==== TABLE OF CONTENTS ====
A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//01-introduction.tex
A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//02-construction.tex
A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//03-filtration.tex
A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//04-bundle.tex
A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//05-rigidity.tex
A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//main.tex
A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//refs.bib
===========================
==== BEGIN A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//01-introduction.tex ====
\section{Introduction}\label{sec:introduction}
Let $k=\C$. An element $f\in k[x_1,\ldots,x_n]$ is a \emph{coordinate} if there are elements $f_2,\ldots,f_n$ such that
\[
k[x_1,\ldots,x_n]=k[f,f_2,\ldots,f_n]
\]
as a polynomial ring. Equivalently, a polynomial automorphism sends $f$ to $x_1$. It is a \emph{one-stable coordinate} if it is a coordinate in $k[x_1,\ldots,x_n,w]$, where $w$ is independent. The stable coordinate conjecture asks whether an element that becomes a coordinate after adjoining finitely many variables was already a coordinate. The one-variable formulation appears in \cite[Conjecture~3]{ShpilrainYu2002}; successive applications give the formulation with finitely many added variables. Over $\C$, the answer is affirmative in two and three variables by \cite[Proposition~1.1(c)]{ShpilrainYu2002}. The case of one original variable is elementary. Thus four variables are the first possible dimension for a counterexample, and one added variable already suffices.
\begin{theorem}\label{thm:main}
In $R=\C[x_1,x_2,x_3,x_4]$, put
\begin{equation}\label{eq:explicit-counterexample}
Q=x_2^2-x_4^2+x_1x_3,
\qquad
f=x_1-2Q\bigl(Q(x_2+x_4)+x_1x_4\bigr).
\end{equation}
There is a $\C$-algebra automorphism of $R[w]$ sending $f$ to $x_1$, but there is no such automorphism of $R$.
\end{theorem}
The example also has a geometric consequence. A closed embedding $\A^{n-1}\hookrightarrow\A^n$ is \emph{rectifiable} if an automorphism of the ambient affine space takes its image to a coordinate hyperplane. The Abhyankar--Sathaye embedding conjecture asserts, in characteristic zero, that a polynomial $g$ with $k[x_1,\ldots,x_n]/(g)\simeq k^{[n-1]}$ is a coordinate; see \cite[p.~8429]{BlancVanSanten2019}. Here $k^{[r]}$ denotes a polynomial ring in $r$ variables. Corollary~\ref{cor:fibers} gives a negative answer to this conjecture in ambient dimension four. In fact, every fiber is affine three-space and its inclusion in $\A^4$ is nonrectifiable.
\begin{corollary}\label{cor:fibers}
For every $\lambda\in\C$, the hypersurface $f^{-1}(\lambda)\subset\A^4$ is isomorphic to $\A^3$ and its embedding in $\A^4$ is not rectifiable.
\end{corollary}
\begin{proof}
Proposition~\ref{prop:fibers} identifies each fiber with $\A^3$; in particular, $(f-\lambda)$ is its prime defining ideal. If an ambient automorphism takes $f^{-1}(\lambda)$ to a coordinate hyperplane, it takes its defining principal ideal to the ideal of a coordinate. The two generators therefore differ by a nonzero constant. This would make $f-\lambda$, and hence $f$, a coordinate, contrary to Theorem~\ref{thm:main}.
\end{proof}
\subsection*{History and related methods}
The coordinate question retains the embedding in affine space: it asks whether a given function can be made into a coordinate projection by an ambient automorphism. Cancellation asks instead whether an affine variety whose product with an affine line is an affine space must itself be an affine space. Makar-Limanov, van Rossum, Shpilrain, and Yu established an implication from the stable coordinate conjecture to cancellation in \cite[Theorem~1.2]{MakarLimanovRossumShpilrainYu2004}. In our example the ambient variety is already $\A^4$, and the obstruction concerns its distinguished polynomial.
The low-dimensional proof in \cite[proof of Proposition~1.1(c)]{ShpilrainYu2002} uses curve cancellation and the Abhyankar--Moh embedding theorem in two variables. In three variables it combines affine-surface cancellation, using work of Fujita and Miyanishi--Sugie, with Kaliman's theorem that a polynomial on $\C^3$ with general fiber $\C^2$ is a coordinate \cite[Main Theorem]{Kaliman2002}.
The four-variable V\'en\'ereau polynomials have been an important testing ground for these coordinate questions. Lewis proved that the second V\'en\'ereau polynomial is a coordinate \cite{Lewis2013}. Blanc and Poloni developed a description in terms of affine bundles and bivariables, revisited that result, and proved one-stability for a broader family of related polynomials \cite[Proposition~2.3]{BlancPoloni2022}. These results illustrate the distinction between exhibiting polynomial coordinates after stabilization and proving that a polynomial cannot occur in any original coordinate system. For the polynomial in Theorem~\ref{thm:main}, the first problem is settled by explicit substitutions; the second requires a derivation obstruction.
\subsection*{The construction and the obstruction}
The proof uses two presentations of the same four-variable polynomial ring. We first write it as a hypersurface $A=P/(H)$ in a five-variable polynomial ring $P$, with a distinguished element $p$. A determinant-preserving matrix transformation makes $H$ linear in one new coordinate. It identifies $A$ with $R$ and carries $p$ to the polynomial in \eqref{eq:explicit-counterexample}. A second construction, using the exponential of a locally nilpotent derivation, shows that $p$ belongs to polynomial coordinates on $A[w]$. These explicit changes of coordinates occupy Section~\ref{sec:construction}.
To show that $p$ is not a coordinate, we put an integer-indexed filtration on $A$. Its associated graded ring has the form
\[
G=B[\tau,I/\tau]\subset B[\tau,\tau^{-1}],
\qquad B=k[x,y,z,u]/(xy-z(z+1)),
\]
where $I$ is an explicitly defined two-generated ideal of $B$ and $\tau$ is the initial form of $p$. The notation $B[\tau,I/\tau]$ means the subalgebra generated by $B$, $\tau$, and all elements $h/\tau$ with $h\in I$. If $p$ were a coordinate, differentiation in a suitable other coordinate would induce a nonzero locally nilpotent derivation $E$ on $B$ whose kernel contains a nonzero element of $I$. Section~\ref{sec:filtration} proves this implication, including the care needed for negative filtration degrees.
