diff --git a/noncoordinate_polynomial_fibre_v2.txt b/noncoordinate_polynomial_fibre_v2.txt
index eda2f4c..31749ec 100644
--- a/noncoordinate_polynomial_fibre_v2.txt
+++ b/noncoordinate_polynomial_fibre_v2.txt
@@ -16,20 +16,20 @@
 
 
 
-# An explicit noncoordinate polynomial\\[2pt]with affine three-space zero fibre
+# An explicit noncoordinate polynomial[2pt]with affine three-space zero fibre
 
 
-\hypersetup{
+hypersetup{
   pdftitle={An explicit noncoordinate polynomial with affine three-space zero fibre}}
 
 
-\maketitle
+maketitle
 
 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
+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
+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.
@@ -38,54 +38,54 @@ every identity in this note can be checked by hand.
 # The question
 
 All polynomials in this note have complex coefficients. Write
-$R_n=\CC[z_1,...,z_n]$ and $\CC^{[m]}=\CC[t_1,...,t_m]$.
+$R_n=CC[z_1,...,z_n]$ 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.
+A polynomial $fin R_n$ is a coordinate if there are
+$f_2,...,f_nin R_n$ for which $Phi=(f,f_2,...,f_n)colonCC^ntoCC^n$ is
+a bijection with polynomial inverse. Such a $Phi$ is a polynomial
+automorphism of $CC^n$, and $f$ is its first component.
 
 
 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
-$(w_1,w_2)\mapsto(w_1-w_2^2,w_2)$.
+$(z_1,z_2)mapsto(z_1+z_2^2,z_2)$ has the polynomial inverse
+$(w_1,w_2)mapsto(w_1-w_2^2,w_2)$.
 
 
-If $f\in R_n$ is a coordinate, then $R_n/(f)\cong\CC^{[n-1]}$.
+If $fin R_n$ is a coordinate, then $R_n/(f)congCC^{[n-1]}$.
 
 
 Proof. 
 In the coordinates in which $f$ is the first coordinate,
-$R_n=\CC[w_1,...,w_n]$ and $R_n/(f)=\CC[w_2,...,w_n]$.
+$R_n=CC[w_1,...,w_n]$ 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.
+$CC^{n-1}$. The question is whether the converse holds.
 
 [Abhyankar--Sathaye]
-If $f\in R_n$ satisfies $R_n/(f)\cong\CC^{[n-1]}$, must $f$ be a coordinate?
+If $fin R_n$ satisfies $R_n/(f)congCC^{[n-1]}$, must $f$ be a coordinate?
 
 
-Geometrically: can every copy of $\CC^{n-1}$, cut out inside $\CC^n$ by one
+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
+automorphism of $CC^n$? For $n=2$ the answer is yes, by the Abhyankar--Moh
 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
+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]. Below we answer the question negatively for
-$n=4$ (Section~sec:fibre) and, by adjoining variables, for all $n\ge4$
+$n=4$ (Section~sec:fibre) and, by adjoining variables, for all $nge4$
 (Section~sec:higher). The three-variable case is not addressed here.
 
 # 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
+$umapsto(u^3,-u^2)$ 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
 
-  x=u^3+hv,\qquad y=-u^2+hw,\qquad
+  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.
 
 Expanding gives $x^2=u^6+2hu^3v+h^2v^2$ and
@@ -97,7 +97,7 @@ The perturbation has turned the cusp equation into a divisible expression.
 
 The second ingredient is an identity in independent indeterminates $x,y,s$. Put
 
-  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.
 
 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
@@ -106,7 +106,7 @@ cancel and
 
 Now define
 
-  F=h-p(x,y,s)-1\in R,
+  F=h-p(x,y,s)-1in R,
 
 with $x,y,s$ the polynomials (eq:xyz); thus
 $F=h-1+2s^2x-3sy^2+3s^3y$ in the four variables $h,u,v,w$.
@@ -118,24 +118,24 @@ in $R$
 
 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
+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.
 
