POLAR QUOTIENTS OF A PLANE CURVE AND THE NEWTON ALGORITHM

Andrzej Lenarcik

IMUJ PREPRINT 2003/12 KRAKÓW

... do not take on a great amount of business ... (Si 11,10)

Abstract

Using the Newton algorithm we show how to compute all the polar quotients and their multiplicities of a plane curve f = 0, where f is a formal power series of two variables over an algebraically closed field k with characteristic zero. The curve is not necessarily reduced.

1 Introduction

Let k be an algebraically closed field with characteristic zero. We use standard notations; k[[X, Y]] is the ring of formal power series, ord f is the order of f ∈ k[[X, Y]] (ord 0 = +∞). For elements a_{1}, …, a_{p} of a given set we define the system A = ⟨a_{1}, …, a_{p}⟩ as the sequence a_{1}, …, a_{p} treated as unordered. Put deg A = p (see [Wh], notion of symetric power).

Let f(X, Y) ∈ k[[X, Y]] be such a series that p = ord f(0, Y) > 1. Recall the polar curve ∂f/∂Y = 0 with its positive-order-roots z_{1}(X), …, z_{p−1}(X) in the ring k[[X]]^{∗} = ⋃_{n≥1} k[[X^{1/n}]] of the Puiseux series. We consider the system

2000 Mathematics Subject Classification: Primary 32S55.

Key words and phrases: plane curve singularity, polar quotients,

Newton polygon, Newton algorithm.

Supported in part by the KBN grant No 2 P03A 022 15.

Q̄(f, X) = ⟨ord f(X, z_{1}(X), …, ord f(X, z_{p−1}(X))⟩  (1)

of polar quotients of f with respect to X. Every polar quotient is either a positive rational or +∞. We omit the bar over Q to denote the system of finite quotients. We have Q̄(f, X) = Q(f, X) if and only if f is reduced.

When X and f are transverse, the polar quotients are topological invariants called polar invariants (k = C). Substracting one from the maximal polar invariant results in the Łojasiewicz exponent of grad f near zero [T].

The study of polar quotients and polar invariants extends over many authors ([T], [M], [E], [Eph], [CA], [LMW1], [LMW2], [Ga], [D], [GP1], [GP2], [LMP]). Examples include: Merle’s [M] description of polar invariants for irreducible series (see also [GP1]); Casas-Alvero’s developement of the case of two branches; and the computation of polar invariants for multi-branched singularity using Egger’s diagrams ([E], [Ga]). The authors of [GP2] give explicit formulae for the polar quotients in terms of characteristics and intersection multiplicities of branches. For a nondegenerate series f the polar invariants can be expressed using the Newton polygon of f [LP,LPM].

This paper aims to compute the system of polar quotients using a version of the Newton algorithm ([W], [Ca], [KP], [L]). We generalize the approach from [LP,LMP].

2 Main result

It is convenient to consider the ring k[[X^{∗}, Y]] = ⋃_{n≥1} k[[X^{1/n}, Y]]. Take f = ∑ f_{αβ}X^{α}Y^{β} ∈ k[[X^{∗}, Y]]. As usual we define the support supp f as {(α, β) : f_{αβ} ≠ 0}, the Newton diagram Δ(f) as conv(supp f + R_{+}^{2}), and the Newton polygon N (f) as the set of compact faces of Δ(f). By δ(f) we denote the distance between Δ(f) and the horizontal axis.

For S ∈ N (f), by |S|_{1} and |S|_{2} we denote the lengths of projections of S onto the horizontal and vertical axies, respectively. We call the ratio |S|_{1}/|S|_{2} the inclination of S. For θ ≥ 0 it will be useful to consider the conditional polygon N^{θ}(f), which consists of all the faces S ∈ N (f) with an inclination strictly greater than θ. Let α(S) denote the abscissa of the point where the line determined by S intersects the horizontal axis. We define the initial form in(f, S) = ∑ f_{αβ}X^{α}Y^{β}, where (α, β) runs over S ∩ supp f. By t(f, S) we denote the number of different linear factors of in(f, S) in k[[X]]^{∗}[Y]. The number ε(S) ∈ {−1, 0} is defined as −1 when S touches the horizontal axis and as 0 otherwise. Put d(f, S) = |S|_{2} + ε(S) − t(f, S) + 1. Note that d(f, S) = 0 if and only if every nonzero root of in(f, S) in k[[X]]^{∗} is single. Then we call the series f nondegenerate on S.

For any φ ∈ k[[X]]^{∗}, ord φ > 0 one can apply the substitution f_{φ}(X, Y) = f(X, φ(X) + Y) ∈ k[[X^{∗}, Y]] ([Ca], [GP1], [KP]). Clearly, f_{φ} = f for φ = 0. Consider ring k[X]^{∗} = ⋃_{n≥1} k[X^{1/n}] of Puiseux polynomials. For φ ∈ k[X]^{∗}, deg φ < +∞. Put deg 0 = −∞. The set T (f, X) ⊂ k[X]^{∗} of the tracks (of the Newton algorithm) for f is defined to be the minimal set satisfying two properties: (I) 0 ∈ T (f, X), (II) for every φ(X) ∈ T (f, X), if there exists S ∈ N^{deg φ}(f_{φ}), then for every nonzero root aX^{θ} of in(f_{φ}, S), φ + aX^{θ} ∈ T (f, X). We will write N_{φ} instead of N^{deg φ}(f_{φ}) when f is fixed.

We call series ψ ∈ k[[X]]^{∗} a continuation of φ ∈ k[X]^{∗} if ord (φ − ψ) > deg φ. Then we write ψ = φ + . . .. Let φ be a track of the Newton algorithm for f. Let Q̄_{φ}(f, X) denote the subsystem of (1) where solutions are continuations of φ. By analogy we use notation Q_{φ}(f, X) when quotients are finite.

Instead of ⟨\underbrace{a_{1}, …, a_{1}}_{m_{1} times}, … \underbrace{a_{p}, …, a_{p}}_{m_{p} times}⟩ we write ⟨a_{1} : m_{1}, … a_{p} : m_{p}⟩. For

A = ⟨a_{1}, …, a_{p}⟩ and B = ⟨b_{1}, …, b_{q}⟩ we have a natural addition A ⊕ B = ⟨a_{1}, … a_{p}, b_{1}, … b_{q}⟩ with the neutral element ⟨⟩. By convention ⟨a : 0⟩ = ⟨⟩.