The last two sections exclude that invariant. Section~\ref{sec:bundle} passes from $\Spec B$ to the complement of the zero section in a line bundle. In its coordinate ring
\[
C=k[a,d,b,c,u]/(ac-bd-1),
\]
the ideal $I$ becomes principal, generated by $v=a^3b+a^2u^2-d^2$. The additive action defined by $E$ lifts to this bundle and commutes with fiber scaling. A nonzero invariant in $I$ would force the lifted derivation to annihilate $v$. In Section~\ref{sec:rigidity}, a second grading reduces this to a derivation fixing $a,d,u$. Such a derivation has a coefficient that would simultaneously have to be a polynomial in $a,d,u$ and have negative fiber weight. This is impossible.
The graded-ring and line-bundle strategy adapts the construction in \cite[Sections~2--6]{Cancellation2026}. That paper studies an affine-space cancellation example. Here the quadratic choice in the defining equation has two different uses: it permits the determinant-preserving change that makes $A$ polynomial, and it makes the highest auxiliary part of $v$ factor as $(au-d)(au+d)$. The latter factorization gives the final rigidity argument without a polynomial $abc$ estimate. The passage to leading derivations follows the filtered methods of Kaliman and Makar-Limanov \cite[Sections~3--5]{KalimanMakarLimanov2007}, and the lift of an additive action uses the line-bundle normalization argument of Brion \cite[Lemma~2.9]{Brion2015}. Every construction and obstruction needed for Theorem~\ref{thm:main} is proved below.
==== END A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//01-introduction.tex ====
==== BEGIN A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//02-construction.tex ====
\section{Polynomial presentation and stabilization}\label{sec:construction}
We construct the counterexample through a hypersurface presentation of a
four-variable polynomial ring. This
presentation makes stabilization explicit and will also supply the filtration
used to obstruct coordinateness.
It adapts the construction of \cite[Section~2]{Cancellation2026}, with
quadratic terms that permit the polynomial change of coordinates below.
Let
\begin{equation}\label{con:presentation}
\begin{gathered}
P=k[p,s,u,F,J],\qquad x=s^2-u^2+pF,\\
H=x^2F-(1+2sx)J-pJ^2-u,\qquad A=P/(H).
\end{gathered}
\end{equation}
The five generators of $P$ are independent. We use the same letters for their
images in $A$; the distinguished element is $p\in A$.
\subsection{A polynomial change of coordinates}
\begin{proposition}\label{prop:polynomial}
There are polynomial coordinates $p',s',F',J$ on $A$ such that
\begin{equation}\label{con:candidate}
x=(s')^2-J^2+p'F',\qquad
p=p'-2x\bigl(x(s'+J)+p'J\bigr).
\end{equation}
Thus, under the identification $(p',s',F',J)=(x_1,x_2,x_3,x_4)$,
the element $p$ is the polynomial $f$ in Theorem~\ref{thm:main}.
\end{proposition}
\begin{proof}
The relation $H$ becomes linear after a change of coordinates preserving $x$.
To exhibit that change, write
\[
M=\begin{pmatrix}F&s-u\\s+u&-p\end{pmatrix},\qquad
e=\binom{x}{-J},\qquad n=\binom{J}{x}.
\]
Then $x=-\det M$, $e^{\mathsf t}n=0$, and
\[
H=e^{\mathsf t}Me-J-u.
\]
Keep $J$ fixed and define $F',s',u',p'$ by
\begin{equation}\label{con:matrix-change}
M'=(I_2-2ne^{\mathsf t})M
=\begin{pmatrix}F'&s'-u'\\s'+u'&-p'\end{pmatrix}.
\end{equation}
Since $(ne^{\mathsf t})^2=0$ and $\det(I_2-2ne^{\mathsf t})=1$,
we have $-\det M'=x$. Consequently $x$ and $J$, and hence $e$ and $n$,
are unchanged by the substitution. Its inverse is the polynomial substitution
\[
M=(I_2+2ne^{\mathsf t})M',\qquad x=-\det M',
\]
so $p',s',u',F',J$ form a coordinate system on $P$.
Taking half the difference of the off-diagonal entries in
Equation~\eqref{con:matrix-change} gives
\[
u'=u-e^{\mathsf t}Me,\qquad H=-J-u'.
\]
Eliminating $u'$ therefore identifies $A$ with $k[p',s',F',J]$.
In this quotient $u'=-J$, so the determinant identity gives the first formula
in Equation~\eqref{con:candidate}. The lower-right entry of the inverse
matrix identity gives
\[
p=p'-2x\bigl(x(s'-u')+p'J\bigr),
\]
which becomes the second formula after substituting $u'=-J$.
\end{proof}
The expression for $p$ in Equation~\eqref{con:candidate} takes the values $0$ and $1$ at
$(p',s',F',J)=(0,0,0,0)$ and $(1,0,0,0)$, respectively. In particular,
$p$ is a nonzero nonunit of $A$.
\subsection{One-variable stabilization}
We next change the defining equation after adjoining one variable, while
keeping $p$ fixed. For this purpose set
\begin{equation}\label{con:xyz}
y=s+x(x+u^2),\qquad z=sx+pJ.
\end{equation}
Expansion gives the identity
\begin{equation}\label{con:quadric-identity}
xy-z(z+1)=p(H+u).
\end{equation}
Moreover,
\begin{equation}\label{con:localized-coordinates}
P[p^{-1}]=k[p,p^{-1}][x,y,z,u],
\end{equation}
with inverse coordinate formulas
\begin{equation}\label{con:localized-inverse}
s=y-x(x+u^2),\qquad
F=\frac{x-s^2+u^2}{p},\qquad
J=\frac{z-sx}{p}.
\end{equation}
\begin{proposition}\label{prop:stable}
For an independent variable $w$, the ring $A[w]$ is a polynomial ring in
five variables with $p$ as one of its coordinates.
\end{proposition}
\begin{proof}
A derivation $\delta$ of a $k$-algebra is \emph{locally nilpotent} if, for
every element $g$, some iterate $\delta^r(g)$ is zero. We use such a derivation
to replace the equation $H=0$ by $H+pw=0$.