 
 The polynomial $F$ of (eq:F) satisfies
-\[
-  R/(F)\cong\CC^{[3]}
-  \qquadand\qquad
-  \grad F(2,0,-\tfrac12,\tfrac12)=0.
-\]
+[
+  R/(F)congCC^{[3]}
+  qquadandqquad
+  grad F(2,0,-tfrac12,tfrac12)=0.
+]
 Consequently $F$ is not a coordinate of $R$.
 
 
-Let $P=(2,0,-\frac12,\frac12)$, in the variable order $(h,u,v,w)$. There
+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
+of $CC^3$. The isomorphism in the theorem is proved in
 Section~sec:fibre, the vanishing of the gradient in
 Section~sec:critical.
 
@@ -144,7 +144,7 @@ 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
+cite[proof of Theorem~3.2, case~(3)]{MaubachMonomial}, but nothing below uses
 that origin: the proof needs only (eq:cusp) and (eq:shift).
 
 
@@ -153,67 +153,67 @@ that origin: the proof needs only (eq:cusp) and (eq:shift).
 One general statement is needed. Throughout, $S$ denotes the polynomial of
 (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
-\[
-  u^3+hv=x,\qquad -u^2+hw=y,\qquad S(h,u,v,w)=s
-\]
+$alpha h+beta y=1$ for some $alpha,betain B$, 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.
 
 
-Let $B$ be a commutative ring and let $h,x,y,s\in B$ satisfy $x^2+y^3=hs$ and
+Let $B$ be a commutative ring and let $h,x,y,sin B$ satisfy $x^2+y^3=hs$ and
 $(h,y)=B$. Then
-\[
-  C=B[U,V,W]/(U^3+hV-x,\ -U^2+hW-y,\ S(h,U,V,W)-s)\cong B[T].
-\]
+[
+  C=B[U,V,W]/(U^3+hV-x, -U^2+hW-y, S(h,U,V,W)-s)cong B[T].
+]
 
 
 Proof. 
-Choose $\alpha,\beta\in B$ with $\alpha h+\beta y=1$. Substitute $G=V+UW$ (so
+Choose $alpha,betain B$ 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 (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
 
-  C\cong B[U,G,W]/(hG-yU-x,\ hW-U^2-y,\ y^2W-s+hG^2-2UyG).
+  Ccong 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 (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
+]
+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
+$(hW-a,,y^2W-b)$ equals $(W-W_0)$, and evaluating at $W_0$ gives $Ccong A$.
+
+It remains to identify $A$ with $B[T]$. The $B$-algebra maps $Ato B[T]$ and
+$B[T]to A$ given by
+[
+  Umapsto hT-beta x,qquad Gmapsto yT+alpha x,qquad
+  Tmapstoalpha 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 $Acong B[T]$, and $Ccong A$.
+
+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 $V=G-UW$.
 
 
@@ -222,7 +222,7 @@ recovers $U$ and $G$ from $T$, then $W$ from $W_0$, then $V=G-UW$.
 We prepare the base ring of Lemma~lem:lift. Let $h,x,y,s$ now be four
 independent indeterminates and set
 
-  B=\CC[h,x,y,s]/(h-1-p(x,y,s),\ x^2+y^3-sh),
+  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 (eq:xyz).
@@ -230,37 +230,37 @@ 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 (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]},
+  BcongCC[X,Y,s]/(X^2+Y^3-s)congCC^{[2]},
 
 with inverse presentation
-\[
-  s=X^2+Y^3,\qquad x=X-s^3,\qquad y=Y+s^2,\qquad h=1+p(x,y,s).
-\]
+[
+  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
 
-  \alpha=1+2s^2x+4s^5,\qquad
-  \beta=(3s^3-3sy)(1+2s^2x)-4s^4y^2,
+  alpha=1+2s^2x+4s^5,qquad
+  beta=(3s^3-3sy)(1+2s^2x)-4s^4y^2,
 
-for which $\alpha h+\beta y=1$ in $B$, because the identity
+for which $alpha h+beta y=1$ in $B$, because the identity
 
-  \alpha h+\beta y-1=(1+2s^2x)(h-1-p(x,y,s))-4s^4(x^2+y^3-sh)
+  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~lem:lift over this $B$. Its algebra $C$ is
-\[
-  C=B[u,v,w]/(u^3+hv-x,\ -u^2+hw-y,\ S(h,u,v,w)-s).
-\]
-In the polynomial ring $\CC[h,x,y,s,u,v,w]$ the last three relations are solved
+[
+  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 (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 (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].
+  R/(F)cong Ccong B[T]congCC[X,Y,T].
 
