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\centerline{INTRODUCTION TO CURVE SINGULARITIES}\medskip
\centerline{Bernard Teissier}\vskip.2in\medskip
\centerline{\it D.M.I., Ecole normale sup\'erieure, U.R.A. 762 du C.N.R.S.}
\centerline{\it 45 rue d'Ulm, F 75005 PARIS}
\centerline{email: teissier@dmi.ens.fr}\vskip.3in\medskip
\noindent{\sevenrm  INTRODUCTION}\vskip.1in
My purpose in the course was to provide a simple introduction to the resolution of
singularities of curves in analytic Geometry and to the basic invariants of these
singularities. However, the course is not elementary or self contained in the usual sense that it
uses only simple notions and results. On the contrary, taking advantage of the
existence nowadays of excellent treatises of complex analytic Geometry, I freely refer to them
for some results, for example on normalization. My hope is to entice the reader to
study them carefully by showing some applications of some of the material they contain in the
fairly intuitive context of singularities of curves. 
\par\vskip.1in The study of singularities of functions begins with that of analytic functions of
one variable, say
$f(x)$. The local study consists in the assertion that in a neighbourhood of a zero of order (also
called multiplicity)
$k$ of a formal or convergent power series $f(x)$, which we may assume to be the
origin after a translation on the variable $x$, one may write $f(x)=x^kg(x)$
with $g(0)\neq 0$, and by a change of variable
$x_1=xg(x)^{1\over k}$ which is well defined near $0$ since $g(0)\neq 0$, we get the function
$f(x')=x'^k$ ; the point is that the lowest exponent of the series $f(x)$
determines completely the local behaviour of the function $f(x)$ up to an
analytic isomorphism. Over an algebraically closed field, say the field $\C$ of
complex numbers, the  global study in the case of polynomials functions of
one variable consists on one hand in the assertion that the total number of
roots counted with multiplicity is the degree of the polynomial and on the
other hand on the residue theorem.
\par  We will begin the study of functions of two variables in the
neighbourhood of a zero. First let us consider the formal situation, that is the
case where we deal with formal power series in two variables without
constant term.  It means that we do not worry about the convergence of the
series which we construct. It is then of course awkward to speak the
''mappings'' defined by formal power series, since its value may be
undefined but it is customary to do this, with a cautionary ''formal'' in front
of everything, or an ''oid'' after.  Thus a formal power series in two
variables $f(x,y)\in \C[[x,y]]$ represents a formal mapping $(\C^2,0)\to
(\C,0)$ and the quotient ring $\C[[x,y]]/(f)$ is known as an
\it algebroid plane curve
\rm although if the series $f(x,y)$ is divergent, there is no germ of curve in
$(\C^2,0)$ defined by $f(x,y)=0$ and such that this ring is the ring of formal
functions on this germ. In any case, the first goal is to try and
\it parametrize
\rm such a curve, that is, find all possible different pairs of power series
$x(t),y(t)\in
\C[[t]]$ without constant term such that $f(x(t),y(t))=0$. This is the geometric
problem of presenting any (formal) germ of curve in $\C^2$ as the union of
the  images of (formal) maps $(\C,0) \to (\C^2,0)$. Let me first describe an
apparently more ''analytic'' problem, which will turn out to be equivalent,
and its solution by Newton, who used a favourite method of analysts:
successive approximations.\par\vskip.2in\noindent{\sevenrm1. NEWTON'S STUDY OF PLANE
CURVE SINGULARITIES}\par\vskip.1in A general reference for this paragraph is [W]. 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\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
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

 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 
$$\eqalign{x=&x_1^h\cr y=&x_1^l(c_0+y_1)}$$ 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 $$\eqalign{x=& x_1\cr
y=&y_1+c_0x_1^{\nu}}$$ 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
$$\bigsqcup_{i=1}^r(\C,0)_i\longrightarrow (\C^2,0).$$
This method was perfected by Hironaka and is the basis for his method of resolution in all
dimensions over a field of characteristic zero.
\par\vskip.2in\noindent { \sevenrm 2. RESOLUTION BY BLOWING UPS}\par\vskip.1in
Let us consider the projective space $\P^n(\C)$ as the space of lines through the origin in
$\C^{n+1}$. If we choose coordinates $x_0,\ldots ,x_n$ on $\C^{n+1}$ the projective space is 
covered by affine charts $U_i$, the points of which correspond to the lines contained in
the open set $x_i\neq 0$.  It is customary to take homogeneous coordinates $(u_0:\cdots :u_n)$
on the projective space, corresponding to the lines given parametrically by $x_i=u_it$, or by the
equations $x_iu_j-x_ju_i=0$, where it is enough to take the $n$ equations for which $j=i+1\ 
\hbox{\rm and}\  i<n$. The term ''homogeneous coordinates '' means that for any $\lambda\in
k*$ the coordinates $(u_0:\cdots :u_n)$ and $(\lambda u_0:\cdots :\lambda u_n)$ define
the same point. \par Now consider the subvariety
$Z$ of the product space
$\C^{n+1}\times \P^n$ defined by these $n$ equations. It is a nonsingular algebraic variety of
dimension $n+1$ and the first projection induces an algebraic morphism $B_0\colon Z\to
\C^{n+1}$. \par The fiber $B_0^{-1}(0)$ is the entire projective space $\P^n(k)$ since when all
$x_i$ are zero, all the equations between the $u_j$ vanish,while the fiber $B^{-1}_0(x)$ for a
point $x\neq 0$ consists of a unique point because then the coordinates $x_i$ determine
uniquely the ratios of the $u_j$ which means a point of $\P^n(k)$. ''Blowing up a point
''replaces the observer at the point by what he sees'', because the observer essentially sees a
projective space (in fact a sphere, but this is just a metaphor).\par 

A basic properties of blowing up is that it separates lines: in fact consider the algebraic map
$\delta \colon \C^{n+1}\setminus \{ 0\}\to \P^n$ which to a point outside the origin associates
the line joining the origin to this point. Of course we cannot extend the definition of this map
through the origin; The graph of $\delta$ however, is an algebraic subvariety of
$(\C^{n+1}\setminus \{ 0\})\times \P^n$, and we may take the closure (for the strong topology
if $k=\C$, or for the Zariski topology) of this graph. It is a good exercise to check that this
closure coincides with $Z$ as defined above. A point of $B_0^{-1}(0)$ is
precisely a direction of line, so the map $\delta\circ B_0$ can be defined there as the map
which to this point associates the direction: in $Z$ we have separated all the lines meeting at
the origin.\par Let us consider in more detail the case $n=1$. Then $Z$ is a surface covered by
two affine charts corresponding to the charts of the projective space: for convenience of
notation set $u_0=u,\ u_1=v,\ x_0=x,\ x_1=y$ so that $Z$ is defined by $vx-uy=0$. On the
open set $U$ of $Z$ where $u\neq 0$ we may taxe as coordinates $x_1=x ,\ y_1={v\over u}$
and then the map induced by $B_0$ on $U$ is described in these coordinates by 
$$\eqalign{x\circ B_0=&x_1\cr y\circ B_0=&x_1y_1\cr}$$
and similarly on the open set $V$ defined by $v\neq 0$, we take as coordinates $x_1={u\over
v},\ y_1=y$ and the map $B_0$ is described by
$$\eqalign{x\circ B_0=&x_1y_1\cr y\circ B_0=&y_1\cr}$$
\noindent Remark that in the first chart the projective space $B_0^{-1}(0)$ is defined by $
x_1=0$ and in the second by $y_1=0$ (remember that they are coordinates on two
distinct charts, and on the intersection of the two charts they define the same subspace). It is a
crucial property of blowing up that it transforms the blown-up subspace (here the origin) into
a subspace defined locally by one equation (called a divisor); it is a good exercise to check that
this is the case in any dimension. The space $B_0^{-1}(0)$ is called the \it exceptionnal
divisor\rm .  We are now able to study the effect on a function
$f(x,y)$ (formal or convergent) of its composition with the map $B_0$. Consider the expansion of
$f$ as a sum of homogeneous polynomials
$$f(x,y)=f_m(x,y)+f_{m+1}(x,y)+\cdots +f_{m+k}(x,y)+\cdots \ ,$$
where $f_j$ is homogeneous of degree $j$. In the chart $U$, we may write
$$f\circ B_0=f(x_1,x_1y_1)= x_1^m\bigl(f_m(1,y_1)+x_1f_{m+1}(1,y_1)+\cdots
+x_1^kf_{m+k}(1,y_1)+\cdots\bigr) $$
and there is a similar expansion in the other chart.
