diff --git a/noncoordinate_polynomial_fibre_v2.txt b/noncoordinate_polynomial_fibre_v2.txt
index 47863e0..8d0378b 100644
--- a/noncoordinate_polynomial_fibre_v2.txt
+++ b/noncoordinate_polynomial_fibre_v2.txt
@@ -26,14 +26,13 @@
 \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.
+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}
@@ -42,30 +41,26 @@ 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
+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}
 
-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 $(w_1,w_2)\mapsto(w_1-w_2^2,w_2)$.
+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 new coordinates $f$ is the first coordinate $w_1$, so
+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, and in particular its zero fibre, is a copy of
+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}
@@ -74,42 +69,37 @@ If $f\in R_n$ satisfies $R_n/(f)\cong\CC^{[n-1]}$, must $f$ be a coordinate?
 
 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.
+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 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
+$u\mapsto(u^3,-u^2)$ 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 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 the independent indeterminates
-$x,y,s$. Put
+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 .
+  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
+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).
@@ -118,63 +108,59 @@ 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
+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
-$R$
+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.
+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
 \[
   R/(F)\cong\CC^{[3]}
   \qquad\text{and}\qquad
-  \grad F\Bigl(2,0,-\frac12,\frac12\Bigr)=0 .
+  \grad F\bigl(2,0,-\tfrac12,\tfrac12\bigr)=0.
 \]
 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
-\[
-  x=u^3+hv=-1,\quad y=-u^2+hw=1,\quad s=1,\quad
-  p=-2s^2x+3sy^2-3s^3y=2,
-\]
-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}.
+$x=u^3+hv=-1$, $y=-u^2+hw=1$, $s=1$, 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} 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
+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 only needs the identities \eqref{eq:cusp} and
-\eqref{eq:shift}.
+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, 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
+\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.
+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
@@ -185,49 +171,50 @@ $(h,y)=B$. Then
 \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
+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, modulo the relation $hW=U^2+y$, the relation $U^3+hV=x$ becomes $hG-yU=x$.
-Hence
+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
-\[
-  a=U^2+y,\qquad b=s-hG^2+2UyG .
-\]
-The last two relations of \eqref{eq:Clinear} read $hW=a$ and $y^2W=b$. They are
+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
 \[
-  y^2a-hb=y^2(U^2+y)-h(s-hG^2+2UyG)=(hG-yU)^2+y^3-hs=x^2+y^3-hs=0 .
+  y^2a-hb=y^2(U^2+y)-h(s-hG^2+2UyG)
+  =(hG-yU)^2+y^3-hs=x^2+y^3-hs=0.
 \]
 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
+$rh+qy^2=(1-\beta y)(1+\beta y)+\beta^2y^2=1$, 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 $C\cong A$.
+$(hW-a,\,y^2W-b)$ 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
 \[
-  U\mapsto hT-\beta x,\qquad G\mapsto yT+\alpha x,\qquad T\mapsto\alpha U+\beta G
+  U\mapsto hT-\beta x,\qquad G\mapsto yT+\alpha x,\qquad
+  T\mapsto\alpha U+\beta G
 \]
 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.
+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}
@@ -267,44 +254,45 @@ 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
+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
-the forward and inverse maps in terms of $h,u,v,w$.
+both the forward and the 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
+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
 \[
-  2X\,dX+3Y^2\,dY=(1+F)\,ds+s\,dF .
+  2X\,dX+3Y^2\,dY=(1+F)\,ds+s\,dF.
 \]
 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.
+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.
+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}
 
@@ -319,8 +307,8 @@ $z_1,\dots,z_n$ 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
+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}
 
@@ -337,15 +325,17 @@ The forward map takes a point of the zero fibre $F=0$ to $\CC^3$ by
 \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, &
+  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 $v=G-UW$. No division occurs, so both
-maps are polynomial on all of $\CC^3$ and on all of the zero fibre.
+$r=\alpha(1+\beta y)$ 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.
 
 \begin{thebibliography}{99}
 \small
@@ -362,8 +352,8 @@ Algebra Colloq. \textbf{27} (2020), no.~3, 587--598.
 
 \bibitem{Ghosh}
 P.~Ghosh,
-\emph{On $\mathbb{A}^1$-contractibility of certain simple birational extensions of
-affine spaces}, arXiv:2506.09166v4 (2026).
+\emph{On $\mathbb{A}^1$-contractibility of certain simple birational extensions
+of affine spaces}, arXiv:2506.09166v4 (2026).
 
 \bibitem{GGP}
 P.~Ghosh, N.~Gupta, and A.~Pal,
@@ -386,8 +376,8 @@ J.~Pure Appl. Algebra \textbf{179} (2003), 159--168.
 
 \bibitem{MaubachMonomial}
 S.~Maubach,
-\emph{Triangular monomial derivations on $k[X_1,X_2,X_3,X_4]$ have kernel generated
-by at most four elements},
+\emph{Triangular monomial derivations on $k[X_1,X_2,X_3,X_4]$ have kernel
+generated by at most four elements},
 J.~Pure Appl. Algebra \textbf{153} (2000), no.~2, 165--170.
 
 \bibitem{SY}