homogeneous element of $C_0=k[a,d,u]$ has nonnegative weight. This contradiction proves the proposition. $\square$

9. Conclusion

Theorem 9.1. Let $k$ be a field of characteristic zero and $A=P/(H)$ as above. Then $A[w]\cong k^{[5]}$ but $A\not\cong k^{[4]}$. So the cancellation problem has a negative answer for affine four-space over $k$.

Proof. The first assertion is Theorem 4.1 of the previous lesson. If $A\cong k^{[4]}$, Proposition 6.1 produces a derivation that Proposition 8.1 excludes. $\square$

Reference: [OpenAI-cancellation, Theorem 1.1].

10. Exercises

Exercise 10.1 (easy). Verify that the auxiliary degrees $(0,3,-3,0,2)$ make $ac-bd-1$ homogeneous and give $N$ degree $6$ and $a^3b$ degree $-3$.

Exercise 10.2 (medium). Show directly that the derivation $E_+$ at the end of Section 6 is locally nilpotent and compute $E_+(v)$.

Exercise 10.3 (medium). In Proposition 8.1, explain where the hypothesis $e\le0$ is used, and why the argument fails for $E_+$.

11. Solutions

10.1. $ac$ has degree $0+0$, $bd$ has degree $-3+3=0$, and $1$ has degree $0$. $d^2$ has degree $6$, $a^2u^3$ has degree $0+6=6$, and $a^3b$ has degree $0-3=-3$.

10.2. On $k[a,d,b,c,u]$, $E_+=a\,\partial/\partial b+d\,\partial/\partial c$ kills $a,d,u$ and lowers the total degree in $b,c$ by one, so it is locally nilpotent; it kills $ac-bd-1$ because $E_+(ac)-E_+(bd)=ad-ad=0$, so it induces a locally nilpotent derivation of $\widetilde R$. Then $E_+(v)=E_+(a^3b)=a^4$ and $E_+^2(v)=0$.

10.3. It is used only at the very end: $h\in k[a,d,u]$ has weight $e-2$, which is negative exactly when $e<2$. For $E_+$, with $e=+2$, one finds $E_+(a)=E_+(d)=E_+(u)=0$ and $h=cE_+(b)-bE_+(c)=ca-bd=1$, of weight zero, and no contradiction arises.

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
[Kaliman–Makar-Limanov] S. Kaliman, L. Makar-Limanov, AK-invariant of affine domains, in Affine Algebraic Geometry, Osaka University Press, 2007, 231–255. Author version: https://www.math.miami.edu/~kaliman/library/KM-L.miyanishi.pdf
[Kaliman–Zaidenberg] S. Kaliman, M. Zaidenberg, Affine modifications and affine hypersurfaces with a very transitive automorphism group, Transform. Groups 4 (1999), 53–95. https://arxiv.org/abs/math/9801076

\(h\)	\(\deg h\)	image of \(\operatorname{in}(h)\) in (1.2)
\(p\)	\(-1\)	\(\tau\)
\(s\)	\(0\)	\(s_R\)
\(u\)	\(0\)	\(u\)
\(F\)	\(2\)	\(f/\tau^2\)
\(J\)	\(2\)	\(g/\tau^2\)

Cancellation fails in dimension four

1. The valuation along \(p=0\)