# Automorphisms of affine Veronese surfaces

###### Abstract.

We prove that every derivation and every locally nilpotent derivation of the
subalgebra $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$, where $n\geq 2$, of the
polynomial algebra $K[x,y]$ in two variables over a field $K$ of
characteristic zero is induced by a derivation and a locally nilpotent
derivation of $K[x,y]$, respectively. Moreover, we prove that every
automorphism of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ over an
algebraically closed field $K$ of characteristic zero is induced by an
automorphism of $K[x,y]$. We also show that the group of automorphisms of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ admits an amalgamated free product
structure.

Bakhyt Aitzhanova 111Department of Mathematics, Wayne State University,
Detroit, MI 48202, USA, e-mail<EMAIL_ADDRESS>and Ualbai
Umirbaev222Department of Mathematics, Wayne State University, Detroit, MI
48202, USA and Institute of Mathematics and Mathematical Modeling, Almaty,
050010, Kazakhstan, e-mail<EMAIL_ADDRESS>
Mathematics Subject Classification (2020): 14R10, 14J50, 13F20.

Key words: Automorphism, derivation, polynomial algebra, affine rational
normal surface, free product.

##  1\. Introduction

Let $K$ be an arbitrary field and let $\mathbb{A}^{n}$ and $\mathbb{P}^{n}$ be
the affine and the projective $n$-space over $K$, respectively. The Veronese
map of degree $d$ is the map

$\displaystyle\nu_{d}:\mathbb{P}^{n}\to\mathbb{P}^{m}$

that sends $[x_{0}:\ldots:x_{n}]$ to all $m+1$ possible monomials of total
degree $d$, where

$\displaystyle m={n+d\choose d}-1.$

It is well known that the image of the Veronese map is a projective variety
and is called the Veronese variety [9].

The rational normal curve $C_{n}\subset\mathbb{P}^{n}$ is a particular case of
the Veronese variety and is defined to be the image of the map

$\displaystyle\nu_{n}:\mathbb{P}^{1}\to\mathbb{P}^{n}$

given by

$\displaystyle\nu_{n}:[x_{0}:x_{1}]\mapsto[x_{0}^{n}:x_{0}^{n-1}x_{1}:\ldots:x_{1}^{n}]=[X_{0}:\ldots:X_{n}].$

It is well known that $C_{n}$ is the common zero locus of the polynomials

(1) $\displaystyle F_{i,j}=X_{i-1}X_{j+1}-X_{i}X_{j}\,\,\textit{for}\,\,1\leq
i\leq j\leq n-1.$

For $n=2$ it is the plane conic $X_{0}X_{2}=X_{1}^{2}$ and for $n=3$ it is the
twisted cubic [9].

Denote by $V_{n}\subset\mathbb{A}^{n+1}$ the common zero locus of the
polynomials (1) in $\mathbb{A}^{n+1}$. The variety $V_{n}$ is the affine cone
of the rational normal curve $C_{n}$ and is called the Veronese cone in [12].
We will call $V_{n}$ the affine Veronese surface of index $n$ in order to
separate it from the Veronese cones of higher dimensions. Veronese surfaces
play an important role in the description of quasihomogeneous affine surfaces
given by M.H. Gizatullin [6] and V.L. Popov [18]. They form one of the main
examples of the so called Gizatullin surfaces [12].

M.H. Gizatullin and V.I. Danilov devoted two papers [7, 8] to the systematic
study of automorphisms of affine surfaces including affine cones of rational
normal curves. In particular, generators of the automorphism group of $V_{n}$
can be deduced from their work along with its amalgamated product structure.
L. Makar-Limanov [15, 16] gave an algebraic description of generators of the
automorpism groups of algebraic surfaces defined by an equation of the form
$x^{n}y=P(z)$. This gives an explicit description of generators of the
automorphism group of $V_{2}$.

It is well known [10, 14] that all automorphisms of the polynomial algebra
$K[x,y]$ in two variables $x,y$ over a field $K$ are tame. The well-known
Nagata automorphism (see [17])

$\displaystyle\sigma=(x+2y(zx-y^{2})+z(zx-y^{2})^{2},y+z(zx-y^{2}),z)$

of the polynomial algebra $K[x,y,z]$ over a field $K$ of characteristic zero
is proven to be non-tame [22].

The automorphism group $\mathrm{Aut}\,K[x,y]$ of this algebra admits an
amalgamated free product structure [14, 21], i.e.,

(2)
$\displaystyle\mathrm{Aut}\,K[x,y]=\mathrm{Aff}_{2}\,(K)\ast_{C}\mathrm{Tr}_{2}\,(K),$

where $\mathrm{Aff}_{2}\,(K)$ is the group of affine automorphisms,
$\mathrm{Tr}_{2}\,(K)$ is the group of triangular automorphisms, and
$C=\mathrm{Aff}_{2}\,(K)\cap\mathrm{Tr}_{2}\,(K)$.

It follows that any algebraic subgroup $G\subseteq\mathrm{Aut}\,K[x,y]$ is
conjugate to a subgroup of one of the factors $\mathrm{Aff}_{2}\,(K)$ and
$\mathrm{Tr}_{2}\,(K)$ [8, 11, 25]. In particular, any reductive subgroup
$G\subseteq\mathrm{Aut}\,K[x,y]$ is linearizable, i.e., is conjugated to a
subgroup of linear automorphisms $GL_{2}(K)$. The first examples of
nonlinearizable actions were given by G.W. Schwarz [23] and a nonlinearizable
action of the symmetric group $S_{3}$ on $\mathbb{C}^{4}$ is given in [5]. It
is still open question if any finite automorphism of $\mathbb{C}^{n}$ for
$n\geq 3$ is linearizable [13].

Recently I. Arzhancev and M. Zaidenberg [1] proved that every automorphism of
the Veronese surface $V_{n}$ can be extended to an automorphism of the plane
using the construction of Cox rings. It is also shown that the automorphism
group of the Veronese surface $V_{n}$ admits an amalgamated product structure
induced by (2) and an analogue of the Kambayashi [11] and Wright [25] result
for $V_{n}$ is proven.