Now if we look at the zero set of $f\circ B_0$ we see that in each chart it contains the
exceptionnal divisor counted $m$ times. If we remove this exceptionnal divisor as many
times as possible, i.e divide $f\circ B_0$ by $x_1^m$ in the first chart and by $y_1^m$ in the
second, we obtain the equation of a curve on the surface $Z$, either formal or defined near
$B_0^{-1}(0)$, which no longer contains the exceptionnal divisor. This curve is called the \it
strict transform \rm of the original curve. We also say that the equation obtained in this way
is the strict transform of $f$). In the first chart it is $x_1^{-m}f(x_1,x_1y_1)$, and in the
second $y_1^{-m}f(x_1y_1,y_1)$. \par By construction, the strict transform meets the
exceptionnal divisor only in finitely many points; let us determine them: in the first chart they
are given by $f_m(1,y_1)=0$ and in the second, by $f_m(1,y_1)=0$. By construction of the
projective space the points we seek are therefore the points in the projective line defined by
the homogeneous equation $f_m(u,v)=0$. The homogeneous polynomial $f_m$ of lowest
degree appearing in $f(x,y)$ is called the \it initial form \rm and $f_m(x,y)=0$ is a union of
$m$ lines (counted with multiplicity) called the \it tangent cone \rm of $f$ at the point $0$. So
we see that the strict transform of $f$ meets the exceptionnal divisor at the points in this
projective space corresponding to the lines which are in the tangent cone at $0$ of our
curve.\par
In particular, if our curve has two components with tangent cones meeting only at the origin,
their strict transforms are disjoint. Consider for example $f(x,y)=(y^2-x^3)(y^3-x^2)$.\par 
In order to analyze in more detail what goes on, we have to \it assume that $k$ is
algebraically closed, which we will do from now on \rm , and introduce the concept of
\it intersection number \rm of two curves at a point. The simplest definition (but not the most
useful for computations) is the following:\par\noindent
Let $f,h\in k[[x,y]]$ be series without constant term and without common irreducible factor.
Let $(f,h)$ denote the ideal generated by $f$ and $h$ in $k[[x,y]]$. Then the dimension 
$$\hbox{\rm dim}k[[x,y]]/(f,h)$$ is finite and is by definition the intersection number of the
two curves at $0$. If $k=\C$ and $f,h$ are in $\C\{ x,y\}$, then the dimension above is also
$$\hbox{\rm dim}\C\{x,y\}/(f,h)$$ where now $(f,h)$ is the ideal generated in $\C\{ x,y\}$.\par
To prove the finiteness we first remark that it is sufficient to prove it after replacing $k$ by its
algebraic closure and then we may use the Hilbert nullstellensatz which tells us that since
$f=0,\ h=0$ meet only at the origin, the ideal $(f,h)$ contains a power of the maximal ideal
${\bf m}=(x,y)$ say ${\bf m}^N$. This implies the finiteness since $k[[x,y]]/(f,h)$ is then a
quotient vector space of $k[[x,y]]/{\bf m}^N$ and also shows that we may without changing the
ideal assume that $f,h$ are polynomials of degree $<N$, so that for example if $f,h$ are
convergent power series the vector spaces
$\C\{x,y\}/(f,h)$ and $\C [[x,y]]/(f,h)$ are equal. \par
The definition of intersection multiplicity at the point $0$, of the two curves $f=0,\ h=0$, say in
the analytic case is then
$$\big( f,h\big)_0 =\hbox{\rm dim}\C\{x,y\}/(f,h).$$
Note that we use large parentheses for the intersection number, small ones for the ideal
generated by $f,g$.\par
 In any case this definition of the intersection multiplicity has the
advantage to suggest the following intuitive interpretation :\par\noindent
Consider a 1-parameter deformation of one of the two functions, say $f+\epsilon$; it is possible
to show that if $f,h$ converge in a nice neighborhood $U$ of $0$, for small enough $\epsilon$,
then the two curves $h=0, f+\epsilon =0$ meet in $U$ transversally at points which are
non-singular on each. Moreover, these points tend to $0$ as $\epsilon$ tends to $0$, and the
number of these points is $\hbox{\rm dim}\C\{x,y\}/(f,h)$. So this number may be thought of
as the number of ordinary intersections (i.e transverse intersection of non-singular curves)
which are concentrated at $0$.\par
There is another way to present this intersection number, which is very useful for
computations:\par\noindent
Suppose that $h(x,y)=uh_1^{e_1}\ldots h_r^{e_r}$ with $u(0)\neq 0$. For each $i$, $1\leq i \leq
r$, let us parametrize the curve $h_i(x,y)=0$ by $x(t_i),y(t_i)$. Now substitute these power series
in $f(x,y)$; we get a series in $t_i$, the order of which we denote by $I_i$. Then we have 
\par\noindent
$$I_i=\hbox{\rm dim}\C\{x,y\}/(f,h_i)\ ,$$
and 
$$\big(f,h\big)_0=\sum_1^re_iI_i.$$\vskip.2truecm
Remark: Given a germ of curve $f=0$, where $f=f_m+f_{m+2}+\cdots $, its multiplicity at the
origin may be defined as the smallest degree $m$ of a monomial appearing in the series
$f$. A better definition is to say that the multiplicity is the intersection number
$\big(f,\ell\big)_0$ for a sufficiently general linear form $\ell$. In fact, we have
$$m\leq \big(f,\ell\big)_0$$ with equality if and only if the line $\ell (x,y) =0$ is not in the
tangent cone defined by $f_m(x,y)=0$.\par
Indeed, we may parametrize the line $\ell =0$ by $x=\alpha t,\ y=\beta t$; then we substitute
in $f$:
\par\noindent
$$f(\alpha t, \beta t)=f_m(\alpha ,\beta )t^m+f_{m+1}(\alpha ,\beta )t^{m+1}+\cdots $$
is of order $\geq m$, and of order $m$ exactly if and only if $f_m(\alpha ,\beta )\neq 0$.\par
It is convenient, given a curve $f(x,y)$ and a point $z$ in the plane, to define the
\it multiplicity of $f$ at $z$ \rm as follows: take coordinates $(x',y,)$ centered at $z$, which
means that they vanish at $z$; if $z=(a,b)$ we may take $x'=x-a, y'=y-b$. Then expand $f$ in
those coordinates (of course we assume that $z$ is in the domain of convergence of $f$). \par
We get $f'(x',y')=f(a+x',b+y')$. Then we compute the lowest degree terms appearing in the
expansion of $f'$ and denote this by $m_z(f)$ or, if $X$ is the curve $f(x,y)=0$, by $m_z(X)$.
We see that $m_z(f)=0$ unless $f(z)=0$, and that if $\ell$ is a line through $z$, we have
$m_z(X)\leq \big(X,\ell\big) _z$ with equality except if $\ell$ is in the tangent cone of $X$ at
$z$. \par
Let us  apply this, in our blowing up as described above, to the line
$x_1=0$ (the exceptionnal divisor) and the strict transform $f_1(x_1,y_1)=0$, at  a point
$x'$ with coordinates $x_1=0, y_1=t_1)$ where $f_m(1,t_1)=0$ i.e a point of intersection of the
strict transform with the exceptionnal divisor. We have
$$f_1=f_m(1,y_1)+x_1f_{m+1}(1,y_1)+\cdots $$ and if we denote by $e_{x'}$ the multiplicity of
$t_1$ as a root of the polynomial $f_m(1,Y)$, it follows from what we saw above that we have
$$e_{x'}\geq m_{x'}(f_1)$$ with equality unless the curve $f_1(x_1,y_1)=0$ is tangent to the
exceptionnal divisor at the point $x'$, in the sense that the tangent at $x'$ to the exceptionnal
divisor is in the tangent cone of $f_1=0$ at the point $x'$. Since the multiplicity of $f_1$ is zero
at points where $f_m(1,y_1)$ does not vanish, we see that if we look at all the points $x'$ in the
blown up surface $Z$ which are mapped to our origin by the projection $Z\to \C^2$, which we
denote by $x'\to 0$, we have 
$$\sum_{x'\to 0}m_{x'}(f_1)\leq \sum_{x'\to 0}e_{x'}=m\ ,$$
so that in particular, if \it  there is a point $x'$ of the strict transform $X'$ of $X$ which is
mapped to
$0$ and is of multiplicity
$m$ on $f_1=0$, then it is the only point of $X'$ mapped to $0$ and $X'$ is transversal to the
exceptionnal divisor at $x'$. \rm This fact and its generalizations play a crucial role in
Hironaka's proof of the resolution of singularities. \par\vskip.2truecm
In order to show that the situation which we have just described cannot persist indefinitely in
a sequence of blowing ups, we have to use the intersection number in another manner,
according to Hironaka:\par 
Given a germ of a plane curve $(X,x)$ with $r$ branches $(X_i,x)_{1\leq i\leq r}$ and a
nonsingular curve
$W$ through the point
$x$, define the \it contact exponent \rm of $W$ with $X$ at $x$ as follows:
$$\delta_x(W,X)=\hbox{\rm min}_{i=1}^r\bigg({\big(X_i,W\big) _x\over m_x(X_i)}\bigg)$$
and the \it contact exponent \rm of $X$ at $x$ as follows
$$\delta_x(X)=\hbox{\rm max}_W\delta_x(W,X),$$
where $W$ runs through the set of germs at $x$ of non-singular
curves.\par\vskip.3truecm\noindent
Lemma.-\it Let $f(x,y)=0$ be an equation for $X$. If the coordinates $(x,y)$ are chosen in such a
way that
$x=0$ is not tangent to
$X$ at $x$ and $W$ is defined by $y=0$, the rational number $\delta_x(W,X)$ is the inclination
of the first side of the Newton polygon of $f(x,y)$.\rm\par\medskip
 By definition of $\delta_x(W,X)$ is enough to prove that for an irreducible $f$, the
inclination of the only side of it Newton polygon is ${\big(X,W\big) _x\over m_x(X)}$, but if we
parametrize $X$ by $x=t^m,\  y=t^q+\cdots $,  we find that the transversality
condition implies
$m\leq q$, and we have $\big(X,W\big)_x=q$; the result follows.\par\vskip.3truecm\noindent
Lemma.- \it Assume that $W$ is the curve $y=0$ and that $f(x,y)$ is in Weierstrass form, i.e.,
 $$f(x,y)=y^n+a_1(x)y^{n-1}+\cdots +a_n(x)\ \ \ \ a_i(x)\in\C\{x\},$$ then the inclination of
the first side of its Newton polygon is $$\delta_x(W,X)=\hbox{\rm min}_{1\leq i\leq n-1}{\nu_0
(a_i)\over i}.$$ Here as usual $\nu_0(a(x))$ denotes the order of vanishing at the origin of
the series $a(x)$.\rm\par\smallskip Indeed, the point
$(0,n)$ is a vertex of the first side of the Newton polygon, and the lemma is just the
observation that if we write
$a_i(x)=\alpha_ix^{c_i}+\cdots$, the other vertices of the Newton polygon are among the points
$(c_i,n-i)$, which follows directly from the definition.\par\medskip
 A nonsingular curve
$W$ such that
$\delta_x(W,X)=\delta_x(X)$ is said to have
\it maximal contact \rm at $x$. 
