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\begin{document}
\title{Monomial ideals, binomial ideals,...}
\author{Bernard Teissier}
\date{}
%\pagestyle{myheadings}
%\markboth{\rm Bernard Teissier}{\rm A polynomial cannot vanish at all its exponents}
%\maketitle
\section{Introduction}
In these lecture, I hope to illustrate some ideas and results of commutative algebra by showing the form which they take for very simple ideals of
polynomial rings:  monomial or binomial, ideals, which can often be understood combinatorially. In the opposite direction, I will show that there are
methods wich allow one to prove results about fairly general ideals by continuously specializing them to monomial or binomial ideals. Topics will include
the theory of multiplicity and mixed multiplicities, integral dependence on ideals, the Brian\c con-Skoda theorem, resolution of singularities.
\par\medskip
Let $k$ be a field, and $k[u_1,\ldots ,u_d]$ (resp $k[[u_1,\ldots ,u_d]]$) the polynomial (resp. power series) ring in $d$ variables. \par\noindent If $d=1$,
given a monomial $u^n$ , if we try to deform it by adding lower degree terms, we get all unitary polynomials of degree $n$.
Adding higher degree terms, say in $k[[u]]$, we add only monomials which are divisible by $u^n$ and therefore obtain a series which can be written as
$u^nI(u)$ where $I(u)$ is a unit in
$k[[u]]$. The series $I(u)$ has an $n$-th root so that after a change of variable in $k[[u]]$, our series can again be written $u^n$. In fact a general result,
the Weierstrass preparation theorem says that any deformation of $u^n$ in $k[[u]]$ or ${\mathbf C}\{u\}$ obtained by adding monomials with
coefficients which are themselves formal or convergent power series \textit {without constant terms} in variables $(t_1,\ldots ,t_s)$ say is, up to a change
of variable in $k[[u]]$ (depending on the variables $(t_1,\ldots ,t_s)$) and multiplication by an invertible series in $(u,t_1,\ldots ,t_s)$, a unitary
polynomial in $u$ of degree $n$.
\par\medskip For $d=2$, working in $k[[u_1,u_2]]$ we meet a serious difficulty:  a series in two variables does not necessarily have a dominant term,
that is, a term which divides all the others. The simplest example is the binomial $u_1^a-cu_2^b$ with $c\in k^*$. As we shall see, if we allow enough
transformations, it is essentially the only example in dimension $2$. So if we want to understand the behavior near the origin of a series $f(u_1,u_2)$, we
cannot reduce it to something like $u_1^au_2^b$.\par\noindent 
In general, for $d>1$, we can try to measure how bad the situation is  with respect to divisibility by considering, for a given $f(u_1,\ldots ,u_d)\in
k[[u_1,\ldots ,u_d]]$, say
$f=\sum_m f_mu^m$, where
$m\in \N^d$ and $u^m=u_1^{m_1}\ldots u_d^{m_d}$, the ideal of $k[[u_1,\ldots ,u_d]]$ or $k[u_1,\ldots ,u_d]$ generated by the monomials
$\{u^m/f_m\neq 0 \}$ which actually appear in $f$. Since both rings are n\oe therian, this ideal is finitely generated in both cases, and we are faced
with the following problem:\par\noindent
\textbf{Problem}: Given an ideal generated by finitely many monomials (a \textit {monomial ideal}) in $k[[u_1,\ldots ,u_d]]$ or $k[u_1,\ldots ,u_d]$,
study how far it is from being principal, or from having the property that given any pair of its generators, one divides the other. \par\noindent
The second property implies principality (exercise), but is stronger (take an ideal in a Principal Ideal Domain, such as $\Z$ or $k[u]$); I shall call it
\textit{strong principality}.
\par\noindent Here we reach a bifurcation point in methodology: \par\noindent 
-one approach is to generalize the notion of divisibility by studying all linear relations,
with coefficients in the ambient ring, between our monomials. This leads to the construction of syzygies for the generators of our monomial ideal
${\mathcal M}$, or free resolutions for the quotient of the ambient ring by ${\mathcal M}$. There are many beautiful results in this direction. See
Eisenbud's lectures and [St]. One is also led to try and compare monomials using \textit{monomial orders} to produce Gr\"obner bases, since as soon as the
ideal is not principal, deciding whether a given element belongs to it becomes arduous in general. 
\par\noindent
\section{Strong principalization of monomial ideals by toric maps}
- Another approach is to try and force the ideal ${\mathcal M}$ to become principal after a change of variables. The first thing to try is changes of
variables which transform monomials into monomials, that is, which are themselves described by monomial functions:
$$
\begin{array}{llrcl}
u_1 &=& {y_1}^{a_{1}^1} \cdot\cdot\cdot\cdot {y_d}^{a_{1}^d} \\
 u_2 &=& {y_1}^{a_{2}^1} \cdot\cdot\cdot\cdot {y_d}^{a_{2}^d} \\
\cdot &&\cdot && \\
\cdot &&\cdot && \\
\cdot &&\cdot && \\
 u_d &=& {y_1}^{a_{d}^1} \cdot\cdot\cdot\cdot {y_d}^{a_{d}^d}
\end{array}
$$
where we can consider the exponents of $y_i$ appearing in the expressions of $u_1,\ldots ,u_d$ as the coordinates of a vector $a^i$ with integral
coordinates. These expressions decribe a monomial, or \textit{toric}, map of $d$-dimensional affine spaces
$$\pi(a^1,\ldots ,a^d)\colon \Aa^d(k)\to \Aa^d(k)$$ in the coordinates $(y_i)$ for the first affine space and $(u_i)$ for the second.\par\noindent
If we compute the effect of the change of variables on a monomial $u^m$, we see that $$u^m\mapsto y_1^{\langle a^1,m\rangle}\ldots y_d^{\langle
a^d,m\rangle}$$
\textbf{Exercise}: Show that the degree of the fraction field extension $k(u_1,\ldots ,u_d)\to k(y_1,\ldots ,y_d)$ determined by $\pi(a^1,\ldots ,a^d)$ is
equal to one (that is, our map $\pi(a^1,\ldots ,a^d)$ is birational) if and only if the determinant of the vectors $(a^1,\ldots ,a^d)$ is $\pm 1$. \par\noindent
\par\medskip\noindent
In view of the form of the transformation on monomials by our change of variables, it makes sense to introduce a copy of $\Z^d$ where the exponents of
our monomials inhabit, and which we will denote by $M$, and a copy of $\Z^d$ in which our vectors $a^j$ inhabit which we will call the \textit{weight
space} and denote by $W$. The lattices $M$ and $W$ are dual and we consider $W$ as the integral lattice of the vector space $\check\Rr^d$ dual to the
vector space $\Rr^d$ in which our monomial exponents live. In this manner, we think of $m\mapsto\langle a^i,m\rangle$ as the linear form on $M$
corresponding to $a^i\in W$.\par Given two monomials $u^m$ and $u^n$, the necessary and sufficient condition for the transform of $u^n$ to divide the
transform of $u^m$ in $k[y_1,\ldots ,y_d]$ is that $\langle a^i,m\rangle \geq\langle a^i,n\rangle\ \ \hbox{\rm for all}\ i,1\leq i\leq d $. If we read this
as
$\langle a^i,m-n\rangle\geq 0\ \hbox{\rm for all}\ i,1\leq i\leq d $ and seek a symmetric formulation, we are led to introduce the rational hyperplane
$H_{m-n}$ in $\check\Rr^d$ which is dual to the vector $m-n\in M$ and obtain the following:\par\noindent
\textit{ A necessary and sufficient condition for the transform of one of the monomials $u^m,u^n$ by the map $\pi (a^1,\ldots ,a^d)$ to divide the other in
$k[y_1,\ldots ,y_d]$ is that all the vectors $a^j$ lie on the same side of the hyperplane $H_{m-n}$ in $\check\Rr_+^d$}.\par\medskip\noindent
The transform of a monomial is nothing but its composition with the map $\pi (a^1,\ldots ,a^d)$ in the coordinates $(y_1,\ldots ,y_d)$.\par\noindent
Remark that the condition is non empty if and only if one of the monomials $u^m,u^n$ does not already divide the other in $k[u_1,\ldots ,u_d]$, because
to say that such divisibility does not occur is to say that the integral equation of the hyperplane $H_{m-n}$ does not have all its coefficients of the same
sign, and therefore meets the first quadrant $\check\Rr_+^d$, where our vectors $a^j$ live.\Par
To force one monomial to divide the other in the affine space $\Aa^d$ with coordinates $(y_i)$ is nice but not extremely useful since it provides
information on the original monomials only in the image of the map $\pi (a^1,\ldots ,a^d)$ in the affine space $\Aa^d$ with coordinates $(u_i)$, which is
contained in the complement of some coordinate hyperplanes. It is much more useful to find a proper and birational (hence surjective)  map
$\pi\colon Z\to
\Aa^d$ such that the compositions with
$\pi$ of our monomials generate a sheaf of ideals in $Z$ which is \textit{locally} principal; if you prefer, $Z$ should be covered by charts $U$ such that if
our monomial ideal ${\mathcal M}$ is generated by $u^{m^1},\ldots , u^{m^q}$, the ideal $(u^{m^1}\circ\pi,\ldots , u^{m^q}\circ\pi)\vert U$ is principal,
or strongly principal.\par\medskip
Toric geometry provides a way to do this: \par\noindent A rational \textit{fan} with 
 $\check\Rr_+^d$ is a finite collection $\Sigma$ of strictly convex rational
cones
$(\sigma_\alpha)_{\alpha\in A}$ with the following properties:\Par
1) The union of all the $(\sigma_\alpha)_{\alpha\in A}$ is the closed first quadrant $\check\Rr_+^d$ of $\check\Rr^d$.\Par
2) Each face of a $\sigma_\alpha\in \sigma$ is in $\Sigma$; in particular $\{0\}\in \Sigma$.\Par
3) each intersection $\sigma_\alpha\cap\sigma_\beta$ is a face of $\sigma_\alpha$ and of $\sigma_\beta$.\par\noindent
Recall that a cone is \textit{strictly convex} if it contains no positive dimensional vector subspace, and cones contained inside the first quadrant will
be strictly convex. A rational convex cone is one bounded by finitely many hyperplanes whose equations have rational (or equivalently, integral)
coefficients. An equivalent condition is that a rational convex cone is the cone positively generated by finitely many vectors with integral
coordinates. In general, the support of a fan is $\bigcup_\alpha\sigma_\alpha$.\par
A fan is sait to be \textit{regular} if each of its $k$-dimensional cones is generated by $k$ integral vectors (it is a simplicial cone) which form part of a
basis of the integral lattice. 
