SJ ¨ALVST ¨ANDIGA ARBETEN I MATEMATIK MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET

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SJ ¨ ALVST ¨ ANDIGA ARBETEN I MATEMATIK

MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET

Puiseux parametrizations and invariants of plane algebroid branches

av

Jens Forsg˚ard

2010 - No 5

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Abstract

The Weierstrass preparation theorem for the ring of germs of holomorphic functions states that any such germ can be, possibly after a linear change of variables, represented as a unit times a Weierstrass polynomial. We give two proofs of the theorem, one using algebra and one using complex analysis.

By use of the preparation theorem we then show that plane branches have Puiseux parametrizations, i.e. parametrizations of the form T 7→ (T^k, φ(T )) where φ is analytic in a neighbourhood of the origin, and we give an algorithm to calculate φ. We define the multiplicity sequence of an analytic (algebroid) plane branch by use of repeated blowups, and show that it is an analytic invariant. By the Puiseux parametrization we can describe an analytic branch as a subring of CJtK, and we can then define its value semigroup and characteristic exponents. Either one of them uniquely determines, and is uniquely determined by, the multiplicity sequence. Hence they are also analytical invariants. Also, we show how to calculate the semigroup of a plane branch given its parametrization.

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Introduction

This paper could be seen as an introduction to the theory of plane analytical branches, with some generalizations. It has of course been a subject of intense study during the history, and we cover only tiny parts of these. We have no main problem which this paper tries to solve, and one could read the three main chapters (2, 3 and 4) as independent pieces.

Given a function f : Cⁿ→ C, assumed to be holomorphic in some neighbourhood of the origin, we are interested in how the curve f (X₁, . . . , X_n) = 0 behaves close to the origin. As our interest is local, we identify functions that coincide in some neighbourhood of the origin and talk about the ring of germs of holomorphic functions. Alternatively, one can consider the ex- pansion of f as a power series in X₁, . . . , X_n and talk about the ring of convergent power series.

One of the most important tools when working with germs of holomorphic functions is Weierstrass preparation theorem, in [5] even called the fundamental theorem of the theory of analytic functions of several variables. It states that every germ can, possibly after a linear change of variables, be represented as a unit times a Weierstrass polynomial, i.e. a monic polynomial in Xn, with the other coefficients being germs of holomorphic functions in X₁, . . . , X_n−1 that vanishes at the origin. This means that when considering curves f (X₁, . . . , X_n) = 0 locally, we can assume f to be a Weierstrass polynomial. In chapter 2 we give two proofs of the theorem, one in the context of germs of holomorphic functions and one in the context of convergent power series. Our main sources are [3] and [4].

In chapter 3 we use the preparation theorem and some ideas from algebraic topology to develop the concept of Puiseux parametrizations for plane branches. Every curve f (X, Y ) = 0 can locally be divided into irreducible branches. For each branch we get a parametrization X 7→ (X, φ(X¹^k)), or equivalently T 7→ (T^k, φ(T )), where φ is a germ of an analytic function. Se- ries of the form φ(X^k¹), i.e. power series with rational exponents, are usually called Puiseux series. We will use the term for φ(T ) as well. Last in chapter

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3 we give an algorithm for calculating the Puiseux series of a branch. Main sources are [3] and [6].

The most interesting case is when the origin is a singular point of the branch. In chapter 4 we will define the multiplicity of a branch, and it should be considered a measure to how singular the branch is. The most important concept is the quadratic transform or blowup of a branch. We think of this as a way to unlock singularities. One defines the multiplicity sequence of a branch as the sequence of multiplicities one gets from making repeated blowups. We will show this is an analytic invariant. It is actually even a topological invariant, and for analytic branches it also determines the topology class. One can also use them to show the classical result that singularities of plane branches (over C) can be resolved.

By use of the Puiseux parametrization we can write a branch as a subring of CJtK. Since this is a DVR, each element of the branch has a value, and from this we can define the value semigroup of the branch. Also we will define the characteristic exponents of the branch, which is a certain subset of the set of exponents in the parametrization of the branch. Both the semigroup and the characteristic exponents uniquely determines, and is uniquely determined by, the multiplicity sequence. Hence they are also topological invariants (and determine the topology class). Finally we will characterize what semigroups and multiplicity sequences that occur for plane branches, and give an algorithm to calculate the semigroup. Our main source in chapter 4 is [1].

1.1 History

One can say algebraic geometry has two sides, one purely algebraic side and one analytic and topological side. Both saw (not independently) much progress during the 19th century. One usually mentions names like Niels Henrik Abel (1802-1829), Augustin-Louis Cauchy (1789-1857), Bernhard Riemann (1826-1866) and Karl Weierstrass (1815-1887).

During this time the focus of the study of holomorphic functions began to shift away from elliptic funtions to power series expansions. Cauchy proved in 1831 that an analytic function can be expanded as a power series, and laid the foundation for the theory on several variables as well [5, p.

149ff]. One of the most important results for functions of several variables is the preparation theorem. Weierstrass proof is from the 1860s, however the attribution to him has been questioned as he was probably not the first to give a proof of the theorem [5, p. 152].

Weierstrass had a way of not publishing his results, but spreading them through his lectures only. Due to this there are also examples of results which was obtained first by Weierstrass, but which are not attributed to him. As in the case of the generalized form of power series expansions

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known as Laurent series, on which Weierstrass wrote a paper in 1841, two years before Pierre Alphonse Laurent (1813-1854) published his. Also, parts of the paper published by Victor Puiseux (1820-1883) in 1850 had been obtained by Weierstrass in 1842 [5, p. 153], though not the use of what is now known as Puiseux series. Calculations with such series was nothing new, the classical way to calculate the Puiseux parametrization of a branch (described in [3, chapter 7]) goes back all the way to Isaac Newton (1643- 1727). But Puiseux’ paper was a breakthrough in the theory of multi-valued analytic functions, where only local methods can be used. Especially he contributed to a better understanding of singular points. For example by making a distinction between poles and critical points, something Cauchy never did [5, p. 189ff].

