# 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](a-fourfold-whose-cylinder-is-affine-space.md), 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:

1. 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;
2. to a line bundle over \(\operatorname{Spec}G\) on which an ideal with two generators becomes principal, generated by one element \(v\);
3. 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](locally-nilpotent-derivations.md) and [Additive actions on line bundles](additive-actions-on-line-bundles.md) 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](course:AG-CA/smooth-algebras-over-a-field-and-the-jacobian-criterion), Krull's intersection theorem from [Noetherian and Artinian rings](course:AG-CA/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.**

1. \(S\) is a domain, \(S\to A\) is injective, and \(S[p^{-1}]=A[p^{-1}]\).
2. \(R\) is a domain, smooth over \(k\).
3. 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](course:AG-CA/smooth-algebras-over-a-field-and-the-jacobian-criterion#2-the-criterion-and-its-open-locus), \(R\) 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](course:AG-CA/noetherian-and-artinian-rings), \(\bigcap_np^nS_{pS}=0\); 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](locally-nilpotent-derivations.md). 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](course:AG-CA/smooth-algebras-over-a-field-and-the-jacobian-criterion#2-the-criterion-and-its-open-locus), \(G\) 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](locally-nilpotent-derivations.md) 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.**

1. \(\widetilde R\) is a domain, and a Laurent bundle algebra over \(R\) in the sense of [Additive actions on line bundles](additive-actions-on-line-bundles.md), 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 \(\widetilde R_0\).
2. 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](additive-actions-on-line-bundles.md) 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](locally-nilpotent-derivations.md)), 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](locally-nilpotent-derivations.md), \(E'\) 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