\Par If we go back to our monomial map, assuming that the determinant of the vectors $(a^1,\ldots ,a^d)$ is $\pm 1$, we can express the $y_j$ as
monomials in the $u_i$; the matrix of exponents will then be the inverse of the matrix $(a^1,\ldots ,a^d)$, and have some negative entries. Monomials
with possibly negative exponents will be called \textit{Laurent monomials} here.\Par If $\sigma=\langle a^1,\ldots ,a^d\rangle$, the cone positively
generated by the vectors $a^1\ldots ,a^d$, then the monomials in
$y_1,\ldots ,y_d$, viewed as Laurent monomials in $u_1,\ldots u_d$ via the expression of the $y_j$ as Laurent monomials in the $u_i$, correspond to the
integral points of the dual convex cone $$\check\sigma=\{x\in \Rr^d/\ell(x)\geq 0\ \forall \ell\in\sigma\}$$ of $\sigma$, i.e., those points $m\in \Z^d$
such that
$\langle a^i,m\rangle
\geq0\
\
\hbox{\rm for all}\ 1\leq i\leq d$. So we can identify the polynomial algebra $k[y_1,\ldots ,y_d]$ with the algebra $k[\check\sigma\cap M] $ of the
semigroup
$\check\sigma\cap M$ with coefficients in $k$. Since $\sigma$ is contained in the first quadrant of $\check\Rr^d$, its convex dual $\check\sigma$
contains the first quadrant of $\Rr^d$, so we have a graded inclusion of algebras
$$k[\Rr_+^d\cap M]=k[u_1,\ldots ,u_d]\subset k[\check\sigma\cap M]=k[y_1,\ldots ,y_d],$$ the inclusion being described by sending each variable $u_i$
to a monomial in $y_1,\ldots ,y_d$ as we did in the beginning.\par\noindent This slightly more abstract formulation has the following use: Given a fan in
$\check\Rr^d$, to each of its cones $\sigma$ we can associate the algebra
$ k[\check\sigma\cap M]$, even if $\sigma$ is not generated by $d$ vectors with determinant $\pm 1$.\Par By a lemma of Gordan [], it is finitely
generated, so it corresponds to an affine algebraic variety
$X_\sigma=\hbox{\rm Spec}k[\check\sigma\cap M]$. This variety is a $d$-dimensional affine space if and only if the cone $\sigma$ is $d$ dimensional
and generated by vectors which form a basis of the integral lattice of $\check\Rr^d$. It is, however, always normal and with rational singularities only
(see [TE]) and moreover it is rational, which means that the field of fractions of $k[\check\sigma\cap M]$ is $k(u_1,\ldots ,u_d)$. If two cones
$\sigma_\alpha$ and
$\sigma_\beta$ have a common face
$\tau_{\alpha\beta}$, the affine varieties $X_{\sigma_\alpha}$ and $X_{\sigma_\beta}$ can be glued up along the open set corresponding to
$X_{\tau_{\alpha\beta}}$ which they have in common. By this process, the fan $\Sigma$ gives rise to an algebraic variety over $\Aa^d$
$$\pi (\Sigma)\colon Z(\Sigma)\to \Aa^d$$
The variety $Z(\Sigma)$ is covered by affine charts corresponding to the $d$-dimensional cones $\sigma$ of $\Sigma$ and in each of these charts the
map $\pi(\Sigma)$ corresponds to the inclusion of algebras $k[u_1,\ldots ,u_d]\subset k[\check\sigma\cap M]$. If $\sigma$ is generated by $d$ vectors
with determinant $\pm 1$, i.e., which form a basis of the integral lattice, the second algebra is a polynomial ring $k[y_1,\ldots ,y_d]$ and the
inclusion is given by the monomial expression we started from.\Par Now, given two monomials
$u^m,u^n$, if we can find a fan
$\Sigma$ which is
\textit{compatible with the hyperplane
$H_{m-n}$} in the weight space in the sense that for each $\sigma\in\Sigma$ the intersection $\sigma\cap H_{m-n}$ is a face of $\sigma$, which implies
that each $\sigma$ is entirely on one side of $H_{m-n}$, or if you prefer that $H_{m-n}$ never meets the interior of a cone $\sigma\in \Sigma$ of
maximal dimension, then in each chart of $Z(\Sigma)$ the transform of one of our monomials will divide the other.\par\medskip\noindent 
Example: In dimension
$d=2$, let us seek to make one of the two monomials
$(u_1,u_2)$ divide the other after a monomial transformation. The hyperplane in the weight space is $w_1=w_2$; its intersection with the first quadrant
defines a fan whose cones are the two cones
$\sigma_1$ and
$\sigma_2$  generated by $a^1=(0,1),a^2=(1,1)$ and by $b^1=(1,1),b^2=(0,1)$ and their faces. The semigroup of integral points of $\check\sigma_1\cap
M$ is generated by $(1,0),(-1,1)$, which correspond respectively to the monomials $y_1=u_1,y_2=u_1^{-1}u_2$ and the semigroup of integral points of 
$\check\sigma_2\cap
M$ is generated by $(0,1),(1,-1)$, which correspond respectively to the monomials $y'_2=u_2,y'_1=u_1u_2^{-1}$. There is a natural isomorphism of the
open sets where $u_1\neq0,u_2\neq0$ and gluing up gives the two-dimensional subvariety of $\Aa^2\times{\mathbf P}^1$ defined by
$t_2u_1-t_1u_2=0$, where $(t_1:t_2)$ are the homogeneous coordinates on $\Pp^1$, with its natural projection to $\Aa^2$: it is the blowing-up of the
origin in $\Aa^2$. \par\medskip\noindent
Now if we have a finite number of distinct monomials $\neq 1$, say $u^{m^1},\ldots,u^{m^q}$, and if we can find a fan $\Sigma$ with support
$\check\Rr^d_+$ and compatible with \textit{all} the hyperplanes $H_{m^s-m^t},\ 1\leq s,t\leq q, s\neq t$, this will give us an algebraic (toric)
variety $Z(\Sigma)$, possibly singular and endowed with a map
$\pi(\Sigma)\colon
Z(\Sigma)\to \Aa^d$ such that the pull back by the map $\pi(\Sigma)$ of the ideal ${\mathcal M}$ generated by our monomials is strongly principal in
each chart.\par
Our collection of hyperplanes through the origin in fact defines a fan $\Sigma_0(F)$ which depends only upon the finite set $F=\{m^1,\ldots ,m^q\}$ of
elements of $\Z^d$: take as cones the connected components of the complement in
$\check\Rr^d_+$ of the union of all the hyperplanes: they are strictly convex rational cones because they lie in the first quadrant and are bounded by
hyperplanes whose equations have integral coefficients. Add all the faces of these cones, and we have a fan, of course not regular in general. To say
that a monomial ideal is locally strongly principal with respect to some functions (the generators of the algebra $k[\check\sigma \cap M]$) is not nearly
as useful when they do not form a system of coordinates as when they do. However, it is important to note that we first make our ideal $\mathcal M$
locally strongly principal by the map $\pi (\Sigma_0)\colon Z (\Sigma_0)\to \Aa^d$, and then resolve the singularities of $Z (\Sigma_0)$ by a toric map.