The algebraic side of algebraic geometry saw much of its progress by a systematic use of projective geometry, both on Riemann surfaces and on higher dimensional varieties. One usually mentions the italian school from 1870 to 1940, as a source to many of the important ideas. However the rigorous threatment had to wait until the 20th century with likes of Oscar Zariski (1899-1986) and Bartel van der Waerden (1903-1996) [2]. One problem classicly studied was if it is possible to resolve singularities. A proof that it is possible in any dimension, over a field of characteristic zero, was given by Heisuke Hironaka (1931-) in 1964.

Semigroups have mainly been studied in the 20th century. Zariski showed in 1932 that the multiplicity sequence determines the topology class of a plane branch, and in 1952 that two branches have the same multiplicity sequence iff they have the same semigroup. The theory has developed much since then, and this is still an active field of research.

1.2 Notation

CJX1, . . . , X_nK is the ring of all formal power series of n variables, and R = ChX1, . . . , Xni is its subring of all convergent power series of n variables.

By R⁰ we will mean the subring R⁰ = ChX1, . . . , X_n−1i of R. Quite often we will use R⁰[X_n] which is the ring of polynomials in X_n with coefficients in R⁰.

When working with power series the variables are most often denoted by X₁, . . . , X_n, but when working with holomorphic functions it is more or less standard to use z1, . . . , zn. In this text we will use both notation. We have chosen to do this since it gives our equations a similiar look to those found in books, and it gives the reader a chance to see from the notation in what context we are currently working. Hopefully, it is more clarifying than confusing.

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Chapter 2

Weierstrass preparation theorem

Working towards Puiseux parametrizations, the Weierstrass preparation theorem will be one of our most important lemmas. Here we will give two proofs of the theorem, one using algebra and one using complex analysis. This is possible because of the fact that a function on Cⁿ that is holomorphic in a neighbourhood of 0 can be locally represented as a convergent power series, and vice versa. Or to put it more technically:

Let U_δ_i(zi) denote an open disc in C around zi with radius δi > 0, and let C_δ_i(z_i) be the (circular) boundary of U_δ_i(z_i). Then U_δ(z) = U_δ₁(z₁) ×

· · · × U_δ_n(zn) is called a polydisc in Cⁿ, and Cδ(z) = Cδ1(z1) × · · · × Cδn(zn) is called its determining set. By [U_δ(z)] we mean the closure of U_δ(z). When z = 0 we will only write U_δ and C_δ. The following claims are true:

i If f ∈ R converges for some z ∈ Cⁿ then f is absolutely convergent for all z ∈ U_δ where δ = (|z₁|, . . . , |z_n|) and it represents a holomorphic function there.

ii If f is a holomorphic function on Uδ then f can be expanded as a power series in z₁, . . . z_n wich converges on U_δ.

Hence R can be considered either as the ring of convergent power series, or as the ring of germs of holomorphic functions at 0, i.e. the ring of all functions holomorphic in some neighbourhood of 0 where two functions are identified if they coincide in some neighbourhood of 0. For a more complete discussion, including proofs, see [4].

For f ∈ R let ¯f = f (0, . . . , 0, X_n). We will call f general in X_n if ¯f 6= 0.

If f ∈ R⁰[Xn] let f = Pk

i=0fiX_nⁱ where fi ∈ R⁰. Then we will call f a Weierstrass polynomial if f₀(0) = · · · = f_k−1(0) = 0 and f_k = 1.

The Weierstrass preparation theorem states that any f ∈ R that is general in Xn has a unique representation f = up where u is a unit and p is a Weierstrass polynomial. One consequence of this is that when studying a

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curve f (z) = 0 where f is general, then locally around 0 this curve coincides with a curve p(z) = 0 where p is a Weierstrass polynomial. Hence we can, by looking at the curve locally, restrict ourselves to such curves. For the sake of universalness we point out that for any f ∈ R there is a linear change of variables that fixes 0 and makes f general in Xn, this will be proved later on.

2.1 The algebraic approach

Thus we now consider elements in R as convergent power series. We will use multi-indices to handle such series more easily. For f ∈ R we write f =P

νaνX^ν where ν ∈ Nⁿ0 and X^ν =Q

iX_i^νⁱ. We will let |ν| = ν1+· · ·+νn. If f ∈ R⁰[X_n] then we have the well known concept of the degree of f . Here we will let deg_n(f ) denote the degree of f in the variable Xn. For arbitrary f ∈ R we will define a similar concept called the order of f , wich is the lowest power that appears in f :

ord(f ) =

∞ if f = 0

min_a_ν6=0(|ν|) else.

We have that ord(f + g) ≥ min(ord(f ), ord(g)) and ord(f g) = ord(f ) + ord(g).

Working with power series, if f is general in Xn, then ord( ¯f ) = k <

∞ and we say that f is general in X_n of order k. Thus assuming f is general gives us more information if we look at f as a power series. Another construction that is only possible if we view f as a power series is the following; whenever k is clear from context we will use the notation f = f + Xˆ _n^kf where ˆ˜ f only have powers of X_n smaller then k. In this case, that f is general of order k means thatf = 0 and ˜¯ˆ f (0) 6= 0.

Working with different constructions of convergent power series, we will often need to prove that the result is a convergent power series as well. In order to handle this we will consider the following family of functions: for δ ∈ Rⁿ with δ > 0 we define k · k_δ : CJX1, . . . , X_nK → [0, ∞] as kf kδ = P

ν|a_ν|δ^ν.

Remark 2.1. For δ, δ⁰ > 0 and f ∈ CJX1, . . . , X_nK we have a) kf k_δ ≥ 0 and kf k_δ= 0 ⇔ f = 0

b) kλf k_δ= |λ|kf k_δ if λ ∈ C c) kf + gk_δ≤ kf k_δ+ kgk_δ d) if δ⁰ ≤ δ then kf k_δ⁰ ≤ kf k_δ

e) |f (x)| ≤ kf k_δ where δ = (|x₁|, . . . , |x_n|).