non singular curves with maximal contact are the nonsingular curves which it is hardest to separate from $X$ by
a succession of blowing ups (in the sense of separating strict transforms), and so when they eventually separate,
something nice should happen; indeed once they separate, there is no point of multiplicity
$m_x(X)$ in the iterated strict transform mapping to $x$. As one says, ''the multiplicity has
dropped''. Hironaka's approach to resolution uses the existence of varieties with maximal
contact to build an induction on the dimension. \par\vskip.3truecm\noindent The next step is to prove the
existence of curves with maximal contact.\par Assume that a non singular curve $W$ defined by $y=0$ does not
have maximal contact with
$X$ at $x$. We way assume that the curve $x=0$ is transversal to $f(x,y)=0$, which means
that $f(0,y)=a_{0,m}y^m+\cdots$, where $m$ is the multiplicity of $f$ at $0$. By a change of
variable $y=(a_{0,m})^{1\over m}y'$, which does not change the contacts, we may assume
that $a_{0,m}=1$. To say that
$\delta_x(W,X)<\delta_x(X)$ means that there is a series
$A(x)$ such that the contact of the curve
$f(x,y)=0$ with $y-A(x)=0$ is greater than its contact with $y=0$. By a change of the
coordinate $x$ which does not affect the contacts, we may assume that $A(x)=\xi x^d$ for
some integer $d$ and $\xi\in\C^*$. Let us now compute the power series expansion in the
coordinates
$x'=x, y'=y-A(x)$; 
$$\displaystyle{f'(x,y')=\sum_{{i\over \delta}+j\geq m} a_{i,j}x^i(y'+\xi x^d)^j=\sum_{{k\over
\delta'}+\ell\geq m}  a'_{k,\ell}x^ky'^\ell }.$$ By expanding the powers of $y'+\xi x^d$ we get,
for each $(i,j)$, and $k\leq j$ the inequality
 ${i+kd\over \delta}+j-k\geq m$ but we know that ${i\over \delta}+j\geq m$. From
this follows the inequality $d\geq \delta$. \par\noindent Isolating the terms which lie on the
first side of the Newton polygon, we get: 
$$\leqno{(*)}\displaystyle{\sum_{{i\over \delta}+j=m}a_{i,j}x^iy^j+\sum_{{i\over
\delta}+j>m}a_{i,j}x^iy^j=\sum_{{k\over \delta'}+\ell\geq m} a'_{k,\ell}x^ky'^\ell},$$ and the
slope of the first side of the Newton polygon of the right-hand side is $\delta'>\delta$. Let us
first assume that $\delta=1$. Remark that all the terms $x^ky'^\ell$ with ${k\over
\delta'}+\ell\geq m$ except $y'^m$ are in the ideal $(x,y')^{m+1}$. Therefore we must have the
equality
$$\displaystyle{\sum_{{i\over \delta}+j=m}a_{i,j}x^iy^=y'^m \ \hbox{\rm mod.}(x,y)^{m+1}}$$
so that the left hand side is the m-th power of $y-\xi x^d$. This implies that $d=1=\delta$
since the left hand side is homogeneous. \par
If $\delta >1$ we follow the same method. Since we know that $d\geq \delta$, it is easy to
check that the ideal of $k[[x,y]]$ generated by the monomials $x^ky'^\ell\ ,\ {k\over
\delta'}+\ell\geq m\ ,\ k\neq 0$ is contained in the ideal $I$ generated by the monomials
$x^iy^j\ ,\ {i\over\delta}+j>m$. Looking at the equation $(*)$ modulo $I$ gives us
$$\sum_{{i\over \delta}+j=m}a_{i,j}x^iy^j=y'^m \ \hbox{\rm mod.}I$$
which again by homogeneity shows that $d=\delta$ and the sum on the left hand side is
$(y-\xi x^d)^m$.\par
Note that this argument also works if $\delta_x(X)=\infty$.
So there are two possibilities:\vskip.1in\Par
1) We have $\delta_x(W,X)<\delta_x(X)$; in this case the sum of the
terms of $f(x,y)$ lying on the first side of the Newton polygon is of the form $(y-\xi
x^d)^m$.\Par
2) The sum of the terms of $f(x,y)$ lying on the first side of the Newton polygon is not of the
form $(y-\xi x^d)^m$.\Par
In the first case, as we have seen, $d=\delta_x(W,X)$. We make the change of variables
$x'=x;\ y'=y-\xi x^d$ and in the new coordinates $x',y'$, if $W'$ is the curve $y'=0$, we have
$\delta_x(W',X)>\delta_x(W,X)$. This follows easily from the computation we have just made;
an effect of the change of variables is that all the terms lying on the first side of the Newton
polygon, of inclination $d$, are transformed into the single term monomial $y'^m$. So the
inclination of the new Newton polygon has to be $>d$; but we know this inclination to be
$\delta_x(W',X)$. If we have not reached $\delta_x(X)$, we continue the same procedure,
and after possibly infinitely many steps, i.e. after a change of variables of the form $$x'=x\ ;\
y'=y-\xi_1x^{d_1}-\xi_2x^{d_2}-\cdots -\xi_rx^{d_r}-\cdots$$
we reach the stage where the sum of terms on the first side of the Newton polygon is not a
$m-th$ power, so $\delta_x(W_s,X)=\delta_x(X)$, with $s$ possibly infinite. Since the
denominators of the $\delta_x(W,X)$'s are bounded, the series is infinite only in the case
where $\delta_x(X)=\infty$. At least formally this series converges, since we have
$d_1>d_2>\cdots >d_r>\dots$, but we can omit the proof of convergence if we work in
$\C\{x,y\}$ since the equality $\delta_x(X)=\infty$ means that in some coordinates $f(x,y)$ is
of the form $u(x,y)y^m$ where $u$ is an invertible element in $k[[x,y]]$; indeed for any other
case, we see from the definition that $\delta_x(X)<\infty$. But the Weierstrass preparation
theorem tells us that if such a presentation exists with formal power series, it also exists with
convergent power series, so that the series defining our final coordinates converges.\par
So in all cases, we can find a nonsingular curve $W$ which has \it maximal contact \rm with
$X$ at $x$, i.e. such that $\delta_x(W,X)=\delta_x(X)$.