The second step corresponds to a refinement of $\Sigma_0$ into a regular fan $\Sigma$,  where refinement means that each cone of the
second fan in contained in a cone of the original.\Par This is always possible in view of the:\par 
\par\medskip\noindent Theorem (Kempf, Knudsen, Mumford, St. Donat, see [E])
\textit{ Given a rational fan, it is always possible to refine it into a regular fan.}\par\medskip\noindent
From this follows\par\medskip\noindent
\begin{theorem}  Let $k$ be a field. Given a monomial ideal ${\mathcal M}$ in $k[u_1,\ldots ,u_d]$, there exists a fan $\Sigma_0$ with support
$\check\Rr^d_+$ such that, given any regular refinement $\Sigma$ of $\Sigma_0$, the associated proper birational toric map of non singular toric
varieties
$$\pi(\Sigma)\colon\Z(\Sigma)\to \Aa^d(k)$$ has the property that the transform of ${\mathcal M}$ is strongly principal in each chart.\end{theorem}
\begin{remark} By construction, for each chart
$Z(\sigma)$ of $Z(\Sigma)$ there is an element of ${\mathcal M}$ whose transform generates the ideal ${\mathcal M}{\mathcal O}_{Z(\sigma)}$. This
element cannot be the same for all charts unless
${\mathcal M}$ is already principal. \end{remark} To see this, assume that there is a monomial $u^n$ whose transform generates ${\mathcal
M}{\mathcal O}_{Z(\Sigma)}$ in every chart. This means that every simplicial cone $\sigma$ of our fan with support $\Rr_+^d$ is in the positive side of all
the hyperplanes $H_{m-n}$ for all the other monomials
$u^m$ generating ${\mathcal M}$. But this is possible only if none of these hyperplanes meets the positive quadrant outside $(0)$, which means that
$u^n$ divides all the other $u^m$.
\section{Integral closure of ideals}
Given a finite set $F=\{m^1,\ldots ,m^q\}$ of elements of $\Z^d$, let us define its \textit{support function} as follows: it is the function
$\hbox{\rm h}_F\colon\check
\Rr^d\to \Rr$ defined by $$\hbox{\rm h}_F(\ell)=\hbox{\rm min}_{1\leq s\leq q}\ell (m^s).$$ It is a classical result on convexity that the convex hull of
$F$ is given by
$$\hbox{\rm Conv}F=\{n\in \Rr^d/\ell (n)\geq \hbox{\rm h}_F(\ell)\ \ \forall \ell\in\check\Rr^d\}.$$ It is often found in convexity books as saying that
a convex set is the intersection of the half-spaces determined by its support hyperplanes. The same proof shows that
$$\hbox{\rm Conv}\bigcup_{1\leq s\leq q}(m^s+\Rr_+^d)=\{n\in \Rr^d/\ell (n)\geq \hbox{\rm h}_F(\ell)\ \ \forall \ell\in\check\Rr_+^d\}.$$
\begin{lemma}\label{lin} The support function $\hbox{\rm h}_F$ is linear in each cone of the fan $\Sigma_0(F)$.\end{lemma}

 Let us choose a strongly principalizing map $\pi(\Sigma)\colon Z(\Sigma)\to\Aa^d(k)$ with $\Sigma$ a refinement of $\Sigma_0(F)$ and let $u^n$ be a
monomial in
$k[u_1,\ldots ,u_d]$. Given a chart $X_\sigma$ of $Z(\Sigma$, corresponding to $\sigma\in \Sigma$, a necessary and sufficient condition for
$u^n\circ\pi(\Sigma)$ to belong in $k[\check\sigma\cap M]$ to the ideal generated by the transforms of the generators of
${\mathcal M}$ is that we have $\ell (n)\geq \hbox{\rm h}_F(\ell)$ for all $\ell\in\sigma$: by Lemma \ref{lin}, we have for some $t\in\{1,\ldots ,q\}$
that
$\hbox{\rm h}_F(\ell)=\ell(m^t)$ for all $\ell\in\sigma$, and then by definition of $\check\sigma$ our inequality means that the quotient of the
transform of $u^n$ by the transform of $u^{m^t}$ is in $k[\check\sigma\cap M]$, which means that $u^nk[\check\sigma\cap M]\subset {\mathcal
M}k[\check\sigma\cap M]$. For this to be true in all charts it is necessary and sufficient, as we saw, that $n\in\hbox{\rm Conv}\bigcup_{1\leq s\leq
q}(m^s+\Rr_+^d)$. So we have finally, using a little sheaf-theoretic language:\begin{lemma} We have $u^n{\mathcal O}_{Z(\Sigma)}\in {\mathcal
M}{\mathcal O}_{Z(\Sigma)}$ if and only if
$n$ is in the convex hull of
$\bigcup_{1\leq s\leq q}(m^s+\Rr_+^d)$.\end{lemma}
Now one defines the integral closure of an ideal (a concept which goes back to Pr\"ufer or even Dedekind) as follows:
\begin{definition} An element $h$ of a commutative ring $R$ is integral over an ideal $I$ of $R$ if it satisfies an algebraic relation
$$h^k+a_1h^{k-1}+\cdots +a_k=0\ \ \ \ \ \hbox{\rm with}\ \ a_i\in I^i\ \ \ \hbox{\rm for}\ 1\leq i\leq k.$$\end{definition}
It is not difficult to see that the set of elements integral over $I$ is an ideal $\overline I$ containing $I$ and contained in $\sqrt I$.
We have the following characterization in algebraic geometry, which follows from the Riemann extension theorem:
\begin{proposition} If $R$ is essentially of finite type over a field, for any proper and birational morphism $t\colon Z\to\hbox{\rm Spec}R$ such that $Z$
is normal and $I{\mathcal O}_Z$ is invertible, we have that $h$ is integral over $R$ if and only if $h{\mathcal O}_Z=h\circ t\in I {\mathcal
O}_Z$.\end{proposition}
From this follows the interpretation of the Lemma:
\begin{proposition} The integral closure of a monomial ideal generated by the monomials $u^{m^1},\ldots ,u^{m^q}$ is the monomial ideal generated by the
monomials with exponents in the convex hull of
$\bigcup_{1\leq s\leq q}(m^s+\Rr_+^d)$. \end{proposition}

\section{The monomial Brian\c con-Skoda Theorem}
\begin{theorem}{\rm{Caratheodory's Theorem:}} Let $E$ be a subset of $\Rr^d$; every point of the convex hull of $E$ is in the convex hull of $d+1$ points
of
$E$.\end{theorem}
\begin{proposition}\label{FHH} {\rm (Fenchel ([F]), Hanner-H\aa dstr\"om ([H-H]))} 
Let $E\subset {\bf R}^d$ be a subset having at most $d$ connected components. Any point of the convex hull of $E$
is in the convex hull of $d$ points of $E$.\end{proposition}
\begin{proof}(after [H-H]): Assume that a point $m$ of the convex hull is {\it not} in the convex hull of any subset of $d$ points
of $E$. By Caratheodory's theorem, $m$ is in the convex hull $\tau\subset {\bf R}^d$ of $d+1$
points of $E$; if these $d+1$ points were not linearly independent, the point $m$ would be in the convex hull of the intersection of $E$ with a hyperplane
and we could apply Caratheodory's theorem in a space of dimension
$d-1$ and contradict our assumption, so the convex hull $\tau$ of the $d+1$ points is a $d$-simplex. Let us choose
$m$ as origin, and denote by
$(q_0,\ldots ,q_d)$ the vertices of
$\tau$. We have therefore
$0=\sum_0^dr_iq_i$ with
$r_i\geq 0$ and $\sum_0^dr_i=1$. Our assumption that $0$ is not the barycenter of $d$ points implies that $0$ is in the interior of $\tau$, that is, $r_i>0$
for $0\leq i\leq d$. Consider the simplex
$-\tau$ and the cones with vertex
$0$ drawn on the faces of
$-\tau$. Since the $r_i$ are $>0$, we can reinterpret the expression of
$0$ as a barycenter of the $q_i$ to mean that each $q_i$ is in the cone with vertex $0$ generated by the vectors $-0q_j$ for $j\neq i$; thus 
 each of these cones drawn from $0$ on the faces of $-\tau$ contains a point of $E$, namely one of the $q_i$. The union of their closures is ${\bf R}^d$
because
$-\tau$ is a $d$-simplex, and no point of $E$ can be on the boundary of one of these cones; if such was the case this point,
together with $d-1$ of the vertices of $\tau$, would define a $(d-1)$-simplex with vertices in $E$ and containing $0$, a
possibility which we have excluded. Therefore these $d+1$ cones divide $E$ into $d+1$ disjoint non empty parts, and
$E$ must have at least $d+1$ connected components.\end{proof}
Now let us remark that, given finitely many points $m^1,\ldots ,m^q$ in the positive quadrant $\Rr_+^d$, the set $E=\bigcup_{s=1}^q(m^s+\Rr_+^d)$ is
connected. Indeed, by definition, each point of this set is connected by a line to at least one of the points $m^s$, and any point of $\Rr_+^d$ having each
of its coordinates larger than the maximum over $s\in\{1,\ldots, q\}$ of the corresponding coordinate of the $m^s$ is in $E$ and connected by lines to all
the points $m^s$, so that any two of the points $m^s$ are connected in $E$.\Par
Now let $\sigma$ be a strictly convex rational cone in $\check\Rr^d$ and $\check\sigma\subset \Rr^d$ its dual. We need not assume that $\sigma$ is
regular, or even simplicial. Let $m^1,\ldots ,m^q$ be integral points in $\check\sigma$, corresponding to monomials $u^{m^1},\ldots ,u^{m^q}$ in the
algebra $k[\check\sigma\cap M]$. The integral closure $\overline{\mathcal M}$ in $k[\check\sigma\cap M]$ of the ideal ${\mathcal M}$ generated by
the monomials $u^{m^s}$ is the monomial ideal generated by the monomials $u^n$ such that $n$ is in the convex hull of the set
$E=\bigcup_{s=1}^q(m^s+\check\sigma)$. What we have just said about the connectedness of $E$ extends immediately.