All proofs are easy. One could say that the next lemma is the reason why we choose to work with this family of functions. It also relates them to the notation U_δ. It is a special formulation of Weierstrass M-test, and we will refer to it by that name.

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Lemma 2.2 (Weierstrass M-test). Let {f_n}^∞_n=0 be a sequence of formal power series. If ∃δ > 0 such thatP∞

n=0kf_nk_δ< ∞, thenP∞

n=0fn converges uniformly on [U_δ].

Proof. For x ∈ [U_δ] by 2.1.e we haveP∞

n=0|f_n(x)| ≤P∞

n=0kf_nk_δ. Hence the sum is absolutely convergent in [U_δ], and since C is a complete metric space it converges pointwise. Also sinceP∞

n=0kf_nk_δ< ∞ we have that ∀ > 0 ∃N such that for n₀ ≥ N we have P∞

n=n0kf_nk_δ < . Hence |P∞

n=n0f_n(x)| < for all x ∈ [Uδ], so the sum is uniformly convergent there.

Knowing this we can formulate four more, less trivial but as important, properties. We will say that f = P

νaνX^ν and g = P

νbνX^ν have no common powers ν if a_ν 6= 0 ⇒ b_ν = 0 and b_ν 6= 0 ⇒ a_ν = 0.

Lemma 2.3. For δ > 0 and f ∈ CJX1, . . . , XnK we have a) f ∈ R ⇔ ∃δ > 0 such that kf k_δ < ∞

b) if f and g have no common powers ν then kf + gk_δ= kf k_δ+ kgk_δ c) if f ∈ R then limδ→0kf k_δ= |f (0)|

d) if f, g ∈ R then kf gk_δ≤ kf k_δ· kgk_δ

Proof. a) (⇒) is one of the remarks in the beginning of this chapter, (⇐) is clear from the preceeding lemma.

b) If kf k_δ= ∞ then since each term in the sum kf k_δappears in kf + gk_δ the relation is clear. The cases kgk_δ = ∞ is shown similarly. If both kf k_δ and kgk_δ are finite, then ∀ > 0 there exists a ν0 ∈ Nⁿ such that

kf k_δ+ kgkδ− ≤ X

ν<ν0

(|aν|δ^ν + |bν|δ^ν) = X

ν<ν0

|a_ν + bν|δ^ν ≤ kf + gk_δ

Let → 0 to get one inequality, the other is immediate.

c) Let > 0. Since f ∈ R, pick a δ₀such that kf k_δ₀ < ∞. Then ∃ν₀such that P

ν≥ν0|a_ν|δ^ν < if δ < δ₀. Now P

ν<ν0|a_ν|δ^ν has only finitely many terms, all but ν = 0 dependent on δ, hence we can decrease δ0 such that

|P

ν<ν0|a_ν|δ^ν− |f (0)|| < if δ < δ₀, hence |kf k_δ− |f (0)|| < 2 if δ < δ₀. d) f g = P

ν(P

ν1+ν2=νa_ν₁b_ν₂)X^ν so kf gk_δ = P

ν|P

ν1+ν2=νa_ν₁b_ν₂|δ^ν. Since kf k_δkgk_δ =P

ν(P

ν1+ν2=ν|a_ν₁||b_ν₂|)δ^ν the relation is clear.

To give an example of how this notation will be used we prove the following well-known fact.

Theorem 2.4. Let f ∈ R. Then f is a unit ⇔ f (0) 6= 0.

Proof. (⇐) Assume f (0) = 1 and let g = 1 − f and h = P∞

n=0gⁿ. Since g(0) = 0 by 2.3.c ∃δ such that kgk_δ< ¹₂, then for this δ we haveP kgⁿk_δ≤ P kgkⁿ_δ <P ₁

2ⁿ = 2 so that h ∈ R by 2.2. Now f h = (1 − g)P gⁿ= 1 so f is a unit. The other direction is trivial from counting with orders of power series.

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Thus f is general of order k if f = ˆf + X_n^kf where˜ f = 0 and ˜¯ˆ f is a unit. We will now prove what is sometimes called the preparation theorems twin sister: Weierstrass division theorem. This is really the hard part of our algebraic proof, the preparation theorem itself will follow as an easy consequence of the division theorem.

Theorem 2.5 (Weierstrass division theorem). Let f, g ∈ R and g be general in X_nof order k. Then there exists a unique representation f = qg +r where q ∈ R and r ∈ R⁰[X_n] with deg_n(r) < k.

Note that if g is general of order 0, then g is a unit, and the theorem is trivial. One intuitive way to think of the theorem is that if g is general in Xn of order k then the tail of g (i.e. the part of g where Xn has a power greater then k) is invertible.

Proof. We have that f = ˆf + X_n^kf and g = ˆ˜ g + X_n^kg where ˜˜ g is a unit. We do the recursive definition:

φ₀= f, φ_i+1= −ˆg˜g⁻¹φ˜_i Then

f = φ0=

∞

X

i=0

(φi− φ_i+1) =

∞

X

i=0

( ˆφi+ X_n^kφ˜i+ ˆg˜g⁻¹φ˜i) =

=

∞

X

i=0

( ˆφ_i+ ˜g⁻¹(ˆg + X_n^k˜g) ˜φ_i) =

∞

X

i=0

φˆ_i+ g˜g⁻¹

∞

X

i=0

φ˜_i = r + qg

where r = P∞

i=0φˆ_i and q = ˜g⁻¹P∞

i=0φ˜_i. We need to show that q, r ∈ R.

By 2.3.e we have kφ_ik_δ= k ˆφ_ik_δ+ δ^k_nk ˜φ_ik_δ. So k ˆφ_ik_δ≤ kφ_ik_δ and δ_n^kk ˜φ_ik_δ≤ kφ_ik_δ, hence using 2.2 it is enough to show that there exists a δ such that P∞

i=0kφ_ik_δ < ∞.