Now, put C_{φ} = ⟨∞ : δ(f_{φ})−1⟩ if δ(f_{φ}) > 1 and C_{φ} = ⟨⟩ in all other cases. For S ∈ N_{φ} we denote A_{φ,S} = ⟨α(S) : t(f_{φ}, S) − 1⟩. We have the following

Theorem 2.1 (main result)

(a) For φ ∈ T (f, X) we have Q̄_{φ}(f, X) = [ ⨁_{S∈N_{φ}}(A_{φ,S} ⊕ B_{φ,S})]⊕C_{φ} , where

B_{φ,S} is a system of quotients strictly greater than α(S), deg B_{φ,S} = d(f, S).

(b) B_{φ,S} = ⨁_{aX^{θ}} Q̄_{φ+aX^{θ}}(f, X), where aX^{θ} runs over all multiple nonzero

roots of in(f_{φ}, S),

(c) Q(f, X) = ⨁_{φ∈T (f,X)} ⨁_{S∈N_{φ}} A_{φ,S}

We prove the theorem in the next section.

Remark 2.2 Clearly α(S) is a polar quotients if and only if t(f_{φ}, S) > 1. This condition is always satisfied for S ∈ N_{φ}, which does not touch the horizontal axis. When α(S) in not a polar quotient then S touches the axis and in(f_{φ}, S) = bX^{ζ}(Y − aX^{θ})^{|S|_{2}}.

Remark 2.3 We can consider the system A_{φ}(f, X) = ⨁_{S∈N^{φ}} A_{φ,S} of finite quotients which are determined by the behaviour of f_{φ} on N_{φ}. Clearly deg A_{φ}(f, X) = ∑_{S∈N^{φ}}[t(f_{φ}, S) − 1].

Remark 2.4 We state in the above theorem that the sum in (c) is in fact finite. We can replace T (f, X) by T_{min}(f, X), which contains all φ ∈ T (f, X) such that there exists S ∈ N_{φ} with t(f_{φ}, S) > 1.

For φ = 0 we have the following corollary. We write A_{S}, B_{S} instead of A_{0,S}, B_{0,S} and C instead of C_{0}.

Corollary 2.5 (a) Q̄(f, X) = [ ⨁_{S∈N (f)}(A_{S} ⊕ B_{S})] ⊕ C ,

(b) B_{S} = ⨁_{aX^{θ}} Q̄_{aX^{θ}}(f, X), where aX^{θ} runs over all possible multiple nonzero

roots of in(f, S).

Corollary 2.6 If δ(f) ≤ 0 and d(f, S) = 0 for every S ∈ N (f) then

Q̄(f, X) = Q(f, X) = A_{0}(f, X) = ⨁_{S∈N (f)}⟨α(S) : |S|_{2} + ε(S)⟩ .

[figure: α_{1}; α_{2}; …; α_{s}]

When S ∈ N (f) touches the horizontal axis and |S|_{2} = 1 then α(S) is not a polar quotient. We call such a face exceptional ([LP]).

Example 2.7 Consider the curve f = Y (Y^{2} − X)^{3} − 2XY^{2}(Y^{2} − X)^{2} + 8X^{5}Y + X^{6}. We have N (f) = {S, T }, where S joins (0, 7) with (3, 1) and T joins (3, 1) with (6, 0). From Corollary 5.2 Q̄(f, X) = A_{S} ⊕B_{S} ⊕A_{T} ⊕B_{T} ⊕C. Clearly in(f, S) = Y (Y^{2} − X)^{3} = Y (Y − X^{1/2})^{3}(Y + X^{1/2})^{3}, α(S) = 3\frac{1}{2}, t(f, S) = 3, d(f, S) = 4; in(f, T) = −X^{3}Y + X^{6} = −X^{3}(Y − X^{3}), α(T) = 6, t(f, T) = 1, d(f, T) = 0 and δ(f) = 0. Obviously A_{T} = B_{T} = C = ⟨⟩.

[figure: 1; −3; 3; −2; 4; −2; 8; 1; −1; Δ(f); S; T; 8; −8; 8; 1; 8; P; R; Δ(f_{X^{1/2}})]

Hence Q̄(f, X) = A_{S} ⊕ B_{S} = ⟨3\frac{1}{2}, 3\frac{1}{2}⟩ ⊕ B_{S}, where B_{S} contains four quotients strictly greater then 3\frac{1}{2}. Consider two multiple nonzero roots X^{1/2} and −X^{1/2} of in(f, S). We have B_{S} = Q̄_{X^{1/2}}(f, X) ⊕ Q̄_{−X^{1/2}}(f, X). Taking X^{1/2} as a track we obtain f_{X^{1/2}} = f(X, X^{1/2} + Y) = (Y + X^{1/2})Y^{3}(Y + 2X^{1/2})^{3} − 2X(Y + X^{1/2})^{2}Y^{2}(Y + 2X^{1/2})^{2} + 8X^{5}(Y + X^{1/2}) + X^{6}. The polygon N_{X^{1/2}} has two faces: P which joins (2, 3) with (3, 2) and R which joins (3, 2) with (5\frac{1}{2}, 0). We have Q̄_{X^{1/2}}(f, X) = A_{X^{1/2},P} ⊕ B_{X^{1/2},P} ⊕ A_{X^{1/2},R} ⊕ B_{X^{1/2},R} ⊕ C_{X^{1/2}}. Note that in(f_{X^{1/2}}, P) = 8X^{2}Y^{3} − 8X^{3}Y^{2} = 8X^{2}Y^{2}(Y − X), α(P) = 5, t(f_{X^{1/2}}, P) = 2, d(f_{X^{1/2}}, P) = 0; in(f_{X^{1/2}}, R) = −8X^{3}Y^{2} + 8X^{11/2} = −8X^{3}(Y − X^{5/4})(Y + X^{5/4}), α(R) = 5\frac{1}{2}, t(f_{X^{1/2}}, R) = 2, d(f_{X^{1/2}}, R) = 0 and δ(f_{X^{1/2}}) = 0. Since B_{X^{1/2},P} = B_{X^{1/2},R} = C_{X^{1/2}} = ⟨⟩ we finish with Q̄_{X^{1/2}}(f, X) = A_{X^{1/2},P} ⊕ A_{X^{1/2},R} = ⟨5, 5\frac{1}{2}⟩. Analogously, Q̄_{−X^{1/2}}(f, X) = ⟨5, 5\frac{1}{2}⟩ and finally Q̄(f, X) = A_{0}(f, X) ⊕ Q̄_{X^{1/2}}(f, X) ⊕ Q̄_{−X^{1/2}}(f, X) = ⟨3\frac{1}{2}, 3\frac{1}{2}, 5, 5, 5\frac{1}{2}, 5\frac{1}{2}⟩. As a result we see that f is reduced. We can compute the Milnor number from Teissier’s formula μ_{0}(f) = ∑ Q(f, X)−ord f(0, Y)+1 = 22. Note that T_{min}(f, X) = {0, X^{1/2}, −X^{1/2}}.