In the coordinates of Equation~\eqref{con:localized-coordinates}, take
\[
\Delta=-p\frac{\partial}{\partial u},
\qquad \Delta(p)=\Delta(x)=\Delta(y)=\Delta(z)=0.
\]
This locally nilpotent derivation of $P[p^{-1}]$ preserves $P$, because
Equation~\eqref{con:localized-inverse} yields
\begin{equation}\label{con:delta}
\begin{gathered}
\Delta(p)=0,\qquad \Delta(u)=-p,\qquad \Delta(s)=2pxu,\\
\Delta(F)=-4sxu-2u,\qquad \Delta(J)=-2x^2u.
\end{gathered}
\end{equation}
Its restriction to $P$ is therefore locally nilpotent as well. Applying
$\Delta$ to Equation~\eqref{con:quadric-identity} gives $\Delta(H)=p$.
Extend $\Delta$ to $P[w]$ by $\Delta(w)=0$. The exponential
\[
\exp(w\Delta)(g)=\sum_{r\geq0}\frac{w^r\Delta^r(g)}{r!}
\]
is a polynomial automorphism: each sum is finite, and its inverse is
$\exp(-w\Delta)$. It fixes $p$ and sends $H$ to $H+pw$, inducing a
$k[p]$-algebra isomorphism
\begin{equation}\label{con:stabilized-hypersurface}
A[w]\simeq P[w]/(H+pw).
\end{equation}
It remains to eliminate one variable from the equation on the right.
Put $x_0=s^2-u^2$ and introduce
\begin{equation}\label{con:linear-coordinates}
\begin{aligned}
L&=x_0^2F-(1+2sx_0)J,\\
N&=(1-2sx_0)F+4s^2J.
\end{aligned}
\end{equation}
The coefficient matrix of this substitution has determinant
$4s^2x_0^2+(1+2sx_0)(1-2sx_0)=1$; explicitly,
\begin{equation}\label{con:linear-inverse}
F=4s^2L+(1+2sx_0)N,\qquad
J=-(1-2sx_0)L+x_0^2N.
\end{equation}
Thus $p,s,u,L,N$ are polynomial coordinates on $P$. Expanding $x=x_0+pF$
in $H$ gives
\begin{equation}\label{con:H-expansion}
H=L-u+pQ_0,\qquad Q_0=2x_0F^2+pF^3-2sFJ-J^2.
\end{equation}
Since $Q_0\in P$, the substitution $w'=w+Q_0$ is another polynomial change
of coordinates, and
\[
P[w]/(H+pw)
=k[p,s,u,L,N,w']/(L-u+pw')
\simeq k[p,s,u,N,w'].
\]
Together with Equation~\eqref{con:stabilized-hypersurface}, this proves the
assertion with $p$ fixed throughout.
Composing with the identification in Proposition~\ref{prop:polynomial} and
relabeling the target variables so that $p$ maps to $x_1$ gives an
automorphism of $R[w]$ sending $f$ to $x_1$.
\end{proof}
The two steps explicitly realize the exponential shift of
\cite[Lemma~3.3]{DuttaLahiri2021} and the case of the transfer construction
in \cite[Theorem~7]{EdoVenereau2001} in which the perturbation $pQ_0$ is
divisible by $p$.
\subsection{The fibers}
The same coordinate computations describe every fiber of $p$. This will
give a geometric consequence once noncoordinateness has been proved.
\begin{proposition}\label{prop:fibers}
For every $\lambda\in k$, the quotient $A/(p-\lambda)$ is a polynomial
ring in three variables over $k$.
\end{proposition}
\begin{proof}
At $\lambda=0$, Equations~\eqref{con:linear-coordinates}--\eqref{con:H-expansion}
give
\[
A/(p)=k[s,u,L,N]/(L-u)=k[s,u,N].
\]
For $\lambda\ne0$, Equations~\eqref{con:localized-coordinates} and
\eqref{con:quadric-identity} give
\[
A/(p-\lambda)
=k[x,y,z,u]/\bigl(xy-z(z+1)-\lambda u\bigr)
\simeq k[x,y,z],
\]
by elimination of $u$.
\end{proof}
Propositions~\ref{prop:polynomial} and~\ref{prop:stable} establish the
polynomial presentation and the stable-coordinate assertion of
Theorem~\ref{thm:main}. The remaining task is to show that $p$ cannot be
a coordinate of $A$ itself.
==== END A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//02-construction.tex ====
==== BEGIN A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//03-filtration.tex ====
\section{A graded obstruction to coordinateness}
\label{sec:filtration}
We now turn to the assertion that $p$ is not a coordinate of $A$.
The first step is to pass to a weighted degeneration of $A$.
If $p$ were a coordinate, differentiation in another coordinate would
then produce a nonzero locally nilpotent derivation of a simpler
quadric ring, with a nonzero invariant in an explicit ideal.
The rest of the proof will exclude that possibility.
The use of leading derivations follows the filtration methods of
\cite[Sections~3--5]{KalimanMakarLimanov2007}; the particular quadric construction is
adapted from \cite[Sections~3--5]{Cancellation2026}.
\subsection{The associated graded ring}
Give the generators of $P=k[p,s,u,F,J]$ the weights
\begin{equation}
\label{gr:weights}
\begin{array}{c|rrrrr}
&p&s&u&F&J\\ \hline
\text{weight}&-1&0&0&1&1.
\end{array}
\end{equation}
For $d\in\mathbb Z$, let $\mathcal F_dP$ be the span of the monomials
of weight at most $d$, and give $A=P/(H)$ the quotient filtration
$\mathcal F_dA=\operatorname{im}(\mathcal F_dP\to A)$.
These are increasing multiplicative filtrations, indexed by all
integers. Their negative indices are essential: $p$ has weight $-1$.
\begin{proposition}
\label{prop:graded}
The filtration on $A$ is exhaustive and separated. Its associated
graded ring is the domain
\begin{equation}
\label{gr:presentation}
G:=\gr A
\simeq k[p,s,u,F,J]/(H_*),
\qquad
H_*=x^2F-(1+2sx)J-pJ^2,
\quad x=s^2-u^2+pF,
\end{equation}
with the grading induced by \eqref{gr:weights}.