As kgk_δ = kˆgk_δ+ δ_n^kk˜gk_δ, we can choose a δ such that kˆgk_δ, k˜g⁻¹k_δ and kf k_δ are all < ∞. Since ˆg(0) = 0 we can pick a δ⁰ such that 0 < δ⁰ < δ and kˆgk_δ⁰ < _k˜_g−1¹kδ. Then

kˆg˜g⁻¹k_δ⁰ ≤ kˆgkδ⁰k˜g⁻¹k_δ⁰ ≤ kˆgkδ⁰k˜g⁻¹k_δ = < 1 thus

kφ_i+1k_δ⁰ = kˆg˜g⁻¹φ˜ik_δ⁰ ≤ kφ_ik_δ⁰ so recursivly we get

kφ_i+1k_δ⁰ ≤ ⁱ⁺¹kf k_δ⁰ and hence

∞

X

i=0

kφ_ik_δ⁰ ≤ kf k_δ⁰

∞

X

i=0

ⁱ = kf k_δ⁰ 1

1 − < ∞

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hence q, r ∈ R. It only remains to show uniqueness, which we do by showing that if f = 0 then q = r = 0. So assume 0 = qg + r. The coefficient of X_n^k in g is 6= 0. If there is an i such that the coefficient of X_nⁱ in q is 6= 0, then since deg_nr < k we get that the coefficient of X_n^k+i in the right hand side is 6= 0. Thus q = 0, and this implies r = 0 as well.

This proof might seem constructive, but it really is not. The φ_i:s con- verge, but not in such a manner that q and r can be easily calculated this way. By the division theorem we can, in some sense, divide with general power series. It can of course not be extended to a division algorithm, since we know that R is not a PID (assuming n ≥ 2 that is). We will however show that R is a UFD later on, using the preparation theorem.

Theorem 2.6 (Weierstrass preparation theorem). Let f ∈ R be general in Xn of order k. Then there exists an unique representation f = up, where u ∈ R is a unit and p ∈ R⁰[X_n] is a Weierstrass polynomial of degree k.

Proof. By assumption f is general in Xnof order k. We divide X_n^kby f and get X_n^k= qf + r where deg_n(r) < k and thus

qf = X_n^k− r = X_n^k−

k−1

X

i=0

a_iX_nⁱ

where ai∈ R⁰. Substituting X1= · · · = Xn−1= 0 we get

q(0, X_n)(cX_n^k+ . . . ) = X_n^k+

k−1

X

i=0

a_i(0)X_nⁱ

and comparing the coefficients gives q(0, 0) = ¹_c and a₁(0) = · · · = a_k−1(0) = 0. So q is invertible, and writing f = q⁻¹·(X_n^k+Pk−1

i=0a_iX_nⁱ) = up where u = q⁻¹ and p = X_n^k+Pk−1

i=0 a_iX_nⁱ completes the proof of existence. Uniqueness follows from uniqueness in the division theorem.

Now we will use this result for a first time to prove one of the lemmas we will need in chapter 3. A polynomial f ∈ R⁰[Xn] is said to be normalized if ¯f (Xn) = X_n^k+ a_k−1X_n^k−1+ · · · + a0.

Lemma 2.7 (Hensel’s lemma). Let f ∈ R⁰[Xn] be normalized and let ¯f = (Xn− c₁)^k¹. . . (Xn− c_r)^k^r where each ci are distinct. Then there are normalized polynomials f₁, . . . , f_r ∈ R⁰[X_n] such that f = f₁· · · f_r, deg_nf_i = k_i and ¯fi= (Xn− c_i)^kⁱ.

Proof. The proof is by induction on r. If r = 1 then f = f₁. Assume it is true for r = m − 1, and let r = m. If ci = 0 for some i then reorder so that i = m. Then g(X) = f (X1, . . . , Xn−1, Xn+ cm) is general in Xn

of order k_m. Hence, g = up where p is a Weierstrass polynomial of degree

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k_m. Setting f_m(X) = p(X₁, . . . , X_n−1, X_n− c_m) we see f_m is normalized and ¯fm= (Xn− c_m)^k^m. Letting f⁰(X) = u(X1, . . . , Xn−1, Xn− c_m) we have f = f⁰fm so ¯f⁰ = (Xn− c₁)^k¹· · · (X_n− c_m−1)^k^m−1 and then we can use the induction hypothesis on f⁰, wich completes the proof.

2.2 The analytic approach

We will now consider R as the ring of germs of holomorphic functions at 0, most calculations will however be made in the ring of all functions that are holomorphic in some neighbourhood of 0. The big advantage working with holomorphic functions, is the strong tools we have from complex analysis.

We will assume the reader to be familier with such tools, as Cauchy’s integral formula, Riemann’s removable singularities theorem and residue calculus, in the one variable case. They can all be found in [7].

Given a function f continous on C_δ(z) we define its integral by Z

Cδ(z)

f (z)dz1. . . dzn:=

Z

C_δ1(z1)

. . . Z

C_δn(zn)

f (z)dz1. . . dzn.

Theorem 2.8 (Cauchy’s integral formula for several variables). Let f be a holomorphic function in a domain D ⊂ Cⁿ. Take a polydisc U_δ(c) ⊂ D such that [U_δ(c)] ⊂ D. Then for z ∈ U_δ(c), f (z) can be represented as

f (z) =

1 2πi

nZ

C_δ(c)

f (w₁, . . . w_n)

(w1− z₁) . . . (wn− z_n)dw1. . . dwn

Proof. The proof is by induction on n. If n = 1 then this is Cauchy’s integral formula, thus true. Assume it is true for n = k − 1, then for n = k we get

Z

Cδ(c)

f (w₁, . . . w_k) Qk

i=1(w_i− z_i)dw₁. . . dw_k=

= Z

C_δ1(c1)

. . . Z

C_δk(c_k)

f (w₁, . . . w_k) Qk

i=1(wi− z_i)dw₁. . . dw_k=

= Z

C_δ1(c1)

. . . Z

C_δk−1(c_k−1)

R

C_δk(ck)

f (w1,...wk) (wk−z_k) dw_k Qk−1

i=1(wi− z_i) dw_k−1. . . dw₁ = by Cauchy’s integral formula in one variable

= Z

C_δ1(c1)

. . . Z

C_δk−1(ck−1)

2πif (w₁, . . . , w_k−1, z_k) Qk−1

i=1(w_i− z_i) dw_k−1. . . dw₁ =

= 2πi Z

C_δ1(c1)

. . . Z

C_δk−1(ck−1)

f (w1, . . . , w_k−1, z_k) Qk−1

i=1(wi− z_i) dw_k−1. . . dw1=

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by the induction hypothesis

= (2πi)^kf (z1, . . . z_k) So it is true for n = k.

