64c3814
==== TABLE OF CONTENTS ====
An-explicit-failure-of-complex-affine-space-cancellation//01-introduction.tex
An-explicit-failure-of-complex-affine-space-cancellation//02-construction.tex
An-explicit-failure-of-complex-affine-space-cancellation//03-degeneration.tex
An-explicit-failure-of-complex-affine-space-cancellation//04-bundle.tex
An-explicit-failure-of-complex-affine-space-cancellation//05-reduction.tex
An-explicit-failure-of-complex-affine-space-cancellation//06-rigidity.tex
An-explicit-failure-of-complex-affine-space-cancellation//07-affine-fibrations.tex
An-explicit-failure-of-complex-affine-space-cancellation//main.tex
An-explicit-failure-of-complex-affine-space-cancellation//preamble.tex
An-explicit-failure-of-complex-affine-space-cancellation//references.tex
===========================
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//01-introduction.tex ====
\section{Introduction}
\label{sec:introduction}
The Zariski cancellation problem asks whether an affine variety is determined
by its cylinder when that cylinder is affine space. Over the complex numbers,
its algebraic formulation is the implication
\begin{equation}
A[w]\simeq \mathbb C\poly{n+1}
\quad\Longrightarrow\quad
A\simeq\mathbb C\poly n,
\label{eq:cancellation-question}
\end{equation}
where $A$ is a finitely generated commutative unital complex algebra, $w$ is
an independent polynomial variable, and all isomorphisms fix the complex
scalars. Here $C\poly r$ denotes a polynomial ring in $r$ variables over a
ring $C$. The hypothesis says that adjoining one variable turns $A$ into a
polynomial ring; the question is whether that variable can be cancelled.
An isomorphism of cylinders need not respect the given projection to the
affine line, so its existence alone does not identify polynomial
coordinates on $A$.
Cancellation for the affine line follows from the work of Abhyankar,
Heinzer, and Eakin~\cite[Corollary~(2.8)]{AEH}. Cancellation for the
complex affine plane was established through the work of Fujita,
Miyanishi, and Sugie~\cite{Fujita,MiyanishiSugie}, using the structure of
affine surfaces and cylinderlike open sets. In positive characteristic,
Asanuma's work on polynomial fiber rings supplied threefolds whose
one-variable cylinders are affine four-space~\cite{Asanuma}. Gupta
proved that suitable members of this family are not affine three-space
over their ground field~\cite{GuptaThreefold}, then established failure
of cancellation in every dimension at least three over every field of
positive characteristic~\cite{GuptaPositive}. The latter result uses a
separate family of examples: adjoining variables to a threefold whose
first cylinder is already polynomial would not produce higher-dimensional
counterexamples.
The characteristic-zero problem in dimensions at least three is still
listed as open in Gaifullin--Petrov's July 2026
preprint~\cite[p.~1]{GaifullinPetrov}. Gupta's
survey~\cite{GuptaSurvey} explains its history and its connections with
coordinates, polynomial fibrations, and the recognition of affine
space. We give a negative answer to \eqref{eq:cancellation-question}
over $\mathbb C$ for $n=4$.
The defining polynomial also gives a negative answer to the Stable
Coordinate Conjecture over $\kk$ in ambient dimension five, already after
adjoining one variable.
Let $p,s,u,F,J$ be algebraically independent over $\kk$, and set
\begin{equation}
\begin{split}
P&=\kk[p,s,u,F,J],\qquad x=s^2+u^3+p^2F,\\
H&=x^2F-(1+2sx)J-p^2J^2-pu,\qquad A=P/(H).
\end{split}
\label{eq:example}
\end{equation}
Thus $x$ is an abbreviation for the displayed polynomial, and $H$ is a
polynomial in exactly the five indicated variables.
\begin{theorem}
\label{thm:main}
The algebra $A$ in \eqref{eq:example} is an integral complex algebra of
dimension four. For an independent variable $w$,
\[
A[w]\simeq\kk\poly5,
\qquad
A\not\simeq\kk\poly4.
\]
Consequently the Zariski cancellation problem over the complex numbers has
a negative answer in dimension four.
\end{theorem}
\paragraph{The polynomial cylinder.}
The stabilization belongs to the theory of residual and stable
coordinates. A polynomial in $\kk[p]\poly4$ is a \emph{residual
coordinate} if its image is a polynomial coordinate over the residue
field of every prime of $\kk[p]$; it is \emph{one-stably a coordinate}
if it becomes a coordinate after adjoining one variable. Our $H$ is a
residual coordinate: it is a coordinate after inverting $p$, while a
linear change in $F,J$ makes its reduction at $p=0$ a coordinate.
Its one-stability therefore follows from the theorem of
El Kahoui--Ouali~\cite[Theorem~1.1]{ElKahouiOuali}, generalized by
Dutta--Lahiri~\cite[Theorem~3.4]{DuttaLahiri}.
The Stable Coordinate Conjecture asks whether a polynomial
$f\in\kk\poly n$ that becomes a coordinate in $\kk\poly{n+1}$ after
adjoining one variable must already be a coordinate in $\kk\poly n$
\cite[p.~342]{MakarLimanovVanRossumShpilrainYu}. The assertion concerns
polynomial automorphisms of the ambient spaces.
\begin{corollary}[Failure of the Stable Coordinate Conjecture]
\label{cor:stable-coordinate}
The polynomial $H$ in \eqref{eq:example} is not a coordinate of
$P=\kk[p,s,u,F,J]$, but it is a coordinate of $P[w]$ for one independent
variable $w$. Thus the Stable Coordinate Conjecture over $\kk$ fails in
ambient dimension five after one stabilization.
\end{corollary}
\begin{proof}
Since $H$ is a residual coordinate of $P=\kk[p]\poly4$, the theorem of
Dutta--Lahiri~\cite[Theorem~3.4]{DuttaLahiri}, applied with $R=\kk[p]$
and four relative variables, gives $G_2,G_3,G_4,G_5\in P[w]$ such that
\[
P[w]=\kk[p][H,G_2,G_3,G_4,G_5]
=\kk[p,H,G_2,G_3,G_4,G_5].
\]
Hence $H$ is a coordinate of $P[w]$ as a $\kk$-algebra. If $H$ were a
coordinate of $P$ as a $\kk$-algebra, quotienting a polynomial
coordinate system containing $H$ by $(H)$ would give
$A=P/(H)\simeq\kk\poly4$, contrary to \Cref{thm:main}.
\end{proof}
A derivation $D$ is \emph{locally nilpotent} if every element is killed
by some iterate of $D$; over $\kk$, its exponential is an algebraic
action of the additive group $\Ga$. We make the stabilization method
explicit. A locally nilpotent derivation of $P$
sends $H$ to $p^3$. Exponentiating it converts $H$ into $H+p^3w$,
and a coordinate calculation identifies the resulting quotient with a
polynomial ring. This exponential-shift argument follows the method of
El Kahoui--Ouali~\cite[Lemma~3.1 and proof of Theorem~1.1]{ElKahouiOuali};
the final coordinate change realizes the transfer criterion of
Edo--V\'en\'ereau~\cite[Theorem~7]{EdoVenereau}. We include its
polynomial inverse and the divisibility that makes it valid at $p=0$.
\paragraph{Affine-space fibrations.}
An $\Aff^3$-fibration is a morphism of complex affine varieties whose
base change to every geometric point of the target is affine three-space.
The affine-fibration assertion of Dolgachev and Weisfeiler predicts that
an affine morphism to a normal locally noetherian integral base is
Zariski-locally a product with affine $r$-space if every fiber is affine
$r$-space over its residue field~\cite[Section~3.8.5]{DolgachevWeisfeiler}.
The residual-coordinate property of $H$ gives two counterexamples in
relative dimension three:
\[
p:\Spec A\longrightarrow\Aff^1,
\qquad
(p,H):\Spec P=\Aff^5\longrightarrow\Aff^2.
\]
Both maps are smooth $\Aff^3$-fibrations, but neither is a
Zariski-locally trivial $\Aff^3$-bundle. \Cref{sec:affine-fibrations}
proves this consequence using the locally-polynomial-algebra theorem
of Bass--Connell--Wright~\cite[Theorem~3(a)]{BassConnellWright} and the
Quillen--Suslin theorem~\cite[Theorem~2(b)]{Suslin}.
\paragraph{From polynomiality to a homogeneous additive action.}
The main work is to prove that $A$ itself is not polynomial.
Additive actions have long served to distinguish affine varieties from
affine space: Makar-Limanov used them to show that the Russell cubic is
not affine three-space~\cite{MakarLimanov}. Our obstruction has three
stages, beginning with the passage from filtered derivations to
homogeneous derivations developed by Kaliman and
Makar-Limanov~\cite[Sections~3--5]{KalimanMakarLimanov}.
The construction in \Cref{sec:construction,sec:degeneration} identifies
an auxiliary subalgebra of $A$ with the presentation
\[
S=\kk[p,x,y,z,u]/\bigl(xy-z(z+1)-p^3u\bigr),
\qquad S[p^{-1}]=A[p^{-1}].
\]
Under this inclusion, $p^2F$ and $p^2J$ belong to $S$. Define
\[
R=S/pS=\kk[x,y,z,u]/\bigl(xy-z(z+1)\bigr),\qquad
I=(\overline{p^2F},\overline{p^2J})\subset R,
\]
where the bars denote residues of elements of $S$ modulo $pS$.
The order $\ord_p$ along $p=0$ in $S$ applies to every nonzero
element of $A$. Its negative defines the \emph{valuation degree}
$\deg h=-\ord_p(h)$ and a filtration of $A$. We prove the exact
identification
\[
G=\gr A=R[\tau,I/\tau^2]\subset R[\tau,\tau^{-1}],
\qquad \deg\tau=-1,\qquad \deg R=0.
\]
Here $\tau$ is an indeterminate representing the initial form of $p$,
and the right-hand side is generated by $R$, $\tau$, and all
$i/\tau^2$ with $i\in I$. The coefficient inclusion is the one induced
by $S\subset A$: the residue of $b\in S$ corresponds to its
degree-zero class. This is an affine modification of $R[\tau]$ in
the sense of Kaliman--Zaidenberg~\cite{KalimanZaidenberg}, and it is
smooth.
If $A$ were polynomial, some polynomial coordinate would have positive
valuation degree: otherwise all of $A$ would have nonpositive degree,
contrary to $\deg F=2$. Differentiation in a different coordinate fixes
that positive-degree coordinate. The exact graded identification then
induces a nonzero homogeneous locally nilpotent derivation of $G$ with
a nonzero homogeneous invariant of positive degree. This is the action
the remaining two stages exclude.
\paragraph{Lifting through the principalizing line bundle.}
The standard torus bundle from $\mathrm{SL}_2$ to the smooth affine
quadric principalizes $I$ after adjoining the unchanged variable $u$.
This quadric and its polynomial embeddings also occur in the work of
Blanc and van Santen~\cite{BlancVanSanten}. The bundle algebra and
the pullback of $G$ are
\[
\begin{gathered}
\Rt=\kk[a,d,b,c,u]/(ac-bd-1),\qquad
x=ab,\quad y=dc,\quad z=db,\\
\Gt=G\otimes_R\Rt
=\Rt[\tau,V]/(\tau^2V-v),\qquad
v=a^3b-a^2u^3-d^2.
\end{gathered}
\]
The valuation degree is zero on $\Rt$, with $\deg\tau=-1$ and
$\deg V=2$. Separately, fiber scaling gives $a,d$ weight $1$,
$b,c$ weight $-1$, and $u,\tau$ weight $0$; thus $v,V$ have
fiber weight $2$.
The morphism $\Spec\Gt\to\Spec G$ is the complement of the zero
section in a line bundle. Every additive action on the smooth affine
base $\Spec G$ lifts uniquely to an action on this line bundle that is
linear on fibers and
normalized at the identity. For the homogeneous action just obtained,
uniqueness makes the lift homogeneous with the same valuation degree
shift, while fiber linearity preserves fiber weights. This applies even
when the original action does not preserve the quadric algebra $R$.
A positive-degree invariant is divisible by $V$, forcing
$V$ itself to be invariant. The homogeneous pieces of $\Gt$, with
separate arguments according to whether $\tau$ is invariant, then
produce a nonzero locally nilpotent derivation $E$ of $\Rt$ with
nonpositive fiber weight and
\[
E^2(v)=0.
\]
Thus the lift converts the geometric action into a constrained
derivation on the explicit determinant ring.
\paragraph{The weighted rigidity obstruction.}
A highest component $E'$ for an auxiliary grading remains a nonzero
locally nilpotent derivation with the same fiber weight. The grading
separates the two parts of $v$ so that $(E')^2(d^2+a^2u^3)=0$.
The polynomial orbit of $d^2+a^2u^3$ therefore has degree at most one.