The filtration degrees of $p,s,u,F,J$ in $A$ are respectively
$-1,0,0,1,1$.
\end{proposition}
\begin{proof}
Exhaustiveness follows because every element of $A$ has a polynomial
representative. For $r>0$, every monomial of weight at most $-r$ is
divisible by $p^r$, so
\[
\mathcal F_{-r}A\subseteq p^rA.
\]
By Proposition~\ref{prop:polynomial}, $A$ is a polynomial ring, and
Equation~\eqref{con:candidate} shows that $p$ is a nonzero nonunit.
Therefore $\bigcap_{r>0}p^rA=0$: in a unique
factorization domain, a nonzero element cannot be divisible by
arbitrarily high powers of an irreducible factor of $p$.
This proves separation. In particular, every nonzero $a\in A$ has a
well-defined degree
\[
\deg a=\min\{d\in\mathbb Z:a\in\mathcal F_dA\}
\]
and a nonzero initial form in $\mathcal F_dA/\mathcal F_{d-1}A$.
The element $x$ is homogeneous of weight zero in $P$, and
$H=H_*-u$ has highest weight part $H_*$, of weight one.
For every nonzero $q\in P$, the highest weight part of $qH$ is
$(\operatorname{in}q)H_*$: the polynomial ring is a graded domain,
so the product of the highest parts does not vanish.
Consequently the ideal of initial relations is exactly $(H_*)$.
For completeness, the natural graded map $P\to\gr A$ sends a homogeneous
polynomial of weight $d$ to its class in
$\mathcal F_dA/\mathcal F_{d-1}A$, which may at this stage be zero.
This map is surjective.
A homogeneous polynomial of weight $d$ lies in its kernel exactly
when its image in $A$ has a representative of weight at most $d-1$;
subtracting that representative gives a relation with the original
polynomial as its highest part. Thus the kernel is the ideal of
initial relations, proving \eqref{gr:presentation}.
To prove that this quotient is a domain, use the polynomial
coordinates $x,y,z,u$ of $P[p^{-1}]$ from the preceding section.
The identity $xy-z(z+1)=p(H+u)$ becomes
\[
pH_*=xy-z(z+1).
\]
The polynomial on the right is irreducible over
$k[p,p^{-1},x,z,u]$: as a polynomial in $y$, it is primitive,
because $x$ and $z(z+1)$ are coprime, and it has degree one over
the fraction field. Thus $H_*$ is irreducible after inverting $p$.
It is not divisible by $p$ before localization, since its reduction
modulo $p$ is
\[
(s^2-u^2)^2F-\bigl(1+2s(s^2-u^2)\bigr)J\ne0.
\]
Any factorization of $H_*$ in $P$ has a factor that becomes a unit after
inverting $p$, hence is a scalar times a power of $p$.
Since $p\nmid H_*$, this factor is already a unit in $P$.
Thus $H_*$ is irreducible in $P$, and hence prime by unique factorization.
Finally, $H_*$ is not associated to any
of the five variables, so their classes in $G$ are nonzero.
These classes are the initial forms of the corresponding elements
of $A$, proving the asserted degrees.
\end{proof}
\subsection{A quadric and its distinguished ideal}
From now on write $\tau$ for the class of $p$ in $G$, to distinguish
the graded ring from $A$. Consider the quadric ring
\begin{equation}
\label{gr:quadric}
B=k[x,y,z,u]/\bigl(xy-z(z+1)\bigr).
\end{equation}
The formulas
\[
x=s^2-u^2+\tau F,\qquad
y=s+x(x+u^2),\qquad z=sx+\tau J
\]
define elements of degree zero in $G$. They satisfy the relation in
\eqref{gr:quadric}; after inverting $\tau$, the inverse coordinate
formulas from the preceding section give an isomorphism
\begin{equation}
\label{gr:localization}
G[\tau^{-1}]=B[\tau,\tau^{-1}],
\qquad \deg B=0,\quad \deg\tau=-1.
\end{equation}
In particular, the induced map $B\to G$ is injective: its composition
with localization is the inclusion of $B$ in this Laurent polynomial
ring. We henceforth regard $B$ as a subring of $G$.
Define elements and an ideal of $B$ by
\begin{equation}
\label{gr:ideal}
\begin{aligned}
s_B&=y-x(x+u^2),\\
f_0&=x-s_B^2+u^2,\qquad
g_0=z-s_Bx,\qquad I=(f_0,g_0).
\end{aligned}
\end{equation}
In \eqref{gr:localization} we have $s=s_B$, $F=f_0/\tau$, and
$J=g_0/\tau$. These identities give the more precise description
\begin{equation}
\label{gr:modification}
G=B[\tau,I/\tau]
=B[\tau,f_0/\tau,g_0/\tau]
\subset B[\tau,\tau^{-1}].
\end{equation}
This is the algebraic construction of an affine modification, with
divisor $\tau=0$ and center $(\tau,I)$ in $B[\tau]$; see
\cite[Definition~1.1 and Proposition~1.1(b)]{KalimanZaidenberg1999}.
We will only need its explicit
homogeneous pieces.
\begin{lemma}
\label{gr:positive-piece}
For each integer $\ell>0$, the degree-$\ell$ piece of $G$ is
\[
G_\ell=\tau^{-\ell}I^\ell.
\]
\end{lemma}
\begin{proof}
A monomial in the generators in \eqref{gr:modification}, with
coefficient in $B$, has the form
\[
b\,\tau^r(f_0/\tau)^i(g_0/\tau)^j,
\qquad r,i,j\ge0.
\]
It has degree $\ell$ exactly when $i+j-r=\ell$.
In that case it equals $\tau^{-\ell}b f_0^i g_0^j$, whose numerator
belongs to $I^{\ell+r}\subseteq I^\ell$.
Since the presentation is homogeneous, such monomials span $G_\ell$.
Conversely, every element of $I^\ell$ is a $B$-linear combination of
the products $f_0^i g_0^j$ with $i+j=\ell$, and dividing these
products by $\tau^\ell$ gives elements of $G_\ell$.