Next we need the concept of analytic continuation, which is based on the following lemma.

Lemma 2.9. If two holomorphic functions f and g in a domain D coincide in some neighbourhood of a point c ∈ D, then f and g coincide in all of D.

Proof. We will prove it by showing that if a holomorphic function and all its partial derivatives vanishes in a point c ∈ D, then it coincides with the zero function in all of D. Expanding f as a power series around c, all coefficients in the power series are 0. Hence f (z) ≡ 0 in a polydisc around c. So the set D0 of all points where f and all its partial derivatives vanish is open. But the set D1 where f or some partial derivative doesn’t vanish is clearly open.

Since D₀∪ D₁= D, D₀∩ D₁ = ∅ and D 6= D₁ we have D = D₀.

This lemma is normally used as follows; let f₁ be holomorphic on D₁ and f2 be holomorphic on D2. If D1∩ D₂ 6= ∅ and open, and f₁ and f2

coincide on D1∩ D₂, then f2 is said to be the analytic continuation of f1

to D₂. One could also say that f₁ can be expanded to a function analytic on D1∪ D₂, by simply letting f1 := f2 on D2. This is reasonable because of our lemma: any analytic function on D1∪ D₂ that agrees with f1 on D1

must agree with f₂ on D₂.

We will now need to distinguish the n:th variable from the other. The following notation will be used: for z = (z1, . . . , zn) ∈ Cⁿ we will let z⁰ = (z₁, . . . , z_n−1) ∈ Cⁿ⁻¹. For δ = (δ₁, . . . δ_n) ∈ Rⁿ let δ⁰ = (δ₁, . . . , δ_n−1) ∈ Rⁿ⁻¹. For f ∈ R we will let fn= _∂z^∂f

n.

Lemma 2.10 (Continuity of the roots). Let f be holomorphic on U_δ. If there exists r ∈ (0, δ_n) such that f (z⁰, z_n) has no zeros for r ≤ |z_n| < δ_n, then there is a s ∈ N such that for each fixed z⁰, f (z⁰, zn) has exactly s zeros, counted with multiplicities, on U_δ_n.

Proof. For each fix z⁰ and r < R < δ_n let s(zn) = 1

2πi Z

CR

fn(z⁰, zn) f (z⁰, z_n)dzn

then, by the argument principle, s(zn) is the number of zeros, with multiplicities, of f (z⁰, z_n) on U_R. Since it depends continously on z⁰it is constant.

Theorem 2.11 (Weierstrass preparation theorem). Let f ∈ R be general in Xn. Then there exists a polydisc Uδ such that in this polydisc there is a unique representation f = up where u ∈ R is non-vanishing and p ∈ R⁰[Xn] is a Weierstrass polynomial.

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Comparing with the earlier formulation, we see that the degree of p can no longer be as easily determined by f . The claim on u is that it should be non-vanishing, but by 2.4 that is equivalent to u being a unit. Also we need to state that this representation is a local property, which is understood without saying when working with power series.

Before we begin the proof we would like to especially remind the reader about the fact that for a one variable analytic function, not identically zero, all zeros are isolated. Again, see [7].

Proof. By assumption, ¯f 6= 0 and f (0) = 0, hence 0 is an isolated zero of f (0, zn) of some order s. Thus taking 2 small enough and 0 < 1 < 2, f (0, zn) is holomorphic and f (0, zn) 6= 0 on 0 < 1 ≤ |z_n| ≤ ₂. Let D = {zn

such that ₁ ≤ |z_n| ≤ ₂} ⊂ C. Then D is compact, so |f(0, zn)| attains its minimum on D, let

m = min

zn∈D(|f (0, zn)|) > 0

Then taking δ⁰ small enough we get |f (z⁰, zn)| ≥ ^m₂ on zn∈ C_δ⁰× D. Hence σ_k(z⁰) = 1

2πi Z

C₁

fn(z⁰, zn)

f (z⁰, z_n) z_n^kdzn k = 0, 1, 2, . . .

are holomorphic functions on U_δ⁰. So they are continous. Looking at the previous lemma, we see that σ0(z⁰) = s for all z⁰. For a fix z⁰, let w1(z⁰), . . . , ws(z⁰) be the (not necessary distinct) zeros of f (z⁰, zn) in the disc |z_n| < ₁. By residue calculus, we get

σ_k(z⁰) = w₁(z⁰)^k+ · · · + w_s(z⁰)^k now let

p(z⁰, zn) =

s

Y

i=1

(zn− w_i(z⁰)) = z^s_n+ a1(z⁰)z^s−1_n + · · · + as(z⁰)

The elementary symmetric functions ai(z⁰) of wi(z⁰) can be expressed as polynomials in σ₀(z⁰), σ₁(z⁰), . . . without constant term. This is known as Newton’s identities or the Newton-Girard formulae. For example

a₁ = σ₁, a₂ = 1

2(σ₁²− σ₂), a₃ = 1

3(σ³₁− 3σ₁σ₂+ 2σ₃),

and so on. This means that i) they are holomorphic in Uδ⁰, and ii) remember- ing that for z⁰ = 0, w1(0) = · · · = ws(0) = 0, we get a1(0) = · · · = as(0) = 0.

