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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 the point
$(2,0,-\frac12,\frac12)$. 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, and adjoining
variables gives counterexamples in every dimension at least four. The proof rests
on one elementary lemma about two compatible linear equations, 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 polynomials
$f_2,\dots,f_n\in R_n$ for which the map
\[
  \Phi=(f,f_2,\dots,f_n)\colon\CC^n\longrightarrow\CC^n
\]
is a bijection with polynomial inverse. Such a map is a \emph{polynomial
automorphism} of $\CC^n$, and $f$ is its first component.
\end{definition}

A coordinate is thus a polynomial that can serve as one of the new coordinates
after an invertible polynomial change of variables. For example $z_1$ 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 new coordinates $f$ is the first coordinate $w_1$, so
$R_n=\CC[w_1,\dots,w_n]$ and $R_n/(f)=\CC[w_2,\dots,w_n]$.
\end{proof}

Hence every fibre of a coordinate, and 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: an embedded line can be straightened, by the
Abhyankar--Moh theorem \cite{AM} and Suzuki \cite{Suzuki}. In dimension four the
answer was known to be yes for equations that are linear in one variable with a
coefficient depending 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 treatments of such linear equations
appear in \cite{EEO,GGPthreefold}; \cite{GGP} imposes additional hypotheses on the
coefficients and their divisors, and \cite{Ghosh} proves an affirmative embedding
theorem for one further family of equations. We show that the answer is no for
$n=4$ in Section~\ref{sec:fibre} and extend this to all $n\ge4$ in
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 cuspidal 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 and
\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 the 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 $s^6$ 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 in \eqref{eq:xyz}, so that
$F=h-1+2s^2x-3sy^2+3s^3y$ 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 is
to produce a third coordinate on that fibre, together with its polynomial inverse.

\begin{theorem}\label{thm:main}
The polynomial $F$ of \eqref{eq:F} satisfies
\[
  R/(F)\cong\CC^{[3]}
  \qquad\text{and}\qquad
  \grad F\Bigl(2,0,-\frac12,\frac12\Bigr)=0 .
\]
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
\[
  x=u^3+hv=-1,\quad y=-u^2+hw=1,\quad s=1,\quad
  p=-2s^2x+3sy^2-3s^3y=2,
\]
so $F=h-p-1=-1$. The critical point therefore lies on the fibre over $-1$, not on
the zero fibre; and the zero fibre is smooth, being a copy of $\CC^3$. The
isomorphism in the theorem is proved in Section~\ref{sec:fibre} and the vanishing
of the gradient in Section~\ref{sec:critical}.

\begin{remark}
The polynomials \eqref{eq:xyz} do not come from nowhere. The triangular derivation
$D=h\partial_u-3u^2\partial_v+2u\partial_w$ of $R$ annihilates each of
$h,x,y,s$, as a direct check shows: $Dx=3hu^2-3hu^2$, $Dy=-2hu+2hu$, and
$Ds=0$. These four invariants appear in Maubach's computation of kernels of
triangular monomial derivations
\cite[proof of Theorem~3.2, case~(3)]{MaubachMonomial}, but nothing below uses
that origin: the proof only needs the identities \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, the lemma says: if a commutative ring $B$ contains
$h,x,y,s$ with $x^2+y^3=hs$, and if $h$ and $y$ have no common factor 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 $G=V+UW$. Direct
expansion of \eqref{eq:xyz} gives
\[
  S(h,U,G-UW,W)=hG^2-2UyG+y^2W,
\]
and, modulo the relation $hW=U^2+y$, 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
\[
  a=U^2+y,\qquad b=s-hG^2+2UyG .
\]
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
$rh+qy^2=(1-\beta y)(1+\beta y)+\beta^2y^2=1$, so $W_0=ra+qb$ satisfies both
equations: using $y^2a=hb$ 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 $C\cong A\cong B[T]$, with parameter $T=\alpha U+\beta(V+UW)$ in the original
letters.
\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 express
$x,y,s$ in terms of $h,u,v,w$ by \eqref{eq:xyz}; the relation $x^2+y^3-sh$ is then
zero by \eqref{eq:cusp}, and the relation $h-1-p(x,y,s)$ is exactly 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
the forward and inverse maps in terms of $h,u,v,w$.

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

Now the second assertion of Theorem~\ref{thm:main}. Read the identity
\eqref{eq:ambient} with $X=x+s^3$ and $Y=y-s^2$ computed from \eqref{eq:xyz}; it
says $X^2+Y^3=s(1+F)$.

At $P=(2,0,-\frac12,\frac12)$, in the variable order $(h,u,v,w)$, the values are
$x=-1$, $y=1$, $s=1$, $p=2$, so $X=0$, $Y=0$ and $F=-1$. Differentiating
\eqref{eq:ambient} 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 $P$.

It remains 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 an invertible matrix. 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, & y&=Y+s^2, & h&=1+p(x,y,s),\notag\\
  u&=hT-\beta x, & g&=yT+\alpha x, &
  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 $v=G-UW$. 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}