\end{proof}
\subsection{What a coordinate would force}
The ideal $I$ records the positive part of the grading. We now show
why a hypothetical coordinate system containing $p$ supplies a nonzero
locally nilpotent derivation of $B$ fixing a nonzero element of $I$.
\begin{proposition}
\label{prop:coordinate-obstruction}
If $p$ is a coordinate of $A$, then there exist a nonzero locally
nilpotent derivation $E$ of the ring $B$ in \eqref{gr:quadric} and
an element $0\ne h\in I$ such that $E(h)=0$.
\end{proposition}
\begin{proof}
Suppose $A=k[p,q_1,q_2,q_3]$ for a polynomial coordinate system.
At least one $q_i$ has strictly positive filtration degree.
Indeed, if all three had degree at most zero, then so would every
polynomial in $p,q_1,q_2,q_3$, contrary to $\deg F=1$.
Relabel so that $\deg q_1>0$, and take
\[
D=\frac{\partial}{\partial q_2}.
\]
This is a nonzero locally nilpotent derivation of $A$ satisfying
$D(p)=D(q_1)=0$.
We give the details of its passage to $G$, because the filtration
has both positive and negative indices.
Write $\omega(t)$ for the assigned weight of a generator
$t\in\{p,s,u,F,J\}$, and put
\begin{equation}
\label{gr:shift}
m=\max_{D(t)\ne0}\bigl(\deg D(t)-\omega(t)\bigr).
\end{equation}
This maximum is defined: a nonzero derivation cannot kill all
algebra generators. The product rule shows that a monomial of
weight at most $d$ has derivative in $\mathcal F_{d+m}A$.
Every element of $\mathcal F_dA$ has a representative that is a sum
of such monomials, by the definition of the quotient filtration.
Thus
\[
D(\mathcal F_dA)\subseteq\mathcal F_{d+m}A
\qquad(d\in\mathbb Z).
\]
It follows that $D$ induces a homogeneous derivation $D_0$ of $G$
of degree $m$.
For $a\in\mathcal F_dA$, write $[a]_d$ for its class in
$\mathcal F_dA/\mathcal F_{d-1}A$. Explicitly,
\begin{equation}
\label{gr:iterate}
D_0([a]_d)=[D(a)]_{d+m},\qquad
D_0^r([a]_d)=[D^r(a)]_{d+rm}\quad(r\ge0).
\end{equation}
The first formula is well-defined because $D$ also sends
$\mathcal F_{d-1}A$ into $\mathcal F_{d+m-1}A$; the second follows
by induction. It proves that $D_0$ is locally nilpotent, first on
homogeneous elements and then on finite sums of them.
Moreover, equality is attained in \eqref{gr:shift} for some
generator $t$, whose initial form has degree $\omega(t)$ by
Proposition~\ref{prop:graded}. Its image under $D_0$ is the nonzero
initial form of $D(t)$, so $D_0\ne0$.
Finally, \eqref{gr:iterate} shows that $D_0$ kills both $\tau$ and
the nonzero initial form $\eta$ of $q_1$. The latter is homogeneous
of degree $\ell=\deg q_1>0$.
Because $D_0(\tau)=0$, extending $D_0$ to
$G[\tau^{-1}]=B[\tau,\tau^{-1}]$ preserves local nilpotence.
Homogeneity implies that there is a derivation $E:B\to B$ with
\begin{equation}
\label{gr:descent}
D_0(b)=\tau^{-m}E(b)\qquad(b\in B).
\end{equation}
Indeed, every degree-$m$ element of the Laurent ring is uniquely
$\tau^{-m}$ times an element of $B$.
The derivation $E$ is nonzero: otherwise $D_0$ would kill both $B$
and $\tau$, hence the whole Laurent ring and its subring $G$.
It is locally nilpotent because
\[
E^r(b)=\tau^{rm}D_0^r(b)
\qquad(b\in B, r\ge0).
\]
Here, as in \eqref{gr:iterate}, the formula follows by induction,
using that $D_0$ fixes $\tau$.
By Lemma~\ref{gr:positive-piece}, write
$\eta=\tau^{-\ell}h$ with $0\ne h\in I^\ell$.
Equations \eqref{gr:descent} and $D_0(\eta)=0$ give
\[
0=D_0(\eta)=\tau^{-m-\ell}E(h).
\]
Thus $E(h)=0$, and $I^\ell\subseteq I$ because $\ell>0$.
\end{proof}
Proposition~\ref{prop:coordinate-obstruction} has removed the
original coordinate system from the problem. It remains to show
that no nonzero locally nilpotent derivation of $B$ can have a
nonzero invariant in $I$. We will make this ideal principal after
pulling it back to the complement of the zero section in a line bundle,
where a second grading gives the contradiction.
==== END A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//03-filtration.tex ====
==== BEGIN A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//04-bundle.tex ====
\section{Lifting the obstruction to a line bundle}\label{sec:bundle}
Proposition~\ref{prop:coordinate-obstruction} reduces noncoordinateness to
excluding a nonzero locally nilpotent derivation $E$ of $B$ with a nonzero
invariant in $I=(f_0,g_0)$. We pull $I$ back to the complement of the zero
section in a line bundle over $\Spec B$; on this complement the ideal
becomes principal. Every additive action on the base lifts linearly to
the line bundle and preserves this complement. Consequently, an invariant
in $I$ will force the lifted action to fix one specific polynomial. This use of the quadric and its line bundle
adapts the construction of \cite[Sections~4--6]{Cancellation2026}; all the
lifting and principalization arguments needed here are proved below.
Put $Y=\Spec B$ and define
\begin{equation}\label{bun:C}
C=k[a,d,b,c,u]/(ac-bd-1).
\end{equation}
The assignment
\begin{equation}\label{bun:base-map}
x=ab,\qquad y=dc,\qquad z=db,\qquad u=u
\end{equation}
defines a homomorphism $B\longrightarrow C$, since
$xy=abdc=(db)(ac)=z(z+1)$. The ring $C$ is a domain: the polynomial
$ac-bd-1$, viewed as a primitive linear polynomial in $c$ over
$k[a,d,b,u]$, is irreducible. We give $C$ the \emph{fiber weights}
\begin{equation}\label{bun:fiber-weights}
\operatorname{wt}(a,d,b,c,u)=(1,1,-1,-1,0).