\begin{theorem}{{\rm (Monomial Brian\c con-Skoda theorem)}} Let $\sigma$ be a strictly convex rational cone in $\Rr_{\geq 0}^d$. Given a monomial ideal
${\mathcal M}$ in
$k[\check\sigma\cap M]$, we have the inclusion of ideals
$$\overline{{\mathcal M}^d}\subset{\mathcal M}.$$
\end{theorem}
\begin{proof} Let $y_1,\ldots, y_N$ be a system of homogeneous generators of the graded $k$-algebra $k[\check\sigma\cap M]$ and let $y^{m^1},\ldots
,y^{m^q}$ be generators of
${\mathcal M}$ in
$k[\check\sigma\cap M]$. Set
$E=\bigcup_{1\leq s\leq q}(m^s+\check\sigma)\subset \check\sigma$. Thanks to Proposition
\ref{FHH} and the fact that $E$ is connected, it suffices to show that any point
$n\in
\check\sigma\cap M$ which is the barycenter of
$d$ points
$x_1,\ldots ,x_d$, each of which is the sum of $d$ points of $E$, is in $E$. But then $\frac{1}{d}n$ is also, as a barycenter of
barycenters of points of $E$, in the convex hull of $E$, and therefore, by Proposition \ref{FHH}, the barycenter of $d$ points of $E$. Let us write
$\frac{1}{d}n=\sum_{i=1}^dr_ie_i$ with $e_i\in E$,
$r_i\geq 0$ and $\sum_{i=1}^dr_i=1$. At least one of the $r_i$, say $r_1$, must be $\geq \frac{1}{d}$, so that $n\in e_1+\check\sigma\subset E$, which
proves the result. \end{proof}
Exercise: Prove by the same method that for each integer $\lambda\geq 1$ we have\par\noindent\centerline{ $\overline{{\mathcal
M}^{d+\lambda-1}}\subset{\mathcal M}^\lambda$.}
\begin{remark} It is not difficult to check that $\hbox{\rm conv}E=\bigcup_{k\in \N}\frac{kE}{k}$, where $kE$ means the Minkowski multiple, i.e., set of
all sums of $k$ elements of $E$, while dividing by $k$ meant homothety of ratio $\frac{1}{k}$. In fact, the inclusion $\bigcup_{k\in
\N}\frac{kE}{k}\subset \hbox{\rm conv}E$ is clear, and by Proposition \ref{FHH},  points with rational barycentric coordinates with respect to $d$ points
of $E$ lie dense in $\hbox{\rm conv}E$, and for each such point $x$ there is an integer $k$ such that $kx\in dE$. Thus, for every element of the convex
hull, there is a multiple $k$ such that
$kx\in kE\subset E$; the combinatorial avatar of the weak form of the Brian\c con-Skoda theorem which states that
$x\in\overline{\mathcal M}\implies x^d\in{\mathcal M}$  is the existence of a uniform bound for the $k$ such that $x\in\hbox{\rm conv}E\implies kx\in
E$, namely
$k=d$. There is a similar interpretation for the form given above. 
\end{remark}
The monomial Brian\c con-Skoda theorem follows from the results of [Li-T] in the case where ${\mathcal M}$ contains an ideal
generated by a regular sequence and with the same integral closure, since $k[\check\sigma\cap M]$ has only rational singularities (see [TE]) and hence is
pseudo-rational.
\section{Polynomial ideals and non degeneracy}
Let $$f=\sum_pf_pu^p$$ be an arbitrary polynomial or power series in $d$ variables with coefficients in the field $k$. Let $\hbox{\rm
Supp}(f)=\{p\in \Rr^d/f_p\neq 0\}$ be its \textit {support}. The affine Newton polyhedron of $f$ in the coordinates $(u_1,\ldots ,u_d)$ is the
boundary ${\mathcal N}(f)$ of the convex hull in
$\Rr^d_+$  of the support of $f$. The local Newton polyhedron is the boundary ${\mathcal N}_+(f)$ of $${\mathcal P}_+(f)=\hbox{\rm convex hull of}\
(\hbox{\rm Supp}(f)+\Rr^d_+).$$
It has finitely many compact faces and its non compact faces of dimension $\leq d-1$ are parallel to coordinate hyperplanes. Note that both
polyhedra depend not only on $f$ but also on the choice of coordinates. Remark also that the local Newton polyhedron has no interest if the
polynomial or series $f$ has a non zero constant term.\par We can define the affine (resp. local) support function associated with the function $f$ as
follows:in the affine case, it is the same definition as that given above, applied to the set of monomials appearing in $f$.
\par\noindent for the affine Newton polygon it is the function defined on $\check\Rr^d$ by
$$\hbox{\rm h}_{\mathcal N (f)}(\ell)=\hbox{\rm min}_{p\in \mathcal N (f)}\ell(p),$$
and for the local Newton polyhedron it is defined on the first quadrant $\check\Rr_+^d$ by
$$\hbox{\rm h}_{\mathcal N_+ (f)}(\ell)=\hbox{\rm min}_{p\in \mathcal N_+ (f)}\ell(p).$$
Both functions are piecewise linear in their domain of definition, which means that there is a decomposition of their respective domains of definition into
convex cones such that the corresponding functions are linear in each cone. These collections of cones are actually fans, in $\check \Rr^d$ and $\check
\Rr^d_+$ respectively. These fans are ''dual'' to the Newton polyhedra in the following sense:\par\noindent
Consider, say for the local polyhedron, the following equivalence relation between linear functions:
$$\ell\equiv \ell'\iff \{p\in \mathcal N_+ (f)\vert \ell (p)=\hbox{\rm h}_{\mathcal N_+ (f)}(\ell)\}=\{p\in \mathcal N_+ (f)\vert \ell' (p)=\hbox{\rm
h}_{\mathcal N_+ (f)}(\ell')\}$$
Then, the equivalence classes form a decomposition of the first quadrant into convex rational fans, and by definition the support function is linear in
each of them, given by $\ell\mapsto \ell (p)$ for any $p$ in the set $\{p\in \mathcal N_+ (f)\vert \ell (p)=\hbox{\rm h}_{\mathcal N_+ (f)}(\ell)\}$.