\smallskip
Remark that all the discussion above is valid on a germ of a non singular surface, since it is
analytically isomorphic to the plane. The definition of the blowing up is independant of
the choice of coordinates, and makes sense on any nonsingular surface.\par The next step is to
study the behavior of the contact under blowing up of the origin. I will leave the proof of this
as an exercise, since it is a direct application of what we have just seen and the definition of
blowing up:\smallskip\noindent {\bf Theorem}(Hironaka).- \it Let
$m$ be an integer, let
$f(x,y)=0$ define a germ of a plane curve, $(X,0)\subset (\C,0)$ of multiplicity $m$ and let
$(W,0)\subset (\C,0)$ be a non singular curve with maximal contact with $X$ at
$0$. If, after blowing up the point $0$ by the map $B_0\colon Z\to\C^2$, there is a point
$x'\in X'$ of multiplicity $m$ in the strict transform $X'\subset Z$ of $X$, then \Par
1) The point $x'$ is the only point of $X'$ mapped to $0$ by $B_0$,\Par
2) The strict transform $W'$ of $W$ by $B_0$ contains the point $x'$, and $W'$ has maximal
contact with $X'$ at $x'$,\Par
3) We have the equality $\delta_{x'}(W',X')=\delta_x(W,X)-1$.\rm
\smallskip\noindent
{\bf Corollary}.- \it The maximal length of a sequence of infinitely near points of multiplicity
$m$ on the strict transforms of $X$, each mapping to its predecessor in successive blowing
ups $$\cdots Z^{(r)}\to Z^{(r-1)}\to \cdots \to Z^{(2)}\to Z^{(1)}\to \C^2$$
is equal to the integral part $[\delta_x(X)]$.\rm
\smallskip
This suffices to show that unless the curve is of the form $y^m=0$,  the multiplicity of its strict
transform in the sequence of blowing ups obtained by blowing up at each step the points of
maximal multiplicity drops after a finite number of steps. By induction on the multiplicity,
this proves the resolution of the singularity of $X$ at $0$ by a finite number of blowing ups of
points on non singular surfaces.\medskip
We should remark that the map $X'\to X$ of the strict transform of $X$ to $X$ is defined by
itself, without any reference to an embedding $(X,0)\subset (\C^2,0)$ (see [K]). \par 
We have proved a local result, but if now we consider any algebraic or analytic curve, it has
finitely many singular points, and the local resolution processes at each point are independant,
so we have:
\smallskip\Par {\bf Theorem}.-\it  Given an algebraic or analytic plane curve $X$ there
exists a finite sequence of point blowing ups such that in the composed map $X'\to X$ the curve
$X'$ has no singularities. \rm\par\smallskip  Actually we can get, by the same method, a better
result, known as
\it embedded resolution
\rm and originally due to Max N\oe ther, as follows:\smallskip\Par {\bf Theorem}.- \it Given a
curve
$X$ on a non singular surface
$S$, there exists a finite sequence of blowing ups of points $$S^{(r)}\to \cdots \to S^{(1)}\to S$$
such that if we denote by $\pi\colon S^{(r)}=S'\to S$ their compositum, then the inverse
image of the singular points of $X$ (the exceptionnal divisor) is a union of non singular curves
meeting transversally on the non singular surface $S'$, and the strict transform $X'$ of $X$ by
$\pi$ is a non singular curve meeting transversally these curves.\par
In analytic terms, if $f(x,y)=0$ is  a local equation for $X$ in $S$, then $f\circ \pi$ is, at every
point $x'$ of $S'$, of the form $(f\circ \pi)_{x'}=u^av^b$ for suitable local coordinates of $S'$ at
$x'$. Of course $a$ and $b$ will be zero unless we have $x'\in \pi^{-1}(X)$.\rm
\smallskip The induced map $\pi\colon X'\to X$ is a resolution of singularities of $X$. If we fix
a singular point
$x\in X$, let $r$ be the number of analytically irreducible components of the germ $(X,x)$.
The number of points in $\pi^{-1}(x)$ is equal to $r$ and for a small enough representative
$X_x$ of the germ $(X,x)$, the part $\pi^{-1}(X_x)$ of $X'$ lying over $X_x$ consists of $r$ non
singular curves ${\bf D}_i$, each marked with one of the points of $\pi^{-1}(x)$. The image
by $\pi$ of each of these non singular curves ${\bf D}_i$ is one of the irreducible components
of $X_x$.\par 
If we choose for each non singular curve ${\bf D}_i$ a coordinate $t_i$ vanishing at the only
point $z_i$ of ${\bf D}_i$ lying over $x$, then ${\bf D}_i$ is described parametrically,
in local coordinates $(u,v)$ on $S'$ centered at $z_i$, by \it convergent \rm power series
$u(t_i),v(t_i)$, because of the implicit function theorem. Since the map $\pi \colon S'\to S$ is a
composition of algebraic maps, $x\circ \pi$ and $y\circ \pi$ are at worst convergent power
series in $(u,v)$, so when we restrict them to ${\bf D}_i$, we get convergent power series in
$t_i$. This shows that each branch of our curve has a convergent parametrization, and from
this we deduce that the formal parametrization constructed by Newton's method converges.

\smallskip Note that this convergence argument works equally well with the first
resolution theorem. The new fact in the resolution result above with respect to the resolution
theorem is the transversality of the strict transform with the exceptionnal divisor, which is
not part of the resolution theorem as we have stated it above. The proof of this improvement
is not difficult: it amounts to resolving singularities, by a sequence of points blowing up, of the
union of the strict transform and the exceptional divisor of the map which resolves the
singularities of
$X$.\Par
As an example, given an integer $m>1$, after one blowing up the strict transform of a curve
with equation $y^m-x^{m+1}=0$, but is not transversal to the exceptionnal divisor. \par It is
the first example of a fundamental fact of analytic or algebraic geometry:\it  you can make
spaces (in fact, their strict transforms) transversal by well chosen sequences of blowing
ups.\rm\par\vskip1truecm\noindent
{\sevenrm 3. RESOLUTION OF SPACE CURVES}\medskip
\Par
1. Integral dependance\par\medskip
To prove a resolution theorem for space curves, one meets the difficulty that their equations
may be complicated (for example to define a curve in $\C^n$ one may need more that $n-1$
equations; those for which $n-1$ equations suffice are called \it complete intersections \rm ,
and also that rather different looking sets of equations may generate the same ideal in
$\C\{x_1,\ldots ,x_n\}$ and therefore define the same curve. In the proofs above we have
used constructions which depend heavily on the equation. Moreover, even to show that a
germ of a complex curve in $\C^d$ has a finite number of irreducible components, which are
analytic germs, is not completely trivial (see [\L ], II.5). There are two possibilities: we can
conceptualize and abstract the proof for plane curves to make it less dependent on the
equation, or try to reduce to the plane curve case. As it happens, the two methods are not so
different, at least for one of the ways of abstracting the ideas. \Par To reduce to the plane
curve case, the natural idea is to project the space curve
$X$ to a plane curve $X_1$. One can then show that a resolution of $X_1$ has to map to $X$,
and that this map is a resolution of the singularities of $X$!.\par The key idea is that of \it
normalization.
\rm The Italian geometers called \it normal \rm a projective variety $Z\in \P^n$  having the
property that any map $Z'\to Z$ presenting $Z$ as a ''general'' projection by a linear map
$\P^{n'}\setminus L\to \P^n$ of an algebraic variety $Z'\subset \P^{n'}$ had to be an
isomorphism. A typical non normal surface in $\P^3$ is therefore a general projection of a non
singular surface in
$\P^4$; such a projection has a curve of double points, on which are finitely many more
complicated singular points, the ''pinch points''. Here the meaning of ''general'' has to be
made precise; \par\noindent 
The variety $Z$ is normal if any map $\pi \colon Z'\to Z$ which \par\noindent
a) is finite-to-one and\par\noindent
b)  induces an isomorphisme $Z'\setminus \pi^{-1}(U)\to U$
over the complement $U$ of a closed algebraic or analytic subset of $Z$
is an isomorphism.\par\noindent
\par The resolution
theorem we saw above shows that a singular curve in
$\P^2$ cannot be normal. \par\medskip
\Par The concept of normalization was ''localized'' and
transfigured into a concept of commutative algebra, as follows: Recall that the total ring of
quotients of a ring $A$ is the ring of equivalence classes of couples $(a,b)$ of elements of $A$,
where $b$ is not a zero divisor in $A$ with addition
$(a,b)+(a',b')=(ab'+ba',bb')$ and componentwise multiplication, the equivalence being
