diff --git a/noncoordinate_polynomial_fibre_v2.txt b/noncoordinate_polynomial_fibre_v2.txt
index 8d0378b..eda2f4c 100644
--- a/noncoordinate_polynomial_fibre_v2.txt
+++ b/noncoordinate_polynomial_fibre_v2.txt
@@ -1,31 +1,31 @@
-\documentclass[11pt]{article}
-\usepackage[T1]{fontenc}
-\usepackage{lmodern}
-\usepackage{amsmath,amssymb,amsthm}
-\usepackage[margin=1.05in]{geometry}
-\usepackage{microtype}
-\usepackage[colorlinks=true,linkcolor=blue,citecolor=blue,urlcolor=blue]{hyperref}
-\numberwithin{equation}{section}
-\newtheorem{theorem}{Theorem}[section]
-\newtheorem{lemma}[theorem]{Lemma}
-\newtheorem{corollary}[theorem]{Corollary}
-\newtheorem{definition}[theorem]{Definition}
-\newtheorem{question}[theorem]{Question}
-\theoremstyle{remark}
-\newtheorem*{remark}{Remark}
-\newcommand{\CC}{\mathbb C}
-\newcommand{\Aff}{\mathbb A}
-\DeclareMathOperator{\grad}{\nabla}
-\title{An explicit noncoordinate polynomial\\[2pt]with affine three-space zero fibre}
-\author{}
-\date{}
+
+
+
+
+
+
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+
+
+# An explicit noncoordinate polynomial\\[2pt]with affine three-space zero fibre
+
+
 \hypersetup{
   pdftitle={An explicit noncoordinate polynomial with affine three-space zero fibre}}
 
-\begin{document}
+
 \maketitle
 
-\begin{abstract}
+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
@@ -33,125 +33,125 @@ 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}
+
+# The question
 
 All polynomials in this note have complex coefficients. Write
-$R_n=\CC[z_1,\dots,z_n]$ and $\CC^{[m]}=\CC[t_1,\dots,t_m]$.
+$R_n=\CC[z_1,...,z_n]$ and $\CC^{[m]}=\CC[t_1,...,t_m]$.
+
+[coordinate]
+A polynomial $f\in R_n$ is a coordinate if there are
+$f_2,...,f_n\in R_n$ for which $\Phi=(f,f_2,...,f_n)\colon\CC^n\to\CC^n$ is
+a bijection with polynomial inverse. Such a $\Phi$ is a polynomial
+automorphism of $\CC^n$, and $f$ is its first component.
 
-\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}
+
+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}
+$R_n=\CC[w_1,...,w_n]$ and $R_n/(f)=\CC[w_2,...,w_n]$.
+
 
 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}
+[Abhyankar--Sathaye]
 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.
+theorem~[cite] and Suzuki~[cite]: 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.
+[cite]. Below we answer the question negatively for
+$n=4$ (Section~sec:fibre) and, by adjoining variables, for all $n\ge4$
+(Section~sec:higher). The three-variable case is not addressed here.
 
-\section{The polynomial}\label{sec:example}
+# The polynomial
 
 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
+
+with $x,y,s$ the polynomials (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
+Combining (eq:cusp), (eq:shift) and (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
+
+The polynomial $F$ of (eq:F) satisfies
 \[
   R/(F)\cong\CC^{[3]}
-  \qquad\text{and}\qquad
-  \grad F\bigl(2,0,-\tfrac12,\tfrac12\bigr)=0.
+  \qquadand\qquad
+  \grad F(2,0,-\tfrac12,\tfrac12)=0.
 \]
 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}.
+Section~sec:fibre, the vanishing of the gradient in
+Section~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
+
+The polynomials (eq:xyz) are annihilated by the triangular derivation
+$D=h∂_u-3u^2∂_v+2u∂_w$ 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}
+that origin: the proof needs only (eq:cusp) and (eq:shift).
+
 
-\section{A lifting lemma}\label{sec:lifting}
+# A lifting lemma
 
 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
+(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
 \[
@@ -162,27 +162,27 @@ 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
 \[
-  C=B[U,V,W]/\bigl(U^3+hV-x,\ -U^2+hW-y,\ S(h,U,V,W)-s\bigr)\cong B[T].
+  C=B[U,V,W]/(U^3+hV-x,\ -U^2+hW-y,\ S(h,U,V,W)-s)\cong B[T].
 \]
-\end{lemma}
 
-\begin{proof}
+
+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
+Modulo the relation $hW=U^2+y$, a direct expansion of (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}
+
+  C\cong B[U,G,W]/(hG-yU-x,\ hW-U^2-y,\ y^2W-s+hG^2-2UyG).
+
 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
+two relations of (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)
@@ -215,23 +215,23 @@ 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
+# The zero fibre is affine three-space
+
+We prepare the base ring of Lemma~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}
+
+  B=\CC[h,x,y,s]/(h-1-p(x,y,s),\ x^2+y^3-sh),
+
 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}.
+$x,y,s$ are independent variables, not the polynomials (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}
+variables $X=x+s^3$, $Y=y-s^2$ of (eq:p) turns this into $X^2+Y^3=s$ by
+(eq:shift). Hence
+
   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).
@@ -240,41 +240,41 @@ with inverse presentation
 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}
+
+  \alpha h+\beta y-1=(1+2s^2x)(h-1-p(x,y,s))-4s^4(x^2+y^3-sh)
+
 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
+Apply Lemma~lem:lift over this $B$. Its algebra $C$ is
 \[
-  C=B[u,v,w]/\bigl(u^3+hv-x,\ -u^2+hw-y,\ S(h,u,v,w)-s\bigr).
+  C=B[u,v,w]/(u^3+hv-x,\ -u^2+hw-y,\ S(h,u,v,w)-s).
 \]
 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 $x=u^3+hv$, $y=-u^2+hw$,
+for $x,y,s$ by (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}
+identically zero by (eq:cusp), and the relation $h-1-p(x,y,s)$ becomes the
+equation $F=0$ of (eq:F). Therefore
+
   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
+
+This proves the first assertion of Theorem~thm:main: $X$, $Y$ and $T$ are
+polynomial coordinates on the zero fibre. Appendix~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}
+# A point where the gradient vanishes
 