Hence p is a Weierstrass polynomial. Now let

˜

u(z) = f (z⁰, zn)

p(z⁰, z_n), v(z) =˜ p(z⁰, zn) f (z⁰, z_n)

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We know that f and p are non-vanishing on U_δ⁰ × D, hence ˜u and ˜v are holomorphic there. Let δn = 2. For each z⁰ let ˜uz⁰(zn) = ˜u(z⁰, zn) and

˜

v_z⁰(zn) = ˜v(z⁰, zn). By construction, for each fix z⁰, f and p have the same zeroes with the same multiplicities in z_n on U_δ_n. Hence, by Riemann’s removable singularities theorem, both ˜uz⁰ and ˜vz⁰ can be extended analytic functions u_z⁰ and v_z⁰ on U_δ_n. Hence by setting u(z⁰, zn) = u_z⁰(zn) and v(z⁰, z_n) = v_z⁰(z_n) we get extensions of ˜u and ˜v which are holomorphic in z_n on Uδ= Uδ⁰ × U_δ_n. We use the integral formula for one variable to see that they are also holomorphic in z⁰. Choose γ as a circle around 0 in D. For each z⁰, we can represent u on U_δ as

u(z⁰, zn) = 1 2πi

Z

γ

u(z⁰, w) z_n− wdw

and similarly for v. This shows that both u and v are holomorphic in z⁰. Since 1 = ˜u˜v on U_δ⁰ × D, by uniqueness in analytic continuation we have that 1 = uv on U_δ, so u is a unit.

It is left to show uniqueness in the preparation theorem. We note that if f = u1p1 = u2p2 then, since u1, u2 is non-vanishing on U_δ, for every fixed z⁰, p₁ and p₂ have the same zeros in z_n. Since a one variable polynomial is uniquely determined by its zeros and the leading coefficient (which is 1 since both are Weierstrass polynomials), p1(z) = p2(z) ∀z ∈ U_δ, hence p1 = p2

as elements in the ring of germs.

This proof is a good example of how far you can get using only standard techiques from complex analysis. The most important part here is getting the double expressions for σi(z⁰), one that shows them being holomorphic and one used for the construction of p. Everything else is very straightfor- ward. Doing the residue calculus, which I have left out, is easy.

2.3 Changing variables

Let Φ : Cⁿ→ Cⁿbe a holomorphic automorphism of Cⁿ, i.e. a holomorphic homeomorphism from Cⁿ to itself, such that Φ(0) = 0. Given an f ∈ CJX1, . . . , X_nK we will say that f ◦ Φ is the function f after the change of variables Φ. Since Φ(0) = 0 and Φ is holomorphic it is clear that f ◦ Φ ∈ CJX1, . . . , XnK.

Lemma 2.12. Let f ∈ CJX1, . . . , X_nK.

i) If f is convergent then so is f ◦ Φ.

ii) If f is irreducible then so is f ◦ Φ.

Proof. i) This is clear since the inverse image of a neighbourhood of 0 under Φ is a neighbourhood of 0. ii) If f ◦ Φ would factor, then (f ◦ Φ) ◦ Φ⁻¹would give a factorization of f .

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As promised in the beginning of this chapter we now show that

Lemma 2.13. If 0 6= f ∈ CJX1, . . . , X_nK has ord(f ) = k, then there is a change of variables Φ such that f ◦ Φ is general in X_n of order k.

Proof. If f is already general in X_n of order k we are done. If not, set Xi = Yi+ ciYn if 1 ≤ i < n

Xn= Yn

Let Φ(Y ) = X, then Φ is a holomorphic automorphism for all choices of c₁, . . . , c_n−1 and Φ(0) = 0. Consider the initial polynomial of f ◦ Φ, i.e. the homogeneous polynomial of f ◦ Φ of degree k. This is not identically zero, since f has order k. Also, since f is not general in Xnof order k, each term in this polynomial has at least one factor Y_i+ c_iY_n. Hence it contains the term Y_n^kwith a coefficient that is a non constant polynomial in c₁, · · · , c_n−1. But then we can choose them so that f ◦ Φ is general in Yn of order k.

When considering the ring of germs we’re only interested in what hap- pends close to the origin, which motivates the following related definition.

Let Ψ be a holomorphic homeomorphism of two neighbourhoods of 0 such that Ψ(0) = 0. Then f ◦ Ψ is said to be the function f after the local change of variables Ψ. It is immediate that 2.12 holds for any local change of variables as well.

We actually know quite much about what a local change of variables looks like only from the definition. If we restrict ourselves to the plane case we can write Ψ(X, Y ) = (ψ₁(X, Y ), ψ₂(X, Y )). Let f (X, Y ) = X, then ψ₁ = f ◦ Ψ ∈ R, and by a similar argument ψ₂ ∈ R. Since Ψ(0) = 0, we have that ord(ψ1) ≥ 1 and ord(ψ2) ≥ 1. The same is of course true for Ψ⁻¹ = (ψ₁⁻¹, ψ⁻¹₂ ), and then since 1 = ord(X) = ord(ψ1(ψ₁⁻¹, ψ⁻¹₂ )) ≥ ord(ψ₁) we get that ord(ψ₁) = 1. By a similar argument ord(ψ₂) = 1. Since Ψ is invertible, at least one of ψ1 and ψ2 must be general in X of order one, and at least one must be general in Y of order one. Hence we can, also using the preparation theorem, always write

Ψ(X, Y ) =

u1· (X + a₁(Y )) u2· (Y + a₂(X))

or Ψ(X, Y ) =

u1· (Y + a₁(X)) u2· (X + a₂(Y ))

where u₁, u₂ are units and a₁(0) = a₂(0) = 0. One should note that this representation need not be unique. It is also clear that any Ψ of one of the above forms is a local change of variables.