3 Proof of the main result

The proof of the main result will be performed in several steps. The first step is the classical Newton-Puiseux theorem, which provides a description of the Puiseux roots of a series in terms of its Newton polygon (Theorem 3.1). The second and fundamental step is the description of the Newton polygon of the derivative. In the final steps of the proof we applicate these methods to the Newton algorithm.

The Newton-Puiseux theorem

Let f ∈ k[[X^{∗}, Y]] be a series such that p = ord f(0, Y) > 1. Let Zer f = ⟨y_{1}(X), …, y_{p}(X)⟩ be the system of all the positive-order-roots of f = 0 in k[[X]]^{∗}. Consider S ∈ N (f) and define the form in(f, S)^{∘} by the equation in(f, S) = X^{a_{S}}Y^{b_{S}}in(f, S)^{∘} where a_{S} and b_{S} are the maximal possible powers. Let us note that deg Zer in(f, S)^{∘} = |S|_{2}. Fix θ > 0. By analogy to the conditional polygon N^{θ}(f), we define the conditional system of zeros Zer^{θ}f, which contains those roots of Zer f which satisfy ord y_{i}(X) > θ. By the heigh of N^{θ}(f) we mean ∑_{S∈N^{θ}(f)} |S|_{2} and we denote it by |N^{θ}(f)|. We need the Newton-Puiseux theorem in the following form (we use convention in 0 = 0).

Theorem 3.1 Let Zer^{θ}f = ⟨y_{1}(X), …, y_{s}(X)⟩. Then

(i) ⟨ord y_{1}(X), …, ord y_{s}(X)⟩ = ⨁_{S∈N^{θ}(f)}⟨|S|_{1}/|S|_{2} : |S|_{2}⟩ ⊕ ⟨+∞ : δ(f)⟩ ,

(ii) ⟨in y_{1}(X), …, in y_{s}(X)⟩ = ⨁_{S∈N^{θ}(f)} Zer in(f, S)^{∘} ⊕ ⟨0 : δ(f)⟩ .

(iii) s = |N^{θ}(f)| + δ(f).

The following property geometrically expresses the orders from (1). For θ > 0 and any closed nonempty subset Z ⊂ R_{+}^{2} we define the number

α(θ, Z) = min {α + βθ : (α, β) ∈ Z} .

Obviously α(θ, Δ_{f}) = α(θ, supp f). This number is the abscissa of the point where the line of inclination θ, supporting Δ(f), intersects the horizontal axis. We have the following Property 3.2 (see [L2] (iii)′, (iv)′ p. 24) Let z(X) ∈ k[[X]]^{∗}, 0 < ord z(X) < +∞. Then

(a) ord f(X, z(X)) ≥ α(ord z, Δ_{f}),

(b) the inequality is sharp if and only if there exists a face S ∈ N (f) such that |S|_{1}/|S|_{2} = ord z(X) and in z(X) is a root of in(f, S).

Polygon of the derivative

In this next step we describe the polygon N (∂f/∂Y). By eliminating the points from supp f that lie on the horizontal axis and by moving all remaining points one unit down, we obtain the support of ∂f/∂Y. In effect, if S ∈ N (f) does not touch the horizontal axis, then T = S−(0, 1) is a face of the polygon N (∂f/∂Y) and

in(\frac{∂f}{∂Y} , T) = \frac{∂}{∂Y} in(f, S) .  (2)

If S touches the horizontal axis then there exists the corresponding family of faces T ∈ N (∂f/∂Y) such that |T |_{1}/|T |_{2} ≥ |S|_{1}/|S|_{2} and ∑ |T |_{2} + δ(∂f/∂Y) = |S|_{2} − 1. If there exists an element T, of the family which is parallel to S, then (2) is also satisfied. As a result we have

Corollary 3.3 (see [L1], Corollary 5.4) Let T ∈ N (∂f/∂Y).

(a) If T is parallel to face S ∈ N (f), then (2) is satisfied.

(b) If T is not parallel to any face of N (f), then

· |T |_{1}/|T |_{2} > |S|_{1}/|S|_{2} for every S ∈ N (f), ·· the polygon N (f) touches the horizontal axis.

We need a more detailed analysis of the relations between roots of in(f, S) and ∂/∂Y in(f, S). Consider the factorization

in(f, S) = bX^{a_{S}}Y^{b_{S}}L_{1}^{r_{1}} … L_{k}^{r_{k}} ,  (3)

where Y, L_{1}, …, L_{k} are different monic linear factors of in(f, S) in k[X]^{∗}[Y] (b ≠ 0, k > 0, r_{i} > 0). Every factor has the form Y − aX^{θ}, where θ = |S|_{1}/|S|_{2}. We have the following Lemma 3.4 (see [LMP], Lemma 4.1)

(i) If S does not touch the horizonatal axis (b_{S} > 0), then

\frac{∂}{∂Y} in(f, S) = b′X^{a_{S}}Y^{b_{S}−1}L_{1}^{r_{1}−1} … L_{k}^{r_{k}−1}L′_{1} … L′_{k} ,

where L′_{1}, …, L′_{k} are monic linear factors different from Y, L_{1}, …, L_{k}.

(ii) If S touches the horizontal axis (b_{S} = 0) and k > 1 then

\frac{∂}{∂Y} in(f, S) = b′X^{a_{S}}L_{1}^{r_{1}−1} … L_{k}^{r_{k}−1}L′_{1} … L′_{k−1} ,

where L′_{1}, …, L′_{k−1} are monic linear factors different from L_{1}, …, L_{k}.

Proof. If b_{S} > 0 then Y -differentiation of in(f, S) = X^{a_{S}}Y^{b_{S}}in(f, S)^{∘} moves the support of in(f, S) by the vector (0, −1). We obtain

\frac{∂}{∂Y} in(f, S) = X^{a_{S}}Y^{b_{S}−1}h(X, Y) ,  (4)

where the powers a_{S} and b_{S} − 1 are the maximal possible. Let us consider a factor of in(f, S) of the form L^{r} = (Y − aX^{θ})^{r}. By substitution aX^{θ} + Y in place of Y, we obtain Y^{r} in place of L^{r}. We use (4) to show that L^{r−1} divides ∂/∂Y in(f, S) and r − 1 is the maximal possible power.