The Mason--Stothers polynomial $abc$ inequality~\cite{Mason,Stothers}
and the nonsquareness of $-u^3$ in $\Frac\Rt$ force $E'$ to fix
$a,d,u$. The determinant relation then forces positive fiber weight,
contradicting the sign obtained in the preceding stage.
The use of polynomial $abc$ for rigidity of additive actions has
precedents in Finston--Maubach~\cite[Section~2]{FinstonMaubach}; here
the nonsquare obstruction requires keeping the leading coefficients
in the original fraction field.
\begin{table}[bp]
\centering
\small
\begin{tabular}{@{}llp{0.48\textwidth}@{}}
\toprule
Ring & Definition & Role in the proof\\
\midrule
$A$ & $P/(H)$ & Coordinate ring of the fourfold.\\
$S$ & \eqref{deg:S-R} & Subalgebra of $A$ defining the order along $p=0$.\\
$R$ & $S/pS$ & Coefficient ring of the degeneration.\\
$G$ & $\gr A=R[\tau,I/\tau^2]$ & Exact graded algebra carrying the induced action.\\
$\Rt$ & \eqref{bun:determinant-ring} & Torus-bundle algebra in which $I$ becomes principal.\\
$\Gt$ & $G\otimes_R\Rt$ & Pullback algebra carrying the lifted action.\\
\bottomrule
\end{tabular}
\caption{The rings in the proof. The valuation degree on $G$ and
$\Gt$ is separate from the fiber weight on $\Rt$ and $\Gt$.}
\label{tab:ring-dictionary}
\end{table}
\paragraph{Organization.}
\Cref{tab:ring-dictionary} records the rings used in the argument.
\Cref{sec:construction} proves the stabilization.
\Cref{sec:degeneration} constructs the filtration and identifies $G$.
\Cref{sec:bundle} describes the torus bundle and proves the lifting
statement. \Cref{sec:reduction} extracts the constrained derivation on
$\Rt$, and \Cref{sec:rigidity} excludes it and completes the proof of
\Cref{thm:main}. \Cref{sec:affine-fibrations} derives the two nontrivial
affine-space fibrations. All locally nilpotent derivations and algebraic group
actions in the proof are defined over $\kk$.
==== END An-explicit-failure-of-complex-affine-space-cancellation//01-introduction.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//02-construction.tex ====
\section{The equation and its polynomial cylinder}\label{sec:construction}
We first prove that the displayed equation defines an integral fourfold,
then construct the five polynomial coordinates on its cylinder. The
localization at $p$ explains the equation; the stabilization argument
will produce coordinates in the original ring, including the fiber
$p=0$.
The construction realizes the residual-coordinate method of
El Kahoui--Ouali~\cite[Lemma~3.1 and Theorem~1.1]{ElKahouiOuali},
generalized by Dutta--Lahiri~\cite[Lemma~3.3 and Theorem~3.4]{DuttaLahiri}.
The coordinate criterion of Maubach~\cite[Theorem~4.5]{Maubach}
applies to $H+p^3w$: the only root of $p^3$ is $p=0$, where
the coordinate change below makes $H$ a coordinate. We retain the
explicit formulas so that the resulting isomorphism is available
without an existence argument.
Recall the polynomial ring and its elements
\begin{equation}\label{con:setup}
\begin{gathered}
P=\kk[p,s,u,F,J],\qquad x=s^2+u^3+p^2F,\\
y=s+x(x-u^3),\qquad z=sx+p^2J,\\
H=x^2F-(1+2sx)J-p^2J^2-pu,\qquad A=P/(H).
\end{gathered}
\end{equation}
Thus \(H\) is a specified polynomial in the five independent variables
\(p,s,u,F,J\). We will use \(w\) for the polynomial indeterminate adjoined
to \(A\). All maps below are \(\kk\)-algebra maps.
\subsection{Localization and integrality}
The definitions give the identity
\begin{equation}\label{con:identity}
xy-z(z+1)=p^2(H+pu).
\end{equation}
Indeed, substituting \(y\) and \(z\) into the left-hand side and cancelling
the terms \(sx\) gives
\[
x^2(x-u^3-s^2)-p^2(1+2sx)J-p^4J^2,
\]
which is the right-hand side because \(x-u^3-s^2=p^2F\).
\begin{lemma}\label{con:domain}
The algebra \(A\) is an integral domain of dimension four, \(p\ne0\)
in \(A\), and
\[
A[p^{-1}]=\kk[p,p^{-1},x,y,z]
\]
is a Laurent polynomial ring in the displayed generators.
\end{lemma}
\begin{proof}
Over \(\kk[p,p^{-1}]\), the elements \(x,y,z,u\) are polynomial
coordinates on \(P[p^{-1}]\). The inverse substitutions are
\begin{equation}\label{con:localized-coordinates}
s=y-x(x-u^3),\qquad
F=p^{-2}(x-s^2-u^3),\qquad J=p^{-2}(z-sx).
\end{equation}
Substitution in either direction gives the identity, so in particular
the new coordinates are algebraically independent. In these coordinates,
\Cref{con:identity} reads
\begin{equation}\label{con:localized-h}
H=p^{-2}\bigl(xy-z(z+1)\bigr)-pu.
\end{equation}
The coefficient of \(u\) is a unit. Replacing \(u\) by \(H\) is therefore
another polynomial coordinate change over \(\kk[p,p^{-1}]\). Consequently
\((H)\) is prime in \(P[p^{-1}]\), and its quotient is the Laurent
polynomial ring asserted in the lemma, with
\(u=p^{-3}(xy-z(z+1))\).
Set \(x_0=s^2+u^3\). Reduction modulo \(p\) gives
\[
H\bmod p=x_0^2F-(1+2sx_0)J\ne0,
\]
so \(p\) does not divide \(H\) in \(P\). To descend irreducibility from
the localization, factor \(H\) in the unique factorization domain \(P\).
Its image in \(P[p^{-1}]\) is irreducible, so all but one of its
irreducible factors must become units there. An irreducible element of
\(P\) that becomes a unit upon inverting \(p\) divides a power of \(p\),
and hence is associated to \(p\). Nondivisibility by \(p\) excludes such
factors. Thus \(H\) is irreducible in \(P\), and is prime because \(P\)
is a unique factorization domain. This proves that \(A\) is a domain.
Its localization computed above is nonzero, which also proves that
\(p\ne0\) in \(A\). This localization has fraction field
\(\kk(p,x,y,z)\), of transcendence degree four over \(\kk\).
The same is true of \(A\), so the dimension formula for finitely
generated domains over a field~\cite[Tag~00P0]{Stacks} gives \(\dim A=4\).
\end{proof}
\subsection{An explicit stabilization}
The main calculation has two stages. An exponential automorphism
replaces $H$ by $H+p^3w$. A linear coordinate change in $F,J$ then
makes this new equation linear modulo $p^2$, which allows a polynomial
root modulo $p^3$ and a fifth coordinate to be written down explicitly.
\begin{proposition}\label{prop:stabilization}
For the algebra \(A\) in \Cref{con:setup}, there is an algebraic
\(\kk\)-isomorphism
\[
A[w]\cong\kk\poly{5}.
\]
\end{proposition}
\begin{proof}
We first change the equation by a polynomial automorphism, and then
exhibit five polynomial coordinates on the resulting quotient.
\smallskip
\noindent\emph{Changing the equation.}
On \(P[p^{-1}]=\kk[p,p^{-1},x,y,z,u]\), define
\(\Delta=-p^2\partial/\partial u\), with \(p,x,y,z\) fixed.
This derivation is locally nilpotent: every polynomial is killed by a
sufficiently high iterate of \(\Delta\). Applying it to
\Cref{con:localized-coordinates} gives
\begin{equation}\label{con:derivation}
\begin{gathered}
\Delta p=0,\qquad \Delta u=-p^2,\qquad
\Delta s=-3p^2xu^2,\\
\Delta F=(6sx+3)u^2,\qquad \Delta J=3x^2u^2.
\end{gathered}
\end{equation}
These values belong to \(P\), so \(\Delta\) restricts to a locally
nilpotent derivation of \(P\). Extend it by \(\Delta w=0\). The formula
\[
\Phi(f)=\exp(w\Delta)(f)
=\sum_{i\ge0}\frac{w^i}{i!}\Delta^i(f)
\qquad(f\in P[w])
\]
is a finite sum for every \(f\). The iterated Leibniz rule makes \(\Phi\)
an algebra homomorphism, and the exponential identity gives its inverse
\(\exp(-w\Delta)\). By \Cref{con:localized-h},
\(\Delta H=p^3\) and \(\Delta^2H=0\). Hence
\begin{equation}\label{con:cylinder-automorphism}
\Phi(H)=H+p^3w,\qquad
A[w]\xrightarrow[\ \Phi\ ]{\ \sim\ }
T:=P[w]/(H+p^3w).
\end{equation}
In particular, \(T\) is a domain by \Cref{con:domain}, and its element
\(p\) is nonzero since \(\Phi\) fixes \(p\).
\smallskip
\noindent\emph{Coordinates adapted to the fiber \(p=0\).}
With \(x_0=s^2+u^3\), put
\begin{equation}\label{con:linear-coordinates}
L=x_0^2F-(1+2sx_0)J,\qquad
M=(1-2sx_0)F+4s^2J.
\end{equation}
The determinant of this linear change in \((F,J)\), over
\(\kk[p,s,u]\), is
\[
4s^2x_0^2+(1+2sx_0)(1-2sx_0)=1.
\]
Its inverse is
\begin{equation}\label{con:linear-inverse}
F=4s^2L+(1+2sx_0)M,\qquad
J=-(1-2sx_0)L+x_0^2M.
\end{equation}
It follows that \(p,s,u,L,M\) are algebraically independent and that
\[
P=B[L],\qquad B:=\kk[p,s,u,M]\cong\kk\poly{4}.
\]
For the rest of the proof, regard \(F,J,H\) as polynomials in \(L\)
over \(B\), using \Cref{con:linear-inverse}. Expansion only in powers
of \(p\) gives
\begin{equation}\label{con:h-expansion}
H(L)=L-pu+p^2Q(L)+p^4F(L)^3,
\qquad Q(L)=2x_0F(L)^2-2sF(L)J(L)-J(L)^2.
\end{equation}
At this point the transfer theorem of
Edo--V\'en\'ereau~\cite[Theorem~7]{EdoVenereau} also applies, with
coefficient ring $B$ and parameter $p^3$: every coefficient of
$H(L)-L$ is divisible by $p$, hence nilpotent modulo $p^3B$.
The following calculation gives a polynomial inverse in this instance.
\smallskip
\noindent\emph{A polynomial root modulo \(p^3\).}
Define elements and a polynomial over \(B\) by
\begin{equation}\label{con:root}
\begin{gathered}
C=Q(0)=M^2(2x_0+6sx_0^2+4s^2x_0^3-x_0^4),\\
Q_1(L)=\frac{Q(L)-C}{L}\in B[L],\qquad
L_*=pu-p^2C\in B.
\end{gathered}
\end{equation}
Since \(Q(L_*)-C=L_*Q_1(L_*)\), \Cref{con:h-expansion} gives the exact
identity
\begin{equation}\label{con:root-divisibility}
H(L_*)=p^3\bigl((u-pC)Q_1(L_*)+pF(L_*)^3\bigr)\in p^3B.
\end{equation}
\smallskip
\noindent\emph{A polynomial representative for the fifth coordinate.}
The quotient $(L-L_*)/p^3$ suggests the remaining coordinate after
localization. The next identity provides a polynomial representative
for it before $p$ is inverted.
In \(B[L,w]\), put
\begin{equation}\label{con:e-polynomial}
h=u-pQ(L)-p^3F(L)^3-p^2w,
\qquad e_0=-w-pF(L)^3-Q_1(L)h.
\end{equation}
Then \(H(L)+p^3w=L-ph\). Using \(Q(L)-C=LQ_1(L)\), we obtain the
polynomial identity
\begin{equation}\label{con:e-certificate}
p^3e_0-(L-L_*)
=(p^2Q_1(L)-1)\bigl(H(L)+p^3w\bigr).
\end{equation}
Thus the image \(e\) of \(e_0\) in \(T\) satisfies
\begin{equation}\label{con:e-relation}
p^3e=L-L_*.
\end{equation}
In particular, \(e\) belongs to the unlocalized ring \(T\).
\smallskip
\noindent\emph{Generation and algebraic independence.}
Inverting \(p\) in the presentation
\(T=B[L,w]/(H(L)+p^3w)\) allows us to eliminate \(w\), giving the
isomorphism
\begin{equation}\label{con:t-localized}
T[p^{-1}]\xrightarrow{\ \sim\ }B[p^{-1}][L],
\qquad w\longmapsto-p^{-3}H(L),
\end{equation}
which is the identity on \(B\) and \(L\). The composite
\(B\to T\to T[p^{-1}]\) is therefore the natural injection
\(B\hookrightarrow B[p^{-1}][L]\). This proves that the displayed copy
of \(B\) embeds in \(T\). Moreover, \Cref{con:e-relation} identifies
\(e\) in this localization with \(p^{-3}(L-L_*)\), an affine linear
coordinate in the independent variable \(L\). Consequently \(e\) is
transcendental over \(B[p^{-1}]\), and hence over \(B\).