These sets are {\it faces} of the Newton polyhedron, and this establishes a one to one decreasing correspondence between the cones of the fan and the
faces of all dimensions of the Newton polyhedron. Note that the cones corresponding to non compact faces of the Newton polyhedron meet coordinate
hyperplanes outside the origin.\par
 \par\medskip\noindent
We have associated to each polynomial $f=\sum_pf_pu^p$ a fan in $\check\Rr^d$ corresponding to the global Newton polyhedron, and another one in
$\check\Rr^d_+$ corresponding to the local Newton polyhedron. The local polyhedron is also defined for a series $f=\sum_pf_pu^p$, and the
combinatorial constructions are the same. Let us for now restrict our attention to the local polyhedron, assuming
that $f_0=0$ and let us choose a regular refinement of the fan associated to it.\par\noindent 
By the definition just given, this means that for each cone $\sigma=\langle a^1,\ldots ,a^k\rangle$ of the fan, the primitive vectors $a^i$ form part of a
basis of the integral lattice, and all the linear forms $p\mapsto \langle a^i,p\rangle$, when restricted to the set $\{p/f_p\neq 0\}$, take their
minimum value on the same subset, which is a face, of the (local) Newton polyhedron of $f=\sum_pf_pu^p$. This face may or may not be compact.\par
Let us examine the behavior of
$f$ under the map
$\pi (\sigma)\colon Z(\sigma)\to
\Aa^d$ corresponding to a cone
$\sigma=\langle a^1,\ldots ,a^d\rangle\subset \check\Rr^d_{\geq 0}$ of a regular fan which is a subdivision of the fan associated to the local polyhedron of
$f$. If we write
$\hbox{\rm h}$ for
$\hbox{\rm h}_{\mathcal N_+(f)}$ we get 
$$\begin{array}{lr}f\circ\pi(\sigma)=\cr \sum_pf_py_1^{\langle a^1,p\rangle}\ldots y_d^{\langle a^d,p\rangle}=y_1^{h(a^1)}\ldots
y_d^{h(a^d)}\sum_pf_py_1^{\langle a^1,p\rangle- h(a^1)}\ldots y_d^{\langle
a^d,p\rangle -h(a^d)}\end{array}$$
The last sum is by definition the strict transform of $f$ by $\pi (\sigma)$.\par
I leave it as an exercise to check that:\par\noindent
a) In each chart $Z(\sigma)$ the exceptional divisor consists (set-theoretically) of the union of those hyperplanes $y_j=0$ such that $a^j$ is not a basis
vector of $\check \Z^d$.\par\noindent
b) Provided that no monomial in the $u_i$ divides $f$, the hypersurface $$\sum_pf_py_1^{\langle a^1,p\rangle- h(a^1)}\ldots y_d^{\langle
a^d,p\rangle -h(a^d)}=0$$ is indeed the strict transform by the map
$\pi (\sigma)\colon Z(\sigma)\to \Aa^d(k)$ of the hypersurface $X\subset \Aa^d(k)$ defined by $f(u_1,\ldots ,u_d)=0$, in the sense that it is the closure in
$Z(\sigma)$ of the image of $X\cap (k^*)^d$ by the isomorphism of the tori of the two toric varieties induced by $\pi (\sigma)$ as well as in the sense that it
is obtained from $f\circ\pi(\sigma)$ by factoring out as many times as possible the equations of the components of the exceptional divisor.
\par\medskip\noindent Let us denote by
$\tilde f$ the strict transform of $f$ and note that by construction it has a non zero constant term: the cone $\sigma$ is of maximal dimension, which means
that there is a unique exponent
$p$ such that
$\langle a,p\rangle=h(a)$ for
$a\in\sigma$. If we consider the map $\pi (\tau)$ associated to a face $\tau$ of $\sigma$, it coincides with the restriction of $\pi (\sigma)$ to an
open set $Z(\tau)\subset Z(\sigma)$ which is of the form $y_k\neq 0$ for $k\in K$ where $K$ depends on $\tau\subset \sigma$.
 \par Now we can, for each cone
$\sigma$ of our regular fan, stratify the space
$Z(\sigma)$ in such a way that
$\pi (\sigma)^{-1}(0)$ is a union of strata. Let
$I$ be a subset of $\{1,2,\ldots ,d\}$ and define $S_I$ to be the constructible subset of $Z(\sigma)$ defined by $y_i=0\ \hbox{\rm for}\ i\in I,\
y_i\neq 0\ \hbox{\rm for}\ i\notin I$. The $S_I$ for $I\subset\{1,2,\ldots ,d\}$ constitute a partition of $Z(\sigma)$ by non singular varieties,
constructible in $Z(\sigma)$. \par\noindent
If we restrict the strict transform
$$\tilde f(y_1,\ldots ,y_d)=\sum_pf_py_1^{\langle a^1,p\rangle- h(a^1)}\ldots y_d^{\langle
a^d,p\rangle -h(a^d)}$$
to a stratum $S_I$, we see that in the sum representing $\tilde f(y_1,\ldots ,y_d)$ only the terms $f_py_1^{\langle a^1,p\rangle- h(a^1)}\ldots
y_d^{\langle a^d,p\rangle -h(a^d)}$ such that $\langle a^i,p\rangle- h(a^i)=0$ for $i\in I$ survive. But these equalities, since our fan
is a subdivision of the dual fan of the Newton polyhedron of $f$, are satisfied if and only if the exponent $p$ belongs to a face, say $\gamma_I$, of that
Newton polyhedron. Moreover, if we define
$$f_{\gamma_I}=\sum_{p\in\gamma_I}f_pu^p$$
to be the partial polynomial associated to the face $\gamma_I$, which is nothing but the sum of the terms of $f$ whose exponent is in the
face $\gamma_I$, we see that we have the fundamental equality
$$\tilde f\vert_{S_I}=\widetilde{f_{\gamma_I}}\vert_{S_I}$$
and we remark moreover that $\widetilde{f_{\gamma_I}}$ is a function on $Z(\sigma)$ which is independent of the coordinates $y_i;\ i\in I$, so that it is
determined by its restriction to $S_I$.\par Now to say that the strict transform $\tilde f=0$ in $Z(\sigma)$ of the hypersurface $f=0$ is transversal to the
stratum $S_I$ and is non singular in a neighborhood of its intersection with it is equivalent to saying that the restriction $\tilde f\vert_{S_I}$ of the
function $\tilde f$ defines,  by the equation $\tilde f\vert_{S_I}=0$, a non singular hypersurface of $S_I$. By the definition of
$S_I$ and what we have just seen, this in turn is equivalent to saying that the equation $\widetilde{f_{\gamma_I}}=0$ defines a non singular hypersurface
in the torus
$(k^*)^d$ of
$Z(\sigma)$, and this finally is equivalent to saying that $ f_{\gamma_I}=0$ defines a non singular hypersurface in the torus $(k^*)^d$ of the affine space
$\Aa^d(k)$ since
$\pi(\sigma)$ induces an isomorphism of the two tori. \par
Given a series $f=\sum_pf_pu^p$ and a weight vector $a\in \Rr_{\geq 0}^d$, the set $\{p\in \Z_{\geq 0}^d/f_p\neq 0\ \hbox{\rm and }\ \langle
a,p\rangle =h(a)\}$ is a face of the local Newton polyhedron of $f$, corresponding to the cone of the dual fan which contains $a$ in its relative interior. If
all the coordinates of the vector
$a$ are positive, this face is compact.\
\begin{definition} The series $f=\sum_pf_pu^p$ is \textit{ non degenerate} with respect to its Newton polyhedron in the coordinates $(u_1,\ldots
,u_d)$ if for every covector $a\in \Rr_+^d$ the function $f_a=\sum_{\{p/\langle a,p\rangle=h(a)\}}f_pu^p$ has no critical point in the torus
$(k^*)^d$.\end{definition}
\begin{remark} By definition of the faces of the Newton polyhedron and of the dual fan, the function $f_a$ is equal to the function $f_\gamma$, where
$\gamma$ is the face of the Newton polyhedron which is dual to the cone of the dual fan which contains the cone $\sigma$ of the regular subdivision
which itself contains the vector
$a$ in its relative interior. Since
$a$ is in $\Rr_+^d$, the non-degeneracy condition concerns only the restrictions $f_\gamma$ of $f$ to the \textit{compact} faces of the Newton
polyhedron, and these in turn correspond in each chart $Z(\sigma)$ to the strata $S_I$ of the canonical stratification which are contained in
$\pi(\sigma)^{-1}(0)$. Each stratum $S_I$ which is not contained in $\pi(\sigma)^{-1}(0)$ contains in its closure strata which are. \end{remark}
Since if $f$ is non-degenerate, its strict transform is non singular and transversal in each chart to the strata contained in $\pi(\sigma)^{-1}(0)$.