This paper is devoted to the study of vector fields and automorphisms of the
affine Veronese surface $V_{n}$ for all $n\geq 2$ by purely algebraic methods.
The algebra of polynomial functions on $V_{n}$ is isomorphic to the subalgebra
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ of $K[x,y]$ (Proposition 1). Thus
the group of automorphisms of $V_{n}$ is anti-isomorphic to the group of
automorphisms of the algebra $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$. We
show that over a field $K$ of characteristic zero every derivation and every
locally nilpotent derivation of the algebra
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ is induced by a derivation and a
locally nilpotent derivation of $K[x,y]$, respectively. Using the proof of the
Rentchler’s Theorem [19] on locally nilpotent derivations of $K[x,y]$ given in
[4, Ch. 5], we prove that every automorphism of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ is induced by an automorphism of
$K[x,y]$ if $K$ is an algebraically closed field of characteristic zero. This
gives an explicit description of generators of the automorphism group of
$V_{n}$ as opposed to papers [1, 8]. We also show that the amalgamated free
product structure of the automorphism group of $K[x,y]$ induces an amalgamated
free product structure on the automorphism group of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$.

The paper is organized as follows. In Section 2 we describe the algebra of
polynomial functions on the affine Veronese surface $V_{n}$. In Section 3 we
recall some necessary results on the structure of the automorphism group of
$K[x,y]$ from [2, 4]. Section 4 is devoted to lifting of derivations of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ to derivations of $K[x,y]$. In
Section 5 we prove that so called $n$-graded derivations of $K[x,y]$ are
triangulable. In Section 6 we prove that every automorphism of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ is induced by an automorphism of
$K[x,y]$. The amalgamated free product structure of the automorphism group of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ is given in Section 7.

##  2\. Polynomial functions on $V_{n}$

Let $K$ be an arbitrary field and let $K[X_{0},X_{1},\ldots,X_{n}]$ be the
polynomial algebra over $K$ in the variables $X_{0},X_{1},\ldots,X_{n}$. The
set of all monomials of the form

(3) $\displaystyle u=X_{0}^{i_{0}}X_{1}^{i_{1}}\ldots X_{n}^{i_{n}},$

where $i_{0},i_{1},\ldots,i_{n}\geq 0$, is a linear basis of
$K[X_{0},X_{1},\ldots,X_{n}]$. Set
$\alpha(u)=(i_{0},i_{1},\ldots,i_{n})\in\mathbb{Z}^{n+1}$. If $u$ and $v$ are
two monomials of the form (3) then set $u\leq v$ if $\alpha(u)\leq\alpha(v)$
with respect to the lexicographical order.

Let $I$ be the ideal of $K[X_{0},X_{1},\ldots,X_{n}]$ generated by all
elements $F_{ij}$ defined in (1).

######  Lemma 1.

The images of all different monomials of the form $X_{k}^{i}X_{k+1}^{j}$,
where $0\leq k\leq n-1$ and $i,j\geq 0$, in $K[X_{0},X_{1},\ldots,X_{n}]/I$
form a linear basis of $K[X_{0},X_{1},\ldots,X_{n}]/I$.

Proof. The leading monomial of $F_{ij}$ with respect to the ordering $\leq$ is
$X_{i-1}X_{j+1}$. Consider the leading monomials $X_{i-1}X_{j+1}$ and
$X_{k-1}X_{l+1}$ of $F_{ij}$ and $F_{kl}$, respectively. Assume $i\leq k$ and
$F_{ij}\neq F_{kl}$. Then monomials $X_{i-1}X_{j+1}$ and $X_{k-1}X_{l+1}$ have
nontrivial intersection in the following cases:

(a) $i=k$ and $j<l$;

(b) $j+1=k-1$;

(c) $i<k$ and $j=l$.

Case (a). We form an $S$-polynomial (see, for example [3])

$\displaystyle
S(F_{ij},F_{il})=(X_{i-1}X_{j+1}-X_{i}X_{j})X_{l+1}-(X_{i-1}X_{l+1}-X_{i}X_{l})X_{j+1}$
$\displaystyle=-(X_{j}X_{l+1}-X_{j+1}X_{l})X_{i}=-F_{j+1,l}X_{i}.$

The leading monomial of $F_{j+1,l}X_{i}$ is equal to $X_{i}X_{j}X_{l+1}$ and
is less than $X_{i-1}X_{j+1}X_{l+1}$.

Case (b). We have $i+2\leq j+2\leq l$. Then

$\displaystyle
S(F_{ij},F_{(j+2)l})=(X_{i-1}X_{j+1}-X_{i}X_{j})X_{l+1}-X_{i-1}(X_{j+1}X_{l+1}-X_{j+2}X_{l})$
$\displaystyle=-X_{i}X_{j}X_{l+1}+X_{i-1}X_{j+2}X_{l}=F_{i(j+1)}X_{l}-X_{i}F_{(j+1)l}.$

The leading term of $F_{i(j+1)}X_{l}$ is $X_{i-1}X_{j+2}X_{l}$ and the leading
term of $X_{i}F_{(j+1)l}$ is $X_{i}X_{j}X_{l+1}$. Both of them are less than
$X_{i-1}X_{j+1}X_{l+1}$.

Case (c) can be handled similar to the case (a).

Consequently, the set of all elements $F_{ij}$ forms a Gröbner basis for the
ideal $I$ [3, Theorem 6, p. 86]. Since the leading monomial of $F_{ij}$ is
$X_{i-1}X_{j+1}$ it follows the statement of the lemma [3, Ch. 5, section 3].
$\Box$

######  Proposition 1.

$K[X_{0},X_{1},\ldots,X_{n}]/I\cong K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$.

Proof. The homomorphism

$\phi:K[X_{0},X_{1},\ldots,X_{n}]\to K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$

defined by $\phi(X_{i})=x^{n-i}y^{i}$ for all $i$ induces the homomorphism

$\bar{\phi}:K[X_{0},X_{1},\ldots,X_{n}]/I\to
K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$

since
$\phi(X_{i-1}X_{j+1}-X_{i}X_{j})=x^{n-(i-1)}y^{i-1}x^{n-(j+1)}y^{j+1}-x^{n-i}y^{i}x^{n-j}y^{j}=0$
for all $1\leq i\leq j\leq n-1$.

