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\title{An explicit noncoordinate polynomial\\[2pt]with affine three-space zero fibre}
\author{}
\date{}
\hypersetup{
  pdftitle={An explicit noncoordinate polynomial with affine three-space zero fibre}}

\begin{document}
\maketitle

\begin{abstract}
We give an explicit polynomial $F$ in four complex variables whose zero fibre is a
copy of $\CC^3$ and whose gradient vanishes at $(2,0,-\tfrac12,\tfrac12)$. A
polynomial whose gradient vanishes somewhere cannot be the first component of a
polynomial automorphism of $\CC^4$. This answers the Abhyankar--Sathaye question
negatively in ambient dimension four; adjoining variables gives counterexamples
in every dimension at least four. The proof rests on one elementary lemma, and
every identity in this note can be checked by hand.
\end{abstract}

\section{The question}\label{sec:question}

All polynomials in this note have complex coefficients. Write
$R_n=\CC[z_1,\dots,z_n]$ and $\CC^{[m]}=\CC[t_1,\dots,t_m]$.

\begin{definition}[coordinate]\label{def:coordinate}
A polynomial $f\in R_n$ is a \emph{coordinate} if there are
$f_2,\dots,f_n\in R_n$ for which $\Phi=(f,f_2,\dots,f_n)\colon\CC^n\to\CC^n$ is
a bijection with polynomial inverse. Such a $\Phi$ is a \emph{polynomial
automorphism} of $\CC^n$, and $f$ is its first component.
\end{definition}

For instance $z_1$ is a coordinate, and so is $z_1+z_2^2$, because
$(z_1,z_2)\mapsto(z_1+z_2^2,z_2)$ has the polynomial inverse
$(w_1,w_2)\mapsto(w_1-w_2^2,w_2)$.

\begin{lemma}\label{lem:necessary}
If $f\in R_n$ is a coordinate, then $R_n/(f)\cong\CC^{[n-1]}$.
\end{lemma}

\begin{proof}
In the coordinates in which $f$ is the first coordinate,
$R_n=\CC[w_1,\dots,w_n]$ and $R_n/(f)=\CC[w_2,\dots,w_n]$.
\end{proof}

Hence every fibre of a coordinate, in particular its zero fibre, is a copy of
$\CC^{n-1}$. The question is whether the converse holds.

\begin{question}[Abhyankar--Sathaye]\label{q:AS}
If $f\in R_n$ satisfies $R_n/(f)\cong\CC^{[n-1]}$, must $f$ be a coordinate?
\end{question}

Geometrically: can every copy of $\CC^{n-1}$, cut out inside $\CC^n$ by one
polynomial equation, be carried to a coordinate hyperplane by a polynomial
automorphism of $\CC^n$? For $n=2$ the answer is yes, by the Abhyankar--Moh
theorem~\cite{AM} and Suzuki~\cite{Suzuki}: an embedded line can be straightened.
In dimension four the answer is yes when the equation is linear in one variable
and the coefficient of that variable depends on only one variable, by Kaliman,
V\'en\'ereau and Zaidenberg~\cite[Theorem~4.2(a)]{KVZ} and
Maubach~\cite[Corollaries~4.2--4.3]{Maubach}; further linear cases appear in
\cite{EEO,GGPthreefold,GGP,Ghosh}. Below we answer the question negatively for
$n=4$ (Section~\ref{sec:fibre}) and, by adjoining variables, for all $n\ge4$
(Section~\ref{sec:higher}). The three-variable case is not addressed here.

\section{The polynomial}\label{sec:example}

The starting point is the cusp. Since $(u^3)^2+(-u^2)^3=0$, the map
$u\mapsto(u^3,-u^2)$ parametrizes the curve $x^2+y^3=0$. We perturb both entries
by independent multiples of a new variable $h$. In $R=\CC[h,u,v,w]$ put
\begin{equation}\label{eq:xyz}
  x=u^3+hv,\qquad y=-u^2+hw,\qquad
  s=S(h,u,v,w):=2u^3v+3u^4w+h(v^2-3u^2w^2)+h^2w^3.
\end{equation}
Expanding gives $x^2=u^6+2hu^3v+h^2v^2$ and
$y^3=-u^6+3hu^4w-3h^2u^2w^2+h^3w^3$; the two $u^6$ terms cancel, so
\begin{equation}\label{eq:cusp}
  x^2+y^3=hs.
\end{equation}
The perturbation has turned the cusp equation into a divisible expression.