$(a,b)\equiv (a',b')$ when $ab'-ba'=0$. The map $a\mapsto (a,1)$ induces an injection of $A$
in $F$ and we indentify $A$ with its image in $F$. If
$A$ is an integral domain,
$F$ is its field of fractions.\par\medskip\noindent {\bf Definition} Let
$A$ be a commutative ring without nilpotent elements, and let $F$ be its total ring of
quotients. \par\noindent {\bf Definition}.- An element $h\in F$ is \it integral \rm over $A$ if
it satisfies an equation
$$h^k+a_1h^{k-1}+\cdots +a_k=0\ \ \hbox{with}\ a_i\in A.$$
\medskip\noindent
{\bf Example}.- Consider the germ of plane curve $X$ in $\C^2$ defined by the equation
$y^p-x^q=0$. The quotient $\O$ of the ring $\C\{x,y\}$ by the ideal generated by $y^p-x^q$ is
the ring of germs of analytic functions on the germ $X$ (the restrictions to $X$ of two analytic
functions on $\C^2$ coincide if and only if their difference is in the ideal). The ring $\O$ is an
integral domain; let $K$ be its field of fractions. If we keep the notations $x,y$, etc.. for the
restictions to $X$ of functions on $\C^2$, we have ${y\over x}\in F$. I claim that if $p\leq q$,
it is integral over $\O$; indeed, we have the relation $$({y\over x})^p-x^{q-p}=0\ .$$
We can remark that the function ${y\over x}$ is defined and analytic on the strict
transform of $X$ by the blowing up of the origin for any sufficiently small representative of
the germ $X$. We remark also that the condition $p\leq q$ is equivalent to saying that the
meromorphic function ${y\over x}$ remains bounded on $X$ for any small
representative. \vskip.2in\noindent
{\bf Proposition}.- \it Given a ring $A$ without nilpotent elements, let $F$ be its total ring of
fractions; the set of elements of $F$ integral over $A$ is a ring for the operations induced by
those of $F$.\rm\vskip.1in
This ring is called the normalization of $A$ (or the integral closure of $A$ in $F$) and often
denoted by $\overline A$. Of course we have $A\subset \overline A$; a ring such that
$A=\overline A$ is said to be integrally closed. Is is not difficult to check that
$\overline A$ is integrally closed.\par
If $A$ is n\oe therian and integrally closed, any injective map $A\to B$ to a subring $B$ of the
total ring of fractions of
$A$ which makes $B$ into a finite $A$-module is an isomorphism; this is the translation of
the original definition of normality. To prove it, check that if $h$ is an element of $B$, the
powers of $h$ cannot all be linearly independant over $A$, so $h$ satisfies an integral
dependance relation, and if $A$ is normal, it is in $A$!\par
 An important theorem is that if
$A$ is an analytic algebra, i.e a quotient of a convergent power series ring by some ideal, then
$\overline A$ is a finite sum of integrally closed analytic algebras, and moreover that the
injection $A\to \overline A$ makes $\overline A$ into a \it finitely generated
\rm $A$-module. Taking a common denominator (in $F$) for a finite set of generators of the
$A$-module
$\overline A$, we see that the (''conductor'') ideal ${\cal C}=\{d\in \overline A,\  d.\overline
A\subset A\}$ is not zero. \Par
Another important fact is that if the analytic algebra of germs of functions on a curve at a
point is normal, the point is non singular on the curve, and the analytic algebra is isomorphic
to a convergent power series ring in one variable $\C\{t\}$. ([\L ], VI.3,
Thm.2)\par
So this abstract idea, normalization, provides us with a proof of the resolution of
singularities of space curves: given $(C,0)\in (\C^d,0)$, the normalization ${\cal
O}\to\overline{\cal O}$ of the (reduced) analytic algebra of germs of functions on $C$ is an
analytic algebra which is a product $\Pi_{i=1}^r\C\{t_i\}$ of a finite number of convergent
power series rings in one variable. If $x_1,\ldots ,x_d$ generate the maximal ideal of ${\cal
O}$, we get $r$ $d$-uples of convergent power series expansions $x_j(t_i)$, which are our
Newton series in this case. They geometrically correspond to a map 
$$\bigsqcup_{i=1}^r (\C,0)_i\to (C,0)$$
which is our resolution of singularities.
However, normalization is geometrically subtle in general, and the finiteness of normalization
is a subtle theorem; in addition, we may seek a more geometric proof, as follows
\vskip.1in\Par We now turn to the definition of plane projections of a space curve.\Par Let
$(C,0)\in (\C^d,0)$ be a germ of a (reduced) space curve defined by an ideal
$I\subset
\C\{x_1,\ldots ,x_d\}$. Let us choose a linear projection $p\colon \C^d\to \C^2$. Let $M$
denote the space of all such projections; think of it as a set of $d\times 2$ matrices of rank
$2$. We endow $M$ with the topology (complex or Zariski) induced by that of the space
of matrices. We wish to consider only the projections such that $p\vert C\colon C\to p(C)$ is
finite to one. If that is not the case, the kernel of $p$, which is a linear subspace of
codimension $2$ of $\C^d$, contains one of the irreducible components of the curve $C$; the
intersection is analytic, so it is either of dimension $0$ or $1$. By looking at the equations of
$C$, it is not too difficult to check that the projections which do not contain a component of
$C$ form a dense open set of $M$. The fact that they are those which induce a finite map $C\to
p(C)$ is a consequence of the Weierstrass preparation theorem. \Par
Assume now that the map $C\to p(C)$ is finite. Again by the Weiertrass theorem, it means
that the map of analytic algebras $\C\{x,y\}\to \O$ defined by $f\mapsto (f\circ p)\vert C$
makes $\O$ a $\C\{x,y\}$-module of finite type. Since $\C\{x,y\}$ is n\oe therian, it means
we have a presentation by an exact sequence of $\C\{x,y\}$-modules:
$$\C\{x,y\}^q\to \C\{x,y\}^p\to \O\to 0$$
An argument detailed in [T1] shows that since $C$ is of dimension $1$, we must have $q=p$, so
the first map is described by a square matrix with entries in $\C\{x,y\}$. Let $\phi (x,y)$ be
the determinant of that matrix. This determinant is, up to an invertible factor, independant
of the choice of the presentation. Then the image
$p(C)$ is the plane curve with equation
$\phi (x,y)=0$.\par\vskip.1in\noindent
On the other hand, let us say that a plane projection $p\colon \C^d\to \C^2$ is \it general \rm
for $C$ if it has the following property:\Par
For any sequence of couples of points $(a_i,b_i)\in (C\setminus \{0\})\times (C\setminus
\{0\})$ tending to $0$, the limit direction of the secant line $\overline{a_i,b_i}$ (for any
subsequence) is \it not \rm contained in the kernel of $p$.\par
We will see in the next paragraph that all general projections of a given space curve are
topologically indistinguishable as curves in $\C^2$. In [T2] it is shown that if $p$ is general for
$C$, then the inclusion of the ring
$\O_1$ of the image $X_1=p(X)$ as defined above into the ring $\O$ (induced by the
composition of functions with $p$) induces an isomorphism of the total rings of fractions of
these two rings, and because $\O$ is a finite $\O_1$-module, every element of $\O$ is integral
over $\O_1$, as we saw above. Therefore $\O$ is contained in the normalization
$\overline{\O_1}$ of
$\O_1$.\Par Therefore $\overline{\O_1}$ is also the normalization of $\O$, and it is a finite
$\O$-module for general reasons ( it suffices to know that
the integral closure of $\O_1$ is a finite $\O_1$-module). Now we can use the universal
property of blowing ups: in $\overline{\O_1}$ all ideals become principal and
generated by a non zero divisor in each $\C\{t_i\}$. By the universal property of blowing up
([\L ], VII.5) if we blow up the origin in $\O$, the resulting algebra is still contained in
$\overline{\O_1}$, and as we repeat blowing up points, we get an increasing sequence of
$\O$-algebras contained in $\overline{\O_1}$, all having the same total ring of fractions.
Since $\overline{\O_1}$ is a finite $\O$-module, this sequence stabilizes after finitely many
steps. We have to show that this limit algebra is $\overline{\O_1}$. But if this were not the
case, the maximal ideal of one of the component local algebras would not be principal, so we
could blow it up and get a strictly bigger algebra, contradicting the stability.
\Par\smallskip
In conclusion, we have shown that any space curve singularity can also be desingularized by
a finite sequence of point blowing ups.\Par\smallskip
One can also prove embedded resolution for space curves; it is not much more difficult than
in the plane curve case. I refer to the lecture of J. Castellanos for facts about space curves.