\end{equation}
The relation is homogeneous of weight zero, and the image of $B$ has weight
zero. These weights describe the action of $\G_m$ that scales the fibers
of the following bundle. Geometrically this is the familiar quotient of
$\mathrm{SL}_2$ by a diagonal torus, times the affine $u$-line; see
\cite[p.~8431]{BlancVanSanten2019}. We use explicit charts throughout.
\begin{lemma}\label{bun:charts}
The map $B\longrightarrow C$ is injective. The morphism
$\Spec C\longrightarrow Y$ is the complement of the zero section in a line
bundle $\mathcal L\longrightarrow Y$. It is trivial over the open cover
\[
U_0=D(z+1),\qquad U_1=D(z)
\]
of $Y$, with respective invertible fiber coordinates $a$ and $d$.
\end{lemma}
\begin{proof}
Over $U_0$, the identity $ac=z+1$ makes $a$ invertible. The formulas
\begin{equation}\label{bun:chart-a}
c=\frac{z+1}{a},\qquad b=\frac{x}{a},\qquad
d=\frac{ay}{z+1}
\end{equation}
give an isomorphism
$C[(z+1)^{-1}]=B[(z+1)^{-1}][a,a^{-1}]$.
In particular the localized map from $B$ is injective, and so is the
original map because $B$ is a domain. Over $U_1$, the corresponding formulas
are
\begin{equation}\label{bun:chart-d}
b=\frac{z}{d},\qquad c=\frac{y}{d},\qquad a=\frac{xd}{z},
\end{equation}
and give $C[z^{-1}]=B[z^{-1}][d,d^{-1}]$.
The opens cover $Y$ because $z$ and $z+1$ generate the unit ideal. On their
overlap the fiber coordinates satisfy
\begin{equation}\label{bun:transition}
d=\frac{y}{z+1}\,a,
\end{equation}
where $y/(z+1)$ is a unit with inverse $x/z$. Gluing the trivial line bundles
by this transition gives $\mathcal L$; removing their zero sections gives
precisely the two charts of $\Spec C$ above. Both fiber coordinates have
weight one, so the scaling agrees with \eqref{bun:fiber-weights}.
\end{proof}
The two generators of $I$ acquire a common factor in $C$. Define
\begin{equation}\label{bun:v}
v=a^2(x+u^2)-d^2=a^3b+a^2u^2-d^2\in C.
\end{equation}
\begin{lemma}\label{bun:principalization}
The extended ideal $IC$ is the principal ideal $(v)$.
\end{lemma}
\begin{proof}
In $C[a^{-1}]$, put $\xi=d/a$ and $j=x+u^2-\xi^2$. The determinant relation
and \eqref{bun:base-map} give
\[
z=\xi x,\qquad y=\xi+\xi^2x,\qquad
s_B=y-x(x+u^2)=\xi-xj.
\]
Consequently
\begin{equation}\label{bun:ideal-chart}
f_0=(1+2\xi x-x^2j)j,\qquad g_0=x^2j,\qquad v=a^2j.
\end{equation}
The two coefficients of $j$ generate the unit ideal, as the identity
\[
(1-2\xi x)(1+2\xi x-x^2j)
+x^2(4\xi^2+j-2\xi xj)=1
\]
shows. Thus $IC[a^{-1}]=(j)=(v)$.
It remains to check primes of $C$ containing $a$. Modulo such a prime,
$bd=-1$, while $x=0$ and $z=-1$. Hence $g_0=z-s_Bx$ has nonzero residue
$-1$, and $v$ has nonzero residue $-d^2$. Both are units in the corresponding
local ring. The ideals $IC$ and $(v)$ therefore agree at every prime, which
proves the assertion.
\end{proof}
Figure~\ref{bun:figure} records the geometric meaning of this ideal
equality: the inverse image of $V(I)$ is cut out by the single function $v$.
\begin{figure}[ht]
\centering
\begin{tikzcd}[column sep=large,row sep=large]
V(v) \arrow[r,hook] \arrow[d] & \Spec C \arrow[d] \\
V(I) \arrow[r,hook] & Y=\Spec B
\end{tikzcd}
\caption{A Cartesian square of closed subschemes. Pulling $I$ to the
complement of the zero section of $\mathcal L$ gives the principal ideal
$(v)$. In particular, every $h\in I$ pulls back to a multiple of $v$.}
\label{bun:figure}
\end{figure}
We next lift additive actions from $Y$ to $\mathcal L$. The elementary
reason that a lift exists is that adjoining an affine-line parameter does
not change line bundles on $Y$.
\begin{lemma}\label{bun:homotopy}
Every line bundle on $\A^1\times Y$ is isomorphic to the pullback of its
restriction to $\{0\}\times Y$.
\end{lemma}
\begin{proof}
The two charts of Lemma~\ref{bun:charts} have coordinate rings
\begin{align*}
B[(z+1)^{-1}]&=k[x,t,u,(xt+1)^{-1}],
&t&=\frac{y}{z+1},\quad z=xt,\\
B[z^{-1}]&=k[x,t,u,(xt-1)^{-1}],
&t&=\frac{y}{z},\quad z=xt-1.
\end{align*}
Here $t$ denotes a separate coordinate on each chart. Each ring, and its
polynomial extension by the affine-line coordinate $T$, is a noetherian
unique factorization domain. A line bundle on the spectrum of such a
ring is trivial: its divisor class is represented by a Cartier divisor,
and unique factorization makes every divisor principal.
Let $\mathcal M$ be a line bundle on $\A^1\times Y$. Choose trivializations
on $\A^1\times U_0$ and $\A^1\times U_1$. Its transition function belongs
to $B[z^{-1},(z+1)^{-1}][T]^*$. A polynomial over a domain is a unit only
when it is a constant unit, so this transition function is independent of
$T$. Its specialization at $T=0$ is therefore the same transition
function. The chosen trivializations identify $\mathcal M$ with the
pullback of its restriction at zero.
\end{proof}
The lifting argument has a useful general form. Its normalization step
is the standard one for linearizing a line bundle under an additive action;
compare \cite[Lemma~2.9]{Brion2015}.
\begin{lemma}\label{bun:action-lift}
Let $X$ be an affine integral $k$-variety such that every line bundle on
$\A^1\times X$ is the pullback of its restriction at zero. Every
$\G_a$-action on $X$ lifts to any line bundle on $X$ by maps linear on
fibers, commuting with fiber scaling.
\end{lemma}
\begin{proof}
Write $R=k[X]$, let $\mathcal M$ be the line bundle, and let
$\alpha:\A^1\times X\longrightarrow X$ be the action. If
$\pi:\A^1\times X\longrightarrow X$ is projection, then
$\alpha^*\mathcal M$ and $\pi^*\mathcal M$ restrict to $\mathcal M$ at
parameter zero. The hypothesis therefore supplies an isomorphism
\[
\eta:\pi^*\mathcal M\longrightarrow\alpha^*\mathcal M.
\]
At zero, $\eta$ is an automorphism of $\mathcal M$, hence multiplication
by a unit of $R$. Multiplying $\eta$ by the inverse pullback of that unit
normalizes it to the identity at zero.
The isomorphism $\eta$ gives a fiberwise linear lift of $\alpha$. To check
the group law, compare, over $\A^2\times X$, the map on fibers obtained
by applying parameters $S$ and then $T$ with the map obtained by applying
$T+S$. Both are global isomorphisms of the same source and target line bundles,
with fiber maps from $\mathcal M_x$ to
$\mathcal M_{\alpha(T+S,x)}$. Their ratio is an automorphism of the source
line bundle. Its endomorphism sheaf is canonically $\mathcal O$, so this
automorphism is multiplication by a global unit in $R[T,S]$.
Since $R$ is a domain, this unit is independent of $T,S$. At $(T,S)=(0,0)$
it equals one by the normalization of $\eta$. Thus the group law holds.
The resulting action is linear on fibers, so it commutes with fiber
scaling and preserves the complement of the zero section.
\end{proof}
We will also use the following elementary property of locally nilpotent
derivations; see \cite[Proposition~2.1(3)]{KalimanMakarLimanov2007}.
\begin{lemma}\label{bun:factorial-closure}
Let $R$ be a domain over a field of characteristic zero and let $\delta$ be a locally
nilpotent derivation of $R$. If $r,s\in R$ are nonzero and $\delta(rs)=0$,
then $\delta(r)=\delta(s)=0$.
\end{lemma}
\begin{proof}
The exponential $\exp(T\delta)$ takes values in $R[T]$. If $rs$ is fixed,
then
\[
\exp(T\delta)(r)\,\exp(T\delta)(s)=rs.
\]
Degrees in $T$ add under multiplication of nonzero polynomials over a
domain. Both factors on the left therefore have degree zero, and their
coefficients of $T$ give the assertion.
\end{proof}
\begin{proposition}\label{prop:lift}
Let $E$ be a nonzero locally nilpotent derivation of $B$ such that
$I\cap\ker E\ne0$. There is a nonzero locally nilpotent derivation
$\widetilde E$ of $C$ that extends $E$, preserves fiber weights, and satisfies
$\widetilde E(v)=0$.
\end{proposition}
\begin{proof}
The exponential of $E$ defines an additive-group action on $Y$.
By Lemmas~\ref{bun:homotopy} and~\ref{bun:action-lift}, it lifts to
$\mathcal L$ by fiberwise linear maps. Restrict to the complement of the
zero section, namely $\Spec C$.
Differentiating this action at the identity gives a derivation
$\widetilde E$ of $C$. The action law expresses its pullback on each
function as $\sum_{r\ge0}T^r\widetilde E^{\,r}/r!$; because the action is
algebraic, the sum is finite for each function. Hence $\widetilde E$ is
locally nilpotent. It extends $E$ and is nonzero, and commutation with
fiber scaling says exactly that it preserves the weights
\eqref{bun:fiber-weights}.
Finally choose $0\ne h\in I\cap\ker E$. The inclusion $B\subset C$ and
Lemma~\ref{bun:principalization} give a factorization $h=vq$ in $C$, with
both factors nonzero. Since $\widetilde E(h)=0$,
Lemma~\ref{bun:factorial-closure} yields $\widetilde E(v)=0$.
\end{proof}
==== END A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//04-bundle.tex ====
==== BEGIN A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//05-rigidity.tex ====
\section{Rigidity and the coordinate obstruction}\label{sec:rigidity}
Proposition~\ref{prop:lift} converts the hypothetical invariant in $I$ into
a locally nilpotent derivation of $C$ that fixes $v$ and preserves fiber
weights. We now rule out such a derivation. An auxiliary grading isolates
the difference of squares $a^2u^2-d^2$ inside $v$. Its two factors force
three generators to be invariant, leaving a coefficient whose required
fiber weight is impossible.
\begin{proposition}\label{prop:rigidity}
There is no nonzero locally nilpotent derivation of
$C=k[a,d,b,c,u]/(ac-bd-1)$ that preserves the fiber weights
\eqref{bun:fiber-weights} and annihilates
$v=a^3b+a^2u^2-d^2$.
\end{proposition}
\begin{proof}
Suppose that $\widetilde E$ is such a derivation. Give $C$ the auxiliary
grading in the second row of the following table:
\begin{equation}\label{rig:gradings}
\begin{array}{c|rrrrr}
&a&d&b&c&u\\ \hline
\text{fiber weight}&1&1&-1&-1&0\\
\text{auxiliary degree}&0&1&-1&0&1
\end{array}
\end{equation}
The two gradings are compatible because $ac-bd-1$ is homogeneous of degree
zero for each. The auxiliary degrees of the two parts of $v$ are
\[
\deg_{\mathrm{aux}}(a^3b)=-1,\qquad
\deg_{\mathrm{aux}}(a^2u^2-d^2)=2.
\]
Decompose $\widetilde E$ into its homogeneous components for auxiliary
degree shift, and let $E'$ be the nonzero component with largest shift,
say $m$. There are only finitely many components: a derivation is
determined by its values on the five homogeneous generators, and each
value is a finite sum of homogeneous terms. Each component is a
derivation by the product rule. Moreover, $E'$ is locally nilpotent.
Indeed, for homogeneous $r\in C$ of auxiliary degree $e$, the component
of auxiliary degree $e+nm$ in $\widetilde E^{\,n}(r)$ is $(E')^n(r)$.
For sufficiently large $n$ the former expression vanishes, so the latter
vanishes as well. Homogeneous elements span $C$, proving local
nilpotence on every element. Since the gradings in
\eqref{rig:gradings} are compatible, $E'$ still preserves fiber weights.
Taking the component of degree $m+2$ in $\widetilde E(v)=0$ gives
\[
E'(a^2u^2-d^2)=0.
\]
The factorization
\[
a^2u^2-d^2=(au-d)(au+d)
\]
and Lemma~\ref{bun:factorial-closure} imply that $E'$ annihilates both
$au-d$ and $au+d$. Thus $E'(d)=E'(au)=0$, and another application of that
lemma gives
\begin{equation}\label{rig:fixed}
E'(a)=E'(d)=E'(u)=0.
\end{equation}
All the factors used here are nonzero; for example, after inverting $a$
one has
\begin{equation}\label{rig:a-chart}
C[a^{-1}]=k[a,a^{-1},d,u][b],\qquad c=\frac{1+bd}{a},
\end{equation}
where their nonvanishing is immediate.
It remains to control the possible values of $E'$ on $b,c$.
Differentiating $ac-bd=1$ and using \eqref{rig:fixed} yields
$aE'(c)=dE'(b)$. Define
\[
q=cE'(b)-bE'(c)\in C.
\]
Multiplication by $a$ and $d$, respectively, then gives
\begin{equation}\label{rig:q}
E'(b)=aq,\qquad E'(c)=dq.
\end{equation}
The element $q$ is nonzero, since otherwise $E'$ would kill every generator.
We claim that $q$ belongs to the polynomial subring $k[a,d,u]$.
The elements $a,d,u$ are algebraically independent by
\eqref{rig:a-chart}. Put $K=k(a,d,u)$ and localize $C$ at the nonzero
elements of $k[a,d,u]$. Because these elements lie in $\ker E'$, the
localized derivation remains locally nilpotent. The resulting algebra is
$K[b]$, with $c=(1+bd)/a$. A locally nilpotent $K$-derivation of $K[b]$
sends $b$ into $K$: if its value $P(b)$ had degree at least one, iterating
$P(b)\,d/db$ on $b$ would never give zero. More explicitly, differentiation
of a nonconstant polynomial multiplies its nonzero leading coefficient
by its positive degree, and multiplication by $P$ then leaves the result
nonconstant. Equation~\eqref{rig:q} therefore gives $q\in K$.
To compute $C\cap K$ inside the fraction field of $C$, use both coordinate
charts:
\[
C[a^{-1}]=k[a,a^{-1},d,u][b],\qquad
C[d^{-1}]=k[a,d,d^{-1},u][c].
\]
An element of the first polynomial ring lying in $K$ has no $b$ terms,
and an element of the second lying in $K$ has no $c$ terms. Hence
\[
C\cap K\subset
k[a,a^{-1},d,u]\cap k[a,d,d^{-1},u]=k[a,d,u].
\]
The last equality follows from unique factorization in $k[a,d,u]$:
a reduced denominator that is both a power of $a$ and a power of $d$
is a unit. The reverse inclusion is immediate, proving the claim.
Finally, preservation of fiber weights makes $E'(b)$ have weight $-1$.
By \eqref{rig:q}, the nonzero polynomial $q$ has weight $-2$. This is
impossible in $k[a,d,u]$, whose generators have weights $1,1,0$.
The contradiction proves the proposition.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
By Proposition~\ref{prop:polynomial}, the algebra $A$ is a polynomial ring
in four variables and its element $p$ becomes the polynomial $f$ stated
in the theorem. Proposition~\ref{prop:stable} makes $p$ a coordinate of
$A[w]$. If $p$ were a coordinate of $A$,
Proposition~\ref{prop:coordinate-obstruction} would produce a nonzero
locally nilpotent derivation $E$ of $B$ with $I\cap\ker E\ne0$.
Proposition~\ref{prop:lift} would then supply a derivation forbidden by
Proposition~\ref{prop:rigidity}. Thus $p$ is not a coordinate of $A$.
Transporting these two properties through the polynomial presentation
proves the assertion for $f$.
\end{proof}
==== END A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//05-rigidity.tex ====
==== BEGIN A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//main.tex ====
\documentclass[11pt]{article}
\usepackage[a4paper,margin=29mm]{geometry}
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\usepackage{amsmath,amssymb,amsthm,mathtools}
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\usepackage{xcolor}
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\usepackage{hyperref}
\hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black,pdftitle={A stable coordinate that is not a coordinate in four variables},pdfauthor={OpenAI}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
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\numberwithin{equation}{section}
\title{A stable coordinate that is not a coordinate\\in four variables}
\author{OpenAI}
\date{October 5, 2026}
\begin{document}
\maketitle
\begin{abstract}
We construct an explicit degree-five polynomial over $\C$ that is not a
coordinate in four variables but becomes one after adjoining a single variable.
This gives a counterexample to the stable coordinate conjecture in four
variables. Every fiber is isomorphic to affine three-space, yet its embedding
in affine four-space is not rectifiable. Thus the example also disproves the
Abhyankar--Sathaye embedding conjecture in ambient dimension four.
\end{abstract}
\input{sections/01-introduction}
\input{sections/02-construction}
\input{sections/03-filtration}
\input{sections/04-bundle}
\input{sections/05-rigidity}
\bibliographystyle{plain}
\bibliography{refs}
\end{document}
==== END A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//main.tex ====
==== BEGIN A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//refs.bib ====
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year = {2026}
}
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}
@article{Kaliman2002,
author = {Kaliman, Shulim},
title = {Polynomials with general {$\mathbf{C}^{2}$}-fibers are variables},
journal = {Pacific Journal of Mathematics},
volume = {203},
number = {1},
year = {2002},
pages = {161--190},
note = {\url{https://doi.org/10.2140/pjm.2002.203.161}}
}
==== END A-stable-coordinate-that-is-not-a-coordinate-in-four-variables//refs.bib ====