 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
+the polynomials (eq:xyz) and (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}, $X^2+Y^3=s(1+F)$,
+Differentiating the ambient identity (eq:ambient), $X^2+Y^3=s(1+F)$,
 gives
 \[
   2X\,dX+3Y^2\,dY=(1+F)\,ds+s\,dF.
@@ -291,105 +291,105 @@ 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
+zero fibre is a copy of a hyperplane. Theorem~thm:main answers that
 question negatively in ambient dimension four.
 
-\section{Higher ambient dimension}\label{sec:higher}
+# Higher ambient dimension
+
 
-\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}
+
+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},
+$z_1,...,z_n$ that does not involve $z_5,...,z_n$. By (eq:fiber),
 \[
-  R_n/(F)=R[z_5,\dots,z_n]/(F)=\CC[X,Y,T][z_5,\dots,z_n],
+  R_n/(F)=R[z_5,...,z_n]/(F)=\CC[X,Y,T][z_5,...,z_n],
 \]
 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}
+$(P,0,...,0)$ and the Jacobian argument of Section~sec:critical applies.
+
 
 \appendix
-\section{Explicit formulas on the zero fibre}\label{app:coordinates}
+# Explicit formulas on the zero fibre
 
-Both directions of the isomorphism \eqref{eq:fiber} are polynomial, and they are
+Both directions of the isomorphism (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}.
+expressions use (eq:xyz) and the pair (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
+  
+
+These are exactly the maps of Lemma~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.
 
-\begin{thebibliography}{99}
+{99}
 \small
 
 \bibitem{AM}
 S.~S.~Abhyankar and T.~T.~Moh,
-\emph{Embeddings of the line in the plane},
-J.~Reine Angew. Math. \textbf{276} (1975), 148--166.
+Embeddings of the line in the plane,
+J.~Reine Angew. Math. 276 (1975), 148--166.
 
 \bibitem{EEO}
 M.~El~Kahoui, N.~Essamaoui, and M.~Ouali,
-\emph{Interpolation in the automorphism group of a polynomial ring},
-Algebra Colloq. \textbf{27} (2020), no.~3, 587--598.
+Interpolation in the automorphism group of a polynomial ring,
+Algebra Colloq. 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).
+On $A^1$-contractibility of certain simple birational extensions
+of affine spaces, arXiv:2506.09166v4 (2026).
 
 \bibitem{GGP}
 P.~Ghosh, N.~Gupta, and A.~Pal,
-\emph{On embedding of linear hypersurfaces}, arXiv:2405.07205v3 (2024).
+On embedding of linear hypersurfaces, arXiv:2405.07205v3 (2024).
 
 \bibitem{GGPthreefold}
 P.~Ghosh, N.~Gupta, and A.~Pal,
-\emph{On the family of affine threefolds $a(x)y=F(x,z,t)$},
-J.~Algebra \textbf{700} (2026), 267--288; arXiv:2403.17397v2.
+On the family of affine threefolds $a(x)y=F(x,z,t)$,
+J.~Algebra 700 (2026), 267--288; arXiv:2403.17397v2.
 
 \bibitem{KVZ}
 S.~Kaliman, S.~V\'en\'ereau, and M.~Zaidenberg,
-\emph{Simple birational extensions of the polynomial algebra $\CC^{[3]}$},
-Trans. Amer. Math. Soc. \textbf{356} (2004), 509--555.
+Simple birational extensions of the polynomial algebra $\CC^{[3]$},
+Trans. Amer. Math. Soc. 356 (2004), 509--555.
 
 \bibitem{Maubach}
 S.~Maubach,
-\emph{The commuting derivations conjecture},
-J.~Pure Appl. Algebra \textbf{179} (2003), 159--168.
+The commuting derivations conjecture,
+J.~Pure Appl. Algebra 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},
-J.~Pure Appl. Algebra \textbf{153} (2000), no.~2, 165--170.
+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 153 (2000), no.~2, 165--170.
 
 \bibitem{SY}
 V.~Shpilrain and J.-T.~Yu,
-\emph{Embeddings of hypersurfaces in affine spaces},
-J.~Algebra \textbf{239} (2001), no.~1, 161--173.
+Embeddings of hypersurfaces in affine spaces,
+J.~Algebra 239 (2001), no.~1, 161--173.
 
 \bibitem{Suzuki}
 M.~Suzuki,
-\emph{Propri\'et\'es topologiques des polyn\^omes de deux variables complexes, et
-automorphismes alg\'ebriques de l'espace $\CC^2$},
-J.~Math. Soc. Japan \textbf{26} (1974), 241--257.
+Propri\'et\'es topologiques des polyn\^omes de deux variables complexes, et
+automorphismes alg\'ebriques de l'espace $\CC^2$,
+J.~Math. Soc. Japan 26 (1974), 241--257.
+
+
 
-\end{thebibliography}
-\end{document}