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Chapter 3

Puiseux parametrizations

In this chapter we will develop the concept of Puiseux parametrizations for plane branches. A good way to introduce them is as a generelization of the implicit function theorem (for plane branches). Let f ∈ ChX, Y i. The implicit function theorem states that if f (0) = 0 and _∂Y^∂f 6= 0 (or in our words: if f is general in Y of order 1), then locally we get a parametrization of {(X, Y ) such that f (X, Y ) = 0} as T → (T, φ(T )) where φ is holomorphic. Puiseux theorem covers f that is general in Y of any order k. If f is also irreducible (any f of order 1 is irreducible), then locally we get a cor- responding parametrization on the form T → (T^k, φ(T )). If f is reducible, we will first get a local division of the zero set of f into branches, and then parametrizations of the branches.

There are some more things we need to know before we turn towards Puiseux theorem.

3.1 Unique factorization in R

Let A be a UFD. Recall that in A every irreducible is also prime, and that any PID is a UFD. The ”unique” factorization makes it possible for us to define the greatest common divisor of elements in A. Then any finite set of elements are called relativly prime if their gcd is a unit. We will not talk so much about gcd:s, but we will do the following definition: a polynomial in A[X] is called primitive if its coefficients are relatively prime. If f is irreducible and nonconstant in A[X], then it is primitive. Also recall that any integral domain A is a subring of its field of fractions, K.

Lemma 3.1. Let A be a UFD. If a is irreducible in A, then it is prime in A[X].

Proof. Assume a | gh, i.e. a divides every coefficient in the polynomial gh.

If a - g and a - h, let gn and hm be the lowest order coefficients in g and h such that a - gn and a - hm. Since a is prime in A, a - gnhm. Then the

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coefficient of X^n+m in gh is a sum in which a divides every term except gnhm, hence a cannot divide this coefficient, contradicting that a | gh.

Lemma 3.2. If f is irreducible in A[X] then it is irreducible in K[X].

Proof. Let f ∈ A[X] factor as f = gh in K[X]. Then there exists a ∈ A such that af = g⁰h⁰ in A[X]. Factor a into irreducibles, then each irreducible is prime in A[X], so each irreducible divides either g⁰ or h⁰. Hence f = g⁰⁰h⁰⁰in A[X]. Then either g⁰⁰ or h⁰⁰ is a unit, assume g⁰⁰. So g⁰⁰∈ A, which implies g⁰ ∈ A as well and hence g is a unit in K[X].

Theorem 3.4. If A is a UFD, then A[X] is a UFD.

Proof. Let f ∈ A[X] and let d = gcd of the coefficients in f . Recall that any nonconstant, irreducible factor of f is primitive. Assume we have two factorizations, which then are also factorizations in K[X]. But K is a field, so K[X] is a PID and hence a UFD, so the factorizations differ only by units in K[X]. Thus the factorizations in A[X] differ only by multiplication with constants, so any nonconstant factor is unique. But by 3.3 the constant factors are factorizations of d and hence unique.

This is a nice result on its own, but we only want it for the proof of 3.6.

Lemma 3.5. If f is an irreducible Weierstrass polynomial in R⁰[Xn] then it is irreducible in R

Proof. Let f be irreducible Weierstrass polynomial of degree k in R⁰[Xn], and assume it factors as f = f₁f₂in R. Then f₁ and f₂must both be general in X_n, say of orders k₁ and k₂, and k = k₁ + k₂. Apply the preparation theorem on f1 and f2 to get f = up1p2 where p1 and p2 are Weierstrass polynomials of degree k₁ and k₂. Then u ∈ R⁰ because otherwise the right hand side has a term with a power of X_n greater then k. Hence, this is a factorization in R⁰[Xn], so p1 or p2 is a unit, so k1 = 0 or k2 = 0, so f1 or f₂ is a unit.

We note especially that this lemma means that any linear Weierstrass polynomial g = X_n− φ(X₁, . . . , X_n−1) is irreducible in R.

Theorem 3.6. R is a UFD.

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Proof. We use induction on n. For n = 0 it is trivial, and for n = 1, R is a PID and hence a UFD. Assume it is true for n = k − 1 and let n = k. Let f ∈ R, and first assume f to be general in Xn. Then every factor of f must also be general in X_n. By the preparation theorem f is the product of a unit and a Weierstrass polynomial. By the induction hyphothesis R⁰is a UFD and by 3.4 R⁰[Xn] is a UFD, so this polynomial factors uniquely into irreducible polynomials in R⁰[X_n]. For any factorization of f into irreducibles in R, apply the preparation theorem to each factor to see that it differs only from the given factorization by multiplication with units. If f is not general in X_n, then there is a change of variables so that it becomes general in X_n. If f would have two different factorizations, this would give two different factorazations after the change of variables as well, a contradiction. Hence R is a UFD.

Knowing that R is a UFD we can make use of the following algebraic construction; given any ring A and two polynomials f (X) = a₀Xⁿ+ · · · + a_n and g(X) = b0X^m+ · · · + bm where a0 6= 0, b₀ 6= 0 and f, g ∈ A[X] we define the resultant of f and g to be the (n + m) × (n + m) determinant

Res_f,g =

a₀ · · · a_n

. .. . ..

a₀ · · · a_n b0 · · · bm

. .. . ..

b0 · · · bm

with the most important property being the following lemma:

Lemma 3.7. If A is a UFD and f, g ∈ A[X] then f and g have a common factor of degree ≥ 1 ⇔ Res_f,g = 0

Proof. Let deg(f ) = n and deg(g) = m. Since A is a UFD, we have that f and g have a common factor of positive degree iff there exists φ and ψ in A[X] such that deg(φ) < m, deg(ψ) < n and φf + ψg = 0.

It is also clear that there exist such φ and ψ in A[X] iff they exists in K[X]. Consider the vectorfield of polynomials in K[X] with degree <

n + m, with the basis X^n+m−1, . . . , X, 1. Then the rows of the resultant are just the vectors X^m−1f, . . . , Xf, f, xⁿ⁻¹g, . . . , Xg, g. The determinant of these vectors are zero iff they are linearly dependent, i.e. iff there exists λ₀, . . . λ_m−1, µ₀, . . . , µ_n−1 ∈ K such that 0 = λ₀f + · · · + λ_m−1X^m−1f + µ0g + · · · + µn−1Xⁿ⁻¹g = φf + ψg.