It remains to prove that \(B\) and \(e\) generate \(T\).
By \Cref{con:e-relation}, \(L=L_*+p^3e\). The polynomial difference
\(H(L_*+p^3Z)-H(L_*)\) is divisible by \(p^3\) in \(B[Z]\).
Together with \Cref{con:root-divisibility}, this shows that
\[
W(Z):=-\frac{H(L_*+p^3Z)}{p^3}\in B[Z].
\]
The defining relation of \(T\) now gives
\(p^3(w-W(e))=0\). Since \(T\) is a domain and \(p\ne0\), it follows
that \(w=W(e)\). Thus \(L,w\in B[e]\), which proves
\begin{equation}\label{con:polynomial-cylinder}
T=B[e]=\kk[p,s,u,M,e]\cong\kk\poly{5}.
\end{equation}
For completeness, the five independent coordinates in \(A[w]\) itself
are the classes of
\[
\Phi^{-1}(p),\quad\Phi^{-1}(s),\quad\Phi^{-1}(u),\quad
\Phi^{-1}(M),\quad\Phi^{-1}(e_0).
\]
These are polynomials by the finite exponential formula. They are
obtained by transporting the five generators in
\Cref{con:polynomial-cylinder} through the inverse of
\Cref{con:cylinder-automorphism}, which proves the proposition.
\end{proof}
==== END An-explicit-failure-of-complex-affine-space-cancellation//02-construction.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//03-degeneration.tex ====
\section{A smooth degeneration and its derivations}
\label{sec:degeneration}
We retain the notation of \Cref{sec:construction}, using the same letters
for elements of $P$ and their images in the domain $A=P/(H)$. The
filtration constructed here will let us replace a locally nilpotent
derivation of $A$ by a homogeneous one on a smooth algebra with a simpler
equation.
\subsection{The valuation and its coefficients}
Consider the presented algebra
\begin{equation}\label{deg:S-R}
S=\kk[p,x,y,z,u]/\bigl(xy-z(z+1)-p^3u\bigr),
\qquad
R=S/pS=\kk[x,y,z,u]/\bigl(xy-z(z+1)\bigr).
\end{equation}
The identity $xy-z(z+1)=p^2(H+pu)$ defines a homomorphism $S\to A$.
\begin{lemma}\label{deg:valuation}
The map $S\to A$ is injective, and
$S[p^{-1}]=A[p^{-1}]$. The ring $R$ is a smooth domain.
On the common fraction field of $S$ and $A$ there is a discrete valuation
$\ord_p$, normalized by $\ord_p(p)=1$, coming from $S_{pS}$.
For $b\in S$ and every integer $i\geq 0$,
\begin{equation}\label{deg:global-divisibility}
\ord_p(b)\geq i \quad\Longleftrightarrow\quad b\in p^iS.
\end{equation}
In particular, each nonzero $h\in S[p^{-1}]$ has a unique expression
$h=p^m b$ with $m=\ord_p(h)\in\mathbb Z$ and $b\in S\setminus pS$.
\end{lemma}
\begin{proof}
After inverting $p$, the defining polynomial of $S$ is linear in $u$
with invertible leading coefficient. It is therefore irreducible in
$\kk[p,p^{-1},x,y,z,u]$. It is not divisible by $p$ in
$\kk[p,x,y,z,u]$, so it is irreducible there as well: any factor becoming
a unit after localization would be a scalar times a power of $p$.
The latter polynomial ring is a unique factorization domain, so $S$ is
a domain and $p$ is nonzero in $S$. The inverse formulas
\[
s=y-x(x-u^3),\qquad
F=\frac{x-s^2-u^3}{p^2},\qquad
J=\frac{z-sx}{p^2}
\]
show that $S\to A$ becomes an isomorphism after inverting $p$.
Its kernel is consequently $p$-torsion, hence zero in the domain $S$.
This also proves equality of the fraction fields.
The polynomial $xy-z(z+1)$ is primitive as a polynomial in $y$ over
$\kk[x,z,u]$, since $x$ and $z(z+1)$ are coprime. It is linear and
irreducible over $\kk(x,z,u)$, so $R$ is a domain. Its hypersurface is
smooth by the Jacobian criterion~\cite[Tags~00T6 and~00T7]{Stacks}:
simultaneous vanishing of the partial
derivatives $y$, $x$, and $-(2z+1)$ would give $x=y=0$, $z=-1/2$,
which does not satisfy its equation.
Thus $pS$ is a nonzero prime ideal, and the noetherian local domain
$T=S_{pS}$ has maximal ideal $pT$. To see directly that $T$ is a
discrete valuation ring, divide a nonzero nonunit repeatedly by $p$.
This process must end: otherwise the principal ideals generated by the
successive quotients form an ascending chain, and stabilization would
imply that $p$ is a unit, by cancellation in the domain $T$.
Every nonzero element of $T$ is therefore a power of $p$ times a unit.
This gives the stated valuation on $\Frac(T)$.
If $b\in S$ has positive order, then $b\in pT$, so $tb\in pS$ for
some $t\in S\setminus pS$. Primality of $pS$ gives $b\in pS$.
Dividing by $p$ and repeating proves
\Cref{deg:global-divisibility}; its converse is immediate.
Finally, write $h=p^{-N}b_0$ with $b_0\in S$, $N\geq 0$, and apply
\Cref{deg:global-divisibility} to $b_0$ to divide out its full order.
The asserted expression follows, and its uniqueness follows from the
valuation and cancellation.
\end{proof}
For $0\ne h\in A$ define its \emph{valuation degree} by
$\deg h=-\ord_p(h)$, and put $\deg 0=-\infty$. In particular,
\[
\deg(hh')=\deg h+\deg h',\qquad
\deg(h+h')\leq\max\{\deg h,\deg h'\}.
\]
These formulas give an exhaustive, separated filtration indexed by all
integers,
\[
\mathcal F_dA=\{h\in A:\deg h\leq d\},\qquad
\gr A=\bigoplus_{d\in\mathbb Z}\mathcal F_dA/\mathcal F_{d-1}A.
\]
Write $[h]_d$ for the class of $h\in\mathcal F_dA$ in the $d$th
summand, and $\inif(h)=[h]_{\deg h}$ for a nonzero $h$.
Give $R[\tau,\tau^{-1}]$ the grading in which $R$ has degree zero
and $\tau$ has degree $-1$. \Cref{deg:valuation} gives a graded
injection
\begin{equation}\label{deg:graded-injection}
\gr A\lhook\joinrel\longrightarrow R[\tau,\tau^{-1}],
\qquad
[h]_d\longmapsto \overline{p^dh}\,\tau^{-d}.
\end{equation}
Indeed $p^dh\in S$ whenever $h\in\mathcal F_dA$, including when
$d<0$, by the last assertion of \Cref{deg:valuation}. Its residue
modulo $pS$ vanishes exactly when $h\in\mathcal F_{d-1}A$.
The displayed maps respect sums and products and are injective on
each summand. This explains why their coefficients lie in $R$
itself.
\subsection{The exact associated graded algebra}
The initial forms of the five generators give a candidate for $\gr A$.
The point requiring proof is that no cancellation among their leading
terms produces an additional generator. We establish the presentation
of the candidate first, then reduce the weight of any representative
whose leading terms cancel.
In $R$ set
\begin{equation}\label{deg:center}
s_R=y-x(x-u^3),\qquad
f=x-s_R^2-u^3,\qquad
g=z-s_Rx,\qquad I=(f,g).
\end{equation}
\begin{proposition}\label{prop:graded}
Under \Cref{deg:graded-injection}, the associated graded algebra is
the following specific subalgebra of $R[\tau,\tau^{-1}]$:
\begin{equation}\label{deg:G}
G:=\gr A
=R[\tau,I/\tau^2]
=R[\tau,f/\tau^2,g/\tau^2].
\end{equation}
It is a smooth domain. Give $P=\kk[p,s,u,F,J]$ the presentation
weights
\begin{equation}\label{deg:presentation-weights}
(\omega(p),\omega(s),\omega(u),\omega(F),\omega(J))
=(-1,0,0,2,2).
\end{equation}
Then every nonzero $h\in A$ has a polynomial representative whose
maximum weight is exactly $\deg h$, and there is a graded isomorphism
\begin{equation}\label{deg:graded-presentation}
G\cong P/(H_{\mathrm{top}}),\qquad
H_{\mathrm{top}}=x^2F-(1+2sx)J-p^2J^2,
\quad x=s^2+u^3+p^2F.
\end{equation}
The images of $p,s,u,F,J$ in this isomorphism are respectively
$\tau,s_R,u,f/\tau^2,g/\tau^2$.
\end{proposition}
\begin{proof}
The residues of $s$ and $u$ in $R$ are the nonzero elements $s_R$
and $u$. Also $f$ and $g$ are nonzero: specializing $x=z=0$ sends
$f$ to $-y^2-u^3$, and specializing $x=0$, $z=-1$ sends $g$ to
$-1$. The identities $p^2F=x-s^2-u^3$ and $p^2J=z-sx$ therefore
give
\begin{equation}\label{deg:initial-generators}
\begin{array}{c|ccccc}
h&p&s&u&F&J\\ \hline
\deg h&-1&0&0&2&2\\
\inif(h)&\tau&s_R&u&f/\tau^2&g/\tau^2.
\end{array}
\end{equation}
Here $s_R$ is nonzero because its specialization at $x=z=0$ is $y$.
Let $\theta:P\to R[\tau,\tau^{-1}]$ substitute the five initial
forms in \Cref{deg:initial-generators}. Its image contains the
specified copy of $R$: the generators of that copy are recovered by
\begin{equation}\label{deg:recover-R}
\begin{split}
x&=s_R^2+u^3+\tau^2(f/\tau^2),\\
y&=s_R+x(x-u^3),\\
z&=s_Rx+\tau^2(g/\tau^2).
\end{split}
\end{equation}
Consequently $\theta(P)$ is exactly the subalgebra on the right of
\Cref{deg:G}. In particular this is an identification over the
given coefficient ring $R$.
For the presentation weights, $x$ has weight zero and
$H_{\mathrm{top}}$ is homogeneous of weight two, with
\begin{equation}\label{deg:top-relation}
H=H_{\mathrm{top}}-pu,
\qquad xy-z(z+1)=p^2H_{\mathrm{top}}.
\end{equation}
After inverting $p$, the coordinate change from
$(s,u,F,J)$ to $(x,y,z,u)$ and its inverse in
\Cref{deg:valuation} identify $\theta$ with the quotient map
\[
\kk[p,p^{-1},x,y,z,u]\longrightarrow R[\tau,\tau^{-1}],
\qquad p\longmapsto\tau.
\]
Its kernel is $(xy-z(z+1))$, or equivalently
$(H_{\mathrm{top}})$ in the original localized coordinates.
This is a prime ideal. Moreover $p$ does not divide
$H_{\mathrm{top}}$, since its reduction modulo $p$ is
\[
x_0^2F-(1+2sx_0)J,\qquad x_0=s^2+u^3,
\]
which is nonzero. The same unique factorization argument as in
\Cref{deg:valuation} shows that $H_{\mathrm{top}}$ is irreducible,
hence prime, in $P$. Its principal ideal is thus saturated with
respect to $p$: if $p^NQ\in(H_{\mathrm{top}})$ then
$Q\in(H_{\mathrm{top}})$. Contracting the localized kernel yields
\begin{equation}\label{deg:kernel}
\ker\theta=(H_{\mathrm{top}}).
\end{equation}
It remains to show that these initial forms generate all of $\gr A$.
For a nonzero polynomial $Q\in P$, let $\omega(Q)$ be its maximum
monomial weight. If $Q$ represents a nonzero $h\in A$, then
$\deg h\leq\omega(Q)$ by \Cref{deg:initial-generators} and the
valuation inequalities. Suppose the inequality is strict, and write
$M=\omega(Q)$. The weight-$M$ part $Q_M$ has
$\theta(Q_M)=0$, since the degree-$M$ class of $h$ is zero.
By \Cref{deg:kernel},
\[
Q_M=H_{\mathrm{top}}C,
\]
where $C$ may be taken homogeneous of weight $M-2$. To justify the
homogeneity, decompose any quotient $C$ into its finitely many
homogeneous parts; multiplication by the nonzero homogeneous
$H_{\mathrm{top}}$ cannot cancel contributions of different weights
in the domain $P$.
The polynomial
\[
Q-HC=Q-Q_M+puC
\]
represents the same element $h$ and has maximum weight strictly less
than $M$, because $puC$ has weight $M-3$.
Repeating this operation terminates: the maximum weights are strictly
decreasing integers bounded below by the fixed finite number $\deg h$.