The second ingredient is an identity in independent indeterminates $x,y,s$. Put
\begin{equation}\label{eq:p}
  p(x,y,s)=-2s^2x+3sy^2-3s^3y,\qquad X=x+s^3,\qquad Y=y-s^2.
\end{equation}
Since $X^2=x^2+2s^3x+s^6$ and $Y^3=y^3-3s^2y^2+3s^4y-s^6$, the two $s^6$ terms
cancel and
\begin{equation}\label{eq:shift}
  X^2+Y^3=x^2+y^3-sp(x,y,s).
\end{equation}
Now define
\begin{equation}\label{eq:F}
  F=h-p(x,y,s)-1\in R,
\end{equation}
with $x,y,s$ the polynomials \eqref{eq:xyz}; thus
$F=h-1+2s^2x-3sy^2+3s^3y$ in the four variables $h,u,v,w$.

Combining \eqref{eq:cusp}, \eqref{eq:shift} and \eqref{eq:F} gives the identity
in $R$
\begin{equation}\label{eq:ambient}
  X^2+Y^3=s(1+F),
\end{equation}
because $x^2+y^3-sp=s(h-p)=s(1+F)$. On the zero fibre $F=0$ it reads
$X^2+Y^3=s$, a surface that is a graph over the two free variables $X$ and $Y$.
That is where the copy of $\CC^3$ on the zero fibre comes from. The work below
produces a third coordinate on that fibre, together with its polynomial inverse.

\begin{theorem}\label{thm:main}
The polynomial $F$ of \eqref{eq:F} satisfies
\[
  R/(F)\cong\CC^{[3]}
  \qquad\text{and}\qquad
  \grad F\bigl(2,0,-\tfrac12,\tfrac12\bigr)=0.
\]
Consequently $F$ is not a coordinate of $R$.
\end{theorem}

Let $P=(2,0,-\frac12,\frac12)$, in the variable order $(h,u,v,w)$. There
$x=u^3+hv=-1$, $y=-u^2+hw=1$, $s=1$, and
$p=-2s^2x+3sy^2-3s^3y=2$, so $F=h-p-1=-1$. The critical point therefore lies on
the fibre over $-1$, not on the zero fibre; the zero fibre is smooth, being a copy
of $\CC^3$. The isomorphism in the theorem is proved in
Section~\ref{sec:fibre}, the vanishing of the gradient in
Section~\ref{sec:critical}.

\begin{remark}
The polynomials \eqref{eq:xyz} are annihilated by the triangular derivation
$D=h\partial_u-3u^2\partial_v+2u\partial_w$ of $R$: a direct check gives
$Dx=3hu^2-3hu^2=0$, $Dy=-2hu+2hu=0$, $Ds=0$. They appear in Maubach's
calculation of kernels of triangular monomial derivations
\cite[proof of Theorem~3.2, case~(3)]{MaubachMonomial}, but nothing below uses
that origin: the proof needs only \eqref{eq:cusp} and \eqref{eq:shift}.
\end{remark}

\section{A lifting lemma}\label{sec:lifting}

One general statement is needed. Throughout, $S$ denotes the polynomial of
\eqref{eq:xyz}. In words: if a commutative ring $B$ contains $h,x,y,s$ with
$x^2+y^3=hs$, and if $h$ and $y$ generate the unit ideal in the sense that
$\alpha h+\beta y=1$ for some $\alpha,\beta\in B$, then the three equations
\[
  u^3+hv=x,\qquad -u^2+hw=y,\qquad S(h,u,v,w)=s
\]
in three new letters $u,v,w$ describe a polynomial ring in one variable. The
hypothesis on $h$ and $y$ is exactly what lets us solve two linear equations
without dividing, which matters because $h$ and $y$ need not be invertible and
$B$ need not be a domain.

\begin{lemma}\label{lem:lift}
Let $B$ be a commutative ring and let $h,x,y,s\in B$ satisfy $x^2+y^3=hs$ and
$(h,y)=B$. Then
\[
  C=B[U,V,W]/\bigl(U^3+hV-x,\ -U^2+hW-y,\ S(h,U,V,W)-s\bigr)\cong B[T].
\]
\end{lemma}

\begin{proof}
Choose $\alpha,\beta\in B$ with $\alpha h+\beta y=1$. Substitute $G=V+UW$ (so
$V=G-UW$; this is an invertible change of generators of the polynomial ring).
Modulo the relation $hW=U^2+y$, a direct expansion of \eqref{eq:xyz} gives
\[
  S(h,U,G-UW,W)=hG^2-2UyG+y^2W,
\]
and the relation $U^3+hV=x$ becomes $hG-yU=x$. Hence
\begin{equation}\label{eq:Clinear}
  C\cong B[U,G,W]/\bigl(hG-yU-x,\ hW-U^2-y,\ y^2W-s+hG^2-2UyG\bigr).
\end{equation}
Put $A=B[U,G]/(hG-yU-x)$ and in $A$ set $a=U^2+y$, $b=s-hG^2+2UyG$. The last
two relations of \eqref{eq:Clinear} read $hW=a$ and $y^2W=b$. They are
compatible, since
\[
  y^2a-hb=y^2(U^2+y)-h(s-hG^2+2UyG)
  =(hG-yU)^2+y^3-hs=x^2+y^3-hs=0.
\]
With $r=\alpha(1+\beta y)$ and $q=\beta^2$ we have
$rh+qy^2=(1-\beta y)(1+\beta y)+\beta^2y^2=1$, where the first step uses
$\alpha h=1-\beta y$. Hence $W_0=ra+qb$ satisfies both equations: from
$y^2a=hb$ one gets $hW_0=(rh+qy^2)a=a$ and $y^2W_0=(rh+qy^2)b=b$. Moreover
\[
  W-W_0=r(hW-a)+q(y^2W-b),
\]
while $hW-a=h(W-W_0)$ and $y^2W-b=y^2(W-W_0)$. So in $A[W]$ the ideal
$(hW-a,\,y^2W-b)$ equals $(W-W_0)$, and evaluating at $W_0$ gives $C\cong A$.

It remains to identify $A$ with $B[T]$. The $B$-algebra maps $A\to B[T]$ and
$B[T]\to A$ given by
\[
  U\mapsto hT-\beta x,\qquad G\mapsto yT+\alpha x,\qquad
  T\mapsto\alpha U+\beta G
\]
are inverse. Indeed, in $A$ the relation $hG-yU=x$ and $\alpha h+\beta y=1$ give
\[
  h(\alpha U+\beta G)=U+\beta x,\qquad
  y(\alpha U+\beta G)=G-\alpha x,
\]
so substituting $U=hT-\beta x$ and $G=yT+\alpha x$ into $\alpha U+\beta G$
returns $T$, and substituting $T=\alpha U+\beta G$ into the first two returns $U$
and $G$. Thus $A\cong B[T]$, and $C\cong A$.

In the original letters, the parameter is $T=\alpha U+\beta(V+UW)$; its inverse
recovers $U$ and $G$ from $T$, then $W$ from $W_0$, then $V=G-UW$.
\end{proof}

\section{The zero fibre is affine three-space}\label{sec:fibre}

We prepare the base ring of Lemma~\ref{lem:lift}. Let $h,x,y,s$ now be four
independent indeterminates and set
\begin{equation}\label{eq:B}
  B=\CC[h,x,y,s]/\bigl(h-1-p(x,y,s),\ x^2+y^3-sh\bigr),
\end{equation}
writing $h,x,y,s$ also for their classes in $B$. This is a new presentation: here
$x,y,s$ are independent variables, not the polynomials \eqref{eq:xyz}.
Eliminating $h$ leaves $x^2+y^3=s(1+p(x,y,s))$, and the triangular change of
variables $X=x+s^3$, $Y=y-s^2$ of \eqref{eq:p} turns this into $X^2+Y^3=s$ by
\eqref{eq:shift}. Hence
\begin{equation}\label{eq:Bplane}
  B\cong\CC[X,Y,s]/(X^2+Y^3-s)\cong\CC^{[2]},
\end{equation}
with inverse presentation
\[
  s=X^2+Y^3,\qquad x=X-s^3,\qquad y=Y+s^2,\qquad h=1+p(x,y,s).
\]

The elements $h$ and $y$ generate the unit ideal of $B$: modulo $(h,y)$ the two
defining relations become $x^2=0$ and $1=2s^2x$, and squaring the second gives
$1=4s^4x^2=0$. An explicit certificate is the pair
\begin{equation}\label{eq:alphabeta}
  \alpha=1+2s^2x+4s^5,\qquad
  \beta=(3s^3-3sy)(1+2s^2x)-4s^4y^2,
\end{equation}
for which $\alpha h+\beta y=1$ in $B$, because the identity
\begin{equation}\label{eq:certificate}
  \alpha h+\beta y-1=(1+2s^2x)\bigl(h-1-p(x,y,s)\bigr)-4s^4(x^2+y^3-sh)
\end{equation}
holds in the free polynomial ring and both terms on the right vanish in $B$.