\section{Resolution of binomial varieties}
Let $k$ be a field. Binomial varieties over $k$ are irreducible varieties of the affine space $\Aa^d(k)$ which can, in a suitable coordinate system, be
defined by a binomial ideal, which means an ideal generated by binomials, i.e., expressions of the form $U^m-\lambda_{mn}U^n$ with $\lambda_{mn}\in
k^*$. Binomial ideals were studied in [E-S] and if the field $k$ is not algebraically closed, the study becomes more complicated. Here I will assume
throughout that $k$ is algebraically closed. It is natural to study the behavior of binomial ideals under toric maps. Let $\sigma=\langle a^1,\ldots
,a^d\rangle$ be a regular cone in $\check \Rr^d_+$. The image of a binomial $u^m-\lambda_{mn}u^n$ by the map $k[u_1,\ldots ,u_d]\to k[y_1,\ldots
,y_d]$ determined by $u_i\mapsto y_1^{a^1_i}\ldots y_d^{a^d_i}$ is given by: $$u^m-\lambda_{mn}u^n\mapsto y_1^{\langle a^1,m\rangle}\ldots
y_d^{\langle a^d,m\rangle}-\lambda_{mn}y_1^{\langle a^1,n\rangle}\ldots y_d^{\langle a^d,n\rangle}.$$
In general this only tells us that the transform of a binomial is a binomial, which is no news since by definition of a toric map the transform of a monomial
is a monomial. However, something interesting happens if we assume that the cone $\sigma$ is \textit{compatible} with the hyperplane $H_{m-n}$ which is
the dual in the space of weights of the vector $m-n$ of the space of exponents, in the following sense:
\begin{definition} A cone $\sigma$ is compatible with a cone $\sigma'$ if $\sigma\cap\sigma'$ is a face of each.\end{definition}
Let us assume that the binomial hypersurface $u^m-\lambda_{mn}u^n=0$ is irreducible; in particular, no variable $u_j$ appears in both monomials, and
the vector $m-n$ is primitive. In the sequel, I will tacitly assume this.\par If the convex cone
$\sigma$ is compatible with the hyperplane
$H_{m-n}$, it is entirely inside one of the closed half-spaces determined by
$H_{m-n}$. This means that all the non zero $\langle a^i,m-n\rangle$ have the same sign, say positive. It also means that, if we renumber the vectors $a^i$
in such a way that $\langle a^i,m-n\rangle=0$ for $1\leq i\leq t$ and $\langle a^i,m-n\rangle>0$ for $t+1\leq i\leq d$, we can write the transform of our
binomial in the following way:
$$u^m-\lambda_{mn}u^n\mapsto y_1^{\langle a^1,n\rangle}\ldots
y_d^{\langle a^d,n\rangle}(y_{t+1}^{\langle a^{t+1},m-n\rangle}\ldots
y_d^{\langle a^d,m-n\rangle}-\lambda_{mn}).$$
And we can see that the strict transform $y_{t+1}^{\langle a^{t+1},m-n\rangle}\ldots
y_d^{\langle a^d,m-n\rangle}-\lambda_{mn}=0$ of our hypersurface in the chart $Z(\sigma)$ is non singular!\par\noindent

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\end{thebibliography}

%-----------------------------------------------
\bigskip\noindent
Institut math\'ematique de Jussieu,\newline 
UMR 7586 du C.N.R.S.,\newline
Universit\'es Paris 6-Pierre et Marie Curie et Paris 7-Denis Diderot,\newline 
Case 7012, 2 Place Jussieu,\newline
75251 Paris Cedex 05, France\newline
teissier@math.jussieu.fr
\end{document}

 Apparently the first person to successfully tackle this difficulty is Newton, who was trying to solve
the equation
$f(u_1,u_2)=0$ in the case where both derivatives vanish at the origin, i.e., you cannot use the implicit function theorem. Newton introduced the Newton
polygon, which is a two-dimensional substitute for the order of a function of one variable. Let 
$$f(u_1,u_2)=\sum_{i,j}a_{i,j}u_1^iu_2^j\in k[[u_1,u_2]],$$
then the 

\section{NEWTON'S STUDY OF PLANE
CURVE SINGULARITIES} A general reference for this paragraph is [W]. In this section, I write $(x,y) $ for $(u_1,u_2)$. Let
$f(x,y)\in \C[[x,y]]$ be a formal power series without constant term. We seek series $y(x)$
without constant term such that $f(x,y(x))=0$. \par Let us first eliminate a marginal case; \par
 If $f(0,y)=0$, it means that $f(x,y)$
is divisible by some power of $x$; let $a$ be the maximum power of $x$
dividing $f(x,y)$, and let us set $f(x,y)=x^af'(x,y)$.
Geometrically, the equality $f(0,y)=0$ means that the curve $f(x,y)=0$
contains the $y$-axis, and the equality above means that this axis should be
counted $a$ times in the curve. This component may be parametrized by 
$x=0, y=t$ and we are left with the problem of parametrizing the rest of the
curve, which is defined by $f'(x,y)=0$. We now have $f'(0,y)\neq 0$, and we
may thus reduce to the case $f(0,y)\neq 0$. From now on we shall assume
that $f(0,y)\neq 0$. \par
Considering series $f(x,y)$ of the form $y^p-x^q$ with
$p,q>1 \ \hbox{and}\  (p,q)=1$ shows that one cannot hope to find series in
powers of $x$. Newton's idea is to seek solutions which are \it fractional
power series \rm in $x$, that is, he seeks series in $x^{1\over m}$ for some
integer $m$, say $\phi (x^{1\over m})\in \C[[x^{1\over m}]]$ such
that $f(x,\phi(x^{1\over m}))=0$.\par\noindent More precisely he seeks
solutions of the form: $$y=x^\nu( c_0+\phi_0(x^{1\over m}))
\hskip1truecm \hbox{with}\ c_0\neq 0,\ \  \nu \in \Q_+,\ \ \phi_0\ 
\hbox{without constant term}.$$ If we write 
$$f(x,y)=\sum_{i,j\in \N}a_{i,j}x^iy^j\hskip1truecm \hbox{with} \
a_{0,0}=0$$ and substitute, we get $$\sum_{i,j}a_{i,j}x^{i+\nu
j}(c_0+\phi_0(x^{1\over m}))^j$$ and we seek $\nu$, $c_0\neq 0$ and
a series $\phi_0(x^{1\over m})$ such that this series
 is zero. In particular, its lowest order terms in $x$ must be zero. Since
$\phi_0$ has no constant term, if we denote by $\mu$ the minimum value
of $i+\nu j$ for $(i,j)$ such that $a_{i,j}\neq 0$, we have 
$$\sum_{i,j}a_{i,j}x^{i+\nu j}(c_0+\phi_0(x^{1\over m}))^j =x^\mu
\sum_{i+\nu j=\mu}a_{i,j}c_0^j +x^{\mu}h(x^{1\over m})
\hskip.2truecm\hbox{where}\ h\ \hbox{has no constant term}.$$
So $c_0$ must satisfy $$\sum_{i+\nu j=\mu}a_{i,j}c_0^j=0$$
For this equation to have a non-zero root in $\C$, it is necessary and sufficient that
the sum has more than one term.\par\noindent
We may then write, since $f(0,y)$ is a formal power series
in $y$, $f(0,y)=y^kg(y),\ \hbox{with}\ g(0)\neq 0$.\par The proof of the
existence of parametrizations proceeds by induction on the integer $k$.
If $k=1$, we have ${\partial f\over \partial y}(0,0)\neq 0$, and by the
implicit function theorem there exists a unique formal power series $y(x)\in
\C[[x]]$ such that $y(0)=0$ and $f(x,y(x))=0$. We now assume that
$k>1$.\par

 Let us consider in the $(i,j)$-plane the set of points $(i,j)$ such that
$a_{i,j}\neq 0$. It is a subset ${\cal N}(f)$ of the first quadrant $\R_0^2=\ \
\{(i,j)\ /i\geq 0,\ j\geq 0\}$, called the Newton cloud of the series $f(x,y)$.
Any two subset $A$ and $B$ of $\R^d$ can be added coordinate-wise, to give
the Minkowski sum $A+B=\{a+b, a\in A,\ b\in B\}$ of $A$ and $B$.  Let us
consider the subset ${\cal N}_+(f)=\ {\cal N}(f)+\R_0^2$ of $\R_0^2$; its
boundary is a sort of staircase with possibly infinite vertical or horizontal
parts. The Newton polygon ${\cal P}(f)$ of $f(x,y)$ is defined as the
boundary of the convex hull of ${\cal N}_+(f)$. It is a broken line with 
infinite horizontal and  vertical sides, possibly different from the coordinate
axis.\par
Note that it is a natural generalization to the case of two variables of the
construction of the ''lowest exponent'' for a function of one variable. \par The
following is a picture of a Newton polygon in the case where the infinite
sides do coincide with the coordinate axis, or equivalently where the area
bounded by the polygon is finite. \par Recall that the convex hull is by
definition the intersection of the half-spaces containing ${\cal N}_+(f)$, so
that the number $\mu = \hbox{min}_{a_{i,j}\neq 0}\{i+\nu j\}$ is the
minimal abscissa of the intersection points with the $ i$-axis of the lines
with slope $-1\over \nu$ meeting ${\cal N}_+(f)$. Let us denote by $L_\nu$
the line which gives this minimum; an example in drawn on the picture. So
the polynomial   $$\sum_{i+\nu j=\mu}a_{i,j}c_0^j$$ corresponds to the sum
of the terms $a_{i,j}x^iy^j$ such that $(i,j)$ lies on the intersection of the line
$L_\nu$ with the Newton polygon. 