Thus it ends with a representative of weight exactly $\deg h$.
No lower bound on the weights of all monomials in $P$ is needed.
The initial form of $h$ is now the image under $\theta$ of the top
part of such an exact representative. This proves
$\gr A=\theta(P)$ and, together with \Cref{deg:kernel}, proves both
\Cref{deg:G} and \Cref{deg:graded-presentation}.
Finally, $G$ is a domain by its embedding in
$R[\tau,\tau^{-1}]$. On the open set where $p$, or equivalently
$\tau$, is invertible, the presentation gives the smooth algebra
$R[\tau,\tau^{-1}]$. On $p=0$, the partial derivatives of
$H_{\mathrm{top}}$ with respect to $F$ and $J$ reduce to
$x_0^2$ and $-(1+2sx_0)$. They cannot vanish simultaneously, as
\[
(1-2sx_0)(1+2sx_0)+4s^2x_0^2=1.
\]
The Jacobian criterion proves smoothness also at every point of
$p=0$, and hence on all of $\Spec G$.
\end{proof}
The weights in \Cref{deg:presentation-weights} compute the valuation
degree on $G$. This degree will remain distinct from the fiber
weight and the auxiliary degree introduced later.
The description $G=R[\tau,I/\tau^2]$ is an affine modification in
the sense of Kaliman--Zaidenberg~\cite[Definition~1.1]{KalimanZaidenberg}:
the base ring is $R[\tau]$, the ideal is $(\tau^2,I)$, and the
element inverted in the construction is $\tau^2$. Here the exact
graded presentation, including its coefficient ring, is what permits
the later bundle pullback.
\subsection{Inducing locally nilpotent derivations}
We now transfer a derivation of $A$ to its associated graded algebra.
The exact representative statement in \Cref{prop:graded} is the needed
input: a bound on the five generators then bounds the degree shift on
every element of $A$.
This is the filtered-to-homogeneous derivation method developed by
Kaliman--Makar-Limanov~\cite[Sections~3--5]{KalimanMakarLimanov}.
We give the argument for the present integer-indexed filtration,
including the possibility of negative degrees.
A $\kk$-derivation of a $\kk$-algebra $C$ is a $\kk$-linear map
$D:C\to C$ satisfying $D(hh')=hD(h')+h'D(h)$.
It is \emph{locally nilpotent} if for each $h\in C$ there is an
integer $N\geq 1$ such that $D^Nh=0$; all subsequent iterates then
vanish as well. We abbreviate this by LND.
On a graded algebra a derivation is \emph{homogeneous of degree
shift $j$} if it sends the component of degree $d$ to that of degree
$d+j$ for every $d\in\mathbb Z$.
\begin{lemma}\label{lem:induced-lnd}
Every nonzero LND $D$ of $A$ induces a nonzero homogeneous LND
$D_0$ of $G$. If $0\ne h\in\ker D$, then
$\inif(h)\in\ker D_0$.
\end{lemma}
\begin{proof}
Let $(h_1,\ldots,h_5)=(p,s,u,F,J)$ be the generators of $A$ and set
\begin{equation}\label{deg:shift}
j=\max_{Dh_i\ne0}\bigl(\deg(Dh_i)-\deg h_i\bigr).
\end{equation}
The maximum is over a finite nonempty set, since $D$ is nonzero.
We claim that
\begin{equation}\label{deg:shift-bound}
\deg(Dh)\leq\deg h+j\qquad(0\ne h\in A).
\end{equation}
By \Cref{prop:graded}, choose a representative of $h$ of maximum
weight exactly $\deg h$. Applying the product rule to each monomial,
every nonzero summand in its derivative has degree at most the
monomial's weight plus $j$: differentiating a factor $h_i$
changes its degree by at most $j$. Summing these bounds gives
\Cref{deg:shift-bound}.
It follows that $D(\mathcal F_dA)\subseteq\mathcal F_{d+j}A$, so
\begin{equation}\label{deg:induced-map}
D_0([h]_d)=[Dh]_{d+j}
\end{equation}
defines a homogeneous derivation of degree shift $j$ on $G$.
It is well defined because $D(\mathcal F_{d-1}A)$ lies in
$\mathcal F_{d+j-1}A$, and the product rule descends to the graded
algebra. For a generator attaining the maximum in
\Cref{deg:shift}, the right side of \Cref{deg:induced-map} applied
to its initial form is nonzero. Thus $D_0\ne0$.
For every $h\in\mathcal F_dA$ and $m\geq 0$, induction gives the
identity in the indicated graded component
\begin{equation}\label{deg:iterate}
D_0^m([h]_d)=[D^mh]_{d+mj}.
\end{equation}
It includes the case when an intermediate class vanishes: the shift
bound ensures that a drop below the indicated degree persists in
all later iterates. Local nilpotence of $D$ therefore kills each
homogeneous class after finitely many iterates of $D_0$.
Every element of $G$ is a finite sum of homogeneous classes, so a
maximum of their nilpotence bounds kills that element. This proves
local nilpotence on all of $G$, without a sign restriction on $j$.
Finally, if $Dh=0$, \Cref{deg:induced-map} gives
$D_0\inif(h)=0$.
\end{proof}
The positive-degree-invariant obstruction is inspired by Gupta's
Derksen-invariant strategy; see \cite[Lemma~2.4 and the proof of
Proposition~3.6]{GuptaThreefold}. Here it is expressed directly through
polynomial coordinates. The characteristic-zero construction and the
subsequent bundle and rigidity arguments are proved in this paper;
Gupta's positive-characteristic theorem is not an input to them.
\begin{proposition}\label{prop:positive-invariant}
If $A\cong\kk\poly{4}$, then $G$ admits a nonzero homogeneous LND
with a nonzero homogeneous invariant of strictly positive valuation
degree.
\end{proposition}
\begin{proof}
Under the hypothesis choose polynomial coordinates
$q_1,q_2,q_3,q_4$ of $A$.
The subset $\mathcal F_0A$ is a $\kk$-subalgebra. Since $F$ has
degree two by \Cref{deg:initial-generators}, this subalgebra is
proper. Therefore at least one coordinate $q_r$ has positive
degree; otherwise the coordinates would generate $A$ inside
$\mathcal F_0A$.
Choose $i\ne r$. The coordinate partial derivative
$D=\partial/\partial q_i$ is a nonzero LND of $A$ and satisfies
$Dq_r=0$. By \Cref{lem:induced-lnd}, its induced derivation $D_0$
on $G$ is nonzero and homogeneous, and it kills the nonzero
homogeneous element $\inif(q_r)$ of degree $\deg q_r>0$.
\end{proof}
Neither \Cref{lem:induced-lnd} nor \Cref{prop:positive-invariant}
asserts that $D_0$ preserves the subalgebra $R\subset G$.
Coefficient preservation can fail even for a homogeneous LND on $G$.
The derivation $\Delta$ in \Cref{con:derivation} has presentation-weight
shift $-2$ and satisfies
$\Delta(H_{\mathrm{top}})=\Delta(H+pu)=p^3-p^3=0$.
It therefore descends through \Cref{deg:graded-presentation} to an LND
on $G$ sending $u\in R$ to $-\tau^2\notin R$.
This example concerns actions on $G$: $\Delta$ does not descend to
$A$, since $\Delta H=p^3$.
The next section lifts the action on $\Spec G$ directly, without
assuming $D_0(R)\subset R$.
==== END An-explicit-failure-of-complex-affine-space-cancellation//03-degeneration.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//04-bundle.tex ====
\section{Principalization and lifted additive actions}
\label{sec:bundle}
The ideal defining $G$ becomes principal after pullback to a line
bundle with its zero section removed. We construct that bundle and then
show that every homogeneous additive action on $G$ lifts to the
pullback. This replaces the two generators of the ideal by one element
while retaining both degree and fiber weight.
We retain the coefficient ring and ideal from \Cref{prop:graded}:
\[
R=\kk[x,y,z,u]/\bigl(xy-z(z+1)\bigr),\qquad
s_R=y-x(x-u^3),\qquad
I=(f,g)=(x-s_R^2-u^3,\ z-s_Rx).
\]
The inclusion $R\subset G=R[\tau,I/\tau^2]$ is the actual coefficient
inclusion in $R[\tau,\tau^{-1}]$. In terms of the initial generators,
it is given by
\begin{equation}
\label{bun:coefficient-inclusion}
x=s_R^2+u^3+\tau^2(f/\tau^2),\qquad
y=s_R+x(x-u^3),\qquad
z=s_Rx+\tau^2(g/\tau^2).
\end{equation}
This is the inclusion used in the base change below. Its elements have
valuation degree zero.
\subsection{A line bundle that principalizes the center}
The bundle used here is the standard quotient of $\mathrm{SL}_2$ by
its diagonal torus, with the coordinate $u$ adjoined; see
Blanc--van Santen~\cite[p.~8431]{BlancVanSanten} for this
description of the affine quadric. We give its charts explicitly
because the transition functions determine the fiber weights.
Set
\begin{equation}
\label{bun:determinant-ring}
\Rt=\kk[a,d,b,c,u]/(ac-bd-1),\qquad
x=ab,\quad y=dc,\quad z=db,
\end{equation}
and let
\begin{equation}
\label{bun:v}
v=a^2(x-u^3)-d^2=a^3b-a^2u^3-d^2.
\end{equation}
The assignments in \Cref{bun:determinant-ring} define a homomorphism
$R\to\Rt$, since $xy=(ac)(bd)=z(z+1)$.
\begin{proposition}
\label{prop:principalization}
The morphism $\Spec\Rt\to\Spec R$ is the complement of the zero
section in a line bundle, with fiber scaling specified by the weights
\[
\wt(a)=\wt(d)=1,\qquad
\wt(b)=\wt(c)=-1,\qquad \wt(u)=0.
\]
The homomorphism $R\to\Rt$ is faithfully flat, and
$I\Rt=(v)$. Moreover, base change gives a faithfully flat inclusion
$G\hookrightarrow\Gt$, where
\begin{equation}
\label{bun:pullback-algebra}
\Gt:=G\otimes_R\Rt
=\Rt[\tau,v/\tau^2]\subset\Rt[\tau,\tau^{-1}],\qquad
\Gt\cong\Rt[\tau,V]/(\tau^2V-v).
\end{equation}
Thus $V$ denotes the element $v/\tau^2$ in the displayed subalgebra.
The ring $\Gt$ is a domain and has the following commuting gradings:
\begin{center}
\begin{tabular}{c@{\qquad}rrrrr}
\toprule
& $a,d$ & $b,c$ & $u$ & $\tau$ & $V$\\
\midrule
Valuation degree & $0$ & $0$ & $0$ & $-1$ & $2$\\
Fiber weight & $1$ & $-1$ & $0$ & $0$ & $2$\\
\bottomrule
\end{tabular}
\end{center}
\end{proposition}
\begin{proof}
First, $\Rt$ is a domain: the polynomial $ac-(bd+1)$ is primitive
as a polynomial in $c$ over the unique factorization domain
$\kk[a,d,b,u]$, and is irreducible over its fraction field because
it is linear. The element $v$ is nonzero, since $u$ is an
indeterminate over the determinant ring and its coefficient in degree
three in $v$ is $-a^2\ne0$.
The principal opens $D(z+1)$ and $D(z)$ cover $\Spec R$.
Over the first open, $ac=z+1$ is invertible, and the formulas
\begin{equation}
\label{bun:first-chart}
c=\frac{z+1}{a},\qquad b=\frac{x}{a},\qquad
d=\frac{ay}{z+1}
\end{equation}
give mutually inverse identifications
\[
\Rt[(z+1)^{-1}]
\cong R[(z+1)^{-1}][a,a^{-1}].
\]
Indeed, they give $ac=z+1$, $ab=x$, $dc=y$, and
$db=xy/(z+1)=z$. Over the second open, the analogous formulas are
\begin{equation}
\label{bun:second-chart}
b=\frac{z}{d},\qquad c=\frac{y}{d},\qquad a=\frac{dx}{z},
\qquad
\Rt[z^{-1}]\cong R[z^{-1}][d,d^{-1}].
\end{equation}
Here $db=z$, $dc=y$, $ab=x$, and $ac=xy/z=z+1$.
On the overlap the fiber parameters satisfy
\[
d=\frac{y}{z+1}a,\qquad
\frac{y}{z+1}\cdot\frac{x}{z}=1.
\]
Thus their transition function is a base unit. Gluing affine lines
by this same transition function gives a line bundle whose complement
of the zero section is $\Spec\Rt$. The indicated weights are exactly
scalar multiplication on the two fiber parameters. Each Laurent
polynomial chart is faithfully flat over its base chart: as a module
it is free with basis the integral powers of its fiber parameter.
Since the base charts cover, $R\to\Rt$ is faithfully flat.