\par\vskip.1in

\noindent{\sevenrm 4. PUISEUX CHARACTERISTIC EXPONENTS}\vskip.3in
Let $$f(x,y)=0\ \ \hbox{with}\ f(x,y)\in \C\lbrace x,y\rbrace$$ be an
equation for a branch $(X,0)\subset (\C^2,0)$, which means that the series $f$
is an irreducible element of $\C\lbrace x,y\rbrace$.\par As we saw, we may
assume thanks to the Weierstrass preparation theorem that $f$ is of the
form $$f(x,y)=y^m +a_1(x)y^{m-1}+\cdots +a_m(x)$$ where $m$ is the
intersection multiplicity at $0$ of the branch $C$ with the axis $x=0$. We
saw also that we may assume that $m$ is equal to the multiplicity
$n=m_0(X)$ of the curve $X$ at the origin.\par \vskip.1in\noindent
\it For this last paragraph we return to the traditional notation; $n$ will be the multiplicity of
the curve, which we have hitherto denoted by $m$, and the curve will be denoted by $X$
instead of $C$.\par \vskip.1in\noindent
\rm As we saw, the branch
$X$ can be parametrized near
$0$ as follows
$$\eqalign{x(t)&=t^n\cr
y(t)&=a_mt^m+a_{m+1}t^{m+1}+\cdots +a_jt^j+\cdots\hskip1truecm
\hbox{with}\ m\geq n}$$
Let us now consider the following grouping of the terms of the series $y(t)$:
set $\beta_0=n$ and let $\beta_1$ be the smallest exponent appearing in
$y(t)$ which is not divisible by $\beta_0$. If no such exponent exists, it
means that $y$ is a power series in $x$, so that our branch is analytically
isomorphic to $\C$, hence non singular. Let us suppose that this is not the
case, and set $e_1=(n,\beta_1)$, the greatest common divisor of these two
integers. Now define $\beta_2$ as the smallest exponent appearing in $y(t)$
which is not divisible by $e_1$. Define $e_2=(e_1,\beta_2)$; we have
$e_2<e_1$, and we continue in this manner. Having defined
$e_i=(e_{i-1},\beta_i)$, we define $\beta_{i+1}$ as the smallest exponent
appearing in $y(t)$ which is not divisible by $e_i$. Since the sequence of
integers
$$n>e_1>e_2>\cdots >e_i >\cdots $$
is strictly decreasing, there is an integer $g$ such that $e_g=1$. At this
point, we have structured our parametric representation as follows:
$$\eqalign{x(t)&=t^n\cr
y(t)&=a_nt^n+a_{2n}t^{2n}+\cdots
+a_{kn}t^{kn}+a_{\beta_1}t^{\beta_1}+a_{\beta_1+e_1}t^{\beta_1+e_1}+\cdots
+a_{\beta_1+k_1e_1}t^{\beta_1+k_1e_1}\cr
&\ +a_{\beta_2}t^{\beta_2}+a_{\beta_2+e_2}t^{\beta_2+e_2}+\cdots
+a_{\beta_q}t^{\beta_q}+a_{\beta_q+e_{q-1}}t^{\beta_q+e_{q-1}}+\cdots \cr
&\ + a_{\beta_g}t^{\beta_g}+a_{\beta_g+1}t^{\beta_g+1}+\cdots  }$$ where
by construction the coefficients of the $t^{\beta_i}; i\geq 1$ are not zero. Let
us define integers $n_i$ and $m_i$ by the equalities
$$\eqalign{e_{i-1}&=n_ie_i \cr \beta_i&=m_ie_i \hskip2truecm \hbox{for}\
1\leq i\leq g}$$ and note that we may rewrite the expansion of $y$ into
powers of $t$ as an expansion of $y$ into fractional powers of $x$ as follows:
$$\eqalign{y=& a_nx+a_{2n}x^2+\cdots
+a_{kn}x^k+a_{\beta_1}x^{m_1\over
n_1}+a_{\beta_1+e_1}x^{{m_1+1}\over n_1}+\cdots
+a_{\beta_1+k_1e_1}x^{{m_1+k_1}\over n_1}\cr &\
+a_{\beta_2}x^{m_2\over n_1n_2}+a_{\beta_2 +e_2}x^{{m_2+1}\over
{n_1n_2}}+\cdots 
+a_{\beta_q}x^{m_q\over
{n_1n_2\cdots n_q}}+a_{\beta_q+e_{q-1}}x^{{m_q+1}\over
{n_1n_2\cdots n_q}}+\cdots \cr &\ +
a_{\beta_g}x^{m_g\over {n_1n_2\cdots n_g}}+a_{\beta_g+1}x^{{m_g+1}\over
{n_1n_2\cdots n_g}}+\cdots }$$ 
The set of pairs of coprime integers $(m_i,n_i)$ are sometimes also called the
Puiseux characteristic pairs. Their datum is obviously equivalent to that
of the characteristic exponents $\beta_i$.
The sequence of integers $B(X) =(\beta_0,\beta_1,\ldots ,\beta_g)$, where $\beta_0=n$, may
be characterized algebraically as follows: let ${\bf \mu}_n$ denote the group of
$n$-th roots of unity. For $\omega \in {\bf \mu}_n$ let us compute the
order in $t$ of the series $y(t)-y(\omega t)$. If we write $\og=e^{{2\pi
ik}\over n}$, we have
$$y(\og t)= a_n\og^nt^n+\cdots 
+a_{\beta_1}\og^{\beta_1}t^{\beta_1}+\cdots$$ and we see that multiplying
$t$ by $\og$ does not affect the terms in $t^{jn}$. The term in $t^{\beta_1}$ is
unchanged if and only if $\og^{\beta_1}=1$, that is $k\beta_1\over n$ is an
integer, i.e; $k\beta_1=ln$ or $km_1=ln_1$ with the notations introduced
above. Since $n_1$ and $m_1$ are coprime, this means that $k$ is a multiple
of $n_1$, which is equivalent to saying that $\og $ belongs to the subgroup
${\bf \mu}_{n\over n_1}$ of ${\bf \mu}_n$ consisting of ${n\over
n_1}=n_2\cdots n_g$-th roots of unity. If this is the case, then the
coefficients of all the terms of the form $t^{\beta_1+je_1}$ in the Puiseux
expansion are also unchanged when $t$ is multiplied by $\og$, and the first
term which may change is $a_{\beta_2}t^{\beta_2}$. An argument similar to
the previous one shows that if $\og \in {\bf \mu}_{n\over n_1}$, then
$\og^{\beta_2}=1$ if and only if $\og \in {\bf \mu}_{n\over {n_1n_2}}$, and
so on. 
\par
Finally, if we denote by $v$ the order in $t$ of an element of $\C\lbrace
t\rbrace$, we see that 
$$v(y(t)-y(\og t))=\beta_i \ \ \hbox{if and only if}\ \  \og \in {\bf
\mu}_{n\over {n_1\cdots n_{i-1}}}\setminus {\bf
\mu}_{n\over {n_1\cdots n_i}}\  \ \hbox{for}\ \  1\leq i\leq g$$
This provides an algebraic characterization, and a sequence of cyclic 
subextensions 
$$\C\lbrace x\rbrace \subset \C\lbrace x^{1\over n_1}\rbrace \subset 
\C\lbrace x^{1\over
{n_1n_2}}\rbrace  \subset
\cdots \subset \C\lbrace x^{1\over
{n_1n_2\cdots n_i}}\rbrace  \subset \cdots \subset \C\lbrace x^{1\over
n}\rbrace $$ corresponding to the nested subgroups ${\bf
\mu}_{n\over {n_1\cdots n_i}}$ of th group ${\bf \mu}_n$. \par This shows
that the sequence $(\beta_0,\beta_1,\ldots \beta_g)$ depends only
upon the ring inclusion $\C\lbrace x\rbrace \subset {\cal O}_{X,0}$.\par We
shall see later in a different way that this sequence does not depend upon the
choice of coordinates $(x,y)$ in which we write the Puiseux expansion \it as
long as the curve $x=0$ is transversal to $X$.\rm If this is not the case, one
still obtains other characteristic exponents, which are related to the
transversal ones by the \it inversion formula \rm which I leave as an exercise.\par For example
consider the curve with equation $y^3-x^2=0$.\par\vskip.2in
I refer to the lectures of L\^e for the Burau-Zariski topological
interpretation of the characteristic sequence $(\beta_0,\beta_1,\ldots
\beta_g)$ as a characteristic of the iterated torus knot that one obtains upon
intersecting the branch $X$ with a sufficiently small sphere in $\C^2$
centered at the origin. \par
Given a germ of a reduced plane curve $X$, it has a decomposition
$X=\bigcup_{i=1}^r X_i$ into branches; each branch has its characteristic
sequence $B(X_i)$, and as numerical characters of $X$, we have also the
intersection numbers $(X_i,X_j)_0$ of distinct branches at
$0$.\par If we remember that these intersection numbers are equal to the
linking numbers in ${\bf S}^3$ of the knots corresponding to
$X_i$ and $X_j$ and are therefore topological characters of the link
$X\cap {\bf S}^3_\epsilon$, since Milnor proved (see L\^e's lectures) that the
curve $X$ is homeomorphic to the cone with vertex $0$ drawn on this link,
we expect that the collection of the characteristic sequences of the branches
and their intersection numbers may be a topological invariant of the curve
$X$. \par
Let us define the local topological type of a germ of subspace of $\C^N$ as
follows:\par\noindent
{\bf Definition}.- Two subspaces $X_1$ and $X_2$ of $\C^N$ are topologically
equivalent at $0$ if there exist neighbourhoods $U$ and $V$ of $0$ in $\C^N$
and an homeomorphism $\psi \colon U\to V$ such that $\psi (X_1\cap
U)=X_2\cap V$. Two germs at $0$ of subspaces are topologically equivalent
if they have representatives which are topologically equivalent at
$0$.\par\vskip.2in
\noindent {\bf Theorem} (Zariski, Lejeune-Jalabert).- \it Two germs of plane
curves $X=\cup_{i\in I}X_i$ and $X'=\cup_{i\in I'}X'_i$ are topologically
equivalent if and only if there exists a bijection $\phi \colon I\to I'$
between their branches which preserves characteristics and
intersection numbers, that is, satisfies 
$$B(X'_{\phi (i)})=B(X_i) \ \ \hbox{for}\ i\in I,\ \ (X'_{\phi (i)},X'_{\phi
(j)})_0=(X_i,X_j)_0 \ \ \hbox{for}\ i\neq j .$$\rm 
Topological equivalence is
less strict a relation than analytic (or even ${\cal C}^1$) equivalence. \par Let
$X_1$ and $X_2$ each consist of four lines through the origin in $\C^2$.
According to the previous theorem, these two germs are topologically
equivalent. However, if there was a germ et $0$ of a ${\cal C}^1$ (and in
particular analytic) isomorphism of $\C^2$ to itself,  sending $X_1$ to $X_2$,
its tangent linear map at $0$ would have to send $X_1$ onto $X_2$. But
two quadruplets of lines through $0$ are linearly equivalent if and only if
they have the same cross-ratio. If the slopes of the lines of $X_1$ are
$a_1,b_1,c_1,d_1$, and similarly for $X_2$, the cross ratios are
$$\big({{a_1-a_3}\over {a_1-a_4}}\big) \big({{a_2-a_4}\over
{a_2-a_3}}\big)$$ and the numbers obtained by permutation. It is therefore
easy to find examples where $X_1$ and $X_2$ are not ${\cal
C}^1$-equivalent.\par In particular, in an analytic family of curves such as
the surface in $\C^3$ with equation
$$(y-x)(y+x)(y-2x)(y+tx)=0$$
for small values of $t$, the fibers are all analytically inequivalent but
topologically equivalent.\Par
\par\vskip.2in \noindent
{\bf Theorem}.- \it Given two reduced germs of plane curves $(X,0)\subset
(\C^2,0)$ and $(X',0)\subset
(\C^2,0)$ the following conditions are equivalent:\par\noindent 
1) $X$ and $X'$ are topologically equivalent,\par\noindent 
2) There exists an integer $d$, a germ of curve $(C,0)\subset (\C^d,0)$ 
and two linear projections $p,p'\colon \C^d\to \C^2$, both general for $C$ at $0$, and such that
$p(C)=X,\ \ p'(C)=X'$,\par\noindent 
3) There exists a one-parameter family of germs of plane curves that is a germ along
$\{0\}\times U$ of a surface in $\C^2\times U$, where $U$ is a disk in $\C$, say
with equation $f(x,y,u)=0$ and $v, v'\in U$ such that the germs of plane curve $f(x,y,v)=0, \
f(x,y,v')=0$ are isomorphic to $X,\ X'$ respectively and all the germs $f(x,y,t)=0$ have the same
topological type for $t\in U$.\rm\par\vskip.2in
In fact, the theory of Lipschitz saturation, summarized in [T2], shows that, given the
topological type of a germ of plane curve $X$, there exists a space curve $X^s\subset \C^N$,
unique up to isomorphism, such that the germs of plane curves having the same topological
type as are exactly the images of $X^s$ by a linear projection which is generic for
$X^s$.\par\vskip.1in\noindent
The $\delta$ invariant of a plane curve singularity.\par\Par
Let $\O$ be the analytic algebra of a germ of curve $(X,0)$ , plane or not, and let $\overline \O$
be its normalization. Since it is an $\O$-module of finite type with the same total ring of
quotients, a version of the Hilbert Nullstellensatz shows that the quotient \it vector space \rm
over $\C$ is finite dimensional. So we may define an invariant to measure how far $\O$ is from
being integrally closed, i.e regular:
$$\delta_X=\hbox{\rm dim}_\C{\overline \O\over \O}$$
In the case of plane curves, this invariant has a geometrical interpretation, (see [T3]) which I will
describe only in the case of a branch, for simplicity:\par\noindent
Let $t^n,y(t)$ be a parametrization of our branch $X$. Consider the product of the normalization
of $X$ with itself, with coordinates $(t't')$ and the two curves in
$(\C^2,0)=(\overline X\times \overline X,0)$ defined by 
$${{t^n-t'^n}\over { t-t'}}=0\ \ \ {{y(t)-y(t')}\over {t-t'}}=0$$
The intersection number of these two curves at the origin is equal to $2\delta_X$; if now we
perturb slightly the parametrization of $X$ by $t^n+v\alpha t,y(t)+v\beta t$ with two ''general
'' complex numbers $\alpha, \ \beta$, we can see that the two curves now have deformed
equations and for small $v$ they now meet transversally in $2\delta_X$ points in $\overline
X\times \overline X$. this means that the curve defined parametrically by $t^n+v\alpha
t,y(t)+v\beta t$ has $\delta_X$ ordinary double points (two branches meeting transversally),
which tend to $0$ as $v$ tends to $0$. So we can view $\delta_X$ as the number of ordinary
double points which have coalesced to form the singularity of $X$ at the origin. Of course, for an
ordinary double point $\delta =1$. 

\par\vskip.2in\noindent
{ \sevenrm 5. THE SEMIGROUP OF A BRANCH}
\vskip.1in\noindent
There is another natural object associated to the inclusion $\O\to \overline \O$; again I will
decribe it only in the case of a branch. \par
Let $\O$ be the analytic algebra of a germ of analytically irreducible curve $X$, and let
$\overline
\O$ be its normalization; we have an injection $\O\to\overline
\O$ which makes $\overline\O$ an $\O$-module of finite type and $\overline\O$ is a
subalgebra of the fraction field of $\O$. Since $\overline\O$ is isomorphic to $\C\{t\}$, the
order in $t$ of the series defines a mapping $\nu\colon\C\{t\}\setminus {0}\to \N$ which
satisfies\Par
i) $\nu(a(t)b(t))=\nu(a(t))+\nu(b(t))$ and \Par ii) $\nu(a(t)+b(t))\geq
\hbox{\rm min}(\nu(a(t)),\nu(b(t)))$ with equality if $\nu (a(t))\neq \nu (b(t))$;\Par in other
words,
$\nu$ is a
\it valuation \rm of the ring 
$\C\{t\}$.\par
We consider the valuations of the elements of the subring $\O$, i.e the image $\Gamma$ of
$\O\setminus \{0\}$ by $\nu$; in view of i), it is a semigroup contained in $\N$. The fact that
$\overline\O$ is a finite $\O$-module implies that $\N\setminus \Gamma$ is finite, and in fact
we have for the $\delta$ invariant of $C$ the equality
$$\delta_X=\#(\N\setminus \Gamma)$$
Now we seek a minimal set of generators of $\Gamma$ as a semigroup: \Par
Let $\overline{\beta_0}$ be the smallest non zero element in $\Gamma$, let
$\overline{\beta_1}$ be the smallest element of $\G$ which is nor a multiple
of $\overline{\beta_0}$, let $\overline{\beta_2}$ be the smallest element of $\G$ which is not a
combination with non negative integral coefficients of $\overline{\beta_0}$ and
$\overline{\beta_1}$, i.e is not in the semigroup $\bigl\langle
\overline{\beta_0},\overline{\beta_1}\bigr\rangle$, and so on. Finally, since $\N\setminus
\Gamma$ is finite, we find in this way a minimal set of generators:
$$\G=\bigl\langle
\overline{\beta_0},\overline{\beta_1},\ldots ,\overline{\beta_g}\bigr\rangle$$
This set is uniquely determined by the semigroup $\G$, and of course determines it.\Par
By a theorem of Apery and Zariski, if $(X,0)$ is a plane branch, the datum of these generators, or
of the semigroup, is equivalent to the datum of the Puiseux characteristic of $(X,0)$, or of its
topological type.\par
Let us take the notations introduced for the Puiseux pairs; it is easy to check that if we set
$\beta_0=n$, the multiplicity, then $\overline{\beta_0}=\beta_0 =n, \overline{\beta_1}
=\beta_1$. After that is becomes more complicated. Zariski ([Z], Th. 3.9) proved the following
formula for
$q=2,\ldots ,g$:
$$ \overline{\beta_q}= (n_1-1)n_2\ldots n_{q-1}\beta_1+(n_2-1)n_3\ldots
n_{q-1}\beta_3+\cdots +(n_{q-1}-1)\beta_{q-1}+\beta_q,$$ which can be summarized in the
following recursive formula:
$$\overline{\beta_q}= n_{q-1}\overline{\beta_{q-1}}-\beta_{q-1}+\beta_q$$
It follows easily from this that the datum of the semigroup is equivalent to the datum of the
multiplicity $n$ and the Puiseux exponents $\beta_i$ of the curve.\par The semigroups coming
from plane branches are characterized among all semigroups of analytically irreducible germs
of curves by the following two properties:\Par
$$\leqno{1)}\ \  n_i\overline{\beta_i}\in \bigl\langle \overline{\beta_0},\ldots
,\overline{\beta_{i-1}}\bigr\rangle$$
$$\leqno{2)}\ \  n_i\overline{\beta_i}<\overline{\beta_{i+1}}$$
That the semigroups of plane branches have these properties follows from the induction
formula and the inequalities $\beta_i<\beta_{i+1}$. The converse can be proved by the
construction outlined below (see [Z], appendix).\par\smallskip
Conversely, given a semigroup $\G$ in $\N$ with finite complement, we can associate to it an
analytic (in fact algebraic) curve, called the \it monomial curve \rm associated to $\G$. 
If $\G=\big\langle
\overline{\beta_0},\overline{\beta_1},\ldots ,\overline{\beta_g}\big\rangle$, the monomial
curve $C^{\G}$ is described parametrically by
$$\eqalign{u_0=&t^{\overline{\beta_0}}\cr
u_1=&t^{\overline{\beta_1}}\cr
.&\cr
.&\cr.&\cr
u_g=&t^{\overline{\beta_g}}\cr}$$
On the other hand, the relations 1) above mean that there exist natural numbers $\ell_i^{(j)}$
such that we have 
$$\eqalign{ n_1\overline{\beta_1}=&\ell_0^{(1)}\overline{\beta_0}\cr
 n_2\overline{\beta_2}=&\ell_0^{(2)}\overline{\beta_0}
+\ell_1^{(2)}\overline{\beta_1}\cr
.&\cr.&\cr
n_j\overline{\beta_j}=&\ell_0^{(j)}\overline{\beta_0}+\cdots
+\ell_{j-1}^{(j)}\overline{\beta_{j-1}}\cr.&\cr.&\cr.&\cr
n_g\overline{\beta_g}=&\ell_0^{(g)}\overline{\beta_0}+\cdots
+\ell_{g-1}^{(j)}\overline{\beta_{g-1}}\cr}$$
These relations translate into equations for the curve $C^\G\subset \C^{g+1}$; since
$u_i=t^{\overline{\beta_i}}$, our curve satisfies the $g$ equations 
$$u_j^{n_i}-u_0^{\ell_0^{(j)}}u_1^{\ell_1^{(j)}}\ldots u_{j-1}^{\ell_{j-1}^{(j)}}=0,\ \ 1\leq j\leq
g,$$
 and it can be shown that they actually define $C^\G\subset \C^{g+1}$, so that if $\G$ is
the semigroup of a plane branch, $C^\G$ is a complete intersection.\par Remark that if we
give to $u_i$ the weight $\overline{\beta_i}$, the $i$-th equation is homogeneous of
degree $n_i\overline{\beta_i}$.\par The connection between a plane curve
$X$ having semigroup
$\G$ and the monomial curve is much more precise and interesting than the formal relation we
have just seen; by small deformations of the monomial curve one obtains all the branches
with the same semigroup. In fact the best way to understand all branches with semigroup
$\G$ is to consider the not necessarily plane curve $C^\G$ ($C^\G$ is plane if and only if
$C$ has only one characteristic exponent).\par
By definition of $\G$, there are elements $\xi_q\in \O$ with
$\nu(\xi_q)=\overline{\beta_q}$. We can write these elements in $\C\{t\}$ as
$$\xi_q=t^{\overline{\beta_q}}+\sum_{j>\overline{\beta_q}}\gamma_{q,j}t^j.$$
Let us consider the one-parameter family of parametrizations
$$\eqalign{u_0=&t^m\cr
u_1=&t^{\overline{\beta_1}}+\sum_{j>\overline{\beta_1}}v^{j-\overline{\beta_1}}\gamma_{1,j}t^j\cr
.&\cr.&\cr
u_g=&t^{\overline{\beta_g}}+\sum_{j>\overline{\beta_g}}v^{j-\overline{\beta_g}}\gamma_{g,j}t^j\cr
}$$
The reader can check that for $v\neq 0$, the curve thus described is isomorphic to our
original curve $C$. (hint: make the change of parameter $t=vt'$ and the change of coordinates
$u_j=v^{-\overline{\beta_j}}v'_j$, and remember the definition of the $\xi_j$). For $v=0$, we
have the parametric description of the monomial curve.\par
So we have in fact described a map $$\C\times \C\to \C^{g+1}\times\C$$ which induces the
identity on the second factors (with coordinate $v$). The image of this map is a surface, which
is the total space of a deformation of the monomial curve, all of its fibers except the one for
$v=0$ being isomorphic to our plane curve $C$. \par
So the monomial curve is a specialization, in this family, of our plane curve. In this
specialization the multiplicity and the semigroup remain constant; in a rather precise sense it
is an equisingular specialization, or one may say that \it the plane curve is an equisingular
deformation of the monomial curve with the same semigroup.\rm \par\smallskip
The same phenomenon can be also observed in the language of equations rather than
parametrizations. Let us consider a one parameter family of equations for curves in
$\C^{g+1}$, of the form
$$\eqalign{&u_1^{n_1}-u_0^{\ell_0^{(1)}}-vu_2=0\cr
 &u_2^{n_2}-u_0^{\ell_0^{(2)}}u_1^{\ell_1^{(2)}}-vu_3=0\cr &
.\cr&.\cr&u_{g-1}^{n_{g-1}}-u_0^{\ell_0^{(g-1)}}u_1^{\ell_1^{(g-1)}}\ldots
u_{g-2}^{\ell_{g-2}^{(g-1)}}-vu_g=0\cr&u_g^{n_g}-u_0^{\ell_0^{(g)}}u_1^{\ell_1^{(g)}}\ldots
u_{g-1}^{\ell_{g-1}^{(g)}}=0\cr}$$
For $v=o$ we get the equations of the monomial curve, and for $v\neq 0$ we get a curve
which has semigroup $\G$; this is a general heuristic principle of equisingularity: we have
added to each equation of the monomial curve, homogeneous of degree $n_i\overline{\beta_i}$,
a perturbation of degree $\overline{\beta_{i+1}}>n_i\overline{\beta_i}$, and this should
not change the equisingularity class (the perturbation is ''small'' compared to the equation).\par
Notice that for each fixed
$v\neq 0$ the curve described by the above equations is a plane curve: for simplicity take
$v=1$; then use the first equation to compute
$u_2=u_1^{n_1}-u_0^{\ell_0^{(1)}}$, substitute this in the next equation, and use this to
compute $u_3$ as a function of $u_0,u_1$, and so on. Finally the last equation gives us the
equation of a plane curve of the form
$${\Bigl(\cdots
{\bigl({(u_1^{n_1}-u_0^{\ell_0^{(1)}})}^{n_2}-u_0^{\ell_0^{(2)}}u_1^{\ell_1^{(2)}}\bigr)}^{n_3}-
\cdots\Big)}^{n_g}-u_0^{\ell_0^{(g)}}u_1^{\ell_1^{(g)}}{(u_1^{n_1}-u_0^{\ell_0^{(1)}})}^{\ell_2^{(g)}}
\cdots =0$$ 
The first consequence is that we can produce explicitely the equation of a plane curve with
given characteristic exponents: compute the semigroup and its generators, and then write the
equation above.\par
A more important fact is that one can show (see [Z], appendix) that any plane curve with a
given semigroup appears up to isomorphism as a fiber in a deformation depending on a
finite number of parameters: it is a deformation of the monomial curve obtained by adding to
the
$j$-th equation a polynomial in the
$u_i$'s of order
$>n_j\overline{\beta_j}$, and these polynomials can in principle be explicitely computed.\par
Finally, all the plane branches with the same semigroup have ''the same'' process of
resolution of singularities: you have to blow up points according to the same rules, the
multiplicities of the strict transforms are the same, and so on. So the resolution of the plane
curve described above shows the structure of the resolutions of all the curves with the same
semigroup.  First you resolve the curve $u_1^{n_1}-u_0^{\ell_0^{(1)}}=0$; when its strict
transform is non singular (after a number of blowing ups which depends on the continued
fraction expansion of the ratio ${\ell_0^{(1)}\over n_1}$, you take it as a coordinate axis: then
you have one parenthesis less in the equation above (the point is that the form of the
equation does not change), and you proceed like this. After $g$ such steps the branch is
resolved.\par
So the deformation to the monomial curve also explains to us how to resolve the singularities,
and it is perhaps the best description. Can we generalize it to higher dimensions?

\vskip1truecm
\centerline{\bf References}
\vskip1truecm
\noindent
[C] A. Chenciner, \it G\'eom\'etrie alg\'ebrique \'el\'ementaire \rm
Publ. Universit\'e Paris 7, No.  \par\noindent
[K] Kaup, L. , Kaup, B., and Barthel, G. \it Holomorphic functions of several variables, \rm De
Gruyter Studies in Math., 003, W. de Gruyter 1983. \par\noindent
[I] I. Newton, A Treatise on
the method of fluxions .
\par\noindent
[\L ] S. \L ojasiewicz, \it Introduction to complex analytic Geometry, \rm Birkh\"auser Verlag,
1991. \par\noindent
[T1] B. Teissier, The hunting of invariants in the Geometry of discriminants, in \it Real and
complex singularities,\rm P. Holm Ed., Sijtjoff and Noordhoff 1977.\par\noindent
[T2] B. Teissier, Vari\'et\'es polaires 2, \it Proc. La Rabida 1981, \rm Springer L.N.M. N0.
961.\par\noindent
[T3] B. Teissier, Simultaneous resolution I, II, in \it S\'eminaire sur les singularit\'es des
surfaces, \rm Demazure, Pinkham, Teissier, Eds., Springer LNM No. 777.\par\noindent
[W] R.J. Walker, \it Algebraic curves \rm Princeton University Press 1950
(Reprinted by Dover, New York 1962).\par\noindent
[Z] O. Zariski \it modules de branches planes (with an appendix by B. Teissier)\rm Hermann,
Paris 1986.
 \end