%\Insertion{120}{75}{Newton scaled 700}
\par  A necessary and sufficient condition for this polynomial to have
more than one term is that $-1\over \nu$ is the slope of one of the sides of
the Newton polygon. For simplicity of notation, let us call $\nu$ the \it
inclination \rm of the line of slope $-1\over \nu$. Let us  denote by $l\over
h$ the inclination of the "first side" of the Newton polygon of $f$, that is, the
side with the smallest inclination. Let  $c_0$ be a non zero root of the
corresponding equation, and let us make the change of variables 
$$\begin{array}{cc} x=&x_1^l\\ y=&x_1^h(c_0+y_1)\end{array}$$ The substitution in $f(x,y)$
gives $$f(x_1^h,x_1^\ell(c_0+y_1))=\sum a_{i,j}x_1^{hi+\ell j}(c_0+y_1)^j\ .$$   By
definition of $\mu$, for each $a_{i,j}\neq 0$, we have $hi+\ell j\geq \mu h$, so
we may factor the series above as $$x_1^{\mu h}f_1(x_1,y_1)\ ,
\hbox{where}\ \ f_1(x_1,y_1)=\sum  a_{i,j}x_1^{hi+\ell j-\mu h}(c_0+y_1)^j\ .$$
We remark that $$f_1(0,y_1)=\sum_{i+\nu j=\mu}a_{i,j}(c_0+y_1)^j\ ,$$
and since $a_{0,k}\neq 0$ by definition of $k$, \it the order in
$y_1$ of $f_1(0,y_1)$ is $\leq k$.\rm\par Since $c_0$ has been chosen as a
root of the polynomial $\sum_{i+\nu j=\mu}a_{i,j}c_0^j$, this order is $\geq 1$. We remark
that \par\noindent\it The order in $y_1$ of $f_1(0,y_1)$ is equal to $k$ if and only if $c_0$ is
a root of multiplicity $k$ of the polynomial $\sum_{i+\nu
j=\mu}a_{i,j}T^j=0$ \par\noindent\rm
But then we must have an equality $$\sum_{i+\nu
j=\mu}a_{i,j}T^j=a_{0,k}(T-c_0)^k$$
which implies by the binomial formula and since $\C$ is a field of
characteristic zero, that the term in $T^{k-1}$ is not zero; this is possible only
if $\nu$ is an integer and then the equality above shows that the "first side
of the Newton polygon" meets the horizontal axis at the point $(\nu k,0)$,
which corresponds to the monomial $x^{\nu k}$ , which has the non zero
coefficient $(-1)^ka_{0,k}c_0^k$, so it is actually the only finite side of the
Newton polygon of $f(x,y)$, which means that we may write in this case
$$f(x,y)=a_{0,k}(y-c_0x^\nu)^k+\sum_{i+\nu
j>\mu}a_{i,j}x^iy^j\ \ \ \hbox{\rm with }\nu\in \N,\ \mu=\nu k.$$
Making the change of variables $$\begin{array}{cc}x=& x_1\cr
y=&y_1+c_0x_1^{\nu}\end{array}$$ the series $f(x,y)$ becomes 
$$f'(x,y)=a_{0,k}y_1^k+\sum_{i+\nu
j>\mu}a_{i,j}x_1^i(y_1+c_0x_1^\nu)^j$$
The monomials which appear are of the form $x_1^{i+\nu l}y_1^{j-l}$, so that
they all satisfy $i+\nu l+\nu (j-l)=i+\nu j >\nu k$.
This means that \it if the order of $f_1(0,y_1)$ is $k$, the Newton polygon of
$f_1(x_1,y_1)$ still contains the point $(0,k)$ and the inclination $\nu _1$
of its first side is strictly greater than $\nu$ \rm.\par
The proof now proceeds as follows, :\par
\noindent
a) If the order in $y_1$ of $f_1(0,y_1)$ is less than $k$, by the induction
hypothesis, there exist an integer $m_1$ and a series $\phi_1(x_1^{1\over
{m_1}})\in \C[[x_1^{1\over
{m_1}}]]$ such that 
$$f_1(x_1,\phi_1(x_1^{1\over
{m_1}}))=0$$
By the definition of $f_1$, this implies that
$$f(x^h_1, x_1^\ell(c_0+\phi_1(x_1^{1\over
{m_1}}))=0$$
If we set $m=m_1h$ and $\phi (x^{1\over m})=x^{l\over
h}(c_0+\phi_1(x^{1\over m}))\in \C[[x^{1\over m}]]$, we have
$f(x,\phi (x^{1\over m} ))=0$ and the result in this case.\par\noindent
b) If the order in $y_1$ of $f(0,y_1)$ is still equal to $k$, we saw that
$\nu$ is an integer and the inclination of the first side of the Newton
polygon of the function $f_1(x_1,y_1)$ obtained from $f(x,y)$ as above is
strictly greater than $\nu$. \par We now set $\nu_0=\nu \in {\bf N}$ and 
repeat the same analysis for $f_1$, defining a function $f_2(x_2,y_2)$. If
again the order of $f_2(0,y_2)$ is $k$, the slope of the first side of the
Newton polygon of $f_1(x_1,y_1)$ is an integer $\nu_1>\nu_0$ and after the
change of variables $x=x_2,y=y_2+c_0x_2^{\nu_0}+c_1x^{\nu_1}$ the slope
of the Newton polygon has become greater than $\nu_1$. \par
There are two possibilities; \par\noindent 
-- either after a finite number of such steps we
get a function $f_p(x_p,y_p)$ such that $f(0,y_p)$ is of order $< k$, and by
the induction hypothesis we have a series $\phi_p(x^{1\over m_p})\in
\C[[x^{1\over m_p}]]$ such that $f_p(x,\phi_p(x^{1\over m_p}))=0$, and so
a series $$y=c_0x^{\nu_0}+c_1x^{\nu_1}+\cdots +c_{p-1}x^{\nu_{p-1}}+
\phi_p(x^{1\over
m_p})$$ such that $f(x,y(x))=0$;\par
Or the order remains indefinitely equal to $k$ and we have an infinite
increasing sequence of integers 
$$\nu_0\ <\nu_1\ <\ldots <\nu_p<\ \ldots $$
and a formal power series
$$\phi_\infty (x)=c_0x^{\nu_0}+c_1x^{\nu_1}+\cdots +c_px^{\nu_p}+\cdots
\in \C[[x]]$$ such that the Newton polygon of the function $f_\infty
(x_\infty,y_\infty)$ obtained from $f(x,y)$ by the change of variables
$x=x_\infty\ ,\ y=y_\infty +\phi_\infty (x)$ has a Newton polygon
containing the point $(0,k)$ and with inclination $0$. This means that
$f_\infty (x_\infty,y_\infty)$ is divisible by $y_\infty^k$, so we may write 
$$f_\infty(x_\infty,y_\infty)=y_\infty
^kg(x_\infty,y_\infty)$$ This implies that the order of $g(0,y_\infty)$ is
zero, so $g(0,0)\neq 0$. Geometrically, our curve is the non singular curve
$y=\phi_\infty (x)$ counted $k$ times. Indeed, for each integer $p$, we
have 
$$f(x,c_0x^{\nu_0}+c_1x^{\nu_1}+\cdots +c_px^{\nu_p})=x^{\nu_0+\nu_1
+\cdots +\nu_p}f_p(x,0),$$ so that by Taylor's expansion theorem,
$f(x,\phi (x))=0$.  This completes in the formal case the
proof of the existence of a fractional power series such that
$f(x,y(x))=0$.\par\smallskip
In order to describe all the solutions of the equation $f(x,y)=0$, it
is convenient to develop a little more the formalism of the Newton
polygon. Let ${\cal P}$ and ${\cal P}'$ be two Newton polygons; we
can define their sum ${\cal P}+{\cal P}'$ as the boundary of the
convex hull of the Minkowski sum of the convex domains in $\R_+^2$
bounded by ${\cal P}$ and ${\cal P}'$ respectively. It is easy to
verify that the following equality holds for $f,\ f'\in \C[[x,y]]$
$${\cal P}(ff')={\cal P}(f)+{\cal P}(f').$$ Any Newton polygon has a
\it length \rm and an \it height \rm which are the length of the
horizontal and vertical projections of its finite part,
respectively.\par We say that a Newton polygon is \it elementary \rm
if it has only one finite side. If it bounds a finite area, it is
then uniquely determined by its length and height. We use the
following notation for such an elementary Newton polygon.
 %\Insertion{60}{45}{elem scaled 800}
\par
We also need a little more algebra, beginning with the following
fundamental theorem: \par\noindent
One says that a holomorphic function $f(x_1,\ldots ,x_n,y)$ defined on a neighborhood of $0$
in $\C^n\times \C$ is $y$-regular (of order $k$) if $f(0,y)$ has a zero of finite order $k$ at
$0\in
\{0\}\times \C$. Geometrically this means that if we consider the germ of hypersurface
$(W,0)\subset \C^n\times \C$ defined by $f(x_1,\ldots ,x_n,y)=0$ and the first projection
$p\colon W\to\C^n$, then for a small enough representative, if $W$ is not empty
(i.e $k\geq 1$), the fiber $p^{-1}(0)$ is the single point ${0}$. In
other words, the fiber is a finite subset of $\{0\}\times \C$. The general idea of the avatars of the Weierstrass
preparation theorem is that finiteness of the fiber over one point $x$ in an analytic map
implies finiteness of the fibers above points sufficiently close to $x$.\par\vskip.1in\noindent
{\bf Weierstrass preparation Theorem}(see [K], [\L ] p.110).-\it If $f(x_1,\ldots ,x_n,y)$ is
regular of order $k$ in $y$, there exist a unique polynomial of the form
$$P(x_1,\ldots ,x_n,y)=y^k+a_1(x_1,\ldots ,x_n)y^{k-1}+\cdots +a_k(x_1,\ldots ,x_n)$$
with $a_i\in \C\{x_1,\ldots ,x_n\}$ and a convergent series $u(x_1,\ldots ,x_n)$ with
$u(0)\neq 0$, i.e invertible in
$\C\{x_1,\ldots ,x_n\}$ such that we have the equality of convergent series
$$f(x_1,\ldots ,x_n,y)=u(x_1,\ldots ,x_n,y)P(x_1,\ldots ,x_n,y).$$\rm
The polynomial $P$ is said to be \it distinguished \rm in $y$, or to be a Weierstrass 
polynomial.\par
If we start with any power series $f$, we have the same result but in the ring of
formal power series.\par\noindent  It can be shown that, given a function $f$, for
almost every choice of coordinates in
$C^n\times\C$, the function $f$ is distinguished with respect to the last
coordinate.\par\vskip.1in\noindent

It follows from the Weierstrass preparation theorem that provided we have
chosen coordinates such that $f(0,y)\neq 0$, say $f(0,y)=a_qy^q+\cdots$ with $a_q\neq 0$, it is equivalent to
seek solutions of $f(x,y)=0$ and of $P(x,y)=0$, where $P(x,y)$ is the Weierstrass
polynomial $$u^{-1}(x,y)f(x,y)=y^q+a_1(x)y^{q-1}+\cdots +a_q(x)=0\ \
\hbox{with}\ a_i(x)\in \C[[x]] $$
Now from an algebraic point of view, we must consider the field of
fractions $\C((x))$ of the integral domain $\C[[x]]$; the
irreducible polynomial $T^m-x\in \C((x))[T]$ defines an algebraic extension
of degree $m$ of $\C((x))$, denoted by $\C((x^{1\over m}))$ which is a
Galois extension with Galois group equal to the group ${\bf \mu}_m$ of
$m$-th roots of unity in $\C$. The action of ${\bf \mu}_m$ is exactly the
change in determination of $x^{1\over m}$, determined by $x^{1\over
m}\mapsto \omega x^{1\over m}$ for $\omega\in {\bf \mu}_m$. A series of
the form $y=\sum a_ix^{i\over m}$ such that the greatest common divisor of
$m$ and all the exponents $i$ which effectively appear is $1$ gives $m$
different series as $\omega$ runs through ${\bf \mu}_m$.\par
Suppose now that our function $f$ is an irreducible element of $\C[[x,y]]$,
and that the order in $y$ of $f(0,y)$ is $k<\infty$. Then the construction
described above provides a series $y(x^{1\over m})$ with $m\leq k$ such
that $f(x,y(x^{1\over m}))=0$. The product
$$\Pi_{\omega \in {\bf \mu}_m}(y-y(\omega x^{1\over m}))$$ is a
polynomial $Q(x,y)\in \C[[x]][y]$ which, by the algorithm of division of
polynomials in $\C((x))[y]$, divides $P(x,y)$; the rest of the division of $P$
by $Q$ is a polynomial of degree $<m-1$ with $m$ different roots; it is
zero.\par\noindent
We have therefore $Q(x,y)=P(x,y)$ and $m=k$ in this case. \par\noindent
We remark that the expansions $y(\omega x^{1\over m})$ all have the
same initial exponent $l\over h$, and by the definition of $Q(x,y)$, only
monomials $x^iy^j$ with ${i\over \ell}+{j\over h}\geq {\mu\over \ell}$ appear, and the
monomial $x^h$ actually appears. So we have
verified:\par\vskip.2truecm\noindent {\bf Proposition}.- \it The Newton
polygon of an irreducible series is elementary, and of the form $\elem(p,k)$,
where $k$ is the order of $f(0,y)$.\rm\par\vskip.1in
Now it is known that rings such as $k[[x,y]]$, where $k$ is a field, or $\C\{ x,y\}$ are \it
factorial\rm ; each element has a decomposition $f=f_1^{a_1}\ldots f_r^{a_r}$ where $f_i$ is
irreducible, which means that it cannot be factored again as a product $f_i=gh$ in a non trivial
way, that is, without $g$ or $h$ being an invertible element in $k[[x,y]]$, (= a series with a
non zero constant term).\par\noindent
My aim now is to prove the following\vskip.1in
\par\noindent {\bf Theorem}.- \it a) Let $k$ be an algebraically closed field of characteristic
zero, and let $f\in k[[x,y]]$ be a power series without constant term such that $f(0,y)\neq 0$.
Consider the decomposition $f=u f_1^{a_1}\ldots f_r^{a_r}$ of $f$ into irreducible Weierstrass
polynomials $f_i^{a_i}$, with a factor $u$ which is invertible in $k[[x,y]]$. For each index $i,\ \
1\leq i\leq r$, there are power series without constant term $x_i(t),y_i(t)\in k[[t]]$ such that
$f(x_i(t),y_i(t))\equiv 0$; we may choose $x_i(t)=t^{m_i}$ where $m_i$ is the degree of the
Weierstrass polynomial $f_i$, and $y_i(t)$ is then uniquely determined. Moreover if we then
write
$y_i(t)= c_it^{l_i}+\ldots
\  with\  c_i\in k^*$, then the Newton polygon of $f$ in the coordinates $(x,y)$ is the
sum
$${\cal N}(f)=\sum_1^r\elem(a_im_i,a_il_i).$$
Here we have to allow the case where for some $i$, $y_i(t)\equiv 0$, that is $l_i=\infty$.
\par\noindent b) If $k=\C$ and $f\in \C\{x,y\}$ is a convergent power series, the series $x_i(t)$
and
$y_i(t)$ are also convergent.\rm
\par\noindent
Remark: if we do not assume $f(0,y)\neq 0$,  a similar result holds, but we may no longer
apply Weiertrass' theorem and we have to allow expansions of the form $x=0,\ y=t$ and the
corresponding Newton polygons appears as summands in ${\cal
N}(f)$.\par\vskip.4truecm\noindent
The geometric interpretation of this result is that if we take any reduced analytic plane curve
$f=uf_1\ldots f_r$ with $f_i$ irreducible, i.e all $a_i=1$, the curve defined by $f=0$ is a
sufficiently small neighborhood of the origin is the analytic image of a representative of a
complex-analytic map-germ $$\bigsqcup_{i=1}^r(\C,0)_i\longrightarrow (\C^2,0).$$
\vskip.2truecm\noindent
Conversely, given two power series $x(t),y(t) \in k[[t]]$ without constant term, one may \it
eliminate \rm $t$ between them to produce an equation $f(x,y)=0$ with the property that 
$f(x(t),y(t)=0$. Indeed, by using the ''natural'' elimination process (see[T1]) we may do this in
such a way that eliminating $t$ beween $x(t^q),y(t^q)$ produces the equation $f^q(x,y)$, so
that we may even represent parametrically a non-reduced
equation.\par
There are several ways to prove this theorem; one is to prove the convergence first, either
directly by providing bounds for the coefficient of the series produced by Newton's method,
which works but is inelegant, or by considering the analytic curve $f(t^m,y)=0$, and proving
that it is a ramified analytic covering of the $t$-axis; it is also the union of $m$ non singular
curves, so each of them is analytic, and this proves the convergence of the series. (see [\L ],
II.6).\par These proofs give no basis for generalizations to higher dimension, so I chose to
present a geometric method of constructing the analytic map
$$\bigsqcup_{i=1}^r(\C,0)_i\longrightarrow (\C^2,0).$$


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