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\markboth{\hfil On plane polynomial vector fields \hfil EJDE--2002/37}
{EJDE--2002/37\hfil M'hammed El Kahoui  \hfil}
\begin{document}

\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2002}(2002), No. 37, pp. 1--23. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  On plane polynomial vector fields and the Poincar\'e problem
 %
\thanks{ {\em Mathematics Subject Classifications:} 34C05, 34A34, 34C14.
\hfil\break\indent
{\em Key words:} Polynomial vector fields, Invariant
algebraic curves, Intersection numbers, \hfil\break\indent
Tjurina number, B\'ezout theorem.
\hfil\break\indent
\copyright 2002 Southwest Texas State University. \hfil\break\indent
Submitted November 20, 2001. Published May 6, 2002.} }
\date{}
%
\author{M'hammed El Kahoui}
\maketitle

\begin{abstract}
  In this paper we address the Poincar\'e problem, on plane
  polynomial vector fields, under some conditions on the nature
  of the singularities of invariant curves. Our main idea
  consists in transforming a given vector field of degree $m$
  into another one of degree at most $m+1$ having its invariant
  curves in projective quasi-generic position. This allows us to
  give bounds on degree for some well known classes of curves
  such as the nonsingular ones and curves with ordinary nodes.
  We also give a bound on degree for any invariant curve in terms
  of the maximum Tjurina number of its singularities and the
  degree of the vector field.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{definition}{Definition}[section]
\numberwithin{equation}{section}

\section{Introduction}

The study of algebraic invariant curves and integrating factors of
plane polynomial vector fields goes back at least to Darboux
\cite{darboux78} and Poincar\'e \cite{poincare28}. We refer the reader
to \cite{schlomiuk93a,schlomiuk93b,christopherllibre99} for an
interesting survey and historical remarks on the problem. For a
given polynomial vector field the question of finding invariant
algebraic curves reduces mainly to the so-called {\it Poincar\'e
problem} which consists in finding an upper bound on the degree of
such curves. Indeed, any time such bound is found for a given
vector field the question of finding its invariant curves can be
algorithmically solved by using linear algebra (see e.g.
\cite{christopher94a,maccallum97a,christopher96a}).

Solving the Poincar\'e problem, and hence finding invariant
curves, yields great advances in the algorithmic study of plane
polynomial vector fields. Darboux \cite{darboux78} showed that the
abundance of invariant algebraic curves of a plane polynomial
vector field ensures its integrability. More precisely, he proved
that a vector field of degree $m$ with a least
$\frac{m(m+1)}{2}+1$ invariant curves has a first integral. Later
on Jouanolou \cite{jouanolou79} showed that any degree $m$ plane
polynomial vector field with at least $\frac{m(m+1)}{2}+2$
invariant curves has a rational first integral. In this direction
Prelle and Singer studied in \cite{prellesinger82} another kind of
first integrals, namely {\it elementary first integrals}. They
proved that the existence of algebraic integrating factors is
necessary for the existence of elementary first integrals, and
that deciding about the existence of algebraic integrating factors
is the main question to be solved in order to decide about the
existence of elementary first integrals. Ten years later, Singer
\cite{singer92} used differential algebra techniques to study a
wider class of first integrals, namely the {\it Liouvillian first
integrals}, and he proved that they have elementary functions as
integrating factors.

It is well known from the work of Jouanolou that a plane
polynomial vector field has either a rational first integral or
finitely many invariant curves. This gives an indirect proof for
the existence of an upper bound for the degree of irreducible
invariant curves of a given vector field. As far as we know there
is actually no effective method to compute such bound for any
given vector field, even more the question promises to be hard.
For example, the related question of deciding whether the closure
of the set of vector fields with given degree and having invariant
curves is an algebraic set is still open (see e.g.
\cite{christopherllibre2000, linsneto2000} for more details on the
question). On the other hand, it is well known that the degree of
the given vector field is not enough in order to get control on
the maximal degree of its irreducible invariant curves. It is for
instance easy to find linear vector fields having rational first
integrals of arbitrarily high degree. Even more, its is
established in \cite{moulin2000} and \cite{christopherllibre2000a}
the existence of quadratic plane vector fields without rational
first integral and having invariant algebraic curves of any given
degree.

Partial answers to the Poincar\'e problem have been given in
recent years. All of them follow the same strategy, which consists
in finding an upper bound in terms of the degree of the vector
field under some additional conditions on its fixed points or on
the nature of the singularities the invariant curves have (see
e.g. \cite{llibrechavarriga2000,
carnicer94,cerveaulinsneto91,tsygvintsev2001,walcher2000a,campillocarnicer97}).

\subsection*{Outline of the paper}

In this paper we study the Poincar\'e problem from ``algebraic
geometry" point of view. For this purpose it is natural to state
the problem in the general setting of a commutative field of
characteristic zero. Our main idea consists in reducing the
problem , by means of projective transformations, to a situation
where invariant curves have no critical points at infinity. This
reduction has a double advantage: first it keeps the geometric
properties of the invariant curves. Secondly, it allows to use
some basic results of projective algebraic geometry such as
B\'ezout theorem.

The paper is structured as follows: in section \ref{sec:1}
we define the concept of curves in projective quasi-generic
position and we show explicitly how to transform projectively any
curve to a curve in such position. Section \ref{sec:2} is devoted
to show how to transform vector fields, without loss of control
on their degree, into vector fields having their invariant curves
in projective quasi-generic position. In section \ref{sec:3} we
apply the techniques developed in sections \ref{sec:1} and
\ref{sec:2} to the Poincar\'e problem. We recover the classical
bound given in the case of nonsingular curves and we give better
bounds than the known ones in the case of curves with ordinary
nodes. A bound in terms of the maximum Tjurina number of the
singularities of an invariant curve is also given in this section.

\subsection*{The setting of a commutative field of characteristic
zero}

Polynomial vector fields and invariant algebraic curves are
objects of algebraic nature. It is hence natural to study their
properties in the general setting of a commutative field of
characteristic zero. This gives as well some flexibility to our
study of vector fields; we shall for example see that this allows
to treat in the same way invariant curves and rational first
integrals (lemma \ref{firstintegral}). Another practical reason
lies in the study of parameterized vector fields, since any given
vector fields ${\cal X}\in \mathbb{R}[u,x,y]$, where
$u=(u_1,\ldots, u_r)$ is a list of parameters, can be viewed as
vector field over $\mathbb{R}(u)[x,y]$.

\subsection*{Notation}
Let $\mathbb{K}$ be a commutative field of characteristic zero and
$\overline{\mathbb{K}}$ its algebraic closure. Let $f$ be a squarefree
polynomial in $\mathbb{K}[x,y]$ and  ${\cal C}(f)$ be the affine plane algebraic
curve, over the field $\overline{\mathbb{K}}$, defined by the equation
$f(x,y)=0$.

 The zeros in $\overline{\mathbb{K}}^{2}$ of the ideal ${\cal I}(f,
{\partial}_yf)$ are called the {\it critical points} of the curve
${\cal C}(f)$ with respect to the projection on the $x$-axis. In the same
way, the zeros of the ideal ${\cal I}(f, {\partial}_xf)$ are
called the critical points of the curve ${\cal C}(f)$ with respect to the
projection on the $y$-axis. A critical point of the curve ${\cal C}(f)$
with respect to one of the projections is simply called a
critical point of the curve.

 The multiplicity of a point $(\alpha,\beta)$ of the curve
${\cal C}(f)$ is defined as the smallest integer $s$ such that
${\partial}^{i+j}_{x^{i}y^{j}}f(\alpha,\beta)=0$ for any $i+j<s$.
When $s\geq 2$ the point is called {\it singular}. The singular
points of the curve ${\cal C}(f)$ are the critical points for both of the
two projections.

If $(\alpha,\beta)$ is a point of multiplicity $s$ in the
curve ${\cal C}(f)$ then the Taylor expansion of $f$ around
$(\alpha,\beta)$ writes as
$$f(x,y)=f_s(x-\alpha,y-\beta)+\ldots +f_n(x-\alpha,y-\beta)$$ where the
$f_i$'s are homogeneous and $f_s\neq 0$. Since $f_s$ is
homogeneous and bivariate it factors over $\overline{\mathbb{K}}$ into a
product of linear polynomials. For each linear factor $\ell(x,y)$
of $f_s$ the equation $$\ell(x,y)=0$$ gives a tangent line to
${\cal C}(f)$ at $(\alpha,\beta)$. Thus a point of multiplicity
$s$ has $s$ tangent lines counted with multiplicities. In the case
where $f_s$ is squarefree the point $(\alpha,\beta)$ is called an
{\it ordinary multiple point} of the curve.

 If ${\cal I}$ is an ideal of $\mathbb{K}[x,y]$ then to each zero
$(\alpha,\beta)$ of ${\cal I}$ corresponds a local ring $\left
(\overline{\mathbb{K}}[x,y]/{\cal I}\right )_{(\alpha,\beta)}$ obtained by
localizing the ring $\overline{\mathbb{K}}[x,y]/{\cal I}$ at the maximal
ideal ${\cal I}(x-\alpha,y-\beta)$. When this local ring is finite
dimensional as $\overline{\mathbb{K}}$-vector space we say that
$(\alpha,\beta)$ is an isolated zero of ${\cal I}$ and the
dimension as vector space of the corresponding local ring is
called the {\it multiplicity} of $(\alpha,\beta)$ as zero of
${\cal I}$.