We prove principalization by global identities. In $\Rt$ put
\[
h=x-u^3,\qquad C_1=c^2-b^2h,\qquad
\alpha=a^2(1-2bd),\qquad
\beta=d^2(3+2bd)+\alpha h.
\]
Since $s_R=dc-abh$ and $ac-bd=1$, we have
\[
as_R-d=-bv,\qquad
g=b(d-as_R)=b^2v.
\]
Expansion also gives
\[
f-C_1v
=h-(dc-abh)^2-(c^2-b^2h)(a^2h-d^2)
=h\bigl(1-(ac-bd)^2\bigr)=0.
\]
Finally, with $q=bd$ the coefficient identity is
\begin{align*}
\alpha C_1+\beta b^2
&= (ac)^2(1-2bd)+3b^2d^2+2b^3d^3\\
&= (1+q)^2(1-2q)+3q^2+2q^3=1.
\end{align*}
Consequently
\begin{equation}
\label{bun:principalization-certificate}
f=C_1v,\qquad g=b^2v,\qquad v=\alpha f+\beta g,
\end{equation}
which proves $I\Rt=(v)$.
Flatness permits us to tensor the injection
$G\hookrightarrow R[\tau,\tau^{-1}]$ with $\Rt$.
The resulting injection has image generated by
$\Rt$, $\tau$, $f/\tau^2$, and $g/\tau^2$.
By \Cref{bun:principalization-certificate}, this image is precisely
$\Rt[\tau,v/\tau^2]$. In particular it is a domain. The map
$G\to G\otimes_R\Rt$ is faithfully flat by base change, and hence
injective.
For completeness, the single displayed relation is the entire
presentation. Let $W$ be an indeterminate and set
$Q=\Rt[\tau,W]/(\tau^2W-v)$.
If $\tau F=(\tau^2W-v)B$ in $\Rt[\tau,W]$, reduction modulo
$\tau$ gives $vB(0,W)=0$. Since $\Rt[W]$ is a domain and $v\ne0$,
we have $B=\tau B_1$; cancellation then gives
$F=(\tau^2W-v)B_1$. Thus $\tau$ is a nonzerodivisor in $Q$.
After inverting $\tau$, the relation eliminates $W$ and identifies
$Q[\tau^{-1}]$ with $\Rt[\tau,\tau^{-1}]$.
It follows that $Q$ injects into that Laurent ring with image
$\Rt[\tau,v/\tau^2]$, proving \Cref{bun:pullback-algebra}.
The two gradings follow from $\deg(v)=0$ and $\wt(v)=2$;
the relation $\tau^2V=v$ is homogeneous for each.
\end{proof}
Write $Y=\Spec R$ and $Z=\Spec G$. The geometric content of
\Cref{bun:pullback-algebra} is the Cartesian square
\[
\begin{tikzcd}[column sep=large,row sep=large]
\Spec\Gt \arrow[r] \arrow[d] & \Spec\Rt \arrow[d] \\
Z \arrow[r] & Y.
\end{tikzcd}
\]
If $L_Y$ is the line bundle just constructed on $Y$, then
$\Spec\Gt=L^\times$, where $L$ is its pullback to $Z$ and
$L^\times$ denotes the complement of its zero section.
\subsection{A normalized lift on a line bundle}
The pullback is a line bundle over $Z=\Spec G$. An additive action on
$Z$ need not descend to $Y=\Spec R$, so a lift obtained by first acting
on $Y$ would not suffice. Instead, we prove a lifting statement for any
line bundle over a smooth integral affine base. The two ingredients
are invariance of the Picard group under adjoining a variable and the
absence of nonconstant polynomial units over a domain.
For a variety $X$, $\Pic(X)$ is the group of isomorphism classes of line
bundles on $X$, with tensor product as its operation. For a line bundle
$L$ on $X$, normalization means that the restriction over
$\{0\}\times X$ is the identity on $L$.
The Picard-group assertion is the regular-ring case of
Traverso's theorem~\cite[Theorem~3.6]{Traverso}. Additive
linearization also follows from the equivariant Picard-group results
of Brion~\cite[Proposition~2.10 and Theorem~2.14]{Brion}. For the normalized
isomorphism and cocycle verification we follow
\cite[Lemma~2.9]{Brion}; we include the argument to make the subsequent
compatibility with valuation degree explicit.
\begin{proposition}
\label{prop:linearization}
Let $X$ be a smooth integral affine variety over $\kk$, and let
$\pi\colon\Aff^1\times X\to X$ be the projection. Then
\[
\pi^*\colon\Pic(X)\xrightarrow{\ \sim\ }
\Pic(\Aff^1\times X).
\]
For every algebraic additive action
$\alpha\colon\Ga\times X\to X$ and every line bundle $L$ on $X$,
there is a unique isomorphism of line bundles
\begin{equation}
\label{bun:linearization-map}
\phi\colon\pi^*L\xrightarrow{\ \sim\ }\alpha^*L
\end{equation}
whose restriction to $\{0\}\times X$ is $\mathrm{id}_L$.
This isomorphism obeys the action cocycle. It therefore defines a
unique action on $L$ that lifts $\alpha$, is linear on its fibers,
and acts as the identity at the group identity. The action preserves
$L^\times$ and commutes with fiber scaling. If $L^\times$ is affine,
the induced derivation of its coordinate ring is locally nilpotent
and preserves the fiber weights.
\end{proposition}
\begin{proof}
Put $C=\Gamma(X,\mathcal O_X)$ and $K=\Frac C$.
Both $X$ and $\Aff^1\times X$ are regular. Their local rings are
factorial, so every Weil divisor is Cartier and their Picard groups
identify with their divisor class groups~\cite[Tags~056S, 0AG0, and~0BE9]{Stacks}.
Let $D$ be a Weil divisor on
$\Spec C[t]$. Its restriction to $\Spec K[t]$ is principal, since
$K[t]$ is a principal ideal domain. Choose $r\in K(t)^*$ giving
that divisor. Subtracting $\operatorname{div}(r)$ from $D$ removes
every prime component whose defining prime ideal has zero
contraction to $C$.
Each remaining component has a height-one prime ideal
$\mathfrak P\subset C[t]$ with nonzero contraction. Choose
$0\ne c\in\mathfrak P\cap C$, and choose a prime
$\mathfrak q$ minimal over $(c)$ and contained in
$\mathfrak P\cap C$. The principal ideal theorem~\cite[Tag~00KV]{Stacks}
and the fact that
$C$ is a domain give $\operatorname{ht}\mathfrak q=1$.
Now $\mathfrak qC[t]$ is a nonzero prime contained in the
height-one prime $\mathfrak P$, so
$\mathfrak P=\mathfrak qC[t]$.
The corresponding divisor is the pullback of the prime divisor
defined by $\mathfrak q$, with multiplicity one: a uniformizer
of the discrete valuation ring $C_{\mathfrak q}$ is also a
uniformizer at the generic point of that pulled-back divisor.
Thus every divisor class on $\Aff^1\times X$ is pulled back
from $X$, proving surjectivity of $\pi^*$. Restriction to
$i_0\colon X\to\Aff^1\times X$, $x\mapsto(0,x)$, is a left
inverse to $\pi^*$ on Picard groups, proving injectivity as well.
Since $\alpha\circ i_0=\pi\circ i_0=\mathrm{id}_X$, the line
bundles $\alpha^*L$ and $\pi^*L$ have the same restriction to
the identity section. The Picard isomorphism just proved implies
that they are isomorphic. Take an isomorphism $\phi_1$ between
them, with the orientation in \Cref{bun:linearization-map}.
Its restriction $i_0^*\phi_1$ is an automorphism of $L$.
Composing on the source with
$\pi^*((i_0^*\phi_1)^{-1})$ produces an isomorphism $\phi$
whose restriction is the identity.
An automorphism of any line bundle on a scheme is multiplication
by a global unit: its endomorphism sheaf is canonically the
structure sheaf. Because $C$ is a domain,
\begin{equation}
\label{bun:parameter-units}
C[t]^*=C^*,\qquad C[s,t]^*=C^*.
\end{equation}
Indeed, total degree of a product of nonzero polynomials over a
domain is the sum of their total degrees, so a polynomial unit
has degree zero. The ratio of two normalized isomorphisms in
\Cref{bun:linearization-map} is a unit of $C[t]$ whose value at
$t=0$ is one. By \Cref{bun:parameter-units}, that ratio is one.
This proves uniqueness, without a restriction on the units of $C$.
To check the cocycle, denote the fiber map by
$\phi_{t,x}\colon L_x\to L_{\alpha(t,x)}$.
Over $\Aff^2\times X$, the two compositions to compare are
\[
\phi_{t,\alpha(s,x)}\circ\phi_{s,x},\quad
\phi_{s+t,x}
\colon L_x\longrightarrow L_{\alpha(s+t,x)}.
\]
Their sources and targets agree by the action law for $\alpha$.
Thus
\[
\phi_{s+t,x}^{-1}\circ
\phi_{t,\alpha(s,x)}\circ\phi_{s,x}
\]
is an automorphism of the pullback of $L$ from $X$, hence a unit
of $C[s,t]$. At $s=0$ it is the identity. By
\Cref{bun:parameter-units}, it is the identity everywhere.
This proves the cocycle. The family of fiber maps consequently
defines an algebraic additive action on the total space of $L$.
Linearity preserves zero and nonzero vectors and commutes with
scalar multiplication, giving the asserted action on $L^\times$.
Finally, suppose $L^\times=\Spec T$ is affine. Write its action
on functions as
\[
\rho(r)=\sum_{n\ge0}\delta_n(r)t^n\in T[t],\qquad r\in T,
\]
where the sum is finite for each $r$. The identity and action laws
give $\delta_0=\mathrm{id}$ and
\[
\delta_i\delta_j=\binom{i+j}{i}\delta_{i+j}.
\]
The coefficient of $t$ in multiplicativity of $\rho$ says that
$\delta_1$ is a derivation. Since the characteristic is zero, the
displayed composition identity gives
$\delta_n=\delta_1^n/n!$. Finiteness of $\rho(r)$ now says that
$\delta_1$ is locally nilpotent. Commutation with fiber scaling
says precisely that this derivation preserves the fiber weights.
\end{proof}
\subsection{Compatibility with valuation degree}
The preceding proposition gives a lift preserving fiber weights. We
still need its valuation degree to agree with that of the original
derivation. The degree action has its own explicit bundle lift;
uniqueness of the normalized additive lift makes the two compatible.
\begin{proposition}
\label{prop:homogeneous-lift}
Let $D_0$ be a homogeneous locally nilpotent derivation of $G$
of valuation degree shift $j$. There is a locally nilpotent
derivation $\Dt$ of $\Gt$ restricting to $D_0$ on $G$, with
fiber weight zero and valuation degree shift $j$. If $D_0\ne0$,
then $\Dt\ne0$.
\end{proposition}
\begin{proof}
By \Cref{prop:graded}, $Z=\Spec G$ is smooth and integral.
Exponentiating $D_0$ gives the algebraic action on functions
\[
\exp(tD_0)(r)=\sum_{n\ge0}\frac{D_0^n(r)}{n!}t^n.
\]
Apply \Cref{prop:linearization} to this action and the line bundle
$L$ on $Z$ whose complement is $\Spec\Gt$.
The resulting locally nilpotent derivation $\Dt$ restricts to
$D_0$ because the action covers the given action on $Z$, and
has fiber weight zero because the lift commutes with fiber
scaling. Since $G\hookrightarrow\Gt$, a nonzero $D_0$ gives a
nonzero $\Dt$.
It remains to establish its valuation degree. For $\lambda\in\kk^*$,
let $\sigma_\lambda$ be the automorphism of $G$ multiplying a
homogeneous element of degree $h$ by $\lambda^h$.
It fixes the coefficient subring $R$ pointwise. Therefore it has
the pullback lift
$\widetilde\sigma_\lambda=\sigma_\lambda\otimes\mathrm{id}_{\Rt}$
on $\Gt$. Explicitly,
\begin{equation}
\label{bun:degree-scaling}
\widetilde\sigma_\lambda(r)=r\quad(r\in\Rt),\qquad
\widetilde\sigma_\lambda(\tau)=\lambda^{-1}\tau,\qquad
\widetilde\sigma_\lambda(V)=\lambda^2V.
\end{equation}
Geometrically this lift acts on $L=Z\times_Y L_Y$ by the degree
action on $Z$ and the identity on $L_Y$. In particular it extends
to a linear bundle automorphism, and its action on the complement
agrees with \Cref{bun:degree-scaling}.
On the coordinate ring of the base, homogeneity of $D_0$ gives
\begin{equation}
\label{bun:base-conjugacy}
\sigma_\lambda D_0\sigma_\lambda^{-1}=\lambda^jD_0.
\end{equation}
Conjugate the lifted additive action by
$\widetilde\sigma_\lambda$, using the expression on coordinate rings
\[
\widetilde\sigma_\lambda\,\exp(t\Dt)\,
\widetilde\sigma_\lambda^{-1}.