Remark 3.5 If S touches the horizontal axis (b_{S} = 0) and k = 1 in (3), then in(f, S) = bX^{a_{S}}L^{r} and ∂/∂Y in(f, S) = b′X^{a_{S}}L^{r−1}.

From the above lemma and the previous statements we have

Proposition 3.6 Let S ∈ N (f).

(a) If t = t(f, S) > 1 then there exists t − 1 solutions ψ′_{1}(X), …, ψ′_{t−1}(X) of ∂f/∂Y = 0 such that ord f(X, ψ′_{i}(X)) = α(S).

(b) If d = d(f, S) > 0 then there exists d solutions ψ_{1}(X), …, ψ_{d}(X) of ∂f/∂Y = 0 such that ord f(X, ψ_{i}(X)) > α(S).

Proof. (a) If S does not touch the horizontal axis, then by using notations of Lemma 3.4, k = t(f, S) − 1. Since T = S − (0, 1) is a face of N (∂f/∂Y), from Theorem 3.1 there exist solutions ψ′_{1}(X), …, ψ′_{k}(X) of ∂f/∂Y = 0 that correspond to the factor L′_{1} … L′_{k} of of ∂/∂Y in(f, S) = in(∂f/∂Y, T).

Clearly ord ψ′_{i}(X) = |T |_{1}/|T |_{2} = |S|_{1}/|S|_{2}. Since L′_{1}, …, L′_{k} are different from Y, L_{1}, …, L_{k}, then in ψ′_{i} is not a root of in(f, S). By Property 3.2 we obtain

ord f(X, ψ′_{i}(X)) = α(|S|_{1}/|S|_{2}, Δ_{f}) = α(S) .  (5)

If S touches the horizontal axis, then k = t(f, S). Let s be the number of appearances of Y in the sequence L′_{1}, …, L′_{k−1}. We can assume that L′_{1}, …, L′_{k−s−1} are different from Y. Analogously, as before, we construct solutions ψ′_{1}(X), …, ψ′_{k−s−1}(X) that satisfy (5). For s > 0, by the Newton-Puieseux theorem, there exist s solutions ψ′_{k−s}(X), …, ψ′_{k−1}(X) of ∂f/∂Y = 0 that correspond to the family of faces T ∈ N (∂f/∂Y), that lie below the line β = s, and to the distance δ(∂f/∂Y). We have ∑ |T |_{2} + δ(∂f/∂Y) = s. If ψ′_{i}(X) comes from a face T of the family, then by Corollary 3.3 we have ord ψ_{1}(X) = |T |_{1}/|T |_{2} > |S|_{1}/|S|_{2}. Since S touches the horizontal axis, it is the lower possible face of N (f). Therefore, the line supporting Δ_{f}, which is parallel to T, meets Δ_{f} at the vertex lying on the horizontal axis. The vertex has the abscissa ord f(X, 0). From Property 3.2 we obtain ord f(X, ψ′_{i}(X)) = α(|T |_{1}/|T |_{2}, Δ_{f}) = ord f(X, 0). If ψ_{i}(X) = 0, then similarly ord f(X, ψ′_{i}(X)) = ord f(X, 0). Since ord f(X, 0) = α(S) we finish with (5) in both cases.

(b) Note that d = d(f, S) = (r_{1} − 1) + … + (r_{k} − 1). If d > 0 then there exists a face T ∈ N (∂f/∂Y) which is parallel to S. By Theorem 3.1 and by Lemma 3.4 there exist solutions ψ_{1}(X), …, ψ_{d}(X) of ∂f/∂Y = 0 corresponding to the factor L_{1}^{r_{1}−1} … L_{k}^{r_{k}−1} of ∂/∂Y in(f, S) = in(∂f/∂Y, T). Clearly ord ψ_{i}(X) = |T |_{1}/|T |_{2} = |S|_{1}/|S|_{2} and in ψ_{i}(X) is a root of in(f, S). From Property 3.2 it follows that ord f(X, ψ_{i}(X)) > α(|S|_{1}/|S|_{2}, Δ_{f}) = α(S), which completes the proof.

The Newton polygon relative to a Puiseux polynomial

Let φ ∈ k[X]^{∗} be an arbitrary Puiseux polynomial of a positive order. We begin by describing the polygon N_{φ} = N^{deg φ}(f_{φ}). Applying the Weierstrass preparation theorem we can write

f(X, Y) = U(X, Y)(Y − y_{1}(X)) … (Y − y_{p}(X)) ,  (6)

where U(X, Y) ∈ k[[X, Y]] is a unit and, as before, Zer f = ⟨y_{1}, …, y_{p}⟩. Hence f_{φ}(X, Y) = f(X, φ(X) + Y) = U′(X, Y)[Y − (y_{1} − φ)] … [Y − (y_{p} − φ)] , where U′(X, Y) ∈ k[[X^{∗}, Y]] is also a unit. Therefore

Zer f_{φ} = ⟨y_{1} − φ, …, y_{p} − φ⟩ .

Using Theorem 3.1 (a) with θ = 0 we obtain

⟨ord(y_{1} − φ), …, ord(y_{p} − φ)⟩ = ⨁_{S∈N (f_{φ})}⟨|S|_{1}/|S|_{2} : |S|_{2}⟩ ⊕ ⟨+∞ : δ(f_{φ})⟩ .

Let Zer_{φ}f = ⟨y_{1}, …, y_{s}⟩ denote the system of such solutions from Zer f that ord(y_{i} − φ) > deg φ (i.e. y_{1}(X), …, y_{s}(X) are continuations of φ). As a consequence of Theorem 3.1 applied with θ = deg φ we have Corollary 3.7

(a) ⟨ord(y_{1} − φ), …, ord(y_{s} − φ)⟩ = ⨁_{S∈N_{φ}}⟨|S|_{1}/|S|_{2} : |S|_{2}⟩ ⊕ ⟨+∞ : δ(f_{φ})⟩

(b) ⟨in(y_{1} − φ), …, in(y_{p} − φ)⟩ = ⨁_{S∈N_{φ}} Zer in(f_{φ}, S)^{∘} ⊕ ⟨0 : δ(f_{φ})⟩

(c) s = |N_{φ}| + δ(f_{φ}). We will need the following simple Property 3.8 Let g ∈ k[[X^{∗}, Y]] be a nonzero series. Fix θ > 0.

(i) There exists a unique representation

g = g_{0} + g_{1} + g_{2} + … such that every g_{i} is a quasi-homogeneous form of the weights (θ, 1).

(ii) If r is the maximal power such that Y^{r} divides g_{0}, then

|N^{θ}(g)| + δ(g) = r .

(iii) If there exists S ∈ N (g) such that |S|_{1}/|S|_{2} = θ, then g_{0} = in(f, S).

Now, we can prove the following

Proposition 3.9 Assume that φ ∈ k[X]^{∗} is such a polynomial that N_{φ} is nonempty. Let S ∈ N_{φ} and let aX^{θ} be a nonzero root of in(f, S). Then

|N_{φ+aX^{θ}}| + δ(f_{φ+aX^{θ}}) = mutiplicity of aX^{θ} as a root of in(f, S) .

Proof. For θ = |S|_{1}/|S|_{2} let us consider the representation

f_{φ} = g_{0} + g_{1} + g_{2} + …

according to Property 3.8. We have g_{0} = in(f_{φ}, S). Let us write in(f_{φ}, S) = (Y − aX^{θ})^{r}h(X, Y), where r is the maximal possible power. We have r > 0. Let us note that f_{φ+aX^{θ}}(X, Y) = f_{φ}(X, aX^{θ} + Y). Hence

f_{φ+aX^{θ}} = g̃_{0} + g̃_{1} + g̃_{2} + …,  (7)

where g̃_{i}(X, Y) = g_{i}(X, aX^{θ} + Y). Since aX^{θ} + Y is the homogeneous form of the weights (θ, 1), (7) is the unique representation guaranteed by Property 3.8. We have g̃_{0}(X, Y) = in(f_{φ}, S)(X, aX^{θ} + Y) = Y^{r}h̃(X, Y). From Property 3.8(b) follows that

r = |N_{φ+aX^{θ}}| + δ(f_{φ+aX^{θ}}) ,

which concludes the proof.

Tracks of the Newton algorithm

We give here two different characterizations of the set T (f, X), of the tracks of the Newton algorithm for f, and prove their equivalency with the definition from Section 2. Recall

Definition 3.10 T (f, X) ⊂ k[X]^{∗} is the minimal subset satisfying two properties: (I) 0 ∈ T (f, X), (II) for every φ(X) ∈ T (f, X), if there exists S ∈ N_{φ}, then for every nonzero root aX^{θ} of in(f_{φ}, S), φ + aX^{θ} ∈ T (f, X).

Let us define

T_{1}(f, X) = {φ ∈ k[X]^{∗} : ∃y(X) ∈ Zer f such that ord(y(X) − φ) > deg φ}

and

T_{2}(f, X) = {φ ∈ k[X]^{∗} : |N_{φ}| + δ(f_{φ}) > 0} .

Proposition 3.11 If Zer f is nonempty then T (f, X) = T_{1}(f, X) = T_{2}(f, X).

Proof. The equality T_{1}(f, X) = T_{2}(f, X) follows directly from Corollary 3.7 (c). Since T (f, X) is the minimal set with properties (I) and (II), it suffices to show that T_{2}(f, X) satisfies both properties in order to verify that T (f, X) ⊂ T_{2}(f, X). Because Zer f is nonempty, the first condition 0 ∈ T_{2}(f, X) is clear. In order to check the second condition, let us consider φ ∈ T_{2}(f, X). Let us ssume that there exists S ∈ N_{φ} and let aX^{θ} be an arbitrary nonzero root of in(f_{φ}, S). We must show that φ+aX^{θ} ∈ T_{2}(f, X), which follows immediately from Proposition 3.9.

To finish the proof it suffices to verify that T_{1}(f, X) ⊂ T (f, X). Let φ = a_{1}X^{θ_{1}} + … + a_{n}X^{θ_{n}} be a non zero element of T_{1}(f, X) i.e. there exists y(X) ∈ Zer f such that ord(y(X) − φ) > deg φ. Let us put φ_{0} = 0 and φ_{k} = a_{1}X^{θ_{1}} + … + a_{k}X^{θ_{k}} for k = 1, …, n − 1. We will show the implication φ_{k} ∈ T (f, X) ⇒ φ_{k+1} ∈ T (f, X). Assume that φ_{k} ∈ T (f, X). We have in(y(X) − φ_{k}) = a_{k+1}X^{θ_{k+1}}. By Corollary 3.7 there exists S ∈ N_{φ_{k}} such that a_{k+1}X^{θ_{k+1}} is a root of in(f_{φ_{k}}, S). From property (II) we obtain φ_{k+1} = φ_{k} + a_{k+1}X^{θ_{k+1}} ∈ T (f, X). Since 0 ∈ T (f, X), by induction we show that φ = φ_{n} ∈ T (f, X), completing the proof.

Completion of the proof

We are in a good position to finish the proof of the main result (Theorem 2.1).

Proof of (a). Let φ ∈ T (f, X). By Proposition 3.11 s = |N_{φ}| + δ(f_{φ}) > 0. By Theorem 3.1 there exist solutions ψ_{1}(X), …, ψ_{s}(X) of f_{φ} = 0 such that deg ψ_{i} > deg φ. By using previous notation we can write Zer^{deg φ}(f_{φ}) = ⟨ψ_{1}, …, ψ_{s}⟩. On the other hand, we have the system Zer_{φ}f = ⟨y_{1}, …, y_{s}⟩ of roots of f = 0 which are continuations of φ. Corollary 3.7 states the oneto-one correspondence between Zer^{deg φ}(f_{φ}) and Zer_{φ}f by ψ ↦ y = ϕ + ψ (opposite: y ↦ ψ = y − φ). Since (∂f/∂Y)_{φ} = ∂f_{φ}/∂Y, this construction can be directly applied to Zer_{φ}(∂f/∂Y). If s = |N_{φ}| + δ(f_{φ}) > 1, then by Y -differentiation and by Theorem 3.1, there exist s−1 solutions ψ_{1}, …, ψ_{s−1} of ∂f_{φ}/∂Y = 0 such that ord ψ_{i} > deg φ. Clearly

Zer_{φ}(∂f/∂Y) = ⟨ϕ + ψ_{1}, …, ϕ + ψ_{s−1}⟩ .  (8)

Consider S ∈ N_{φ}. If t = t(f_{φ}, S) > 1, then by Proposition 3.6 (a) there exist t−1 solutions ψ′_{1}, …, ψ′_{t−1} of ∂f_{φ}/∂Y = 0 such that ord f_{φ}(X, ψ′_{i}(X)) = α(S).

Then for z′_{i}(X) = φ(X) + ψ′_{i}(X) ∈ Zer_{φ}(∂f/∂Y) we have

ord f(X, z′_{i}(X)) = ord f(X, φ(X) + ψ′_{i}(X)) = ord f_{φ}(X, ψ′_{i}(X)) = α(S) . If d = d(f, S) > 0 then by Proposition 3.6 (b) we analogously construct solutions z_{1}(X), …, z_{d}(X) ∈ Zer_{φ}(∂f/∂Y) such that ord f(X, z_{i}(X)) > α(S). We finish with the observation that if δ(f_{φ}) > 1, then φ(X) is a common root of f = 0 and ∂f/∂Y = 0 with multiplicity δ(f_{φ}) − 1 and ord f(X, φ(X)) = ord f_{φ}(X, 0) = +∞.

Proof of (b). Consider roots z_{1}(X), …, z_{d}(X) ∈ Zer_{φ}(∂f/∂Y) such that

B_{φ,S} = ⟨ord f(X, z_{1}(X)), …, ord f(X, z_{d}(X))⟩ .

Fix i ∈ {1, …, d}. We must show the fact that z_{i} has the form φ + aX^{θ} + …, where aX^{θ} is a multiple root of in(f_{φ}, S). Let ψ_{i} = z_{i} − φ. Since inψ_{i} = aX^{θ}, this fact follows immediately from the construction described in Proposition 3.6.

Proof of (c). We begin with an observation that T_{min}(f, X) is finite (Remark 2.4). It is enough to consider the tracks generated by solutions of f = 0 with infinite number of termes. Let us fix such a solution y(X) = a_{1}X^{θ_{1}} + a_{2}X^{θ_{2}} + … ∈ Zer f. Let us define the sequence of tracks φ_{0} = 0, …, φ_{j} = a_{1}X^{θ_{1}} + … + a_{j}X^{θ_{j}}, … and the sequences of series f^{(j)} := f_{φ_{j−1}} and polygons N^{(j)} := N_{φ_{j−1}}. For any j = 1, 2, …, according to Corollary 3.7, there exists a face S^{(j)} ∈ N^{(j)} with the inclination θ_{j}, such that aX^{θ_{j}} is a root of in(f^{(j)}, S^{(j)}) with multiplicity r_{j} > 0. We write l_{j} = deg_{Y} in(f^{(j)}, S^{(j)}). Of course, l_{j} ≥ r_{j}. From Proposition 3.9 we have r_{j} = |N^{(j+1)}|+δ(f^{(j+1)}) ≥ l_{j+1}. As the result we have the infinite sequence l_{1} ≥ r_{1} ≥ l_{2} ≥ r_{2} ≥ … of positive integers. Therefore, there exists j such that r_{j} = l_{j+1} = r_{j+1} = . . .. The first equality means that S^{(j+1)} is the highest face of N^{(j)}. The second equality means that in(f^{(j+1)}, S^{(j+1)}) has a unique nonzero root, hence t(f^{(j+1)}, S^{(j+1)}) = 1. Since S^{(j+1)} touches the horizontal axis (Remark 3.5), N^{(j+1)} has only one face.

To conclude the proof we need to show that for every solution z(X) ∈ Zer(∂f/∂Y), such that

ord f(X, z(X)) < +∞  (9)

there exists a track φ ∈ T (f, X) and a face S ∈ N_{φ} such that ord f(X, z(X)) = α(S). According to (6) and (9) we have ord(z(X) − y(X)) < +∞ for every y(X) ∈ Zerf. Let us choose a solution y(X) with the longest common track φ(X) of both series z(X) and y(X). We have z(X) = φ(X) + ψ_{z}(X) and y(X) = φ(X) + ψ_{y}(X), where ord ψ_{z} > deg φ, ord ψ_{y} > deg φ and in ψ_{y} ≠ in ψ_{z}. Since 0 = (∂f/∂Y)(X, z(X)) = (∂f_{φ}/∂Y)(X, ψ_{z}(X)), therefore in ψ_{z} can be described by N′_{φ} := N^{deg φ}(∂f_{φ}/∂Y) by virtue of Corollary 3.7.

If ord ψ_{z} = +∞, then z(X) = φ(X) and according to (9) we have ord f(X, z(X)) = ord f(X, φ(X)) = ord f_{φ}(X, 0) < +∞. Hence, there exists S ∈ N_{φ} which touches the horizontal axis and ord f_{φ}(X, 0) = α(S). If ord ψ_{z}(X) is finite, but does not appear as an inclination of any face of N_{φ}, then according Corollary 3.3 (b) the line supporting Δ(f_{φ}), of inclination ord ψ_{z}, intersects the horizontal axis at the point (ord f_{φ}(X, 0), 0). Hence, by Property 3.2 we have ord f(X, z(X)) = ord f_{φ}(X, ψ_{z}(X)) = ord f_{φ}(X, 0). Therefore, in both cases ord f(X, z(X)) = α(S).

Let us assume that ord ψ_{z} = |S|_{1}/|S|_{2} for a certain S ∈ N_{φ} but that in ψ_{z} is not a root of in(f_{φ}, S). By Property 3.2 we have ord f(X, z(X)) = α(S) as before. If in ψ_{z} is a root of in(f_{φ}, S), then by Theorem 3.1 there exists y(X) ∈ Zer f of the form y(X) = φ(X) + ψ_{z}(X) + …, which contradicts the definition of y(X). Because of this contradiction, the last possibility cannot happen and we conclude the prrof of the main result.

4 Eggers’ example

The following example shows that curves which are not equisingular can have the same polar invariants, counting their multiplicities. Let us consider the following four Puiseux series

\begin{matrix}y_{1} = X + X^{3/2} + X^{15/4}, & y_{3} = X + X^{5/2} + X^{11/4}, \\ y_{2} = 2X + X^{5/2} + X^{13/4}, & y_{4} = 2X + X^{3/2} + X^{17/4}. \end{matrix}

Let f_{1}, f_{2}, f_{3}, f_{4} ∈ k[[X, Y]] be the minimal series respectively for y_{1}, y_{2}, y_{3}, y_{4}. We put f = f_{1}f_{2} and f′ = f_{3}f_{4}. Let us begin with f. We have the following cycle for y_{1}

y_{1}^{(i)} = ε^{4i}X^{4/4} + ε^{6i}X^{6/4} + ε^{15i}X^{15/4}

as well as for y_{2}

y_{2}^{(i)} = 2ε^{4i}X^{4/4} + ε^{10i}X^{10/4} + ε^{13i}X^{13/4} ,

where i = 0, 1, 2, 3 and where ε is a primitive root of unity of degree four. On the basis of the table provided below we can analyze tracks 0, X, X + X^{3/2} and X − X^{3/2}

\begin{matrix}f(X, Y) & f(X, X + Y) & f(X, X + X^{3/2} + Y) & f(X, X − X^{3/2} + Y) \\ y_{1}^{(0)} = X + X^{3/2} + X^{15/4} & X^{3/2} & X^{15/4} & 2X^{3/2} \\ y_{1}^{(1)} = X − X^{3/2} − εX^{15/4} & −X^{3/2} & −2X^{3/2} & −εX^{15/4} \\ y_{1}^{(2)} = X + X^{3/2} − X^{15/4} & X^{3/2} & −X^{15/4} & 2X^{3/2} \\ y_{1}^{(3)} = X − X^{3/2} + εX^{15/4} & −X^{3/2} & −2X^{3/2} & εX^{15/4} \\ y_{2}^{(0)} = 2X + X^{5/2} + X^{13/4} & X & X & X \\ y_{2}^{(1)} = 2X − X^{5/2} + εX^{13/4} & X & X & X \\ y_{2}^{(2)} = 2X + X^{5/2} − X^{13/4} & X & X & X \\ y_{2}^{(3)} = 2X − X^{5/2} − εX^{13/4} & X & X & X \end{matrix}

The first column presents all the roots of f = 0; the second – the initial forms of solutions of f(X, X + Y) = 0; the third – f(X, X + X^{3/2} + Y) = 0 and the fourth – f(X, X − X^{3/2} + Y) = 0. This information allows us to reconstruct the relative polygons as well as their initial forms.

[figure: (0, 8); (8, 0); (4, 4); (7, 2); (10, 0); (29/2, 0); A; B; C]

Polygon N (f) has one face A, which joins points (0, 8) and (8, 0), in(f, A) = (Y − X)^{4}(Y − 2X)^{4}, thus t = t(f, A) = 2; from this we obtain invariant α(A) = 8 with multiplicity t − 1 = 1. Polygon N_{X} has one face B, which joins points (4, 4) and (10, 0), in(f_{X}, B) = X^{4}(Y − X^{3/2})^{2}(Y + X^{3/2})^{2}, thus t = t(f_{X}, B) = 2; we obtain invariant α(B) = 10 with multiplicity t − 1 = 1. Finally polygon N_{X+X^{3/2}} has a single face C, which joins (7, 2) and (29/2, 0), in(f_{X+X^{3/2}}, C) = 4X^{7}(Y − X^{15/4})(Y + X^{15/4}), thus t = t(f_{X+X^{3/2}}, C) = 2; therefore we obtain invariant α(C) = 29/2 with multiplicity t−1 = 1. Taking into account track X −X^{3/2} we again obtain α(C) = 29/2 with multiplicity 1.

With the help of the second table we analyze tracks 2X, 2X + X^{5/2} as well as 2X − X^{5/2}.

\begin{matrix}f(X, Y) & f(X, 2X + Y) & f(X, 2X + X^{5/2} + Y) & f(X, 2X − X^{5/2} + Y) \\ y_{1}^{(0)} = X + X^{3/2} + X^{15/4} & −X & −X & −X \\ y_{1}^{(1)} = X − X^{3/2} − εX^{15/4} & −X & −X & −X \\ y_{1}^{(2)} = X + X^{3/2} − X^{15/4} & −X & −X & −X \\ y_{1}^{(3)} = X − X^{3/2} + εX^{15/4} & −X & −X & −X \\ y_{2}^{(0)} = 2X + X^{5/2} + X^{13/4} & X^{5/2} & X^{13/4} & 2X^{5/2} \\ y_{2}^{(1)} = 2X − X^{5/2} + εX^{13/4} & −X^{5/2} & −2X^{5/2} & εX^{13/4} \\ y_{2}^{(2)} = 2X + X^{5/2} − X^{13/4} & X^{5/2} & −X^{13/4} & 2X^{5/2} \\ y_{2}^{(3)} = 2X − X^{5/2} − εX^{13/4} & −X^{5/2} & −2X^{5/2} & −εX^{13/4} \end{matrix}

Polygon N_{2X} has one segment D, which joins (4, 4) and (14, 0), in(f_{2X}, D) = X^{4}(Y − X^{5/2})^{2}(Y + X^{5/2})^{2}, thus t = t(f_{2X}, D) = 2; we obtain invariant α(D) = 14 with multiplicity t − 1 = 1.

[figure: (4, 4); (9, 2); 14; 31/2; D; E]

Polygon N_{2X+X^{5/2}} contains one face E, which joins (9, 2) and (31/2, 0), in(f_{2X+X^{5/2}}, E) = 4X^{9}(Y − X^{13/4})(Y + X^{13/4}), thus t = t(f_{2X+X^{5/2}}, E) = 2; we obtain invariant α(E) = 31/2 with multiplicity t − 1 = 1. Considering track 2X − X^{5/2} we again obtain α(E) = 31/2 with multiplicity 1.

The remaining tracks do not lead to new solutions. The complete system of invariants is as follows

⟨8, 10, 14, 29/2, 29/2, 31/2, 31/2⟩ .

An analogous analysis of series f′ results in the same system. The idea behind Eggers’ example becomes visible when we simultaneously mark faces A, B, C, D, E. Dotted lines signify new faces, which take the place of C and E for f′.

[figure: (8, 0); (4, 4); (7, 2); (10, 0); \frac{29}{2}; (9, 2); 14; \frac{31}{2}]

5 The case of one branch

Let us consider an instance when f ∈ k[[X, Y]] is an irreducible series Y −regular of order p (p > 1). Let us fix Puiseux solution

y(X) = a_{1}X^{ν_{1}/p} + a_{2}X^{ν_{2}/p} + …  (10)

of equation f = 0 (a_{j} ≠ 0, 0 < ν_{1} < ν_{2} < integers, GCD(p, ν_{1}, ν_{2}, …) = 1). This solution generates other solutions in the form of a cycle

y_{i}(X) = a_{1}ε^{ν_{1}i}X^{ν_{1}/p} + a_{2}ε^{ν_{2}i}X^{ν_{2}/p} + …, i = 0, …, p − 1 ,  (11)

where ε is a primitive root of unity of degree p. Let θ_{j} = ν_{j}/p. Let us consider a sequence of tracks of the Newton algorithm for f constructed from solution (10): φ_{0}(X) = 0, …, φ_{j}(X) = a_{1}X^{θ_{1}} + … + a_{j}X^{θ_{j}}, …. Let us put f^{(j)} := f_{φ_{j−1}} and N^{(j)} := N_{φ_{j−1}}. It is convenient to denote ν_{0} = p, θ_{0} = 0. We have

Property 5.1

(i) Polygon N^{(j)} consists of one face S^{(j)}, with inclination ν_{j}/p, which touches the horizontal axis.

(ii) deg_{Y} in(f^{(j)}, S^{(j)}) = GCD(ν_{0}, …, ν_{j−1}) for j = 1, 2, . . ..

(iii) Every root of in(f^{(j)}, S^{(j)}) has the multiplicity GCD(ν_{0}, …, ν_{j}).

(iv) t(f^{(j)}, S^{(j)}) = \frac{GCD(ν_{0}, …, ν_{j−1})}{GCD(ν_{0}, …, ν_{j})} ,

(v) α(S^{(j)}) = ∑_{j′=1}^{j}(GCD(ν_{0}, …, ν_{j′−1})−GCD(ν_{0}, …, ν_{j′}))\frac{ν_{j′}}{p} +GCD(ν_{0}, …, ν_{j})\frac{ν_{j}}{p} .

Recall the definition of a generalized characteristic (b_{0}, …, b_{h}):

b_{0} = ν_{0}, …, b_{k} =min {ν_{j} > b_{k−1}| GCD(ν_{0}, …, ν_{j−1}) > GCD(ν_{0}, …, ν_{j}) } ,
…, b_{h} =min {ν_{j}| GCD(ν_{0}, …, ν_{j}) = 1 } .

If p = ν_{0} ≤ ν_{1}, then the generalized characteristic coincides with the topological characteristic (in a complex case). We call b_{1}/p, …, b_{h}/p characteristic exponents. From Property 5.1 we obtain

Corollary 5.2 t(f^{(j)}, S^{(j)}) > 1 if and only if ν_{j}/p is a characteristic exponent.

By j_{1}, …, j_{h} let us denote the values of index j for characteristic exponents and let α_{k} = α(S^{(j_{k})}), k = 1, …, h. Furthermore let

e_{k} = GCD(ν_{0}, …, ν_{j_{k}}) = GCD(b_{0}, …, b_{k}) .

We have

Corollary 5.3 All polar quotients obtained on the tracks of solution (10) have the form

α_{k} = ∑_{k′=1}^{k}(e_{k′−1} − e_{k′})\frac{b_{k′}}{p} + e_{k} \frac{b_{k}}{p} , k = 1, …, h ,

and appear with multiplicity e_{k−1}/e_{k}.

Let n_{k} = e_{k−1}/e_{k}. To take into consideration the multiplicity deriving from the remaining tracks, let us write cycle (11) in the form

y_{i}(X)=ψ_{0}(X) + ψ_{1}(ε_{1}^{i_{1}}X^{1/n_{1}}) + ψ_{2}(ε_{2}^{i_{1}+n_{1}j_{2}}X^{1/(n_{1}n_{2})}) + …
… + ψ_{h}(ε_{h}^{i_{1}+n_{1}i_{2}+...+n_{1}n_{2}...n_{h−1}i_{h}}X^{1/(n_{1}n_{2}...n_{h})}) ,

where ψ_{0}, …, ψ_{h−1} ∈ k[X], ψ_{h} ∈ k[[X]], ε_{h} = ε, ε_{k} = ε^{n_{k+1}...n_{h}} (k = 1, …, h− 1) and i = i_{1} + n_{1}i_{2} + … + n_{1}n_{2} … n_{h−1}i_{h} is the unique decomposition of i ∈ {0, …, p − 1} such that 0 ≤ i_{1} ≤ n_{1} − 1, …, 0 ≤ i_{h} ≤ n_{h} − 1. Therefore the track

φ_{j_{k}}(X) = ψ_{0}(X) + ψ_{1}(X^{1/n_{1}}) + … + ψ_{k−1}(X^{1/(n_{1}...n_{k−1})})

generates the cycle composed of n_{1} … n_{k−1} elements. Hence the multiplicity of polar quotient α_{k} is

n_{1} … n_{k−1}(n_{k} − 1) .

Commentary. From Property 5.1 result the common relations between faces S^{(j)} and S^{(j+1)} depending whether or not j stands in the characteristic position. If j ∉ {j_{1}, …, j_{h}}, then GCD(ν_{0}, …, ν_{j−1}) = GCD(ν_{0}, …, ν_{j}) and faces S^{(j)}, S^{(j+1)} have a common upper end.

[figure: nwp(ν_{0}, …, ν_{j−1}); nwp(ν_{0}, …, ν_{j}); S^{(j)}; θ_{j−1}; S^{(j+1)}]

This means that polygon N^{(j+1)} does not contain a segment parallel to S^{(j)}. If j = j_{k}, then GCD(ν_{0}, …, ν_{j−1}) = e_{k−1} > e_{k} = GCD(ν_{0}, …, ν_{j}).

[figure: θ_{j−1}; nwp(ν_{0}, …, ν_{j−1}); nwp(ν_{0}, …, ν_{j}); S^{(j)}; S^{′(j)}; S^{(j+1)}]

Let us divide face S^{(j)} into n_{k} = e_{k−1}/e_{k} equal parts. Face S^{(j+1)} appears in place of the lowest part. Polygon N^{(j+1)} contains a face S^{′(j)} that is parallel to S^{(j)}. The length of the projection of S^{′(j)} onto the vertical axis is

GCD(ν_{0}, …, ν_{j−1}) − GCD(ν_{0}, …, ν_{j}) = e_{k−1} − e_{k} .

The sequence of polygons N^{(1)}, N^{(2)}, … approaches the polygon of series f(X, y(X) + Y), which is composed of segments S^{′(j_{1})}, …, S^{′(j_{h})}.

[figure: p = b_{0} = e_{0}; e_{1}; e_{h−1}; 1 = e_{h}; b_{1}/p; b_{h}/p; α_{1}; α_{h}]

The numbers α_{1}, …, α_{h} represent the abscissae of the points where the lines determined by the faces intersect the horizontal axis (see: [GP1]).

Acknowledgements

The initial version of this paper emerged during the author’s stay at the University of Bordeaux 1. The author wishes to express his thanks to Pierrette Cassou-Noguès for her invitation and hospitality as well as for many stimulating conversations.

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Department of Mathematics, Technical University, AL. 1000 L PP 7, 25-314 Kielce, Poland

e-mail: ztpal@tu.kielce.pl