 An important particular case is when two curves ${\cal C}(f)$ and
${\cal C}(g)$ meet at a point $(\alpha,\beta)$ and have no
one-dimensional common component passing through this point. In
this case $(\alpha,\beta)$ is an isolated zero of the ideal
${\cal I}(f,g)$ and its multiplicity is called the {\it
intersection number} of the two curves at this point and denoted
by $\mathop{\rm I}(f,g,(\alpha,\beta))$. The intersection number
satisfies the inequality
\begin{equation}\label{internumber}\mathop{\rm I}(f,g,(\alpha,\beta))\geq
st\end{equation}
where $s$ (resp. $t$) is the multiplicity of $(\alpha,\beta)$ as
point of the curve ${\cal C}(f)$ (resp. ${\cal C}(g)$). The
equality holds if and only if the two curves have no common
tangent line at the considered point (see \cite{fulton} for more
details).

 When the two curves have no one-dimensional common component
then they meet at finitely many points. In this case the vector
space $\mathbb{K}[x,y]/{\cal I}(f,g)$ is finite dimensional and its
dimension is the sum of all the intersection numbers at the
common points of ${\cal C}(f)$ and ${\cal C}(g)$ in the affine plane. A
fundamental result of algebraic geometry, namely B\'ezout theorem
asserts that
\begin{equation}\label{bezout}
\dim_{\mathbb{K}}\mathbb{K}[x,y]/{\cal I}(f,g)\leq \deg(f)\deg(g)
\end{equation}
and the equality holds if and
only if the two curves have no common points at infinity.

 Another important case is the so-called {\it Milnor number}:
given a singular point $(\alpha,\beta)$ of a curve ${\cal C}(f)$,
it is easy to see that it is an isolated zero of the ideal ${\cal
I}({\partial}_xf,{\partial}_yf)$. Its multiplicity as zero of this
ideal is called the Milnor number of $(\alpha,\beta)$ as singular
point of ${\cal C}(f)$ and is denoted by $\mu(f,(\alpha,\beta))$. For any
singular point $(\alpha,\beta)$ we have the relation
\begin{equation}\label{milnor}
  \mathop{\rm I}(f,{\partial}_y f,(\alpha,\beta))=\mu(f,(\alpha,\beta))+r-1
\end{equation}
where $r$ is the multiplicity of $\beta$ as root of the univariate
polynomial $f(\alpha,y)$, see e.g. \cite{le73}.

 The multiplicity of a singular point $(\alpha,\beta)$ of the
curve ${\cal C}(f)$ as zero of the ideal ${\cal
I}(f,{\partial}_xf,{\partial}_yf)$
is called the {\it Tjurina number} of the singular point and is denoted by
$\tau(f,(\alpha,\beta))$.

 By {\it plane polynomial vector field} with coefficients in
the field $\mathbb{K}$, or often a vector field over $\mathbb{K}[x,y]$, we mean
a
$\mathbb{K}$-derivation ${\cal X }$ of the algebra $\mathbb{K}[x,y]$. Any plane
polynomial vector field ${\cal X}$ with coefficients in $\mathbb{K}$ can
be uniquely written in the form ${\cal
X}=p{\partial}_x+q{\partial}_y$, with $p,q \in \mathbb{K}[x,y]$. The
maximum of the degrees of $p$ and $q$ is called the degree of
${\cal X}$.

 Let us recall here that any vector field ${\cal X}$ extends
uniquely to the fractions field $\mathbb{K}(x,y)$ and as well to any
algebraic extension of it. In particular ${\cal X}$ extends
uniquely as $\overline{\mathbb{K}}$-derivation of $\overline{\mathbb{K}}[x,y]$.

 In the sequel we shall also be concerned with transcendent
extensions of the field $\mathbb{K}$. In such cases, given a transcendent
extension $\mathbb{F}$ of $\mathbb{K}$ the vector field ${\cal X}$ do not extend
in a unique way to $\mathbb{F}[x,y]$. However, it extends uniquely as
$\mathbb{F}$-derivation, and in all the rest this will be our default
extension of ${\cal X}$.

\subsection*{Invariant curves}

Let ${\cal X}=p{\partial}_x+q{\partial}_y$ be a plane polynomial
vector field with coefficients in $\mathbb{R}$ or $\mathbb{C}$ and
let
\begin{equation}\label{diffsystem}
\dot{x}=p(x,y), \qquad \dot{y}=q(x,y)
\end{equation}
be the differential system associated with it. We say that an
algebraic curve ${\cal C}(f)$ is an {\it invariant curve} of ${\cal X}$ if
any solution of the differential system (\ref{diffsystem})
starting out at a point of ${\cal C}(f)$ lies entirely in ${\cal C}(f)$.
Following
the famous Hilbert Nullstellenzatz this can be expressed in terms
of algebraic identities as
\begin{equation}\label{algebraicinvariance}
p{\partial}_x f+q{\partial}_y f=kf
\end{equation}
This last identity gives a way to define the concept of algebraic
invariant curves for polynomial vector fields with coefficients
in an abstract field $\mathbb{K}$.

\begin{definition} \rm
Let ${\cal X}=p{\partial}_x+q{\partial}_y$ be a plane
polynomial vector field with coefficients in $\mathbb{K}$. A nonconstant
polynomial $f\in {\mathbb{K}}'[x,y]$, where $\mathbb{K}'$ is a
field extension of $\mathbb{K}$, is called an algebraic invariant curve
of ${\cal X}$ if there exists a polynomial $k\in
\mathbb{K}'[x,y]$, called the cofactor of $f$, such that
$${\cal X}(f)=kf.$$
\end{definition}

 In the same way, a nonconstant rational function
$h=f/g$ with coefficients in an extension
$\mathbb{K}'$ of $\mathbb{K}$ is called a rational first integral of the
vector field ${\cal X}$ if
$${\cal X}(h)=0.$$ In this case it is easy to verify that $f$ and $g$ are
invariant algebraic curves of ${\cal X}$ with the same cofactor.

 The following lemma shows how rational first integrals can
be viewed as invariant algebraic curves over transcendent
extensions of the field of coefficients of the vector field.
\begin{lemma}\label{firstintegral} Let ${\cal X}$ be a plane polynomial vector
field with coefficients in the field $\mathbb{K}$. Then the following are
equivalent:

 i) The vector field ${\cal X}$ has a rational first integral.

 ii) There exists an invariant algebraic curve $f$ of ${\cal
X}$ such that the field generated over $\mathbb{K}$ by the coefficients
of $f$ is transcendent over $\mathbb{K}$, and $f$ has no nonconstant
factor belonging to $\mathbb{K}[x,y]$.

 In particular, any vector field ${\cal X}$ with coefficients
in a field $\mathbb{K}$ having a first integral has a first integral with
coefficients in $\mathbb{K}$.
\end{lemma}
\paragraph{Proof:} ``$i\Rightarrow ii$" Let $h=f/g$ be a
rational first integral, with coefficients in
$\mathbb{K}'\supseteq \mathbb{K}$, of the vector field ${\cal X}$ and
suppose that $\gcd(f,g)=1$. Then we have ${\cal X}(h)=0$ which
gives the relation $$g{\cal X}(f)-f{\cal X}(g)=0.$$ According to
the fact that $\gcd(f,g)=1$ and using Gauss lemma we get
$$\begin{array}{ll}{\cal X}(f)=kf\\{\cal X }(g)=kg\end{array}
$$
with $k\in \mathbb{K}'[x,y]$.

 Let $u$ be an indeterminate over $\mathbb{K}'[x,y]$ and let
$h_1=f-ug$. Then an easy computation shows that ${\cal
X}(h_1)=kh_1$. On the other hand, since $u$ is transcendent over
$\mathbb{K}'$ the field generated over $\mathbb{K}$ by the coefficients
of $h_1$ is transcendent over $\mathbb{K}$. The fact that $h_1$ has no
nonconstant factor in $\mathbb{K}[x,y]$ follows immediately from the
irreducibility of $h_1$ in $\mathbb{K}'[u,x,y.]$

 ``$ii \Rightarrow i$" Let $f$ be an invariant algebraic
curve of the vector field ${\cal X}$ such that the field
$\mathbb{K}'$ generated over $\mathbb{K}$ by the coefficients of $f$ is
transcendent over $\mathbb{K}$.

 Since $\mathbb{K}'$ is finitely generated over $\mathbb{K}$ we can
find $u_1,\ldots, u_s\in \mathbb{K}'$ such that
$\mathbb{K}(u_1,\ldots,u_s)$ is purely transcendent over $\mathbb{K}$ and
$\mathbb{K}'$ is algebraic of finite degree over
$\mathbb{K}(u_1,\ldots,u_s)$.

 On the other hand, let us remark that for any
$\mathbb{K}(u_1,\ldots,u_s)$-isomorphism $\sigma$ from $\mathbb{K}'$ into
its algebraic closure $\overline{\mathbb{K}'}$ the curve
$\sigma(f)$ is also invariant for the vector field ${\cal X}$.
Moreover, there are only finitely many such
$\mathbb{K}(u_1,\ldots,u_s)$-isomorphisms and their number equals the
dimension of $\mathbb{K}'$ as $\mathbb{K}(u_1,\ldots,u_s)$-vector space.

 By considering the product of all the $\sigma(f)$'s we get a
polynomial $h$ with coefficients in the field
$\mathbb{K}(u_1,\ldots,u_s)$ which is an invariant algebraic curve of
${\cal X}$. Without loss of generality we may suppose that $h$
has its coefficients in the ring $\mathbb{K}[u_1,\ldots,u_s]$. Indeed, it
suffices to multiply $h$ by the common denominator of its
coefficients.

 Let  $$h=\sum_{\vert \alpha \vert \leq
d}h_{\alpha}(x,y)u^{\alpha}$$
with $\alpha=(\alpha_1,\ldots,\alpha_s)$, $\vert \alpha
\vert=\alpha_1+\ldots +\alpha_s$ and
$u^{\alpha}=u_1^{\alpha_1}\ldots u_s^{\alpha_s}$.
  We have then $${\cal X}(h)=\sum_{\vert \alpha \vert \leq
d}{\cal X}(h_{\alpha})u^{\alpha}$$ and thus $$\sum_{\vert \alpha
\vert \leq d}{\cal X}(h_{\alpha})u^{\alpha}=k\sum_{\vert \alpha
\vert \leq d}h_{\alpha}u^{\alpha}$$ where $k$ is the cofactor of
$h$ as invariant curve of ${\cal X}$. This last equation should
show that $k$ depends on the $u_i$'s, but in fact by comparing the
total degrees with respect to the $u_i$'s in this last equality
it is easy to see that $k$ must be of degree $0$ as polynomial in
the $u_i$'s. Thus $k$ is a polynomial in $\mathbb{K}[x,y]$ and moreover we
have $${\cal X}(h_{\alpha})=kh_{\alpha}$$ for any $\alpha$.

 To find a rational first integral of the vector field ${\cal
X}$ it is enough to prove that for some $\alpha\neq
\alpha'$ the rational function
$h_{\alpha}/ h_{\alpha'}$ is
nonconstant. This last fact is true, otherwise the polynomial $h$
will be of the form $h=ch'$ with $c\in \mathbb{K}(u_1,\ldots,
u_s)$ and $h'\in \mathbb{K}[x,y] $, and thus $f$ will have a
nonconstant factor in $\mathbb{K}[x,y]$ and this will also be the case
for $f$.

 Let us remark that the rational first integral we have
constructed has its coefficients in the field $\mathbb{K}$. This same
proof shows that starting from a first integral of ${\cal X}$ we
can always construct another one with coefficients in the field
$\mathbb{K}$.
\hfill $\Box$

\begin{remark}\label{remark}\rm
When a vector field ${\cal X}$ over $\mathbb{K}[x,y]$
has no rational first integral then essentially the field
generated over $\mathbb{K}$ of any algebraic invariant curve is algebraic
of finite degree over $\mathbb{K}$, i.e. any invariant algebraic curve
$f$ of ${\cal X}$ writes as $f=cf_1$ where $c$ is a constant
lying in an extension of $\mathbb{K}$ and the coefficients of $f_1$
generate over $\mathbb{K}$ a finite degree extension. As by-product, in
any process of computation of invariant curves there is in
general no need to extend the field of coefficients $\mathbb{K}$.
\end{remark}

\section{Curves in quasi-generic position}\label{sec:1}

In this section we define the concept of curves in quasi-generic
position  and give some of their properties that will be needed
for our purpose. Several aspects of such curves are studied in
\cite{elkahoui96,elkahouithesis,elkahoui2001a,elkahoui2001b}.

\begin{definition} \rm
An affine plane algebraic curve ${\cal C}(f)$ is said in quasi-generic position
with
respect to the projection on the $x$-axis if the following
conditions hold:
\begin{description}
\item{i)} $ \deg (f) = \deg_y(f)$,

\item{ii)} the curve ${\cal C}(f)$ has no vertical tangent line at its
singular points,

\item{iii)} the curve ${\cal C}(f)$ has no inflexion point with vertical
tangent line.
\end{description}
Moreover if the curve ${\cal C}(f)$ has no critical points at infinity
with respect to the projection on the $x$-axis then we say that
${\cal C}(f)$ is in projective quasi-generic position  with respect to the
projection on the $x$-axis. A curve ${\cal C}(f)$ is said in quasi-generic
position  (resp. in projective quasi-generic position) with
respect to the projection on the $y$-axis if the curve defined by
the polynomial $f(y,x)$ is in quasi-generic position  (resp. in
projective quasi-generic position) with respect to the projection
on the $x$-axis.
\end{definition}

 When the curve ${\cal C}(f)$ is in quasi-generic position  (resp. in
projective quasi-generic position) with respect to both of the
projections we simply call it a curve in quasi-generic position
(resp. in projective quasi-generic position). Let us notice that
if ${\cal C}(f)$ is in projective quasi-generic position then for
any nonconstant factor $f_1$ of $f$ the curve ${\cal C}(f_1)$ is
also in projective quasi-generic position.

 The previous definition slightly differs from the one given
in \cite{elkahoui96,elkahouithesis}, and the curves defined there
are called in generic position. The concept of curves in generic
position is cooked up exactly so that no overlapping occurs in
the projections, with respect to the coordinates axes, of the
critical points. More precisely, in addition to the conditions
given in the previous definition we add the condition that any two
distinct critical points have distinct coordinates. This last
condition will not be needed for our purpose and adding it will
make somewhat involved the proof of theorem
\ref{genericprojective}. For these reasons we did not include it
in our definition and changed the ``generic position" terminology
into ``quasi-generic position".

 The following lemma will be useful to make more explicit
projective quasi-generic position.

\begin{lemma}\label{dethom}
Let $n$ be a positive integer and $f_0, \ldots, f_n\in \mathbb{K}[x,y]$
be homogeneous polynomials such that $\deg (f_i) = n_i\leq n$.
Let $M(x)$ be the matrix of the $f_i$'s in the canonical basis
$\{1, y, \ldots, y^{n}\}$ of the $\mathbb{K}[x]$-module $\mathbb{K}[x]_n[y]$.
Then $${\det}(M(x)) = {\det}(M(1))x^N$$ where $N = (\sum_{i =
0}^n{n_i}) - \frac 12 n(n+1)$.
\end{lemma}

\paragraph{Proof:}
Let $f_j = \sum_{i = 0}^n{a_{i,j}x^{n_j - i}y^i}$,  with $a_{i,j}
= 0$ if $n_j < i$. The determinant of the matrix $M(x)$ can be
written in the form $${\det} (M(x)) = \sum_{\sigma\in S_{n +
1}}{\varepsilon(\sigma)a_{0,\sigma(0)} \ldots
a_{n,\sigma(n)}x^{n_{\sigma(0)}}x^{n_{\sigma(1)} - 1}\ldots
x^{n_{\sigma(n)} - n}}.$$ If we let $N = (\sum_{i = 0}^n{n_i}) -
\frac 12 n(n + 1)$ then we get ${\det}(M(x)) =
{\det}(M(1))x^N$.
\hfill $\Box$ \smallskip

 As consequence of lemma \ref{dethom} we have the
following proposition.

\begin{proposition} \label{progen} Let ${\cal C}(f)$ be an affine plane
algebraic curve
given by a degree $n$ polynomial $f$ and write $f = f_n + \ldots
+ f_p$, where $f_i$ is homogeneous of degree $i$. Then the
following are equivalent:
\begin{description}
\item{i)} the curve ${\cal C}(f)$ has no critical points at infinity.

\item{ii)} The polynomial $f_n$ is squarefree and does not have $x$ or
$y$ as factors.

\item{iii)} The discriminant ${\mathop{\rm Disc}}_y(f)$ (resp. $Disc_x(f)$) of
the polynomial $f$ with respect to $y$ (resp. $x$) has degree
$n(n-1)$.
\end{description}
\end{proposition}

\paragraph{Proof:} It is sufficient to prove the result for the critical
points with respect to the projection on the $x$-axis.

 Since $f$ is squarefree the polynomial $\mathop{\rm Disc}_y(f)$ vanishes
identically if and only if $x$ is a factor of $f$. Moreover, if
this is not the case the degree of ${\mathop{\rm Disc}}_y(f)$ equals the
number of critical points, with respect to the projection on the
$x$-axis, in the affine plane of the curve ${\cal C}(f)$ counted with
multiplicities of intersection. Thus the assertions i) and iii)
are equivalent according to B\'ezout theorem.

\noindent ``$i \Longrightarrow ii $" Since the curve ${\cal C}(f)$ has no
critical points at infinity the polynomials $f$ and $f_n$ are
monic with respect to the variable $y$, i.e. $\deg(f)=\deg_y(f)$
and $\deg(f_n)=\deg_y(f_n)$. One has then
$${\mathop{\rm Disc}}_y(f) = {\det}\left(y^{n-2}f, \ldots, yf, f,
y^{n-1}{\partial}_y f,
\ldots, {\partial}_y f\right)$$
and
 $${\mathop{\rm Disc}}_y(f_n) = {\det}\left( y^{n-2}f_n, \ldots, yf_n, f_n,
y^{n-1}{\partial}_y f_{n}, \ldots, {\partial}_y f_{n}\right).
$$
Using multilinearity of the function ${\det}$ and lemma
\ref{dethom} we obtain
\begin{equation}\label{equa}
\mathop{\rm Disc}_y(f) = {\mathop{\rm Disc}}_y(f_n) + R=
\mathop{\rm Disc}_y(f_n(1,y))x^{n(n-1)}+R
\end{equation}
where $\deg (R)\leq n(n-1)-1$.
 Since $\mathop{\rm Disc}_y(f)$ has degree $n(n-1)$ it is so for $f_n$
and this means that $f_n$ is squarefree. The fact that $x$ does
not divide $f_n$ follows from $\deg(f_n)=\deg_y(f_n)$.