\]
This is a normalized linear lift of the base action
$\exp(\lambda^jtD_0)$, by \Cref{bun:base-conjugacy}.
The action $\exp(\lambda^jt\Dt)$ is another normalized linear
lift of that same base action. Uniqueness in
\Cref{prop:linearization} therefore gives equality of these
two families. Taking their coefficients of $t$ yields
\[
\widetilde\sigma_\lambda\Dt
\widetilde\sigma_\lambda^{-1}=\lambda^j\Dt
\qquad(\lambda\in\kk^*).
\]
For a homogeneous $r$ of degree $h$, this says
$\widetilde\sigma_\lambda(\Dt r)=\lambda^{h+j}\Dt r$
for every $\lambda$. The distinct integral characters of
$\Gm$ are linearly independent, so $\Dt r$ is homogeneous of
degree $h+j$ whenever it is nonzero. This proves the required
degree shift.
\end{proof}
The additive lift in \Cref{prop:homogeneous-lift} is constructed
on the line bundle over $Z=\Spec G$. It does not require
$D_0(R)\subset R$, and it does not assert $\Dt(\Rt)\subset\Rt$.
The invariance of coefficient rings used in the proof belongs to
the separate degree action: $\sigma_\lambda$ fixes $R$, and
$\widetilde\sigma_\lambda$ fixes $\Rt$.
==== END An-explicit-failure-of-complex-affine-space-cancellation//04-bundle.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//05-reduction.tex ====
\section{Extracting a derivation on the determinant ring}
\label{sec:reduction}
We now pass from a homogeneous additive action on the degeneration to a
locally nilpotent derivation of
\[
\Rt=\kk[a,d,b,c,u]/(ac-bd-1)
\]
with a prescribed sign for its fiber weight. Throughout this section,
\emph{degree} means the valuation degree, so that
\[
\deg \Rt=0,\qquad \deg\tau=-1,\qquad \deg V=2.
\]
The separate fiber weights are
\[
\wt(a)=\wt(d)=1,\quad \wt(b)=\wt(c)=-1,\quad
\wt(u)=\wt(\tau)=0,\quad \wt(V)=2.
\]
By \Cref{prop:principalization}, we work in the domain
\begin{equation}\label{red:ring}
\Gt=\Rt[\tau,V]\subset\Rt[\tau,\tau^{-1}],
\qquad \tau^2V=v=a^3b-a^2u^3-d^2.
\end{equation}
In particular, $\tau$, $V$, and $v$ are nonzero. The notation
$\Rt[\tau,V]$ here denotes the indicated subalgebra, with the displayed
relation.
The positive-degree invariant supplied by polynomiality will be
divisible by $V$. The elementary order of a locally nilpotent
derivation makes its kernel factorially closed, so this divisibility
will force $V$ itself to be invariant. We recall the argument,
including the order estimates needed when $\tau$ moves; compare
Finston--Maubach~\cite[Lemma~1.1]{FinstonMaubach}.
\begin{lemma}[Order of a locally nilpotent derivation]
\label{lem:lnd-order}
Let $T$ be a domain over $\kk$ and let $\delta$ be a locally nilpotent
derivation of $T$. For $0\ne h\in T$, define
\[
\nu_\delta(h)=\max\{n\ge0:\delta^n(h)\ne0\}.
\]
Then for nonzero $f,g\in T$,
\begin{equation}\label{red:order-product}
\nu_\delta(fg)=\nu_\delta(f)+\nu_\delta(g).
\end{equation}
If $\delta(h)\ne0$, then
$\nu_\delta(\delta(h))=\nu_\delta(h)-1$.
Consequently, $\ker\delta$ is factorially closed in $T$: if
$0\ne fg\in\ker\delta$, then $f,g\in\ker\delta$.
Moreover, $\delta(h)\ne0$ implies $\delta(h)\notin hT$.
\end{lemma}
\begin{proof}
Local nilpotence makes the defining maximum finite. Set
$r=\nu_\delta(f)$ and $s=\nu_\delta(g)$. The iterated Leibniz rule gives
\[
\delta^n(fg)=\sum_{i=0}^n\binom ni\delta^i(f)\delta^{n-i}(g).
\]
For $n>r+s$ every term is zero. For $n=r+s$ only the term with $i=r$
can be nonzero, and it equals
\[
\binom{r+s}{r}\delta^r(f)\delta^s(g)\ne0.
\]
Here we use both characteristic zero and the domain property. This
proves \Cref{red:order-product}. The assertion about $\delta(h)$ follows
directly from the definition. Since a nonzero element has order zero
exactly when it is invariant, product additivity proves factorial
closure. Finally, if $0\ne\delta(h)=hq$, then $q\ne0$ and
\[
\nu_\delta(h)-1
=\nu_\delta(\delta(h))
=\nu_\delta(h)+\nu_\delta(q)
\ge\nu_\delta(h),
\]
which is impossible.
\end{proof}
We will also use the exact homogeneous pieces of \Cref{red:ring}.
Writing $\Gt_e$ for its degree-$e$ component, they are
\begin{equation}\label{red:pieces}
\Gt_e=
\begin{cases}
\tau^{-e}\Rt, & e<0,\\
V^\ell\Rt, & e=2\ell,\quad \ell\ge0,\\
\tau V^{\ell+1}\Rt, & e=2\ell+1,\quad \ell\ge0.
\end{cases}
\end{equation}
Indeed, a degree-$e$ element is an $\Rt$-linear combination of monomials
$\tau^iV^h$ with $i,h\ge0$ and $2h-i=e$. Using $\tau^2V=v$,
these monomials become, respectively,
\[
\tau^{-e}v^h,\qquad
V^\ell v^{h-\ell},\qquad
\tau V^{\ell+1}v^{h-\ell-1}.
\]
The exponents of $v$ in the relevant cases are nonnegative. Conversely,
the multipliers displayed in \Cref{red:pieces} belong to $\Gt$ and have
the required degrees. This proves the equalities, including that their
coefficients lie in $\Rt$. Since the multipliers are nonzero and $\Gt$
is a domain, each such coefficient is unique.
\begin{proposition}\label{prop:extracted-lnd}
If $A\cong\kk\poly{4}$, there are a nonzero locally nilpotent derivation
$E$ of $\Rt$ and an integer $q\le0$ such that $E$ is homogeneous of
fiber weight shift $q$ and
\[
E^2(v)=0.
\]
Here homogeneity means that $E(r)$ has fiber weight $\wt(r)+q$ whenever
$r$ is fiber homogeneous and $E(r)\ne0$.
\end{proposition}
\begin{proof}
By \Cref{prop:positive-invariant}, polynomiality of $A$ supplies a nonzero
homogeneous LND $D_0$ of $G$ and a nonzero homogeneous
$h_0\in\ker D_0$ of positive degree. Apply
\Cref{prop:homogeneous-lift} to obtain an LND $\Dt$ of $\Gt$ lifting
$D_0$, homogeneous of some degree shift $j\in\mathbb Z$ and of fiber
weight shift zero. The injection $G\hookrightarrow\Gt$ in
\Cref{prop:principalization} ensures both that $\Dt\ne0$ and that the
image $h$ of $h_0$ is nonzero. This image is a homogeneous invariant
of the same positive degree. By \Cref{red:pieces}, $h$ is divisible
by $V$ in $\Gt$. Factorial closure in \Cref{lem:lnd-order} therefore
gives
\begin{equation}\label{red:V-invariant}
\Dt(V)=0.
\end{equation}
We distinguish whether $\tau$ is also invariant. If it is, the
derivation restricts after removing a power of $\tau$. If it is not,
the homogeneous pieces force a positive factor of $V$ into its values
on $\Rt$; removing this factor produces a derivation of negative
fiber weight. The LND order proves local nilpotence in this second
case without assuming that $\Dt$ preserves $\Rt$.
\medskip
\noindent\textbf{Case 1: $\Dt(\tau)=0$.}
In this case $\Dt$ extends to a locally nilpotent derivation of
\[
\Gt[\tau^{-1}]=\Rt[\tau,\tau^{-1}].
\]
Indeed, for $f\in\Gt$ and $m\ge0$, all iterates on
$f\tau^{-m}$ equal $\tau^{-m}\Dt^n(f)$. The degree-$j$ component of
this Laurent polynomial ring is exactly $\tau^{-j}\Rt$.
Since every $r\in\Rt$ has degree zero, there is thus a unique
$E(r)\in\Rt$ satisfying
\begin{equation}\label{red:fixed-tau}
\Dt(r)=\tau^{-j}E(r).
\end{equation}
Linearity and the Leibniz rule for $\Dt$, followed by cancellation of
$\tau^{-j}$, show that $E:\Rt\to\Rt$ is a $\kk$-derivation.
It has fiber weight shift zero because $\tau$ has fiber weight zero
and $\Dt$ preserves fiber weight. Since $\Dt(\tau)=0$, induction in
\Cref{red:fixed-tau} gives
\[
\Dt^n(r)=\tau^{-nj}E^n(r)\qquad(n\ge0).
\]
Local nilpotence of $\Dt$ therefore proves local nilpotence of $E$.
By \Cref{red:V-invariant},
\[
0=\Dt(\tau^2V)=\Dt(v)=\tau^{-j}E(v),
\]
so $E(v)=0$. Finally, $E=0$ would make $\Dt$ vanish on $\Rt$, $\tau$,
and $V$, which generate $\Gt$; hence $E\ne0$. This proves the desired
conclusion with $q=0$.
\medskip
\noindent\textbf{Case 2: $\Dt(\tau)\ne0$.}
Write $\nu=\nu_{\Dt}$ on the nonzero elements of $\Gt$, and put
$m=\nu(\tau)\ge1$. By \Cref{lem:lnd-order},
$\Dt(\tau)\notin\tau\Gt$. On the other hand, this element is
homogeneous of degree $j-1$. \Cref{red:pieces} shows that
every component of negative degree, and every component of positive
odd degree, is contained in $\tau\Gt$. We must therefore have
\begin{equation}\label{red:odd-shift}
j-1=2\ell\quad\text{for some }\ell\ge0,
\qquad j=2\ell+1.
\end{equation}
For $r\in\Rt$, the value $\Dt(r)$ belongs to the degree-$j$ piece.
By \Cref{red:pieces} there is a unique element $E(r)\in\Rt$ such that
\begin{equation}\label{red:moving-tau}
\Dt(r)=\tau V^{\ell+1}E(r).
\end{equation}
Thus the values of $E$ lie in $\Rt$ before any iteration or nilpotence
argument is used. Cancellation of the common nonzero multiplier in
the linearity and Leibniz identities again proves that $E$ is a
$\kk$-derivation of $\Rt$. Since that multiplier has fiber weight
$2(\ell+1)$, the fiber weight shift of $E$ is
\begin{equation}\label{red:negative-weight}
q=-2(\ell+1)<0.
\end{equation}
We prove local nilpotence inside $\Gt$ using its LND order. Whenever
$E(r)\ne0$, \Cref{red:moving-tau}, \Cref{red:V-invariant}, and
\Cref{lem:lnd-order} imply
\begin{equation}\label{red:order-decrease}
\nu(E(r))=\nu(r)-1-m.
\end{equation}
In fact, the order of the left side of \Cref{red:moving-tau} is
$\nu(r)-1$, whereas that of its right side is
$m+(\ell+1)\nu(V)+\nu(E(r))=m+\nu(E(r))$.
Every iterate $E^n(r)$ still lies in $\Rt$. Consequently, as long as
it is nonzero, successive applications of \Cref{red:order-decrease}
give
\[
\nu(E^n(r))=\nu(r)-n(m+1).
\]
This nonnegative integer cannot decrease indefinitely, proving that
$E$ is locally nilpotent.
Differentiating the relation in \Cref{red:ring} gives
\begin{equation}\label{red:moving-v}
\tau V^{\ell+1}E(v)=\Dt(v)=2\tau V\Dt(\tau)\ne0.
\end{equation}
The last inequality uses characteristic zero and the domain property.
Thus $E(v)\ne0$, and in particular $E\ne0$.
Product additivity and \Cref{red:V-invariant} give
$\nu(v)=\nu(\tau^2V)=2m$; hence \Cref{red:order-decrease} yields
\[
\nu(E(v))=m-1.
\]
If $E^2(v)$ were nonzero, a second application would yield
\[
\nu(E^2(v))=(m-1)-1-m=-2,
\]
an impossibility. Therefore $E^2(v)=0$, as required.
\end{proof}
\begin{remark}\label{red:sign-essential}
The restriction to nonpositive fiber weight is essential to the
obstruction proved in \Cref{sec:rigidity}. Indeed, the derivation
\[
E_+(b)=a,\qquad E_+(c)=d,\qquad
E_+(a)=E_+(d)=E_+(u)=0
\]
annihilates $ac-bd-1$ and so defines a derivation of $\Rt$.
It is locally nilpotent, since on the polynomial ring it lowers total
degree in $b,c$, and it has fiber weight shift $+2$. Nevertheless,
$E_+(v)=a^4$ and $E_+^2(v)=0$.
\end{remark}
==== END An-explicit-failure-of-complex-affine-space-cancellation//05-reduction.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//06-rigidity.tex ====
\section{The rigidity obstruction}\label{sec:rigidity}
We now exclude the derivation supplied by \Cref{prop:extracted-lnd}.
Recall that
\[
\Rt=\kk[a,d,b,c,u]/(ac-bd-1),\qquad
v=a^3b-a^2u^3-d^2,
\]
and that the fiber weights of $(a,d,b,c,u)$ are
$(1,1,-1,-1,0)$. A derivation has fiber shift $e$ if it sends
each fiber-weight-$i$ component into the fiber-weight-$(i+e)$
component.
The proof first passes to the highest part for a third, auxiliary
grading. It then studies the polynomial orbit of
$N=d^2+a^2u^3$. The polynomial $abc$ inequality forces this highest
derivation to fix $u$, and a nonsquare in the original fraction field
forces it to fix $a,d$ as well. The determinant relation leaves only derivations
whose fiber weight is positive.
The use of polynomial $abc$ to constrain locally nilpotent
derivations has precedents in
Finston--Maubach~\cite[Section~2]{FinstonMaubach}. We include the
Wronskian proof of the Mason--Stothers inequality
\cite{Mason,Stothers}, followed by the elementary one-variable fact
needed at the end of the argument.
\begin{lemma}[Mason--Stothers polynomial \(abc\)]\label{rig:abc}
Let $F$ be a field of characteristic zero, and let
$X,Y,Z\in F[q]$ be nonzero, pairwise relatively prime polynomials
satisfying $X+Y=Z$. If they are not all constant, then
\[
\max\{\deg X,\deg Y,\deg Z\}
\le r(XYZ)-1,
\]
where $r(XYZ)$ is the number of distinct roots of $XYZ$ in an
algebraic closure of $F$.
\end{lemma}
\begin{proof}
Primes in this proof denote differentiation with respect to $q$.
Consider the Wronskian
\[
W=X'Y-XY'.
\]
It is nonzero. Indeed, if $W=0$, relative primality implies that
$X$ divides $X'$. In characteristic zero this is impossible for
nonconstant $X$, so $X$ is a nonzero constant. Then $Y'=0$, hence
$Y$ and $Z$ are constant as well, contrary to the hypothesis.
The same Wronskian, up to sign, is obtained from either of the
other two pairs, because $Z=X+Y$. Work over an algebraic closure
only to count roots. A root of multiplicity $h$ of one of the
three polynomials is not a root of either of the others. Computing
$W$ from a pair containing that polynomial shows that $W$ vanishes
at the root to order at least $h-1$. These roots are distinct
across the three polynomials, so, on writing
$T=\deg X+\deg Y+\deg Z$, we obtain
\[
\deg W\ge T-r(XYZ).
\]
On the other hand, compute $W$ from the two polynomials other
than one of maximum degree $M$. Differentiation lowers the degree
of every nonconstant polynomial by one, and therefore
\[
\deg W\le T-M-1.
\]
Together these inequalities give $M\le r(XYZ)-1$.
The nonconstant hypothesis is necessary: for three nonzero
constants the right side would be $-1$. The nonzero hypothesis
also causes no missing nonconstant case: if one member of a
pairwise relatively prime triple $X+Y=Z$ is zero, the other two
must be nonzero constants. The all-zero triple is not pairwise
relatively prime.
\end{proof}
\begin{lemma}[Derivations in one variable]\label{rig:one-variable}
Let $F$ be a field of characteristic zero. If $\delta$ is a locally
nilpotent $F$-derivation of $F[t]$, then $\delta(t)\in F$.
\end{lemma}
\begin{proof}
Write $\delta=P(t)\,d/dt$. There is nothing to prove if $P=0$.
Otherwise let $m=\deg P$ and let $\gamma\ne0$ be its leading
coefficient. If $m\ge1$, induction shows that, for every $n\ge1$,
the polynomial $\delta^n(t)$ has degree $1+n(m-1)$ and leading
coefficient
\[
\gamma^n\prod_{j=0}^{n-1}\bigl(1+j(m-1)\bigr).
\]
Every factor is nonzero in characteristic zero, including when
$m=1$. Thus no iterate vanishes, contradicting local nilpotence.
Consequently $m=0$.
\end{proof}
\begin{proposition}\label{prop:rigidity}
There is no nonzero locally nilpotent $\kk$-derivation $E$ of
$\Rt$ with a fixed fiber shift $e\le0$ and $E^2(v)=0$.
\end{proposition}
\begin{proof}
Suppose such an $E$ exists. We introduce an \emph{auxiliary
grading}, distinct from both the valuation degree and the fiber
weight:
\begin{equation}\label{rig:auxiliary-grading}
\begin{array}{c|rrrrr}
&a&d&b&c&u\\ \hline
\text{valuation degree}&0&0&0&0&0\\
\text{fiber weight}&1&1&-1&-1&0\\
\text{auxiliary degree}&0&3&-3&0&2.
\end{array}
\end{equation}
The relation $ac-bd-1$ is homogeneous of auxiliary degree zero,
so this is an actual $\mathbb Z$-grading on $\Rt$. Its components
need not vanish in negative degrees. The generators are
homogeneous for both this grading and the fiber grading, and the
relation is homogeneous for both; thus the two gradings commute.
\medskip\noindent
\emph{The highest auxiliary component.}
For an auxiliary-homogeneous $r\in\Rt$ of degree $i$, define
$E_j(r)$ to be the degree-$(i+j)$ component of $E(r)$, and extend
linearly to $\Rt$. Projecting the product rule to each degree
shows that every $E_j$ is a derivation. Only finitely many of
these derivations are nonzero: their values on the five generators
are obtained from the finitely many homogeneous components of
their images under $E$, and a derivation vanishing on all five
generators is zero. We therefore have a finite decomposition
\[
E=\sum_j E_j.
\]
Let $\lambda$ be its greatest nonzero shift and set $E'=E_\lambda$.
Then $E'\ne0$. For every auxiliary-homogeneous $r$ of degree $i$
and every $n\ge1$, the component of $E^n(r)$ of degree
$i+n\lambda$ equals $(E')^n(r)$: in the expansion of the iterate,
each chosen shift is at most $\lambda$, and their sum can equal
$n\lambda$ only when every chosen shift equals $\lambda$.
Local nilpotence of $E$ therefore implies local nilpotence of
$E'$ on every homogeneous $r$, and hence on every element, which
is a finite sum of such elements. This reasoning uses finite
shift sets and finite sums, so the negative degrees in
\Cref{rig:auxiliary-grading} cause no difficulty. Projection
to auxiliary degree preserves fiber weight, and consequently
$E'$ still has fiber shift $e$.
Set
\[
N=d^2+a^2u^3.
\]
Both its summands have auxiliary degree $6$, whereas $a^3b$ has
auxiliary degree $-3$. In $E^2(v)=E^2(a^3b-N)$, the first term
has auxiliary degrees at most $-3+2\lambda$. The degree
$6+2\lambda$ component of the second term is $(E')^2(N)$.
Taking that component of $E^2(v)=0$ gives
\begin{equation}\label{rig:linear-orbit}
(E')^2(N)=0.
\end{equation}
\medskip\noindent
\emph{The orbit polynomials fix $a,d,u$.}
The localization obtained by solving for $c$ is
\begin{equation}\label{rig:a-chart}
\Rt[a^{-1}]=\kk[a,a^{-1},d,u][b],
\qquad c=\frac{1+bd}{a}.
\end{equation}
In particular, $a,d,b,u$ are algebraically independent,
$N\ne0$, and the fraction field is
\begin{equation}\label{rig:original-field}
K=\Frac(\Rt)=\kk(a,d,b)(u).
\end{equation}
Since $E'$ is locally nilpotent, exponentiating it gives
polynomials
\[
a(q)=\exp(qE')(a),\quad
d(q)=\exp(qE')(d),\quad
u(q)=\exp(qE')(u)
\quad\text{in }\Rt[q]\subset K[q].
\]
Each is nonzero, since its value at $q=0$ is the corresponding
nonzero element of $\Rt$. The exponential is multiplicative
by the iterated product rule. Hence \Cref{rig:linear-orbit}
implies
\begin{equation}\label{rig:orbit-sum}
d(q)^2+a(q)^2u(q)^3
=N(q):=\exp(qE')(N)=N+qE'(N),
\qquad \deg_q N(q)\le1.
\end{equation}
Here $N(q)\ne0$ because $N(0)=N\ne0$. Thus neither summand nor
their sum is the zero polynomial.
The three polynomials in \Cref{rig:orbit-sum} are pairwise
relatively prime in $K[q]$. To see this, a common root of the
two summands over $\overline K$ would have multiplicity at least
two in each summand. For the second summand this follows
whether the root belongs to $a(q)$, to $u(q)$, or to both.
Their sum would then have a root of multiplicity at least two,
which a nonzero polynomial of degree at most one cannot have.
Thus the summands are relatively prime, and the identity
$X+Y=Z$ implies relative primality of the other two pairs.
No relative primality between $a(q)$ and $u(q)$ is needed.
Write
\[
D_a=\deg_q a(q),\qquad D_d=\deg_q d(q),\qquad
D_u=\deg_q u(q).
\]
Suppose at least one summand in \Cref{rig:orbit-sum} is
nonconstant. A nonconstant summand has degree at least two.
Since the sum has degree at most one, both summands must then
have the same degree
\[
M=2D_d=2D_a+3D_u\ge2.
\]
The number of distinct roots of their product with $N(q)$ is
at most $D_d+D_a+D_u+\deg_q N(q)$. Applying \Cref{rig:abc}
therefore yields
\begin{equation}\label{rig:degree-bound}
2D_d=2D_a+3D_u
\le D_d+D_a+D_u+\deg_q N(q)-1
\le D_d+D_a+D_u.
\end{equation}
This also covers constant $N(q)$; in that case the first bound
is stronger. The last inequality gives $D_d\le D_a+D_u$.
Combining it with $2D_d=2D_a+3D_u$ gives $3D_u\le2D_u$,
so $D_u=0$. Consequently $u(q)=u$ and
$D_d=D_a>0$.
Let $\alpha,\delta\in K^*$ be the leading coefficients of
$a(q)$ and $d(q)$, respectively. Cancellation of the
degree-$M$ terms in \Cref{rig:orbit-sum} requires
\[
\delta^2+\alpha^2u^3=0,
\qquad\text{hence}\qquad
\left(\frac{\delta}{\alpha}\right)^2=-u^3.
\]
This equality would hold in the original field $K$ of
\Cref{rig:original-field}, not merely in $\overline K$.
It is impossible: put $K_0=\kk(a,d,b)$, so that $K=K_0(u)$.
For a nonzero rational function of $u$, the multiplicity of $u$
in its numerator minus its multiplicity in its denominator
defines an integer-valued order. A square has even order,
while $-u^3$ has order three. Passing to $\overline K$ above
was used only for counting roots and has not changed the field
of the leading coefficients.
We conclude that both summands in \Cref{rig:orbit-sum} are
constant. Their nonvanishing gives $2D_d=0$ and
$2D_a+3D_u=0$. All three degrees are nonnegative, so
$D_a=D_d=D_u=0$. Taking coefficients of $q$ in their
exponentials now gives
\begin{equation}\label{rig:fixed-coefficients}
E'(a)=E'(d)=E'(u)=0.
\end{equation}
\medskip\noindent
\emph{The remaining coefficient has an impossible fiber weight.}
Differentiating $ac-bd=1$ and using
\Cref{rig:fixed-coefficients} gives $aE'(c)=dE'(b)$. Define
\[
h=cE'(b)-bE'(c)\in\Rt.
\]
Using both determinant identities, without dividing by either
$a$ or $d$, gives
\begin{equation}\label{rig:h-formulas}
a h=(ac-bd)E'(b)=E'(b),\qquad
d h=(ac-bd)E'(c)=E'(c).
\end{equation}
Let $C_0=\kk[a,d,u]$ and $Q=\Frac(C_0)=\kk(a,d,u)$.
By \Cref{rig:fixed-coefficients}, every nonzero element of
$C_0$ is invariant. Thus $E'$ extends to a locally nilpotent
$Q$-derivation of
\[
(C_0\setminus\{0\})^{-1}\Rt=Q[b],
\]
where the equality follows by solving for $c$ as in
\Cref{rig:a-chart}. Local nilpotence is preserved because
$(E')^n(r/s)=(E')^n(r)/s$ for $s\in C_0\setminus\{0\}$.
By \Cref{rig:one-variable}, $E'(b)\in Q$; hence
\Cref{rig:h-formulas} implies $h\in Q$.