Let $u=X_{k}^{i}X_{k+1}^{j}$ and $v=X_{s}^{p}X_{s+1}^{q}$ where $k\leq s$. We
get

$\displaystyle\phi(u)=(x^{n-k}y^{k})^{i}(x^{n-k-1}y^{k+1})^{j}=x^{(n-k)i+(n-k-1)j}y^{ki+(k+1)j}$

and, similarly,

$\displaystyle\phi(v)=x^{(n-s)p+(n-s-1)q}y^{sp+(s+1)q}.$

Consequently, $\phi(u)=\phi(v)$ if and only if

$\displaystyle(n-k)i+(n-k-1)j=(n-s)p+(n-s-1)q,$ (4) $\displaystyle
ki+(k+1)j=sp+(s+1)q.$

By adding the both sides of these equalities we get $n(i+j)=n(p+q)$, i.e.,
$i+j=p+q$. Then (2) gives that

$\displaystyle k(p+q)+j=s(p+q)+q,$

i.e.,

(5) $\displaystyle(s-k)(p+q)=j-q.$

We get $j-q\geq 0$ since $s\geq k$. Then (5) is possible only if $s=k$ or
$s-k=1$ and $p+q=j-q$. If $s=k$ then (5) gives $j=q$. Then $i=p$ since
$i+j=p+q$. This gives $u=v$. Suppose that $s-k=1$ and $p+q=j-q$. Since
$i+j=p+q$ it follows that $q=0,i=0,p=j$. Then $u=v=X_{k+1}^{j}$.

Thus we proved that the images of different monomials of the form
$X_{k}^{i}X_{k+1}^{j}$ under $\phi$ are different monomials in $x,y$.
Consequently, the images of different monomials of the form
$X_{k}^{i}X_{k+1}^{j}$ are linearly independent.

By Lemma 1, the images of different monomials of the form
$X_{k}^{i}X_{k+1}^{j}$ gives a linear basis for
$K[X_{0},X_{1},\ldots,X_{n}]/I$. Consequently, $\bar{\phi}$ is an injection.
Obviously, $\bar{\phi}$ is a surjection, i.e., $\bar{\phi}$ is an isomorphism.
$\Box$

##  3\. Automorphisms of $K[x,y]$

Let $K[x,y]$ be the polynomial algebra in the variables $x,y$ over a field $K$
and let $\mathrm{Aut}\,K[x,y]$ be the group of automorphisms of $K[x,y]$.
Denote by $\phi=(f,g)$ the automorphism of $K[x,y]$ such that $\phi(x)=f$ and
$\phi(y)=g$, where $f,g\in K[x,y]$. If $\phi=(f_{1},g_{1})$ and
$\psi=(f_{2},g_{2})$ then the product in $\mathrm{Aut}\,K[x,y]$ is defined by

$\phi\circ\psi=(f_{2}(f_{1},g_{1}),g_{2}(f_{1},g_{1})).$

An automorphism $\phi\in\mathrm{Aut}\,K[x,y]$ is called elementary if it has
the form

$\phi=(x,\alpha y+f(x))$

or

$\phi=(\alpha x+g(y),y),$

where $f(x)\in K[x]$, $g(y)\in K[y]$, and $0\neq\alpha\in K$. The subgroup of
$\mathrm{Aut}\,K[x,y]$ generated by all elementary automorphisms is called the
tame subgroup. Elements of this subgroup are called tame automorphisms of
$K[x,y]$.

An automorphism $\phi\in\mathrm{Aut}\,K[x,y]$ is called affine if it has the
form

$\phi=(\alpha_{1}x+\beta_{1}y+\gamma_{1},\alpha_{2}x+\beta_{2}y+\gamma_{2})$

where $\alpha_{1}\beta_{2}\neq\beta_{1}\alpha_{2}$ and
$\alpha_{1},\alpha_{2},\beta_{1},\beta_{2},\gamma_{1},\gamma_{2}\in K$. The
subgroup $\mathrm{Aff}_{2}\,(K)$ of $\mathrm{Aut}\,K[x,y]$ generated by all
affine automorphisms is called the affine subgroup. If
$\gamma_{1},\gamma_{2}=0$ then the affine automorphism $\phi$ is called
linear. The subgroup $\mathrm{GL}_{2}\,(K)$ of $\mathrm{Aff}_{2}\,(K)$
generated by all linear automorphisms is called the linear subgroup.

An automorphism $\phi\in\mathrm{Aut}\,K[x,y]$ is called triangular if it has
the form

(6) $\displaystyle\phi=(\alpha x+f(y),\beta y+\gamma),$

where $0\neq\alpha,\beta\in K$ and $f(y)\in K[y]$. The subgroup
$\mathrm{Tr}_{2}\,(K)$ of $\mathrm{Aut}\,K[x,y]$ generated by all triangular
automorphisms is called the triangular subgroup.

The well known Jung-van der Kulk Theorem [10, 14] says that all automorphisms
of the polynomial algebra $K[x,y]$ in two variables $x,y$ over a field $K$ are
tame. Moreover, van der Kulk and Shafarevich [14, 21] proved that the
automorphism group $\mathrm{Aut}\,K[x,y]$ of this algebra admits an
amalgamated free product structure, i.e.,

$\displaystyle\mathrm{Aut}\,K[x,y]=\mathrm{Aff}_{2}\,(K)\ast_{C}\mathrm{Tr}_{2}\,(K),$

where $C=\mathrm{Aff}_{2}\,(K)\cap\mathrm{Tr}_{2}\,(K)$.

We fix a grading

(7) $\displaystyle K[x,y]=K[x,y]_{0}\oplus K[x,y]_{1}\oplus
K[x,y]_{2}\oplus\ldots\oplus K[x,y]_{n-1}$

of the polynomial algebra $K[x,y]$, where $K[x,y]_{i}$ is the linear span of
all homogeneous monomials of degree $i+ns$, $i=0,1,\ldots,n-1$, and $s$ is an
arbitrary nonnegative integer. This is a $\mathbb{Z}_{n}$-grading of $K[x,y]$,
i.e.,

$K[x,y]_{i}K[x,y]_{j}\subseteq K[x,y]_{i+j},$

where $i,j\in\mathbb{Z}_{n}=\mathbb{Z}/n\mathbb{Z}$. For shortness we will
refer to this grading as $n$-grading.

An automorphism $\phi\in\mathrm{Aut}\,K[x,y]$ is called a graded automorphism
with respect to grading (7) if $\phi(x),\phi(y)\in K[x,y]_{1}$. A graded
automorphism is called graded tame if it is a product of graded elementary
automorphisms.

Recently A. Trushin [24] studied graded automorphisms of polynomial
automorphisms. But his gradings do not include gradings of type (7).

A graded automorphism of $K[x,y]$ with respect to grading (7) will be called
an n-graded automorphism for shortness. Obviosly, every $n$-graded
automorphism induces an automorphism of the algebra
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$.

A derivation $D$ of $K[x,y]$ will be called an n-graded derivation if
$D(x),D(y)\in K[x,y]_{1}$. Recall that every derivation $D$ of $K[x,y]$ can be
uniquely written in the form

$\displaystyle D=f\partial_{x}+g\partial_{y},$

where $D(x)=f$, $D(y)=g$, and $\partial_{x}=\frac{\partial}{\partial x}$ and
$\partial_{y}=\frac{\partial}{\partial y}$ are partial derivatives with
respect to $x$ and $y$, respectively.

##  4\. Derivations of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$

Let $K$ be an arbitrary field of characteristic zero. Let $A$ be any algebra
over $K$. A derivation $D$ of $A$ is called locally nilpotent if for every
$a\in A$ there exists a positive integer $n=n(a)$ such that $D^{n}(a)=0$.

If $D$ is a locally nilpotent derivation of $A$ then

$\displaystyle\exp{D}=\sum_{p\geq 0}\frac{1}{p!}D^{p}$

is an automorpism of $A$ and is called an exponential automorphism.

Moreover, if $D$ is any derivation of $A$ then

$\displaystyle\exp{TD}=\sum_{i=0}^{\infty}\frac{1}{i!}D^{i}T^{i}$

is an automorpism of the formal power series algebra $A[[T]]$. If $D$ is
locally nilpotent then $\exp{TD}$ is an automorphism of $A[T]$.

Consider the grading (7) of $K[x,y]$. A derivation $D$ of $K[x,y]$ will be
called an n-graded derivation if $D(x),D(y)\in K[x,y]_{1}$. Obviously, every
$n$-graded derivation of $K[x,y]$ induces a derivation of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$. The reverse is also true.

######  Lemma 2.

Every derivation of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ can be uniquely
extended to an $n$-graded derivation of $K[x,y]$.

Proof. Let $D$ be a derivation of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$.
Denote by $T$ the unique extension of $D$ [26, p. 120] to a derivation of the
field of fractions $K(x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n})$ of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$. Obviously, the field extension

$K(x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n})\subseteq K(x,y)$

is algebraic. This extension is separable since $K$ is a field of
characteristic zero. By Corollaries 2 and 2’ in [26, pages 124–125], every
derivation of the field $K(x^{n},x^{n-1}y,\ldots,xy^{n-1},$
$y^{n})$ can be uniquely extended to a derivation of $K(x,y)$. Let $S$ be the
unique extension of $T$ to a derivation of $K(x,y)$. Suppose that

(8) $\displaystyle S(x)=\frac{f_{1}}{g_{1}},\,S(y)=\frac{f_{2}}{g_{2}},$

where $f_{1},f_{2}\in K[x,y]$, $0\neq g_{1},g_{2}\in K[x,y]$, and the pairs
$f_{1},g_{1}$ and $f_{2},g_{2}$ are relatively prime. We have

$\displaystyle
D(x^{n-i}y^{i})=S(x^{n-i}y^{i})=(n-i)x^{n-i-1}y^{i}\frac{f_{1}}{g_{1}}+ix^{n-i}y^{i-1}\frac{f_{2}}{g_{2}}$

for all $0\leq i\leq n$.

Since $D(x^{n}),D(x^{n-1}y),\ldots,D(xy^{n-1}),D(y^{n})\in
K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ it follows that

$\displaystyle
g_{1}g_{2}|(n-i)x^{n-(i+1)}y^{i}f_{1}g_{2}+ix^{n-i}y^{i-1}f_{2}g_{1}$

for all $0\leq i\leq n$. Consequently,

$\displaystyle g_{1}|(n-i)x^{n-(i+1)}y^{i}$

and

$\displaystyle g_{2}|ix^{n-i}y^{i-1}$

for all $0\leq i\leq n$.

This means that $g_{1}|x^{n-1}$ and $g_{1}|y^{n-1}$ and, consequently, we may
assume that $g_{1}=1$. Similarly, $g_{2}|y^{n-1}$ and $g_{2}|x^{n-1}$ give
that $g_{2}=1$. Obviously, $f_{1},f_{2}\in K[x,y]_{1}$. $\Box$

For any derivation $D$ of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ denote by
$\widetilde{D}$ its unique extension to a derivation of $K[x,y]$ determined by
Lemma 2. Obviously $D$ is locally nilpotent if $\widetilde{D}$ is locally
nilpotent. The reverse statement is also true.

######  Lemma 3.

If $D$ is a locally nilpotent derivation of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ then $\widetilde{D}$ is a locally
nilpotent $n$-graded derivation of $K[x,y]$.

Proof. Suppose that $D$ is a locally nilpotent derivation of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$. Then $\exp{TD}$ is an automorphism
of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}][T]$. Recall that
$\exp{T\widetilde{D}}$ is an automorphism of $K[x,y][[T]]$. We have

$\displaystyle\exp{TD}(x^{n})=\exp{T\widetilde{D}}(x^{n})=\exp{T\widetilde{D}}(x)^{n}.$

This implies that $\exp{T\widetilde{D}}(x)\in K[x,y][T]$ since
$\exp{TD}(x^{n})\in K[x,y][T]$. Similarly, $\exp{T\widetilde{D}}(y)\in
K[x,y][T]$. This means that there exist natural numbers $m$ and $n$ such that
$\widetilde{D}^{m}(x)=0$ and $\widetilde{D}^{n}(y)=0$. Therefore
$\widetilde{D}$ is locally nilpotent. $\Box$

##  5\. Triangulation of locally nilpotent $n$-graded derivations

A derivation $D$ of $K[x,y]$ is called triangular if

$\displaystyle D(x)=f(y)\in K[y],\ \ D(y)=\alpha\in K.$

A derivation $D$ of $K[x,y]$ is called triangulable if there exists an
automorphism $\alpha\in\mathrm{Aut}\,K[x,y]$ such that $\alpha^{-1}D\alpha$ is
triangular.

Every triangular derivation, and hence every triangulable derivation, is
locally nilpotent. In 1968 R. Rentschler [19] proved that every locally
nilpotent derivation of the polynomial algebra $K[x,y]$ over a field of
characteristic zero is triangulable.

In this section we adopt the proof of this result given in [4, Ch. 5] to prove
that every locally nilpotent $n$-graded derivation of $K[x,y]$ is triangulable
by a tame $n$-graded automorphism.

First recall some necessary definitions from [4].

Let $0\neq w=(w_{1},w_{2})\in\mathbb{Z}^{2}$. Then $w$-degree of the monomial
$x^{a_{1}}y^{a_{2}}$ is defined by
$w(x^{a_{1}}y^{a_{2}})=a_{1}w_{1}+a_{2}w_{2}$. This degree function leads to
the $w$-grading

$\displaystyle K[x,y]=\sum_{d}W_{d}$

of $K[x,y]$, where $W_{d}$ is the span of all monomials of $w$-degree $d$.

Let $T=cx^{a_{1}}y^{a_{2}}\partial_{i}$ be a monomial derivation of $K[x,y]$,
where $i=1,2$. Set $(s,t)=(a_{1},a_{2})-e_{i}$, where $e_{i}$ is the $i$-th
vector of the standard basis of $K^{2}$. Then

$T(x^{m_{1}}y^{m_{2}})\in Kx^{m_{1}+s}y^{m_{2}+t}$

for all $m_{1},m_{2}$. We call $(s,t)$ the strength of $T$.

Every derivation $D$ is a linear combination of monomial derivations. Set

$\displaystyle\mathrm{supp}\,D=\\{(s,t)\in\mathbb{Z}^{2}\,|\,D\,\textit{contains
a term of strength}\,(s,t)\\}.$

Let us denote by $D(s,t)$ the sum of all terms in $D$ of strength $(s,t)$ and
set

$\displaystyle D_{p}=\sum_{sw_{1}+tw_{2}=p}D(s,t).$

Obviously,

$\displaystyle D=\sum_{p}D_{p}$

and this decomposition is called the w-homogeneous decomposition of $D$. If
$p$ is maximal with $D_{p}\neq 0$ then $p$ is called the $w$-degree of $D$ and
is denoted by $wdeg\,D$. When $w=(1,1)$ $p$ is called the degree of D and is
denoted by $deg\,D$.

It is easy to check [4] that $D_{p}W_{d}\subset W_{p+d}$ for all
$p,d\in\mathbb{Z}$.

Consider the grading (7) of $K[x,y]$. Set $K[y]_{1}=K[x,y]_{1}\cap K[y]$.
Every triangular $n$-derivation of $K[x,y]$ can be written as
$f\partial_{x}+\alpha\partial_{y}$ where $f\in K[y]_{1}$ and $\alpha\in K$.

######  Proposition 2.

Let $D$ be a locally nilpotent $n$-graded derivation of $K[x,y]$. Then there
exists a tame $n$-graded automorphism $\alpha$ of $K[x,y]$ and $f(y)\in
K[y]_{1}$ such that

$\displaystyle\alpha^{-1}D\alpha=f(y)\partial_{x}.$

Proof. Let $D$ be a locally nilpotent $n$-graded derivation of $K[x,y]$.
According to Corollary 5.1.16 in [4, p. 91], the following three cases are
possible:

$(i)$ $D=f(y)\partial_{x}$, for some $f(y)\in K[y]$;

$(ii)$ $D=f(x)\partial_{y}$, for some $f(x)\in K[x]$;

$(iii)$ there exist $s_{0},t_{0}\geq 0$ such that $(s_{0},-1)$ and
$(-1,t_{0})$ belong to $supp\,D$ and, furthermore, $\mathrm{supp}\,D$ is
contained in the triangle with vertices $(s_{0},-1)$, $(-1,-1)$, $(-1,t_{0})$.

Case $(i)$. If $D=f(y)\partial_{x}$ with $f(y)\in K[y]_{1}$ then set
$\alpha=\mathrm{id}$. Obviously, the identity automorphism is an $n$-graded
automorphism.

Case $(ii)$. If $D=f(x)\partial_{y}$ with $f(x)\in K[x]_{1}$ then set
$\alpha=(y,x)$. Obviously $\alpha$ is a $n$-graded automorphism of $K[x,y]$
and $\alpha^{-1}D\alpha=f(y)\partial_{x}$ with $f(y)\in K[y]_{1}$.

Case $(iii)$. Suppose that we have $s_{0},t_{0}\geq 0$ such that
$(s_{0},-1),(-1,t_{0})\in\mathrm{supp}\,D$. This implies that $D$ contains
differential monomials of the form $x^{s_{0}}\partial_{y}$ and
$y^{t_{0}}\partial_{x}$ with nonzero coefficients. Hence $s_{0}=1+nk$,
$t_{0}=1+nl$, $k,l\in\mathbb{Z}$ since $x^{s_{0}},y^{t_{0}}\in K[x,y]_{1}$.

Let $L$ be the line passing through the points $(1+nk,-1)$ and $(-1,1+nl)$.
The defining equation of $L$ is

$(nl+2)x+(nk+2)y=n^{2}kl+nk+nl=nM.$

Set $w=(nl+2,nk+2)$ and $p=n^{2}kl+nk+nl$. Obviously $\mathrm{wdeg}\,D=p$ and
$D_{p}$ is the highest homogeneous part of $D$ with respect to the $w$-degree.
It is well known that the highest homogeneous part of a locally nilpotent
derivation is locally nilpotent (see, for example [4, p. 90]). Consequently,
$D_{p}$ is a locally nilpotent $n$-graded derivation.

We can write $D_{p}=gD_{1}$, where $D_{1}=a\partial_{x}+b\partial_{y}$ with
$gcd(a,b)=1$. By Corollary 1.3.34 in [4, p. 29], $D_{1}$ is locally nilpotent
and $D_{1}(g)=0$. By Proposition 1.3.46 in [4, p. 31], $D_{1}$ has a slice in
$K[x,y]$, i.e., there exists $s\in K[x,y]$ such that $D_{1}(s)=1$. This
implies that $a(0,0)\neq 0$ or $b(0,0)\neq 0$. Assume that $a(0,0)\neq 0$.
This means that $D_{1}$ has a term of the form $c\partial_{x}$, where $c\in
K^{*}$. Since $(1+nk,-1)\in\mathrm{supp}\,D_{p}$ and $D_{p}=gD_{1}$ it follows
that $D_{1}$ also has a term of the form $dx^{r}\partial_{y}$ with $r\geq 0$
and $d\in K^{*}$. Moreover, $g$ and $D_{1}$ are $w$-homogeneous since $D_{p}$
is $w$-homogeneous. Therefore $\mathrm{supp}\,D_{1}$ is on the line passing
through the points $(-1,0)$ and $(r,-1)$. Notice that this line does not
contain any other points with integer coordinates. Hence
$D_{1}=c\partial_{x}+dx^{r}\partial_{y}$. Since $D_{p}$ is an $n$-graded
derivation it follows that $g\in K[x,y]_{1}$ and $n|r$.

We have $g\in\mathrm{Ker}\,D_{1}=K[y-\frac{d}{(r+1)c}x^{r+1}]$ since
$D_{1}(g)=0$. Consequently, $g=a(y-\frac{d}{(r+1)c}x^{r+1})^{N}$ for some
$a\in K^{*}$ and $N\in\mathbb{N}$ since $g$ is $w$-homogeneous. So

$\displaystyle
D_{p}=a(y-\frac{d}{(r+1)c}x^{r+1})^{N}(c\partial_{x}+dx^{r}\partial_{y}),$

where $a,c,d\in K^{\ast}$, $r\geq 0$, and $N\in\mathbb{N}$. Obviously,
$t_{0}=N$ and $s_{0}=(r+1)N+r$.

Let $\alpha$ be the automorphism given by

$\displaystyle\alpha(x)=x,\,\alpha(y)=y-\frac{d}{(r+1)c}x^{r+1}.$

This is an elementary $n$-graded automorphism since $n|r$. Direct calculations
give that

$\displaystyle\alpha^{-1}D_{1}\alpha=c\partial_{x}$

and

$\displaystyle\alpha^{-1}D_{p}\alpha=acy^{t_{0}}\partial_{x}.$

Since $\alpha$ is $w$-homogeneous, $\alpha^{-1}D_{p}\alpha$ is the highest
$w$-homogeneous part of $\alpha^{-1}D\alpha$. Thus $\alpha$ turns all points
of $\mathrm{supp}\,D_{p}$ to one point $(-1,t_{0})$. Consequently,
$s_{0}(\alpha^{-1}D\alpha)<s_{0}(D)$. Leading an induction on
$s_{0}(D)+t_{0}(D)$ we can conclude that the statement of the proposition is
true. $\Box$

##  6\. Automorphisms of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$

As we noticed above, every $n$-graded automorphism of $K[x,y]$ induces an
automorphism of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$. In this section we
prove the reverse of this statement.

######  Lemma 4.

Let $p\in K[x,y]$. If $p^{n}\in K[x,y]_{0}$ then $p\in K[x,y]_{i}$ for some
$i\in\mathbb{Z}_{n}=\mathbb{Z}/n\mathbb{Z}$.

Proof. Consider the standard grading

$\displaystyle K[x,y]=A_{0}\oplus A_{1}\oplus\ldots\oplus A_{k}\oplus\ldots,$

where $A_{i}$ is the linear span of monomials of degree $i$ for all $i\geq 0$.
For any $f\in K[x,y]$ denote by $f_{i}\in A_{i}$ its homogeneous part of
degree $i$. Let

$\displaystyle p=p_{i_{1}}+p_{i_{2}}+\ldots+p_{i_{k}},\ \ 0\neq p_{i_{j}}\in
A_{i_{j}},\ \ i_{1}<i_{2}<\ldots<i_{k}.$

Suppose that $p_{i_{1}},p_{i_{2}},\ldots,p_{i_{s}}\in K[x,y]_{i}$ for some
$i\in\mathbb{Z}_{n}$ and $p_{i_{s+1}}\notin K[x,y]_{i}$. Set
$q=p_{i_{1}}+p_{i_{2}}+\ldots+p_{i_{s}}$. Obviously, $q^{n}\in K[x,y]_{0}$.
Set $t=(n-1)i_{1}+i_{s+1}$. Then $t\not\equiv 0\mod{n}$. We get

$\displaystyle(p^{n})_{t}=(q^{n})_{t}+np_{i_{1}}^{n-1}p_{i_{s+1}}=np_{i_{1}}^{n-1}p_{i_{s+1}}\notin
K[x,y]_{0}$

since $q^{n}\in K[x,y]_{0}$. This contradicts to $p^{n}\in K[x,y]_{0}$. $\Box$

######  Theorem 1.

Every automorphism of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ over an
algebraically closed field $K$ of characteristic zero is induced by an
$n$-graded automorphism of $K[x,y]$.

Proof. Consider the derivation $D=y\partial_{x}$ of $K[x,y]$. Let
$\overline{D}$ be the derivation of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$
induced by $D$.

Let $\alpha$ be an arbitrary automorphism of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$. Set
$T=\alpha\overline{D}\alpha^{-1}$. This derivation is locally nilpotent since
$D$ is locally nilpotent. Let $\widetilde{T}$ be the extension of $T$ to a
derivation of $K[x,y]$ that uniquely defined by Lemma 2. By Lemma 3,
$\widetilde{T}$ is a locally nilpotent $n$-graded derivation of $K[x,y]$. By
Proposition 2, there exists an $n$-graded tame automorphism $\beta$ of
$K[x,y]$ such that $S=\beta^{-1}\widetilde{T}\beta$ is a triangular $n$-graded
derivation of $K[x,y]$. Let

$\displaystyle S=\beta^{-1}\widetilde{T}\beta=g(y)\partial_{x},$

where $g(y)\in K[y]_{1}$. We get

$\displaystyle S(f)=g(y)\frac{\partial f}{\partial x},\ \ f\in K[x,y].$

Let $\overline{\beta}$ be the automorphism of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ induced by $\beta$. Then $S$ induces
the derivation
$\overline{S}=\overline{\beta}^{-1}T\overline{\beta}=\overline{\beta}^{-1}\alpha\overline{D}\alpha^{-1}\overline{\beta}$
of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$.

Let $\phi=\overline{\beta}^{-1}\alpha$. Assume that
$\phi(x^{n-i}y^{i})=f_{i}$, where $0\leq i\leq n$. Applying the equation
$\phi\overline{D}=\overline{S}\phi$ to $x^{n-i}y^{i}$ for all $i$, we get

$\displaystyle(n-i)f_{i+1}=g(y)\frac{\partial f_{i}}{\partial x},$

i.e.,

$\displaystyle 0=g(y)\frac{\partial f_{n}}{\partial
x},f_{n}=g(y)\frac{\partial f_{n-1}}{\partial
x},\ldots,(n-1)f_{2}=g(y)\frac{\partial f_{1}}{\partial
x},nf_{1}=g(y)\frac{\partial f_{0}}{\partial x}.$

These equalities immediately give that

$\displaystyle\deg_{x}f_{n}=0,\deg_{x}f_{n-1}=1,\ldots,\deg_{x}f_{n-i}=i,\ldots,\deg_{x}f_{0}=n.$

In particular, $f_{n}\in K[y]$.

We have

(9)
$\displaystyle\frac{f_{0}}{f_{1}}=\frac{f_{1}}{f_{2}}=\ldots=\frac{f_{n-1}}{f_{n}}$

since the generators $x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}$ of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ satisfy the relations

$\displaystyle\frac{x^{n}}{x^{n-1}y}=\frac{x^{n-1}y}{x^{n-2}y^{2}}=\ldots=\frac{xy^{n-1}}{y^{n}}=\frac{x}{y}.$

Let $\frac{f_{0}}{f_{1}}=\frac{p}{q}$, where $p,q\in K[x,y]$ are relatively
prime. Then $\frac{f_{0}}{f_{n}}=\frac{p^{n}}{q^{n}}$ by (9). Since $p^{n}$
and $q^{n}$ are relatively prime it follows that $f_{0}=p^{n}u$ and
$f_{n}=q^{n}u$ for some $u\in K[x,y]$. Moreover, (9) implies that
$f_{i}=p^{n-i}q^{i}u$ for all $i$. From this we get

$\displaystyle K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]\subseteq K+(u),$

where $(u)$ is the ideal of $K[x,y]$ generated by $u$. This is possible if the
leading word of $u$ divides all of the words
$x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}$. Consequently, $u\in K^{*}$. Over an
algebraically closed field we can write $u=v^{n}$ for some $v\in K^{*}$.
Replacing $vp$ with $p$ and $vq$ with $q$, we may assume that $u=1$ and
$f_{i}=p^{n-i}q^{i}$ for all $i$.

We have $q\in K[y]$ since $f_{n}=q^{n}\in K[y]$. We also have $\deg_{x}(p)=1$
since $p^{n}=f_{0}$ and $\deg_{x}(f_{0})=n$. Set $p=xa(y)+b(y)$. We get

$\displaystyle K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]\subseteq
K[f_{n}]+(p)\subseteq K[y]+(p),$

where $(p)$ is the ideal of $K[x,y]$ generated by $p$. Consequently,

$\displaystyle x^{n}=(xa(y)+b(y))h+f(y).$

Introducing a monomial order with prioritized $x$, we get that it is possible
only if $a(y)=a\in K^{*}$. Consequently, $p=ax+b(y)$. By Lemma 4, it implies
that $p\in K[x,y]_{1}$ since $p^{n}\in
K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$. Set $\gamma=(ax+b(y),y)$. Then
$\gamma$ is an elementary $n$-graded automorphism of $K[x,y]$. Set
$\psi=\overline{\gamma}^{-1}\phi$. Then $\psi(x^{n-i}y^{i})=x^{n-i}q^{i}$ for
all $i$. We have

$\displaystyle K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]\subseteq K[q^{n}]+(x),$

where $(x)$ is the ideal of $K[x,y]$ generated by $x$. It is possible only if
$q^{n}=cy^{n}$ for some $c\in K^{*}$. Consequently, $q=ey$ for some $e\in
K^{*}$ since $K$ is algebraically closed.

Let $\delta=(x,ey)$. Then $\overline{\delta}^{-1}\psi=\mathrm{id}$, i.e.,
$\overline{\delta}^{-1}\overline{\gamma}^{-1}\overline{\beta}^{-1}\alpha=\mathrm{id}$.
Consequently,
$\alpha=\overline{\beta}\overline{\gamma}\overline{\delta}=\overline{\beta\gamma\delta}$
is induced by a tame $n$-graded automorphism of $K[x,y]$. $\Box$

##  7\. Amalgamated free product structure of
$\mathrm{Aut}\,K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$

Let $G_{n}$ be the group of all $n$-graded automorphisms of the polynomial
algebra $K[x,y]$.

######  Lemma 5.

The subgroup $G_{n}$ of $\mathrm{Aut}\,K[x,y]$ is generated by all linear
automorphisms and all automorphisms of the type $(x-\alpha y^{m},y)$, where
$m=1+ns$ is a positive integer and $\alpha\in K$.

Proof. For any $f\in K[x,y]$ denote by $\bar{f}$ its highest homogeneous part
with respect the standard degree function $\deg$. Let $\phi=(f,g)$ be a
$n$-graded automorphism of the algebra $K[x,y]$ and suppose that $\deg\,f=k$
and $\deg\,g=l$. If $k+l=2$ then $\phi$ is a linear automorphism.

Suppose that $k+l\geq 3$. It is well known that $k|l$ or $l|k$ (see, for
example [2, 4]). Assume that $l|k$. In this case we have
$\bar{f}=\alpha\bar{g}^{m}$ for some $\alpha\in K^{*}$ and $m\in\mathbb{N}$.
Since $\bar{f},\bar{g}\in K[x,y]_{1}$ it follows that $m=1+ns$ for some $s\geq
0$. In fact, let $\deg(\bar{f})=1+np$ and $\deg(\bar{g})=1+nq$. Then

$\displaystyle 1+np=m(1+nq).$

Consequently, $m-1=np-mnq=ns$.

Therefore $(x-\alpha y^{m},y)$ is an elementary $n$-graded automorphism of
$K[x,y]$. We have

$(f,g)\circ(x-\alpha y^{m},y)=(f-\alpha g^{m},g)=(f^{\prime},g),$

where $\deg(f^{\prime})<\deg(f)$. Leading an induction on $k+l$ we may assume
that $(f^{\prime},g)$ satisfies the statement of the lemma. Then $(f,g)$ also
satisfies the statement of the lemma. $\Box$

######  Corollary 1.

Every $n$-graded automorphism of $K[x,y]$ is $n$-graded tame.

An automorphism $\phi\in\mathrm{Aut}\,K[x,y]$ is called n-graded triangular if
it has the form

$\displaystyle\phi=(\alpha x+f(y),\beta y),$

where $0\neq\alpha,\beta\in K$ and $f(y)\in K[y]_{1}$.

Let $T_{n}$ be the group of all $n$-triangular automorphisms of the polynomial
algebra $K[x,y]$.

######  Corollary 2.

$G_{n}=\mathrm{GL}_{2}\,(K)\ast_{B}T_{n}$, where $B=\mathrm{GL}_{2}\,(K)\cap
T_{n}$.

Proof. Lemma 5 says that $G_{n}$ is generated by $\mathrm{GL}_{2}$ and
$T_{n}$. Consider (2). We have $\mathrm{GL}_{2}\subseteq\mathrm{Aff}_{2}$,
$T_{n}\subseteq\mathrm{Tr}_{2}\,(K)$, and $B\subseteq C$. This means that
every decomposition of an element of $G_{n}$ in the form

$\displaystyle g_{1}g_{2}\ldots g_{k},$

where $g_{i}\in\mathrm{GL}_{2}\cup T_{n}$ for all $i$ and $g_{i}$ and
$g_{i+1}$ do not belong together to $\mathrm{GL}_{2}$ or $T_{n}$ for all
$i<k$, determined by the amalgated free product structure (2). Consequently,

$\displaystyle
G_{n}=\mathrm{GL}_{2}\,(K)\ast_{B}T_{n}\subseteq\mathrm{Aff}_{2}\,(K)\ast_{C}\mathrm{Tr}_{2}\,(K).\
\ \ \Box$

######  Corollary 3.

Let $E=\\{\epsilon\mathrm{id}|\epsilon^{n}=1,\epsilon\in K\\}$. Then

$\displaystyle\mathrm{Aut}\,K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]\cong
G_{n}/E.$

Proof. Consider the homomorphism

(10)
$\displaystyle\psi:G_{n}\to\mathrm{Aut}\,K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$

defined by $\psi(\alpha)=\overline{\alpha}$, where $\overline{\alpha}$ is the
automorphism of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ induced by the
$n$-graded automorphism $\alpha$ of $K[x,y]$.

By Theorem 1, $\psi$ is an epimorphism. Let $\alpha\in\mathrm{Ker}\,\psi$.
Then

$\displaystyle\alpha(x)^{n-i}\alpha(y)^{i}=x^{n-i}y^{i}$

for all $0\leq i\leq n$. This implies that $\alpha(x)=\epsilon
x,\alpha(y)=\epsilon y$ for some $n$th root of unity $\epsilon\in K$.
Consequently, $\alpha\in E$. Obviously, $E\subseteq\mathrm{Ker}\,\psi$. $\Box$

Let

$\displaystyle\overline{\mathrm{GL}_{2}\,(K)}=\mathrm{GL}_{2}\,(K)/E,\overline{T_{n}}=T_{n}/E,\overline{B}=B/E.$

######  Theorem 2.

$\mathrm{Aut}\,K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]\cong\overline{\mathrm{GL}_{2}\,(K)}\ast_{\overline{B}}\overline{T_{n}}$.

Proof. By Corollaries 2 and 3, the group
$\mathrm{Aut}\,K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ is generated by
$\overline{\mathrm{GL}_{2}\,(K)}$ and $\overline{T_{n}}$.

Let $G$ be any group and $\psi_{1}:\overline{\mathrm{GL}_{2}\,(K)}\to G$ and
$\psi_{2}:\overline{T_{n}}\to G$ be any homomorphisms with
$\psi_{1}|_{\overline{B}}=\psi_{2}|_{\overline{B}}$.

Let $\alpha:{\mathrm{GL}_{2}\,(K)}\to\overline{\mathrm{GL}_{2}\,(K)}$ and
$\beta:T_{n}\to\overline{T_{n}}$ be natural homomorphisms. Set
$\phi_{1}=\psi_{1}\alpha:{\mathrm{GL}_{2}\,(K)}\to G$ and
$\phi_{2}=\psi_{2}\beta:\mathrm{Tr}_{n}\to G$. Obviously,
$\phi_{1}|_{B}=\phi_{2}|_{B}$. By the universal property of the amalgamated
free products of groups [20, Ch. 1], there exists a unique homomorphism
$\phi:{\mathrm{GL}_{2}\,(K)}\ast_{B}T_{n}\to G$ such that
$\phi|_{{\mathrm{GL}_{2}\,(K)}}=\phi_{1}$, $\phi|_{T_{n}}=\phi_{2}$. Since
$E\subseteq\mathrm{Ker}(\phi)$, $\phi$ induces the homomorphism
$\overline{\phi}:({\mathrm{GL}_{2}\,(K)}\ast_{B}T_{n})/E\to G$. Obviously,
$\overline{\phi}|_{\overline{\mathrm{GL}_{2}\,(K)}}=\psi_{1}$ and
$\overline{\phi}|_{\overline{T_{n}}}=\psi_{2}$. By the definition of the
amalgamated free product [20], we get

$\displaystyle({\mathrm{GL}_{2}\,(K)}\ast_{B}T_{n})/E\cong\overline{\mathrm{GL}_{2}\,(K)}\ast_{\overline{B}}\overline{\mathrm{Tr}_{n}}.$

Corollary 2 finishes the proof of the theorem. $\Box$

Recall that an automorphism
$f\in\mathrm{Aut}\,K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ is called
linearizable if there exists
$\phi\in\mathrm{Aut}\,K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ such that
$\phi^{-1}f\phi$ is linear.

######  Corollary 4.

Any automorphism of $K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ of finite order
is linearizable.

Proof. By Corollary 1 in [20, page 6] every element of
$\mathrm{Aut}\,K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ of finite order is
conjugate to an element of $\overline{\mathrm{GL}_{2}\,(K)}$ or
$\overline{T_{n}}$. Since $\overline{T_{n}}$ has no elements of finite order
over a field of characteristic zero, any automorphism of
$K[x^{n},x^{n-1}y,\ldots,xy^{n-1},y^{n}]$ of finite order is conjugate to an
element of $\overline{\mathrm{GL}_{2}\,(K)}$. $\Box$

Acknowledgments

The second author is grateful to Max-Planck Institute für Mathematik for
hospitality and excellent working conditions, where part of this work has been
done.

The second author is supported by the grant of the Ministry of Education and
Science of the Republic of Kazakhstan (project AP09261086).

We are grateful to Professors M. Gizatullin, L. Makar-Limanov, and M.
Zaidenberg for their helpful suggestions and comments.

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