 This proves the first assertion of Theorem~thm:main: $X$, $Y$ and $T$ are
 polynomial coordinates on the zero fibre. Appendix~app:coordinates records
@@ -268,127 +268,127 @@ both the forward and the inverse maps in terms of $h,u,v,w$.
 
 # A point where the gradient vanishes
 
-At $P=(2,0,-\frac12,\frac12)$, in the variable order $(h,u,v,w)$, the values of
+At $P=(2,0,-frac12,frac12)$, in the variable order $(h,u,v,w)$, the values of
 the polynomials (eq:xyz) and (eq:p) are
-\[
-  x=-1,\quad y=1,\quad s=1,\quad p=2,\quad
-  X=x+s^3=0,\quad Y=y-s^2=0,\quad F=-1.
-\]
+[
+  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 (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 $P$.
 
 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
+polynomial automorphism $Phi=(F,g_2,g_3,g_4)$ of $CC^4$ with polynomial
+inverse $Psi$. The chain rule gives $JPhi(P)cdot JPsi(Phi(P))=I$, so
+$JPhi(P)$ 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
+cite[S1]{SY}, who asked whether the gradient must be nowhere vanishing when the
 zero fibre is a copy of a hyperplane. Theorem~thm:main answers that
 question negatively in ambient dimension four.
 
 # Higher ambient 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.
+For every integer $nge4$ there is a polynomial $fin R_n$ such that
+$R_n/(f)congCC^{[n-1]}$ and $f$ is not a coordinate.
 
 
 Proof. 
 For $n=4$ take $f=F$. For $n>4$ take the same $F$, regarded as a polynomial in
 $z_1,...,z_n$ that does not involve $z_5,...,z_n$. By (eq:fiber),
-\[
-  R_n/(F)=R[z_5,...,z_n]/(F)=\CC[X,Y,T][z_5,...,z_n],
-\]
+[
+  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
+added variables vanish, and $grad F(P)=0$, so the gradient vanishes at
 $(P,0,...,0)$ and the Jacobian argument of Section~sec:critical applies.
 
 
-\appendix
+appendix
 # Explicit formulas on the zero fibre
 
 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 (eq:xyz) and the pair (eq:alphabeta).
 
-The forward map takes a point of the zero fibre $F=0$ to $\CC^3$ by
+The forward map takes a point of the zero fibre $F=0$ to $CC^3$ by
 
-  X=x+s^3,\qquad Y=y-s^2,\qquad T=\alpha u+\beta(v+uw).
+  X=x+s^3,qquad Y=y-s^2,qquad T=alpha u+beta(v+uw).
 
 The inverse map takes independent $X,Y,T$ to a point of $F=0$ by
 
-  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).
+  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).
   
 
 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
+$r=alpha(1+beta y)$ 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.
 
 {99}
-\small
+small
 
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+bibitem{AM}
 S.~S.~Abhyankar and T.~T.~Moh,
 Embeddings of the line in the plane,
 J.~Reine Angew. Math. 276 (1975), 148--166.
 
-\bibitem{EEO}
+bibitem{EEO}
 M.~El~Kahoui, N.~Essamaoui, and M.~Ouali,
 Interpolation in the automorphism group of a polynomial ring,
 Algebra Colloq. 27 (2020), no.~3, 587--598.
 
-\bibitem{Ghosh}
+bibitem{Ghosh}
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-\bibitem{GGP}
+bibitem{GGP}
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-\bibitem{GGPthreefold}
+bibitem{GGPthreefold}
 P.~Ghosh, N.~Gupta, and A.~Pal,
 On the family of affine threefolds $a(x)y=F(x,z,t)$,
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+bibitem{KVZ}
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 Trans. Amer. Math. Soc. 356 (2004), 509--555.
 
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