We spell out the coefficient intersection that improves this
to $h\in C_0$. Besides \Cref{rig:a-chart}, solving for $b$
gives
\[
\Rt[d^{-1}]=\kk[a,d,d^{-1},u][c],
\qquad b=\frac{ac-1}{d}.
\]
Since $b$ and $c$ are each transcendental over $Q$, comparison
of polynomial coefficients in these two presentations gives
\[
\Rt\cap Q
\subset C_0[a^{-1}]\cap C_0[d^{-1}]
=C_0.
\]
The last intersection is taken in $Q$: in the unique
factorization domain $C_0$, the reduced denominator of an
element belonging to both rings must divide a power of $a$
and a power of $d$. As $a$ and $d$ are relatively prime, that
denominator is a unit. Conversely $C_0\subset\Rt\cap Q$, so
the intersection is exactly $C_0$. In particular $h\in C_0$.
The derivation $E'$ is nonzero and fixes $a,d,u$; therefore
\Cref{rig:h-formulas} implies $h\ne0$. Since $E'$ has fiber
shift $e$, the definition of $h$ shows that it is homogeneous
of fiber weight $e-2<0$. But every monomial in
$C_0=\kk[a,d,u]$ has nonnegative fiber weight, because the
weights of $a,d,u$ are $1,1,0$. A nonzero element of $C_0$
cannot have negative fiber weight. This contradiction proves
the proposition.
\end{proof}
\begin{proof}[Proof of \Cref{thm:main}]
The explicitly defined algebra $A=P/(H)$ is a finitely
generated unital $\kk$-algebra. By \Cref{con:domain} it is
integral, with fraction field $\kk(p,x,y,z)$, and hence has
dimension four. Moreover, \Cref{prop:stabilization}
gives a $\kk$-algebra isomorphism
\[
A[w]\cong\kk\poly{5}.
\]
Suppose that $A\cong\kk\poly{4}$. By
\Cref{prop:graded,prop:positive-invariant}, its associated
graded ring $G$ admits a nonzero homogeneous locally nilpotent
derivation $D_0$ with a nonzero homogeneous invariant of
positive valuation degree. The principal bundle and its
coordinate ring $\Gt$ are given by
\Cref{prop:principalization}. By
\Cref{prop:homogeneous-lift}, $D_0$ lifts to a locally
nilpotent derivation $\Dt$ on $\Gt$ with the same valuation
degree shift and fiber shift zero. The reduction of this lift
and invariant in \Cref{prop:extracted-lnd} produces
a nonzero locally nilpotent derivation $E$ of $\Rt$ with a
fixed nonpositive fiber shift and $E^2(v)=0$. This contradicts
\Cref{prop:rigidity}. Thus $A\not\cong\kk\poly{4}$, while
its cylinder is $\kk\poly{5}$. Taking $n=4$ proves the
claimed counterexample over $\kk$.
\end{proof}
==== END An-explicit-failure-of-complex-affine-space-cancellation//06-rigidity.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//07-affine-fibrations.tex ====
\section{Nontrivial affine-space fibrations}
\label{sec:affine-fibrations}
The residual-coordinate property of $H$ determines the fibers of two
natural morphisms. We first use it to prove that the morphisms are smooth,
then use the nonpolynomiality of $A$ to rule out Zariski local triviality.
\begin{corollary}[Nontrivial smooth $\Aff^3$-fibrations]
\label{cor:affine-fibrations}
Let $P,H,A$ be as in \eqref{eq:example}, and put $X=\Spec A$. Write
$\Aff^1_p=\Spec\kk[p]$ and
$\Aff^2_{p,q}=\Spec\kk[p,q]$, where $q$ is independent of $p$.
Consider the morphisms
\[
\pi:X\longrightarrow\Aff^1_p,
\qquad
\rho=(p,H):\Spec P=\Aff^5\longrightarrow\Aff^2_{p,q},
\]
induced respectively by $\kk[p]\to A$ and by
$\kk[p,q]\to P$ with $q\mapsto H$.
Both are smooth surjective morphisms of relative dimension three.
For either morphism and every point $y$ of its base, the fiber over $y$
is isomorphic to $\Aff^3_{\kappa(y)}$; in particular, every geometric
fiber is affine three-space. Neither morphism is a
Zariski-locally trivial $\Aff^3$-bundle.
Thus the Dolgachev--Weisfeiler affine-fibration conjecture fails in
relative dimension three over each of the smooth complex affine bases
$\Aff^1$ and $\Aff^2$.
\end{corollary}
\begin{proof}
Set $\Lambda=\kk[p]$ and write $\mathbf z=(s,u,F,J)$. For every prime
$\mathfrak p$ of $\Lambda$, let $H_{\mathfrak p}$ denote the image of $H$ in
$\kappa(\mathfrak p)[\mathbf z]$. The coordinate changes in
\Cref{con:localized-coordinates,con:localized-h} show that $H$ is a
coordinate over $\kk[p,p^{-1}]$. At $\mathfrak p=(p)$, the determinant-one
change in \Cref{con:linear-coordinates,con:linear-inverse} gives
$H_{\mathfrak p}=L\bmod p$. Every other prime of $\Lambda$ avoids $p$.
It follows that
for every $\mathfrak p$ there are three polynomials $G_1,G_2,G_3$ over
$\kappa(\mathfrak p)$ such that
\begin{equation}\label{fib:residual-coordinate}
\kappa(\mathfrak p)[\mathbf z]
=\kappa(\mathfrak p)[H_{\mathfrak p},G_1,G_2,G_3]
\end{equation}
is a polynomial coordinate system.
Let $y$ be any point of $\Aff^2_{p,q}$, set
$\mathfrak p=\ker(\Lambda\to\kappa(y))$, and let $b$ be the image of $q$ in
$\kappa(y)$. The field $\kappa(y)$ is an extension of
$\kappa(\mathfrak p)$. Tensoring \eqref{fib:residual-coordinate} with
$\kappa(y)$ and then setting the coordinate $H_{\mathfrak p}$ equal to
$b$ gives
\[
\rho_y
=\Spec\bigl(\kappa(y)[\mathbf z]/(H_{\mathfrak p}-b)\bigr)
\simeq\Aff^3_{\kappa(y)}.
\]
Here $H_{\mathfrak p}$ in the quotient denotes its image after this
field extension. Thus every fiber of $\rho$ is affine three-space over
its residue field, and the same is true after extension to an algebraic
closure. In particular, $\rho$ is surjective. Moreover,
\[
A=P/(H)\simeq P\otimes_{\Lambda[q]}\Lambda,
\qquad q\longmapsto0,
\]
so $\pi$ is the base change of $\rho$ along the closed immersion
$\Aff^1_p\hookrightarrow\Aff^2_{p,q}$ defined by $q=0$. The fiber
assertions and surjectivity follow for $\pi$ as well.
We next check smoothness without assuming local triviality. Let
\[
\mathfrak j=
\left(\frac{\partial H}{\partial s},
\frac{\partial H}{\partial u},
\frac{\partial H}{\partial F},
\frac{\partial H}{\partial J}\right)\subset P.
\]
For every $\mathfrak p$, differentiation with respect to the coordinate
$H_{\mathfrak p}$ in \eqref{fib:residual-coordinate} gives a
$\kappa(\mathfrak p)$-derivation $\delta$ with
$\delta(H_{\mathfrak p})=1$. The chain rule expresses $1$ as a linear
combination of the four partial derivatives of $H_{\mathfrak p}$.
Consequently the image of $\mathfrak j$ generates the unit ideal in
$\kappa(\mathfrak p)[\mathbf z]$. If a prime of $P$ contained
$\mathfrak j$, localization and reduction over its contraction to $\Lambda$
would give a proper prime of this fiber algebra containing that unit
ideal. Hence $\mathfrak j=P$.
As a $\Lambda[q]$-algebra, $P$ has the presentation
\[
P\simeq \Lambda[q][\mathbf z]/(q-H).
\]
The principal open sets on which one of the four displayed partial
derivatives is invertible cover $\Spec P$. On each such open set this
presentation has one relation with an invertible Jacobian minor, so it
is standard smooth of relative dimension $4-1=3$. Thus $\rho$ is smooth
of relative dimension three. Smoothness and relative dimension are
preserved by base change, giving the same assertion for $\pi$.
Suppose now that one of the two morphisms is Zariski-locally trivial,
and put
\[
(R_0,B)=(\kk[p],A)
\quad\hbox{or}\quad
(R_0,B)=(\kk[p,q],P),
\]
respectively, with the algebra structures specified in the statement.
In either case, $B$ is a finitely presented $R_0$-algebra, and local triviality
implies
$B_{\mathfrak m}\simeq R_{0,\mathfrak m}\poly3$
for every maximal ideal $\mathfrak m$ of $R_0$.
Bass--Connell--Wright's theorem
\cite[Theorem~3(a)]{BassConnellWright}, in the formulation recalled by
Gupta~\cite[Theorem~4.1]{GuptaSurvey}, therefore gives an
$R_0$-algebra isomorphism
\[
B\simeq\operatorname{Sym}_{R_0}(M)
\]
for a projective $R_0$-module $M$. This conclusion allows arbitrary
polynomial transition automorphisms in the assumed local trivializations;
no linear structure is assumed beforehand.
The module $M$ is finitely generated. Indeed, if $\mathfrak a$ is the
augmentation ideal of $\operatorname{Sym}_{R_0}(M)$, then
$M\simeq\mathfrak a/\mathfrak a^2$; finite generation follows because
$B$ is a finite-type augmented $R_0$-algebra. Tensoring with each residue
field and comparing the dimensions of the resulting polynomial rings
shows that $M$ has rank three. For the freeness step, Suslin's
Theorem~2(b)~\cite{Suslin} states that a finitely generated projective
module of rank greater than $\dim C$ over $C[t_1,\ldots,t_m]$ is extended
from $C$ when $C$ is regular and noetherian. Here $C=\kk$ has dimension
zero, giving the needed Quillen--Suslin statement: $M$ is extended from
a three-dimensional complex vector space.
Thus $M$ is free and
\[
B\simeq R_0\poly3.
\]
For $\pi$ this would give
$A\simeq\kk[p]\poly3\simeq\kk\poly4$, contrary to \Cref{thm:main}.
For $\rho$ an $R_0$-algebra isomorphism would give a polynomial coordinate
system
$P=\kk[p,H,G_1,G_2,G_3]$ for suitable $G_1,G_2,G_3\in P$.
In particular, $H$ would be a coordinate of $P$, contrary to
Corollary~\ref{cor:stable-coordinate}. Neither morphism is therefore
Zariski-locally trivial.
\end{proof}
The two bases are smooth, normal, and affine over $\kk$. The fiber
calculation verifies the residue-field hypothesis in the original
affine-fibration assertion of Dolgachev--Weisfeiler, not only its
geometric-fiber version. These counterexamples have relative dimension
three; they do not address the corresponding assertions in relative
dimensions one or two.
==== END An-explicit-failure-of-complex-affine-space-cancellation//07-affine-fibrations.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//main.tex ====
\documentclass[11pt]{article}
\input{preamble}
\title{An explicit failure of complex affine-space cancellation}
\author{OpenAI}
\date{September 23, 2026}
\begin{document}
\maketitle
\begin{abstract}
We construct an explicit integral complex affine fourfold $X$ with
$X\times\mathbb A^1\cong\mathbb A^5$ but
$X\not\cong\mathbb A^4$. This gives a negative answer to Zariski's
affine-space cancellation problem over $\mathbb C$ in dimension four.
The same construction disproves the Stable Coordinate Conjecture in
ambient dimension five and yields smooth $\mathbb A^3$-fibrations over
$\mathbb A^1$ and $\mathbb A^2$ that are not Zariski-locally trivial.
These fibrations disprove the Dolgachev--Weisfeiler affine-fibration
conjecture over these bases.
\end{abstract}
\input{sections/01-introduction}
\input{sections/02-construction}
\input{sections/03-degeneration}
\input{sections/04-bundle}
\input{sections/05-reduction}
\input{sections/06-rigidity}
\input{sections/07-affine-fibrations}
\input{references}
\end{document}
==== END An-explicit-failure-of-complex-affine-space-cancellation//main.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//preamble.tex ====
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\usepackage{lmodern}
\usepackage{microtype}
\usepackage{amsmath,amssymb,amsthm,mathtools}
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\hypersetup{colorlinks=true,linkcolor=MidnightBlue,citecolor=MidnightBlue,urlcolor=MidnightBlue,pdfauthor={OpenAI},pdftitle={An explicit failure of complex affine-space cancellation}}
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\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\newcommand{\kk}{\mathbb C}
\newcommand{\Aff}{\mathbb A}
\newcommand{\Ga}{\mathbb G_a}
\newcommand{\Gm}{\mathbb G_m}
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==== END An-explicit-failure-of-complex-affine-space-cancellation//preamble.tex ====
==== BEGIN An-explicit-failure-of-complex-affine-space-cancellation//references.tex ====
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\par\endgroup
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