\noindent ``$ii \Longrightarrow i$" If $f_n$ is squarefree and does not
have $x$ as factor then $\mathop{\rm Disc}_y(f_n)\neq 0$. This proves
that $\deg(\mathop{\rm Disc}_y(f))=n(n-1)$ according to the relation
(\ref{equa}).
\hfill $\Box$

\subsection{Transformation into curves in projective quasi-generic position}

In this subsection we shall show that for a curve ${\cal C}(f)$ and a
sufficiently "generic" projective transformation $T\in PGL(3,\mathbb{K})$
of the projective plane $\mathbb{P}^{2}\mathbb{K}$ the curve defined by
$F\circ T(x,y,1)$ is in projective quasi-generic position. In
fact the set of projective transformations $T$ such that the
curve defined by $F\circ T(x,y,1)$ is not in projective
quasi-generic position is a projective variety in $PGL(3,\mathbb{K})$,
but this result will not be needed in the sequel. For our purpose
we shall consider a generic projective transformation which is
defined in the following way:

Let $u_1,u_2,v,w$ be new indeterminates,
$\mathbb{F}=\mathbb{K}(u_1,u_2,v,w)$ and $\Gamma: \mathbb{F}[x,y]
\mapsto\mathbb{F}[x,y]$
be the map defined by
$$
 f(x,y)  \rightarrow
  F(u_1x+u_2y,-u_2x+u_1y,v(u_1x+u_2y)+w(u_1y-u_2x)+1)
 $$
The map $\Gamma$ is bijective and multiplicative (i.e
$\Gamma(fg)=\Gamma(f)\Gamma(g)$). Moreover, for any polynomial $f$
with coefficients in $\mathbb{F}$ the curves defined by $f$ and
$\Gamma(f)$, over an algebraic closure of $\mathbb{F}$, are projective
transformations one of the other. This shows in particular that
the two curves have the same geometric invariants (such as
multiplicities, Milnor and Tjurina numbers of singularities). On
the other hand, the geometric invariants of a curve defined by a
polynomial $f\in \mathbb{K}[x,y]$ do not depend on the algebraically
closed extension of $\mathbb{K}$ over which we view this curve. Thus, for
any polynomial $f$ with coefficients in $\mathbb{K}$ the curve defined by
$\Gamma(f)$ conserves the same geometric invariants as ${\cal C}(f)$
(defined over $\overline{\mathbb{K}}$) even though these two curves are
not defined over the same algebraically closed field.

 The following theorem relates the main feature of
considering the above generic projective transformation.

\begin{theorem}\label{genericprojective}
Let $f\in \mathbb{K}[x,y]$ be a squarefree polynomial
of degree $n$ defining a curve ${\cal C}(f)$, and $u_1,u_2,v,w$ be new
indeterminates. Then:
\begin{description}
\item{i)} The curve ${\cal C}(g)$ defined, over an algebraic
closure of $\mathbb{K}(u_1,u_2)$, by the polynomial
$g(x,y)=f(u_1x+u_2y,-u_2x+u_1y)$ is in quasi-generic position.
Moreover, if the leading homogeneous term of $f$ is squarefree
then the curve ${\cal C}(g)$ is in projective quasi-generic
position.

\item{ii)} The curve defined, over an algebraic closure of $\mathbb{F}$, by
the polynomial $\Gamma(f)$ is in projective quasi-generic
position.
\end{description}
\end{theorem}

\paragraph{Proof:} i) Let us write $f=\sum_i f_i$ where $f_i$ is homogneous of
degree
$i$. Then the polynomial $g$ writes as $g=\sum_ig_i$ where
$g_i=f_i(u_1x+u_2y,-u_2x+u_1y)$ is the homogeneous term of degree
$i$ of $g$.

 In particular, the leading homogeneous term of $g$ is
$f_n(u_1x+u_2y,-u_2x+u_1y)$. It is easy to see that this last
polynomial has degree $n$ with respect to both $x$ and $y$.

 Let us now prove that the singular points of ${\cal C}(g)$
have neither horizontal nor vertical tangent lines. For this, let
$[(\alpha_1,\beta_1),\ldots,(\alpha_s,\beta_s)]$ be the list of
singular points of the curve ${\cal C}(f)$.

 Then the list of singular points of the curve ${\cal C}(g)$
is given as $$
[\displaystyle{\frac{1}{u_1^{2}+u_2^{2}}}(u_1\alpha_1-u_2\beta_1,u_2\alpha_1+u_1\beta_1)
,\ldots,\displaystyle{\frac{1}{u_1^{2}+u_2^{2}}}(u_1\alpha_s-u_2\beta_s,u_2\alpha_s
+u_1\beta_s)].$$ On the other hand, let $(\alpha,\beta)$ be a
singular point of multiplicity $p$ in the curve ${\cal C}(f)$ and
let $(\alpha',\beta')=
\displaystyle{\frac{1}{u_1^{2}+u_2^{2}}}
(u_1\alpha-u_2\beta,u_2\alpha+u_1\beta)$.

 Let $f(x,y)=\sum_{i\geq p}h_i(x-\alpha,y-\beta)$ by the
Taylor expansion of $f$ around $(\alpha,\beta)$. Then the Taylor
expansion of $g$ around $(\alpha',\beta')$ writes
as
$$g(x,y)=\sum_{i\geq
p}h_i(u_1(x-\alpha')+u_2(y-\beta'),-u_2(x-\alpha')+
u_1(y-\beta')).
$$
 Since the homogeneous term
$h_p(u_1(x-\alpha')+u_2(y-\beta'),-u_2(x-\alpha')+
u_1(y-\beta'))$ is not divisible by $x-\alpha'$ or
$y-\beta'$ we deduce that the curve ${\cal C}(g)$ has
neither horizontal nor vertical tangent lines at the singular
point $(\alpha',\beta')$.

 Using similar arguments to the ones used for singular points
we can prove that the curve ${\cal C}(g)$ has no inflexion point
with vertical tangent line. Indeed, if
$[(\gamma_1,\delta_1),\ldots, (\gamma_s,\delta_s)]$ is the list of
inflexion points of the curve ${\cal C}(f)$ then the list of
inflexion points of the curve ${\cal C}(g)$ is given as $$
[\displaystyle{\frac{1}{u_1^{2}+u_2^{2}}}(u_1\gamma_1-u_2\delta_1,u_2\gamma_1+
u_1\delta_1)
,\ldots,\displaystyle{\frac{1}{u_1^{2}+u_2^{2}}}(u_1\gamma_s-u_2\delta_s,u_2\gamma_s
+u_1\delta_s)].
$$
 Suppose now that $f_n$ is squarefree. Since
$f_n(u_1x+u_2y,-u_2x+u_1y)$ is the leading homogeneous term of $g$
and $u_1,u_2$ are algebraically independent over $\mathbb{K}[x,y]$ we
deduce that its is squarefree and not divisible by $x$ or $y$.
Therefore, the curve ${\cal C}(g)$ is in projective quasi-generic
position according to proposition \ref{progen}.

 ii) To prove that ${\cal C}(\Gamma(f))$ is in projective
quasi-generic position it suffices to show that the leading
homogeneous term of $F(x,y,vx+wy+1)$ is squarefree.
 This leading homogeneous term is $$h_n=F(x,y,vx+wy).$$
According to the fact that $F$ is squarefree and $u_1,u_2,v,w$ are
algebraically independent over $\mathbb{K}[x,y]$ we deduce that $h_n$ is
squarefree and does not have $x$ or $y$ as factors. This proves
that ${\cal C}(\Gamma(f))$ has no critical points at infinity
according to proposition \ref{progen}.
\hfill $\Box$

\section{Transformation into vector fields with Invariant
curves in projective quasi-generic position}\label{sec:2}

In this section we shall construct a $\mathbb{K}$-linear map
$\Gamma^{\star}_{m}$
from the $\mathbb{K}$-vector space of vector fields of degree $\leq m$
over $\mathbb{K}[x,y]$ into the space of vector fields of degree $\leq
m+1$ over $\mathbb{F}[x,y]$ in such a way that a given vector field
${\cal X}$ has $f\in \mathbb{K}[x,y]$ as invariant curve if and only if
$\Gamma(f)$ is an invariant curve of the vector field
$\Gamma^{\star}_{m}({\cal X})$. As by-product all the invariant
algebraic curves of the vector field $\Gamma^{\star}({\cal X})$
are in projective quasi-generic position.

 For this aim. we need to define the family of maps
${\Gamma}_{n}: \mathbb{F}_{n}[x,y]  \mapsto\mathbb{F}_n[x,y]$,
$$ f(x,y)  \longrightarrow
 F_{n}(u_1x+u_2y,-u_2x+u_1y,v(u_1x+u_2y)+w(u_1y-u_2x)+1)
$$
where  $\mathbb{F}_{n}[x,y]$ denotes the $\mathbb{F}$-vector space of
polynomials
of degree at most $n$ and $F_n$ stands for the degree $n$
homogenization of $f$.
 Contrary to the map $\Gamma$ all the $\Gamma_n$'s are
$\mathbb{F}$-linear. Moreover, an easy computation shows that
$$\Gamma_n(f)=(v(u_1x+u_2y)+w(-u_2x+u_1y)+1)^{n-\deg(f)}\Gamma(f).
$$
This last relation shows in particular that
\begin{equation}\label{equationmap}\Gamma_{n_1+n_2}(f_1f_2)=\Gamma_{n_1}
(f_1)\Gamma_{n_2}(f_2)\end{equation} for any polynomials
$f_1,f_2$ with $\deg(f_1)\leq n_1$ and $\deg(f_2)\leq n_2$.

 We are now able to define in an explicit way the map
$\Gamma^{\star}_{m}$. Let ${\cal X}=p{\partial}_x+q{\partial}_y$
be a vector field of degree $\leq m$ over $\mathbb{K}[x,y]$. Then we
define $\Gamma^{\star}_{m}$  by $\Gamma^{\star}_{m}({\cal
X})=r{\partial}_x+ s{\partial}_y$ with
\begin{gather*}
r=\frac{(vu_1^2x+u_1+u_2^{2}vx)\Gamma_m(p)} {u_2^2+u_1^2}
+\frac{(wu_2^2x-u_2+u_1^2wx)\Gamma_m(q)} {u_2^2+u_1^2}\\
s=\frac{(vu_2^2y+u_2+u_1^2vy)\Gamma_m(p)} {u_2^2+u_1^2}
+\frac{(wu_1^2y+u_1+u_2^2wy)\Gamma_m(q)} {u_2^2+u_1^2}
\end{gather*}
 The exact relation concerning invariant curves between
$\Gamma^{\star}_{m}({\cal X})$ and ${\cal X}$ is given in the
following theorem.

\begin{theorem}\label{genericsituation}
Let ${\cal X}=p{\partial}_x+q{\partial}_y$
 be a vector field of degree $m$ over $\mathbb{K}[x,y]$. Then the
following assertions hold:
\begin{description}
\item{i)} The vector field $\Gamma^{\star}_{m}({\cal X})$ has
degree $m+1$ and for any polynomial $f$ of degree at most $n$ we
have
\begin{equation}\label{basicidentity}\Gamma^{\star}_{m}({\cal X})(\Gamma_n(f))=
\Gamma_{m+n-1}({\cal
X}(f))+n\Gamma_{m}(vp+wq)\Gamma_n(f)\end{equation}

\item{ii)} A rational function $\displaystyle{\frac{f}{g}}$ is a
first integral of ${\cal X}$ if and only if
$\displaystyle{\frac{\Gamma_n(f)}{\Gamma_n(g)}}$ is a first
integral of the vector field $\Gamma^{\star}_{m}({\cal X})$, where
$n=\max(\deg(f),\deg(g))$.

\item{iii)} A polynomial $f$ is an invariant curve of ${\cal X}$ if
and only if $\Gamma(f)$ is an invariant curve of
$\Gamma^{\star}_{m}({\cal X})$. In particular, if ${\cal X}$ has
no rational first integral then all the invariant algebraic
curves of $\Gamma^{\star}_{m}({\cal X})$ are in projective
quasi-generic position.
\end{description}\end{theorem}

\paragraph{Proof:} i) Since $u_1,u_2,v,w$ are indeterminates over
$\mathbb{K}[x,y]$ the map
$\Gamma_m$ sends $p$ and $q$ to polynomials of degree $m$. Thus
the vector field $\Gamma^{\star}_{m}({\cal X})$ has degree
exactly $m+1$.
 To simplify the proof of the identity we introduce the
following abbreviations
\begin{gather*}
X=u_1x+u_2y,\quad Y=-u_2x+u_1y, \\
Z=v(u_1x+u_2y)+w(-u_2x+u_1y)+z,\\
 G=F_n(X,Y,Z), \quad
A={({\partial}_{x}F_n})(X,Y,Z),\\
B={({\partial}_y F_n})(X,Y,Z), \quad
C={({\partial }_{z}} F_n)(X,Y,Z),\\
P^{\star}=P_m(X,Y,Z),\quad  Q^{\star}=Q_m(X,Y,Z)\\
R={{{(vu_1^2x+u_1z+u_2^{2}vx)P^{\star}}\over{u_2^2+u_1^2}}
+{{(wu_2^2x-u_2z+u_1^2wx)Q^{\star}}\over{u_2^2+u_1^2}}},\\
S={{{(vu_2^2y+u_2z+u_1^2vy)P^{\star}}\over{u_2^2+u_1^2}}
+{{(wu_1^2y+u_1z+u_2^2wy)Q^{\star}}\over{u_2^2+u_1^2}}}
\end{gather*}
Now let us consider the expression
$R{\partial}_xG+S{\partial}_yG$. On the first hand, by letting
$z=1$ in the last expression we get
$$(R{\partial}_xG+S{\partial}_yG)(x,y,1)=r{\partial}_x\Gamma_n(f)+s{\partial}_y\Gamma_n(f)
=\Gamma_{m}^{\star}({\cal X})(\Gamma_n(f)).$$ On the other hand,
we have
\begin{gather*}
    {\partial}_xG=u_1A-u_2B+(u_1v-u_2w)C \\
    {\partial}_yG=u_2A+u_1B+(u_2v+u_1w)C
\end{gather*}
which gives after computation (This factorization
can be obtained using a symbolic computation software such as Maple)
\begin{eqnarray*}
\lefteqn{ R{\partial}_xG+S{\partial}yG-zP^{\star}A-zQ^{\star}B}\\
&=&(vP^{\star}+wQ^{\star})\Big((wu_1y-u_2wx+u_1vx+u_2vy+z)C\\
&&+(u_1x+u_2y)A+(u_1y-u_2x)B\Big)
\end{eqnarray*}
The Euler formula for the homogeneous polynomial $G$ writes as
$$ nG=(u_1x+u_2y)A+(-u_2x+u_1y)B+((u_1x+u_2y)v+(-u_2x+u_1y)w+z)C.
$$
Thus $$R{\partial}_xG+S{\partial}yG-zP^{\star}A-zQ^{\star}B=
n(vP^{\star}+wQ^{\star})G.$$ By letting $z=1$ in the last equality
and taking into account the linearity of the $\Gamma_n$'s and the
equation (\ref{equationmap}) we finally get
$$\Gamma_{m}^{\star}({\cal X})(\Gamma_n(f))=\Gamma_{m+d-1}({\cal
X}(f))+\Gamma_{m}(k)\Gamma_n(f),$$ where $k=n(vp+wq)$.

\noindent ii) Without loss of generality we may suppose that
$n=\deg(f)$, we have thus $\Gamma_n(f)=\Gamma(f)$. Let
$n_1=\deg(g)$ and let us simplify the expression
$$\Gamma_{m}^{\star}({\cal X
})\left(\displaystyle{\frac{\Gamma_{n}(f)}{\Gamma_{n}(g)}}\right)
(\Gamma_{n}(g))^{2}.$$
 Using elementary rules of derivations we have
$$
\Gamma_{m}^{\star}({\cal X
})\left(\displaystyle{\frac{\Gamma_{n}(f)}{\Gamma_{n}(g)}}\right)
(\Gamma_{n}(g))^{2}=\Gamma_n(g)\Gamma_{m}^{\star}({\cal
X})(\Gamma(f))- \Gamma(f)\Gamma_{m}^{\star}({\cal
X})(\Gamma_n(g)).
$$
Now taking into account the identities
(\ref{equationmap}) and (\ref{basicidentity}) we obtain
\begin{eqnarray*}
\lefteqn{\Gamma_{m}^{\star}({\cal X
})\left(\displaystyle{\frac{\Gamma_{n}(f)}{\Gamma_{n}(g)}}\right)
(\Gamma_{n}(g))^{2} }\\
&=&\Gamma_n(g)\Gamma_{m+n-1}({\cal X}(f))-
\Gamma(f)\Gamma_{m+n_1-1}({\cal X}(g))\Gamma_{n-n_1}(1)\\
&&+\Gamma(fg)
((n-n_1)\Gamma_{m}(vp+wq)\Gamma_{n-n_1}(1)-\Gamma_{m}^{\star}({\cal
X })(\Gamma_{n-n_1}(1)) )
\end{eqnarray*}
On the other hand, using the identity (\ref{equationmap}) it is
easy to verify that
$$
\Gamma_n(g)\Gamma_{m+n-1}({\cal
X}(f))-\Gamma(f)\Gamma_{m+n_1-1}({\cal
X}(g))\Gamma_{n-n_1}(1)
=\Gamma_{m+2n-1}\big(g^{2}{\cal X}\big({\frac{f}{g}}\big)\big).
$$
Moreover, a direct computation shows that
$$
(n-n_1)\Gamma_{m}(vp+wq)\Gamma_{n-n_1}(1)-\Gamma_{m}^{\star}({\cal X
})(\Gamma_{n-n_1}(1))=0
$$
and then
$$
\begin{array}{ll}\Gamma_{m}^{\star}({\cal
X})\left(\displaystyle{\frac{\Gamma_{n}(f)}{\Gamma_{n}(g)}}\right)
(\Gamma_{n}(g))^{2}=\Gamma_{m+2n-1}\big(g^{2}{\cal
X}\big(\displaystyle{\frac{f}{g}}\big)\big).
\end{array}
$$
Thus $\Gamma_{m}^{\star}({\cal X}
)\left(\displaystyle{\frac{\Gamma_{n}(f)}{\Gamma_{n}(g)}}\right)=0$
if and only if ${\cal
X}\left(\displaystyle{\frac{f}{g}}\right)=0$ according to the fact
that $\Gamma_{m+2n-1}$ is one to one.

\noindent iii) Suppose that $f$ is a degree $n$ algebraic invariant
curve of the vector field ${\cal X}$. Then we have the equality
${\cal X}(f)=kf$ with $\deg(k)\leq m-1$, and applying
$\Gamma_{m+n-1}$ to it we get
$$\Gamma_{m+n-1}({\cal X}(f))=\Gamma_{m-1}(k)\Gamma(f).$$ Combining this last
equality with the identity (\ref{basicidentity}) we obtain
$$\Gamma^{\star}_{m}({\cal X})(\Gamma(f))=k_1\Gamma(f).$$ Conversely, suppose
that $\Gamma(f)$ is an invariant algebraic curve of the vector
field $\Gamma^{\star}_{m}({\cal X})$ and let us write
$\Gamma^{\star}_{m}({\cal X})=k_2\Gamma(f)$ with $\deg(k_2)\leq
m$.
 Since $\Gamma_m$ is bijective we can write
$k_2=\Gamma_m(k_3)$ with $\deg(k_3)\leq m$, and taking into
account the identity (\ref{basicidentity}) we get
$$ \Gamma_{m+n-1}({\cal X}(f))=\Gamma_m(k_4)\Gamma(f)
$$
with
$\deg(k_4)\leq m$.
 Since $\deg(\Gamma_{m+n-1}({\cal X}(f))\leq m+n-1$ and
$\deg(\Gamma(f))=n$ we have the bound $\deg(\Gamma_m(k_4))\leq
m-1$, and hence $\Gamma_m(k_4)=\Gamma_{m-1}(k_5)$ with
$\deg(k_5)\leq m-1$.
 This gives the relation $$\Gamma_{m+n-1}({\cal
X}(f))=\Gamma_{m+n-1}(k_5f)$$ and then $${\cal X}(f)=k_5f.$$

 Now assume that the given vector field ${\cal X}$ has no
rational first integral. Then the vector field has finitely many
invariant curves. Moreover, following remark \ref{remark}, and
without loss of generality, we may assume that all these invariant
curves have their coefficients in $\mathbb{K}$. Therefore, all their
transforms under $\Gamma$ are in projective quasi-generic
position, and these are exactly the invariant curves of the vector
field $\Gamma^{\star}({\cal X})$.
\hfill $\Box$

\subsection*{A property of the transformed vector field}

A remarkable property of the transformed vector field lies in the
fact that it has the straight line $v(u_1x+u_2y)+w(-u_2x+u_1y)+1$
as invariant curve. In this subsection we clear up this question
and give some of its consequences.

\begin{lemma}\label{straightline}
Let ${\cal X}=p{\partial}_x +q{\partial}_y$
be a degree $m$ vector field over $\mathbb{K}[x,y]$ and
$\Gamma^{\star}_m({\cal X})=r{\partial}_x+s{\partial}_y$. Then
the following holds:
\begin{description}
\item{i)} $yr_{m+1}-xs_{m+1}=0$.

\item{ii)} The straight line
$\ell(x,y)=v(u_1x+u_2y)+w(-u_2x+u_1y)+1$ is an invariant curve of
$\Gamma^{\star}_m({\cal X})$. If moreover $yp_m-xq_m=0$ then
$\ell$ is a common factor of $r$ and $s$.
\end{description}\end{lemma}

\paragraph{Proof:}i) An easy computation shows that
$$yr-xs=(-u_2x+u_1y)\Gamma_m(p)-(u_1x+u_2y)\Gamma_m(q).$$ This implies in
particular that $yr-xs$ is of degree at most $m+1$ and thus
$yr_{m+1}-xs_{m+1}=0$.

\noindent ii) The fact that $\ell$ is invariant follows immediately
from the relation (\ref{basicidentity}) by taking $f=1$ and $n=1$.
 Suppose now that $yp_m-xq_m=0$. First let us note that
\begin{gather*}
\Gamma_m(p)=\Gamma_m(p_m)+\ell g\\
\Gamma_m(q)=\Gamma_m(q_m)+\ell h
\end{gather*}
where $g,h\in \mathbb{K}[x,y]$. This gives the relation
\begin{eqnarray*}
(u_1^2+u_2^2)r=&{{{(vu_1^2x+u_1+u_2^{2}vx)\Gamma_m(p_m)}}}\\
&+{{(wu_2^2x-u_2+u_1^2wx)\Gamma_m(q_m)}}+g_1\ell.
\end{eqnarray*}
 On the other hand, a direct computation gives (Here
again we used Maple to carry out this division)
\begin{eqnarray*}
\lefteqn{(-u_2x+u_1y)(\displaystyle{{{(vu_1^2x+u_1+u_2^{2}vx)\Gamma_m(p_m)}}
+{{(wu_2^2x-u_2+u_1^2wx)\Gamma_m(q_m)}}}) }\\
&=&x\Gamma_m(q_m)(u_1^2+u_2^2)\ell+
(vu_1^2x+u_1+vu_2^2x)((u_1x+u_2y)\Gamma_m(q_m)\\
&&-(-u_2x+u_1y)\Gamma_m(p_m))
\end{eqnarray*}
Applying $\Gamma_{m+1}$ to the identity $yp_m-xq_m=0$ we get
$$
(u_1x+u_2y)\Gamma_m(q_m)-(-u_2x+u_1y)\Gamma_m(p_m)=0
$$
and thus $\ell$ divides the product
$$
(-u_2x+u_1y)(\displaystyle{{{(vu_1^2x+u_1+u_2^{2}vx)\Gamma_m(p_m)}}
+{{(wu_2^2x-u_2+u_1^2wx)\Gamma_m(q_m)}}}).
$$
Since on the other hand it is prime with $-u_2x+u_1y$ it must divide
the other factor. As by product the polynomial $\ell$ is a factor of $r$.
The case of $s$ can be done in the same way.
\hfill $\Box$ \smallskip

 As consequence of theorem \ref{genericsituation} and lemma
\ref{straightline} we have the following corollary which will be
very useful in the sequel.

\begin{corollary}\label{transformation}
Let ${\cal X}=p{\partial}_x +q{\partial}_y$
be a degree $m$ vector field over $\mathbb{K}[x,y]$. Let $\mathbb{K}'$ be
a field extension of $\mathbb{K}$ and $f_1,\ldots, f_t \in
\mathbb{K}'[x,y]$ be squarefree polynomials defining invariant
curves of the vector field ${\cal X}$. Then there exists a field
extension $\mathbb{F}$ of $\mathbb{K}'$, a projective transformation over
$\mathbb{F}$ sending $f_1,\ldots, f_t$ to $f'_1,\ldots,
f'_t$ and a vector field ${\cal Y}$ of degree $m+1$ over
$\mathbb{F}[x,y]$ such that:
\begin{description}
\item{i)} the curves defined by the $f'_i$'s are in
projective quasi-generic position,

\item{ii)} the curves defined by the $f'_i$'s are invariant
for the vector field ${\cal Y}$.
\end{description}
 Moreover, if $yp_m-xq_m=0$ then we can choose ${\cal Y}$ of
degree $m$.
\end{corollary}

\paragraph{Proof:}
Let ${\cal X}'$ be the extension of ${\cal X}$ to $\mathbb{K}'[x,y]$
obtained by viewing the elements of $\mathbb{K}'$ as constants.
We can then take $\mathbb{F}=\mathbb{K}'(u_1,u_2,v,w)$, $f'_i=\Gamma(f_i)$ and
${\cal Y}=\Gamma^{\star}_m({\cal X}')$.

 Suppose now that $yp_m-xq_m=0$ and let ${\cal
Y}=r{\partial}_x+s{\partial}_y$. Then following lemma
\ref{straightline} the linear polynomial
$\ell=v(u_1x+u_2y)+w(-u_2x+u_1y)+1$ is a common factor of $r$ and
$s$, say $r=\ell r_1$ and $s=\ell s_1$. By taking ${\cal
Y}'=r_1{\partial}_x+s_1{\partial}_y$ we obtain a vector
field fulfilling the required conditions.
\hfill $\Box$

\section{Application to the Poincar\'e problem}\label{sec:3}

In this section we shall apply the results and tools developed in sections
\ref{sec:1} and \ref{sec:2} to the study, under some assumptions
on invariant curves, of the Poincar\'e problem. The main idea in
this respect is to reduce the question to the case of invariant
curves in projective quasi-generic position, and then use
B\'ezout theorem. We first state some lemmas that we will need in
the sequel.

 Let ${\cal C}(f)$ be a plane algebraic curve given by a
squarefree polynomial $f$ and suppose that $x$ and $y$ are not
factors of $f$. In this case the ideals ${\cal I}(f,{\partial}_x
f)$ and ${\cal I}(f,{\partial}_y f)$ are zero-dimensional. For a
given point $(\alpha,\beta)$ of the curve ${\cal C}(f)$ if we
define $L_x$ (resp. $L_y$) to be the $\mathbb{K}$-linear map of the
multiplication by ${\partial}_xf$ (resp. ${\partial}_yf$) in the
$\mathbb{K}$-algebra $\left(\mathbb{K}[x,y]/{\cal I}(f,{\partial}_y f)\right
)_{(\alpha,\beta)}$ (resp. $\left(\mathbb{K}[x,y]/{\cal I}(f,{\partial}_x
f)\right )_{(\alpha,\beta)}$) its range is $\left({\cal I
}(f,{\partial}_x f,{\partial}_y f)/{\cal I}(f,{\partial}_y
f)\right )_{(\alpha,\beta)}$ (resp. $\left({\cal I
}(f,{\partial}_x f,{\partial}_y f)/{\cal I}(f,{\partial}_x
f)\right )_{(\alpha,\beta)}$), and $\left({\cal I }(f,{\partial}_y
f):{\partial}_x f/{\cal I}(f,{\partial}_y f)\right
)_{(\alpha,\beta)}$ (resp. $\left({\cal I }(f,{\partial}_x
f):{\partial}_y f/{\cal I}(f,{\partial}_x f)\right
)_{(\alpha,\beta)}$) is its kernel.

 Now applying the dimension formula, of vector spaces, to
$L_x$ (resp. $L_y$) we get the relations
\begin{gather}\label{lastequation1}
\mathop{\rm I}(f,{\partial}_x
f,(\alpha,\beta))=\dim_{\mathbb{K}}\left(\mathbb{K}[x,y]/{\cal I}(f,{\partial}_x
f):{\cal I}({\partial}_y f)\right
)_{(\alpha,\beta)}+\tau(f,(\alpha,\beta)) \\
\label{lastequation2}
\mathop{\rm I}(f,{\partial}_y
f,(\alpha,\beta))=\dim_{\mathbb{K}}\left(\mathbb{K}[x,y]/{\cal I}(f,{\partial}_y
f):{\cal I}({\partial}_x f)\right
)_{(\alpha,\beta)}+\tau(f,(\alpha,\beta))
\end{gather} This
implies in particular that any singular point of the curve ${\cal
C}(f)$ is a zero of both of the ideals ${\cal I}(f,{\partial}_x
f):{\cal I}(f,{\partial}_x f ,{\partial}_y f)$ and ${\cal
I}(f,{\partial}_y f):{\cal I}(f,{\partial}_x f ,{\partial}_y f)$.

\begin{lemma}\label{fixedpoints}
Let ${\cal X}=p{\partial}_x+q{\partial}_y$ be a
vector field of degree $m$ over $\mathbb{K}[x,y]$ and let ${\cal C}(f)$
and invariant curve of ${\cal X}$. Then the following hold:
\begin{description}
\item{i)} Any singular point of the curve ${\cal C}(f)$ is a fixed
point of ${\cal X}$, i.e. a zero of the ideal ${\cal I}(p,q)$.

\item{ii)} If $\gcd(p,q,f)=1$ then the number of singular points of
${\cal C}(f)$ do not exceed $(m+1)^{2}$. If moreover
$yp_m-xq_m=0$ then it is bounded by $m^{2}$.
\end{description} \end{lemma}

\paragraph{Proof:} i)
Without loss of generality we may assume that $x$ and $y$ are
not factors of $f$. Let $(\alpha,\beta)$ be a singular point of
${\cal C}(f)$. Then $(\alpha,\beta)$ is a zero of both of the
ideals ${\cal I}(f,{\partial}_x f):{\cal I}(f,{\partial}_x f
,{\partial}_y f)$ and ${\cal I}(f,{\partial}_y f):{\cal
I}(f,{\partial}_x f ,{\partial}_y f)$. Since $p\in {\cal
I}(f,{\partial}_y f):{\cal I}(f,{\partial}_x f ,{\partial}_y f)$
and $q\in {\cal I}(f,{\partial}_x f):{\cal I}(f,{\partial}_x f
,{\partial}_y f)$ we have $p(\alpha,\beta)=q(\alpha,\beta)=0$.

\noindent ii) The number of singular points of a curve is a projective
invariant. Thus according to corollary \ref{transformation} we
may assume that ${\cal C}(f)$ has no critical points at infinity
and ${\cal X}$ has degree $m+1$.
 Let $g$ be the $\gcd$ of $p$ and $q$ and let $p=gp_1$,
$q=gq_1$. From the relation $p{\partial}_x f+q{\partial}_y f$ we
deduce that $g$ divides $kf$. Moreover, according to the
assumption $\gcd(p,q,f)=1$ we have $\gcd(g,f)=1$ and then $g$
divides $k$.
 Therefore, the curve ${\cal C}(f)$ is invariant by the
vector field ${\cal X }_1=p_1{\partial}_x+q_1{\partial}_y$, and
following $i)$ the singular points of ${\cal C}(f)$ are zeros of
the zero-dimensional ideal ${\cal I}(p_1,q_1)$.

 Taking into account the inequalities $\deg(p_1)\leq m+1$,
$\deg(q_1)\leq m+1$ and according to B\'ezout theorem we get the
bound $(m+1)^{2}$. If moreover $yp_m-xq_m=0$ then following
corollary \ref{transformation} we can choose ${\cal X}$ of degree
$m$ and this gives the bound $m^{2}$.
\hfill $\Box$

\subsection*{The case of nonsingular invariant curves}\label{subsec:3.1}

In this subsection we prove that the degree of a nonsingular invariant
curve does not exceed $m+1$, where $m$ is the degree of the vector
field. This result has been obtained for the first time by Cerveau
and Lins Neto in \cite{cerveaulinsneto91} using foliations of the
projective plane. Other proofs, more or less involved, of the same
result can be found in
\cite{tsygvintsev2001,chavallibremoulin2001}. It is included here
because we can supply an elementary and purely algebraic proof.

\begin{theorem}\label{nsingularcase}
Let ${\cal X}=p{\partial}_x+q{\partial}_y$ be a degree $m$ vector
field over $\mathbb{K}[x,y]$ and suppose that in some extension
$\mathbb{K}'$ of $\mathbb{K}$ the vector field has a nonsingular
invariant curve of degree $n$ given by a squarefree polynomial
$f\in \mathbb{K}'[x,y]$. Then the following holds:
\begin{description}
\item{i)} The degree of $f$ do not exceed $m+1$. If moreover
$yp_m-xq_m=0$ then $n\leq m$.

\item{ii)} If the leading homogeneous term of $f$ is squarefree
then $n\leq m$.
\end{description} \end{theorem}

\paragraph{Proof:}
i) It is harmless to assume that $\mathbb{K}=\mathbb{K}'$. Indeed, we can
extend the vector field ${\cal X}$ to $\mathbb{K}'[x,y]$ by
viewing the elements of $\mathbb{K}'$ as constants. On the other
hand, According to corollary \ref{transformation} we can reduce
to the case where ${\cal X}$ has degree $m+1$ and ${\cal C}(f)$
is in projective quasi-generic position.

 From the relation $$p{\partial}_x f+q{\partial}_y f=kf$$ we
deduce that $p\in {\cal I}(f,{\partial}_y f): {\cal
I}(f,{\partial}_x f,{\partial}_y f)$. Moreover, the fact that
${\cal C}(f)$ is nonsingular implies that ${\cal I}(f,{\partial}_x
f,{\partial}_y f)=\mathbb{K}[x,y]$ and hence $$p\in {\cal
I}(f,{\partial}_y f).$$ Let $g$ be the greatest common divisor of
$p$ and ${\partial}_y f$ and let $p=gp_1$ and ${\partial}_y
f=gh$. We have then $$p_1\in {\cal I}(f,{\partial}_y f):g.$$ Now
let us prove that ${\cal I}(f,{\partial}_y f):g={\cal I}(f,h)$.
First, it is obvious that ${\cal I}(f,{\partial}_y f):g \supseteq
{\cal I}(f,h)$. On the other hand, let $h_1\in {\cal
I}(f,{\partial}_y f):g$ and let $a,b\in\mathbb{K}[x,y]$ such that
$$
gh_1=af+b{\partial}_y f.
$$
According to the last relation and to the fact
that $g$ divides ${\partial}_y f$ we deduce that $g$ divides the
product $af$. Since $f$ and ${\partial}_y f$ have no common
factors the polynomial $g$ must divide $a$, say $a=ga_1$.

 We have thus the relation $h_1=a_1f+bh$ which means that
$h_1\in {\cal I}(f,h)$. Therefore we have the inclusion
$${\cal I}(p_1,h)\subseteq {\cal I}(f,h)$$ which gives
$$
\dim_{\mathbb{K}}\mathbb{K}[x,y]/{\cal I}(p_1,h)\geq
\dim_{\mathbb{K}}\mathbb{K}[x,y]/{\cal I}(f,h).
$$
On
the other hand, according to the fact that ${\cal C}(f)$ is in
projective quasi-generic position and that $h$ is a factor of
${\partial}_y f$ we deduce that $f$ and $h$ have no common zeros
at infinity. Thus, by using B\'ezout theorem we get
$$\begin{array}{ll}\dim_{\mathbb{K}}\mathbb{K}[x,y]/{\cal I}(f,h)=n\deg(h),\\
\dim_{\mathbb{K}}\mathbb{K}[x,y]/{\cal I}(p_1,h)\leq
\deg(p_1)\deg(h).\end{array}
$$
this finally gives $n\leq
\deg(p_1)$ and then $n\leq m+1$ according to $\deg(p_1)\leq
\deg(p)=m+1$.

 if $yp_m-xq_m=0$ then according to corollary
\ref{transformation} we can reduce to the case of a curve in
projective quasi-generic position but with a vector field of
degree $m$. This gives the bound $n\leq m$.

\noindent ii) Suppose that the leading homogeneous term of $f$ is
squarefree. Then applying theorem \ref{genericsituation} with
$v=w=0$ and according to theorem \ref{genericprojective} we
reduce to the case where ${\cal X}$ has degree $m$ and ${\cal C}(f)$ is in
projective quasi-generic position. This gives the bound $n\leq m$.
\hfill $\Box$

\subsection*{The case of invariant curves with ordinary nodes}
\label{subsec:3.2}

This subsection deals with invariant algebraic
curves having only ordinary singular points. A bound in the case
of ordinary double points is given in \cite{llibrechavarriga2000}
and also in \cite{tsygvintsev2001}, and it is of order $2m$ where
$m$ is the degree of the vector field. Here we give a better bound
for this category of curves. A notable feature of the results we
present is that no irreducibility assumptions are needed.

\begin{theorem}\label{doublepointscase}
Let ${\cal X}=p{\partial}_x+q{\partial}_y$  be a degree $m$ vector
field over $\mathbb{K}[x,y]$ and suppose that in some extension
$\mathbb{K}'$ of $\mathbb{K}$ the vector field has an invariant curve of
degree $n$ given by a squarefree polynomial $f\in
\mathbb{K}'[x,y]$. If the curve ${\cal C}(f)$ has only ordinary double
points as singularities then the following holds:
\begin{description}
\item{i)} the degree of $f$ does not exceed $m+2$. If moreover
$yp_m-xq_m=0$ then $n\leq m+1$.

\item{ii)} If the leading homogeneous term of $f$ is squarefree
then $n\leq m+1$.
\end{description} \end{theorem}

\paragraph{Proof:}
As in the proof of theorem \ref{nsingularcase} we can reduce to
the case where ${\cal C}(f)$ is in projective quasi-generic position  and
${\cal X}$ has degree $m+1$. If moreover $yp_m-xq_m=0$ then
according to corollary \ref{transformation} we can choose ${\cal
X}$ of degree $m$. To treat the two cases at the same time let
$m_1=m$ if $yp_m-xq_m=0$ and $m_1=m+1$ otherwise.

 Let $g$ be the $\gcd$ of $f$ and $p$ and let $p=gp_1$ and
$f=gf_1$. From the relation $p{\partial}_x f+q{\partial}_y f=kf$
we deduce that $g$ divides $q{\partial}_y f$. Since $g$ is a
factor of $f$ and $\gcd(f,{\partial}_y f)=1$ the polynomial $g$
divides $q$, say $q=gq_1$. Therefore ${\cal C}(f_1)$ is an
invariant curve of the vector field ${\cal
X}_1=p_1{\partial}_x+q_1{\partial}_y$.

 On the other hand, since $p_1\in {\cal I}(f_1,{\partial}_y
f_1):{\cal I}(f_1,{\partial}_x f_1,{\partial}_y f_1)$ we have the
inclusion
$${\cal I}(p_1,f_1)\subseteq {\cal I}(f_1,{\partial}_y f_1):{\cal
I}(f_1,{\partial}_x
f_1,{\partial}_y f_1).$$ Let us note that ${\cal
I}(f_1,{\partial}_y f_1)$ and ${\cal I}(f_1,{\partial}_y
f_1):{\cal I}(f_1,{\partial}_x f_1,{\partial}_y f_1)$ have the
same zeros in $\overline{\mathbb{K}}^{2}$ but with eventually different
multiplicities.

 Let $(\alpha,\beta)$ be a zero of ${\cal I}(f_1,{\partial}_y
f_1)$. Then one of the following cases occurs:
\\
Case 1: The point $(\alpha,\beta)$ is nonsingular in the curve
${\cal C}(f_1)$. In such situation $\beta$ is a multiplicity $2$
root of the univariate polynomial $f_1(\alpha,y)$ according to the
fact that ${\cal C}(f_1)$ has no inflexion point with vertical
tangent line. This gives $\mathop{\rm I}(f_1,{\partial}_y
f_1,(\alpha,\beta))=1$ and then
$$\mathop{\rm I}(f_1,p_1,(\alpha,\beta))\geq\mathop{\rm I}(f_1,{\partial}_y
f_1,(\alpha,\beta))$$ since ${\cal C}(p_1)$ and ${\cal C}(f_1)$
meet at $(\alpha,\beta)$.
\\
Case 2:  $(\alpha,\beta)$ is an ordinary double point of the curve
${\cal C}(f_1)$. In this case, according to the fact that the
curve has no vertical tangent line at its singular points, we
have $\mathop{\rm I}(f_1,{\partial}_y f_1,(\alpha,\beta))=2$.

 On the other hand, the curves ${\cal C}(f_1)$ and ${\cal
C}(p_1)$ meet at the point $(\alpha,\beta)$ and from relation
(\ref{internumber}) we have $\mathop{\rm I}(f_1,p_1,(\alpha,\beta))\geq
2t\geq 2$, where $t$ is the multiplicity of $(\alpha,\beta)$ as
point of ${\cal C}(p_1)$. This gives as in the first case
$$\mathop{\rm I}(f_1,p_1,(\alpha,\beta))\geq\mathop{\rm I}(f_1,{\partial}_y
f_1,(\alpha,\beta)).
$$
 We have thus $$\sum_{(\alpha,\beta)}\mathop{\rm I}(f_1,p_1,
(\alpha,\beta))\geq \sum_{(\alpha,\beta)}\mathop{\rm I}(f_1,{\partial}_y
f_1,(\alpha,\beta))$$ where $(\alpha,\beta)$ ranges on the zeros
of ${\cal I}(f_1,{\partial}_y f_1)$ in the affine plane. Since the
curve ${\cal C}(f_1)$ has no critical points at infinity we have
by using relation (\ref{bezout})
$$\sum_{(\alpha,\beta)}\mathop{\rm I}(f_1,{\partial}_y f_1,
(\alpha,\beta))=\deg(f_1)(\deg(f_1)-1).
$$
 As ${\cal I}(p_1,f_1)$ may have other zeros than those of
${\cal I}(f_1,{\partial}_y f_1)$ we have by using once again
relation (\ref{bezout})
$$
\deg(p_1)\deg(f_1)\geq\sum_{(\alpha,\beta)}\mathop{\rm I}
(f_1,p_1, (\alpha,\beta)).
$$
This gives $\deg(p_1)+1\geq \deg(f_1)$ and then $m_1+1\geq n$
after adding $\deg(g)$ to both sides of the inequality.

\noindent ii) Suppose that the leading homogeneous term of $f$ is
squarefree. Then applying theorem \ref{genericsituation} with
$v=w=0$ and according to theorem \ref{genericprojective} we
reduce to the case where ${\cal X}$ has degree $m$ and ${\cal C}(f)$ is in
projective quasi-generic position. This gives the bound $n\leq
m+1$.
\hfill $\Box$

\begin{remark} \rm
Using theorem \ref{doublepointscase} we recover a
part of a result given in \cite{christopher94a} (see also
\cite{chrisllibrepantazi2000} theorem 1) concerning the case where
the invariant curves satisfy genericity conditions so that their
product gives a curve having no critical points at infinity and
only ordinary nodes as singularities.
\end{remark}

\subsection*{A bound in terms of the Tjurina number}\label{subsec:3.3}

This subsection concerns a bound on degree for an invariant curve in
terms of the degree of the vector field and an upper bound of the
Tjurina numbers of the singularities the curve has.

\begin{theorem}\label{Tjurinabound}
Let ${\cal X}=p{\partial}_x+q{\partial}_y$ be a degree $m$ vector
field over $\mathbb{K}[x,y]$ and suppose that in some extension
$\mathbb{K}'$ of $\mathbb{K}$ the vector field has an invariant curve of
degree $n$ given by a squarefree polynomial $f\in
\mathbb{K}'[x,y]$. Let $K$ be the maximum of the Tjurina numbers
of the singularities the curve has in the projective plane. Then
$$n\leq \displaystyle{\frac{(1+\sqrt{(1+4K)})(m+2)}{2}}$$
\end{theorem}

\paragraph{Proof:}
The Tjurina number is a projective invariant of the singularity.
Thus we can reduce to the case where ${\cal C}(f)$ is in
projective quasi-generic position and ${\cal X}$ has degree $m+1$.

 Let $g$ be the $\gcd$ of $p$ and $f$ and let $p=gp_1$,
$f=gf_1$. In the proof of theorem \ref{doublepointscase} we have
shown that $g$ divides $q$, say $q=gq_1$, and that $f_1$ is an
invariant curve of the vector field ${\cal
X}_1=p_1{\partial}_x+q_1{\partial}_y$.

 On the other hand, since $f_1$ is a factor of $f$ any upper
bound of the Tjurina numbers of the singularities of ${\cal
C}(f)$ is also an upper bound for those of ${\cal C}(f_1)$.
Moreover, the curve ${\cal C}(f_1)$ is also in projective
quasi-generic position.

From the inclusion ${\cal I}(p_1,f_1)\subseteq
{\cal I}(f_1,{\partial}_y f_1):{\cal I}(f_1,{\partial}_x
f_1,{\partial}_y f_1)$ we have
\begin{eqnarray*}
\deg(p_1)\deg(f_1)&\geq& \dim_{\mathbb{K}}\mathbb{K}[x,y]/{\cal I}(p_1,f_1)\\
&\geq& \dim_{\mathbb{K}}\mathbb{K}[x,y]/{\cal I}(f_1,{\partial}_y f_1):{\cal
I}({\partial}_x f_1).
\end{eqnarray*}
 According to the fact that ${\cal C}(f_1)$ is in projective
quasi-generic position and following equations
(\ref{lastequation2}) and (\ref{bezout}) we get
$$\
dim_{\mathbb{K}}\mathbb{K}[x,y]/{\cal I}(f_1,{\partial}_y f_1):{\cal
I}({\partial}_x
f_1)=\deg(f_1)(\deg(f_1)-1)-\sum_{(\alpha,\beta)}\tau(f_1,(\alpha,\beta)).
$$
 As $\gcd(p_1,q_1,f_1)=1$ then using lemma \ref{fixedpoints}
$ii)$ we obtain
$$\sum_{(\alpha,\beta)}\tau(f_1,(\alpha,\beta))\leq
(m+1-\deg(g))^{2}K.$$
 We have then the inequality $$(m+1-\deg(g))\deg(f_1)\geq
\deg(f_1)(\deg(f_1)-1)-(m+1-\deg(g))^{2}K .$$
 This gives the inequality
 $$\deg(f_1)\leq
\displaystyle{\frac{(1+\sqrt{(1+4K)})(m+2-\deg(g))}{2}}.
$$
Finally, by adding $\deg(g)$ to both sides of the inequality and
taking into account the fact that
$(1+\sqrt{1+4K})/2\geq 1$ we get the
required inequality.
\hfill $\Box$

\begin{remark} \rm
If we consider the case of a curve with ordinary
nodes as singularities then the previous theorem gives
$\frac{(1+\sqrt 5)}{2}(m+2)$ as bound while theorem
\ref{doublepointscase} gives $m+2$. This shows that the bound
given in theorem \ref{Tjurinabound} is not sharp.
\end{remark}

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\noindent\textsc{M'hammed El Kahoui}\\
 Department of Mathematics, Faculty of Sciences Semlalia\\
Cadi Ayyad University\\
P.O Box 2390, Marrakech, Morocco\\
e-mail {\tt elkahoui@ucam.ac.ma}

\end{document}