Apply Lemma~\ref{lem:lift} over this $B$. Its algebra $C$ is
\[
  C=B[u,v,w]/\bigl(u^3+hv-x,\ -u^2+hw-y,\ S(h,u,v,w)-s\bigr).
\]
In the polynomial ring $\CC[h,x,y,s,u,v,w]$ the last three relations are solved
for $x,y,s$ by \eqref{eq:xyz}: they read $x=u^3+hv$, $y=-u^2+hw$,
$s=S(h,u,v,w)$. Under this substitution the relation $x^2+y^3-sh$ becomes
identically zero by \eqref{eq:cusp}, and the relation $h-1-p(x,y,s)$ becomes the
equation $F=0$ of \eqref{eq:F}. Therefore
\begin{equation}\label{eq:fiber}
  R/(F)\cong C\cong B[T]\cong\CC[X,Y,T].
\end{equation}
This proves the first assertion of Theorem~\ref{thm:main}: $X$, $Y$ and $T$ are
polynomial coordinates on the zero fibre. Appendix~\ref{app:coordinates} records
both the forward and the inverse maps in terms of $h,u,v,w$.

\section{A point where the gradient vanishes}\label{sec:critical}

At $P=(2,0,-\frac12,\frac12)$, in the variable order $(h,u,v,w)$, the values of
the polynomials \eqref{eq:xyz} and \eqref{eq:p} are
\[
  x=-1,\quad y=1,\quad s=1,\quad p=2,\quad
  X=x+s^3=0,\quad Y=y-s^2=0,\quad F=-1.
\]
Differentiating the ambient identity \eqref{eq:ambient}, $X^2+Y^3=s(1+F)$,
gives
\[
  2X\,dX+3Y^2\,dY=(1+F)\,ds+s\,dF.
\]
At $P$ the left-hand side vanishes, $1+F=0$ and $s=1$, hence $dF(P)=0$: all four
partial derivatives of $F$ vanish at $P$.

To see why this excludes a coordinate, suppose $F$ were the first component of a
polynomial automorphism $\Phi=(F,g_2,g_3,g_4)$ of $\CC^4$ with polynomial
inverse $\Psi$. The chain rule gives $J\Phi(P)\cdot J\Psi(\Phi(P))=I$, so
$J\Phi(P)$ is invertible. But its first row is $\grad F(P)=0$, so the matrix is
not invertible. This contradiction proves that $F$ is not a coordinate.

The obstruction uses a critical point of the defining polynomial, not of its zero
fibre; that is the ambient critical locus emphasized by Shpilrain and Yu
\cite[\S1]{SY}, who asked whether the gradient must be nowhere vanishing when the
zero fibre is a copy of a hyperplane. Theorem~\ref{thm:main} answers that
question negatively in ambient dimension four.

\section{Higher ambient dimension}\label{sec:higher}

\begin{corollary}\label{cor:dimension}
For every integer $n\ge4$ there is a polynomial $f\in R_n$ such that
$R_n/(f)\cong\CC^{[n-1]}$ and $f$ is not a coordinate.
\end{corollary}

\begin{proof}
For $n=4$ take $f=F$. For $n>4$ take the same $F$, regarded as a polynomial in
$z_1,\dots,z_n$ that does not involve $z_5,\dots,z_n$. By \eqref{eq:fiber},
\[
  R_n/(F)=R[z_5,\dots,z_n]/(F)=\CC[X,Y,T][z_5,\dots,z_n],
\]
a polynomial ring in $n-1$ variables. All partial derivatives of $F$ in the
added variables vanish, and $\grad F(P)=0$, so the gradient vanishes at
$(P,0,\dots,0)$ and the Jacobian argument of Section~\ref{sec:critical} applies.
\end{proof}

\appendix
\section{Explicit formulas on the zero fibre}\label{app:coordinates}

Both directions of the isomorphism \eqref{eq:fiber} are polynomial, and they are
spelled out here so that the zero fibre can be parametrized directly. All
expressions use \eqref{eq:xyz} and the pair \eqref{eq:alphabeta}.

The forward map takes a point of the zero fibre $F=0$ to $\CC^3$ by
\begin{equation}\label{eq:fibercoords}
  X=x+s^3,\qquad Y=y-s^2,\qquad T=\alpha u+\beta(v+uw).
\end{equation}
The inverse map takes independent $X,Y,T$ to a point of $F=0$ by
\begin{align}
  s&=X^2+Y^3, & x&=X-s^3, \notag\\
  y&=Y+s^2, & h&=1+p(x,y,s), \notag\\
  u&=hT-\beta x, & g&=yT+\alpha x, \notag\\
  v&=g-uw, & w&=\alpha(1+\beta y)(y+u^2)+\beta^2(s-hg^2+2uyg).
  \label{eq:explicitinverse}
\end{align}
These are exactly the maps of Lemma~\ref{lem:lift}: the pair $(u,g)$ is the pair
$(U,G)$ of the lemma, the formula for $w$ is $W_0=ra+qb$ with
$r=\alpha(1+\beta y)$ and $q=\beta^2$, and $v=g-uw$ is $V=G-UW$. No division
occurs, so both maps are polynomial on all of $\CC^3$ and on all of the zero
fibre.

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\end{document}