For a polynomial f ∈ A[X] we define the discriminant of f as Df = Res_f,f⁰ where f⁰ is the formal derivative of f . By 3.7 we have that f has a repeated zero in A ⇔ D_f = 0.

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Remark 3.8. We showed that R⁰is a UFD so the above result is valid for any f ∈ R⁰[Xn]. From the definition we see that Df is a holomorphic function of X⁰ and if D_f(X⁰) ≡ 0 then D_f = 0 and by 3.7 f has a repeated factor as a polynomial over R⁰.

One can also use the resultant to show an important property of the zero set of a holomorphic function. A set M ⊂ U_δ is called a principal analytic set if there exists f ∈ R such that M = {x ∈ U_δ such that f (x) = 0}. Two principal analytic sets is called equivalent if there exists a Uδ for which they coincide, and an equivalent class of principal analytic sets is called a germ of a principal analytic set, denoted by V (f ). We say that V (f ) ⊂ V (g) if there exists a Uδ and representatives f and g for which the inclusion of principal analytic sets holds.

Lemma 3.9 (Study’s lemma). Let f, g ∈ R and f be irreducible, then V (f ) ⊂ V (g) ⇔ f is a divisor of g in R.

Proof. (⇐) is clear, we prove (⇒). Let U_δ be a polydisc in Cⁿfor which the inclusion of the zero sets hold. By the preparation theorem we can assume f and g to be polynomials in R⁰[Xn].

If n = 1 then f and g are holomorphic funtions of one variable. Three cases can occur: i) V (f ) = ∅, but then f (0) 6= 0 so f is a unit and hence not irreducible. ii) V (f ) = {0} but the only such irreducible f is f = X.

Since V (f ) ⊂ V (g) we have g(0) = 0, so X divides g. iii) V (f ) ⊃ U_δ for all U_δ, i.e. f (X) ≡ 0, then g(X) ≡ 0 as well.

If n > 1, by 3.7 and since f is irreducible it is enough to show that Res_f,g = 0. For each X⁰ ∈ U_δ⁰ let f_X⁰ = f (X⁰, Xn) ∈ C[Xn] and similarly define g_X⁰. Then considering U_δ_n we see that V (f_X⁰) ⊂ V (g_X⁰) so by the reverse implication for an irreducible factor of f and then by the case n = 1, f_X⁰ and g_X⁰ must have a common irreducible factor. Hence Res_f,g(X⁰) = Res_f_X0_,g_X0 = 0 so Res_f,g = 0.

The germ of a principal analytic set is called reducible if we can find f1

and f₂ such that V (f ) = V (f₁) ∪ V (f₂), V (f_i) 6= ∅ and V (f₁) 6= V (f₂). It is called irreducible if it is not reducible.

Lemma 3.10. V (f ) is irreducible ⇔ there exists an irreducible g ∈ R such that f = g^k.

Proof. (⇒): Consider two irreducible factors g1 and g2 of f . Since V (f ) is irreducible, V (g1) = V (f ) = V (g2). Then since g1 and g2 both are irreducible, Study’s lemma gives that g₁ = g₂.

(⇐) If V (h) ⊂ V (f ) is nonempty then every irreducible factor of h divides f = g^k hence h = g^l so V (h) = V (f ).

Theorem 3.11. Each germ V (f ) of a principal analytic set has a decomposition V (f ) = V (f1) ∪ · · · ∪ V (fr) where each V (fi) is irreducible. The decomposition is unique up to reordering.

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Proof. f has a unique factorization f = f₁ⁿ¹· · · f_rⁿ^r where the f_i:s are distinct and irreducible. The decomposition is immediate using Study’s lemma and 3.10.

This decomposition is what we refer to as a decomposition of V (f ) into branches. The important part here is that each branch can be assumed to be defined by an irreducible f .

Example 3.12. A standard example is the nodal cubic f (X, Y ) = Y² − X²− X³. It is irreducible in C[X, Y ], but in ChX, Y i is splits into linear factors as f = (Y −X√

1 + X)(Y +X√

1 + X). Here we see its two branches around the origin.

3.2 Covering spaces

For topological spaces S and T , a continous mapping p : S → T is called a covering map if for all t ∈ T there exists a neighbourhood V 3 t such that each for each connected component W of p⁻¹(V ) the map p|W : W → V is a homeomorphism. If p is a covering map, then (S, p) is called a covering space of T . One calls the V :s and W :s from above the fundamental neighbourhoods of T and S. Whenever we use the term neighbourhood in the context of covering spaces we will mean a fundamental neighbourhood.

This section is devoted to a classification theorem from algebraic topology: if Uδ \ {0} is a punctured disc, then it has up to homeomorphism an unique k-sheeted, path-connected and locally path-connected covering space. Since to use this property is the main idea of our proof of Puiseux theorem, we will take a moment to go through some details. However, since we know where we are aiming, we can allow ourself to sometimes make stronger assumptions than what is necessary. A full treatment can be found in [6, chapter 5].

By a path in a topological space T we mean a continous function γ : I → T , where I denotes the unit interval in R. We say that two paths γ0 and γ₁ are homotopic if there exists a continuos function h : I² → T such that

h(0, x₂) = γ₀(x₂) h(1, x₂) = γ₁(x₂)

h(x₁, 0) = γ₀(0) = γ₁(0) h(x₁, 1) = γ₀(1) = γ₁(1)

and denote this by γ0∼ γ₁. Homotopy is an equivalence relation on the set of all paths γ : I → T . Fixing a t ∈ T , a path that begins and ends in t will be called a loop through t. The set of all homotopy classes of loops through t is a group under the operation

(γ₀+ γ₁)(x) =

γ₀(2x) if 0 ≤ x < ¹₂ γ₁(2x − 1) if ¹₂ ≤ x ≤ 1

SJ ¨ALVST ¨ANDIGA ARBETEN I MATEMATIK MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET