Higher order polars of quasi-ordinary singularities

TL;DR

This paper investigates higher order derivatives of quasi-ordinary polynomials, providing new factorizations and generalizations of existing results, extending to irreducible cases and quasi-ordinary power series.

Contribution

It generalizes the factorization of higher order polars for quasi-ordinary polynomials, including irreducible cases and power series, advancing the theoretical understanding of these singularities.

Findings

Derived factorizations of higher order polars.

Extended results to irreducible quasi-ordinary polynomials.

Generalized factorization to quasi-ordinary power series.

Abstract

A quasi-ordinary polynomial is a monic polynomial with coefficients in the power series ring such that its discriminant equals a monomial up to unit. In this paper we study higher derivatives of quasi-ordinary polynomials, also called higher order polars. We find factorizations of these polars. Our research in this paper goes in two directions. We generalize the results of Casas-Alvero and our previous results on higher order polars in the plane to irreducible quasi-ordinary polynomials. We also generalize the factorization of the first polar of a quasi-ordinary polynomial (not necessary irreducible) given by the first-named author and Gonz\'alez-P\'erez to higher order polars. This is a new result even in the plane case. Our results remain true when we replace quasi-ordinary polynomials by quasi-ordinary power series.

Equations

$$\Bigl{\{}\begin{picture}(3.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){3.0}} \put(0.0,0.85){\line(1,0){3.0}} \put(1.5,1.3){\makebox(0.0,0.0)[b]{${\bf q}$}} \put(1.5,0.7){\makebox(0.0,0.0)[t]{$k$}} \end{picture}\Bigr{\}}:=\mbox{convex hull}\;\bigl{(}\{\,(q_{1},\dots,q_{d},0),(0,\dots,0,k)\,\}+\mathbf{R}_{\geq 0}^{d+1}\bigr{)}\;.$$

$$\Delta=\sum_{i=1}^{r}\left\{\begin{picture}(2.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){2.0}} \put(0.0,0.85){\line(1,0){2.0}} \put(1.0,1.3){\makebox(0.0,0.0)[b]{${\bf q}_{i}$}} \put(1.0,0.7){\makebox(0.0,0.0)[t]{$k_{i}$}} \end{picture}\right\},$$

$$\alpha \mathbf{K}[[{\underline{x}}^{1/\mathbf{N}}]]=\{\alpha w:w\in \mathbf{K}[[{\underline{x}}^{1/\mathbf{N}}]]\}$$

$${\alpha }_i-{\alpha }_j={\underline{x}}^{{\mathbf{q}}_{ij}}{v}_{ij}(\underline{x}),for\; some{\mathbf{q}}_{ij}\in {\mathbf{Q}}^d\mathrm{and}{v}_{ij}(0)\ne 0.$$

$$O({\alpha }_i,{\alpha }_j)\ge \min \{O({\alpha }_i,{\alpha }_k),O({\alpha }_j,{\alpha }_k)\}.$$

$$g({\lambda }_B(\underline{x})+z{\underline{x}}^{h(B)})=G_B(z){\underline{x}}^{q(g,B)}+\cdots$$

$$f({\lambda }_B+z{\underline{x}}^{h(B)})=f(z{\underline{x}}^{\left(\frac{3}{2},1\right)})=z(z^2-1)x_1^{9/2}{x}_2^3+\cdots$$

$$f(\gamma )=F_B(\text{mbox}{\mathrm{lc}}_B\gamma ){\underline{x}}^{q(f,B)}+\cdots$$

$$F_B(z)=\text{mbox}const\prod _{i:{\alpha }_i\in B}(z-\text{mbox}{\mathrm{lc}}_B{\alpha }_i)$$

$$q(f,B)=\sum _{i=1}^n\min (O({\lambda }_B,{\alpha }_i),h(B)).$$

$$q(p,B^{\prime })-q(p,B)=♯(\mathrm{missing}Zerp\cap B^{\prime })[h(B^{\prime })-h(B)].$$

$$q(p,B^{\prime })-q(p,B)$$

$$g({\lambda }_B(\underline{x})+z\omega x^{h(B)})=G_B(z){\omega }^k{x}^{q(g,B)}+\cdots$$

$$p({\lambda }_{B^{\prime }}+z{\omega }^{-1}{\underline{x}}^{h(B)})=P_B(z)\theta {\underline{x}}^{q(p,B)}+\cdots$$

$$F(z)=(z^n-c{)}^l\in K[z],$$

$$F^{(k)}(z)={\mathcal{F}}^{\oplus }(z){\mathcal{F}}^{\ominus }(z)$$

$$F_B^{(k)}(z)={\mathcal{F}}_B^{\oplus }(z){\mathcal{F}}_B^{\ominus }(z),$$

$${\bar{g}}^{[\mathbf{r}]}(u,y):=g(u^{r_1},\dots ,u^{r_d},y).$$

$${\bar{g}}^{[\mathbf{r}]}({\lambda }_{\bar{B}}(u)+z{u}^{h(\bar{B})})=G_B(z)u^{⟨\mathbf{r},q(g,B)⟩}+\cdots ,$$

$$l(\vec{r},\Delta )=l(\vec{r},\Delta (g)).$$

$$⟨\vec{r},\alpha ⟩=⟨(1,r_{d+1}),{\pi }_{\mathbf{r}}(\alpha )⟩$$

$$\sum_{\genfrac{}{}{0.0pt}{2}{B\in T(f)}{t_{k}(B)\neq 0}}\left\{\begin{picture}(13.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){13.0}} \put(0.0,0.85){\line(1,0){13.0}} \put(6.5,1.3){\makebox(0.0,0.0)[b]{$t_{k}(B)q(p,B)$}} \put(6.5,0.7){\makebox(0.0,0.0)[t]{$\rule{0.0pt}{10.76385pt}t_{k}(B)$}} \end{picture}\right\}.$$

$$\mathrm{Res}{}_y(f^{(k)}(y),p(y)-T)=\pm \prod _{j=1}^{n-k}(p({\beta }_j)-T).$$

$$\Delta(\bar{R}^{[{\bf r}]})=\sum_{\genfrac{}{}{0.0pt}{2}{B\in T(f)}{t_{k}(B)\neq 0}}\left\{\begin{picture}(14.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){14.0}} \put(0.0,0.85){\line(1,0){14.0}} \put(7.0,1.3){\makebox(0.0,0.0)[b]{$t_{k}(\bar{B})q(\bar{p}^{[{\bf r}]},\bar{B})$}} \put(7.0,0.7){\makebox(0.0,0.0)[t]{$\rule{0.0pt}{10.76385pt}t_{k}(\bar{B})$}} \end{picture}\right\}.$$

$$\left\{\begin{picture}(14.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){14.0}} \put(0.0,0.85){\line(1,0){14.0}} \put(7.0,1.3){\makebox(0.0,0.0)[b]{$t_{k}(\bar{B})q(\bar{p}^{[{\bf r}]},\bar{B})$}} \put(7.0,0.7){\makebox(0.0,0.0)[t]{$\rule{0.0pt}{10.76385pt}t_{k}(\bar{B})$}} \end{picture}\right\}=\left\{\begin{picture}(16.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){16.0}} \put(0.0,0.85){\line(1,0){16.0}} \put(8.0,1.3){\makebox(0.0,0.0)[b]{$t_{k}(B)\langle{\bf r},q(p,B)\rangle$}} \put(8.0,0.7){\makebox(0.0,0.0)[t]{$\rule{0.0pt}{10.76385pt}t_{k}(B)$}} \end{picture}\right\}=\pi_{\bf r}\left(\left\{\begin{picture}(13.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){13.0}} \put(0.0,0.85){\line(1,0){13.0}} \put(6.5,1.3){\makebox(0.0,0.0)[b]{$t_{k}(B)q(p,B)$}} \put(6.5,0.7){\makebox(0.0,0.0)[t]{$\rule{0.0pt}{10.76385pt}t_{k}(B)$}} \end{picture}\right\}\right).$$

$${\mathcal{F}}_B^{\oplus }(z)=\prod _{B{\perp }_{z_i}{B}_i}(z-z_i{)}^{n_k(B_i)}.$$

$$G_B(z)=a{z}^l\mathrm{or}{G}_B(z)=a(z^{n(B)}-c{)}^l$$

$$I_{\bar{B}}=\{i:{\bar{\beta }}_i\in \bar{B},P_B(\text{mbox}{\mathrm{lc}}_{\bar{B}}{\bar{\beta }}_i)\ne 0\},$$

$$J_{\bar{B}}=\{i:{\bar{\beta }}_i\in \bar{B},P_B(\text{mbox}{\mathrm{lc}}_{\bar{B}}{\bar{\beta }}_i)=0\}.$$

$$R(T)$$

Full text

1 Introduction

In [Me] Merle gave a decomposition theorem of a generic polar curve of an irreducible plane curve singularity, according to its topological type. The factors of this decomposition are not necessary irreducible. Merle‘s decomposition was generalized to reduced plane curve germs by Kuo and Lu [K-Lu], Delgado de la Mata [D], Eggers [Eg], García Barroso [GB] among others. In [GB-GP], García Barroso and González Pérez obtained decompositions of the polar hypersurfaces of quasi-ordinary singularities. On the other hand, Casas-Alvero in [Ca] generalized the results of Merle to higher order polars of an irreducible plane curve. In [GB-Gw4] we improved his results giving a finer decomposition in such a way that we are able to determine the topological type of some irreducible factors of the polar as well as their number.

Our research in this paper goes in two directions. We generalize the results of [Ca] and [GB-Gw4] on higher order polars to irreducible quasi-ordinary singularities (see Theorem 10.11 and Proposition 10.12). We also generalize the factorization of the first polar of a quasi-ordinary singularity (not necessary irreducible) from [GB-GP] to higher order polars (see Theorem 10.4). This is a new result even in the plane case.

Our approach is based on Kuo-Lu trees, Eggers trees, Newton polytopes and resultants. As it was remarked in [Po] and [GB-GP], the irreducible factors of the polar of a quasi-ordinary singularity are not necessary quasi-ordinary. For that reason, we mesure the relative position of these irreducible factors and those of the quasi-ordinary singularity using a new notion called the P-contact, which plays in our situation the role of the logarithmic distance introduced by Płoski in [Pł].

The paper is organized as follows.

In Section 2 we recall the notion of the Newton polytope of a Weierstrass polynomial $f\in\mathbf{K}[[\underline{x}]][y]$ and we use it together with the Rond-Schober irreducibility criterium [R-S], in order to give sufficient conditions for the reducibility of $f$. The most important result in this section is Corollary 2.6, which allows us to characterize, in Theorem 9.1, the irreducible factors of the higher order polars of the polynomial $f$.

In Section 3 we present the notion of the Kuo-Lu tree of a quasi-ordinary Weierstrass polynomial. Then in Section 4 we identify the bars of a Kuo-Lu tree with certain sets of fractional power series called pseudo-balls and we introduce the notion of compatibility of a Weierstrass polynomial with a pseudo-ball. Every quasi-ordinary Weierstrass polynomial is compatible with every pseudo-ball associated with its Kuo-Lu tree. Moreover if a Weierstrass polynomial is compatible with a pseudo-ball then any factor of it is compatible too (see Corollary 4.5). In Lemma 4.6 we prove that, under some conditions, the normalized higher derivatives inherit the compatibility property. In Section 5 we introduce, using Galois automorphisms, an equivalence relation in the set of pseudo-balls, called conjugacy, and we explore the compatibility property for conjugate pseudo-balls. We generalize the Kuo-Lu Lemma [K-Lu, Lemma 3.3] to higher derivatives in Section 6. In Section 7 we introduce our main tool, monomial substitutions, that allows us to reduce several questions to the case of two variables. In particular, if $f$ and $g$ are power series in $d+1$ variables such that after generic monomials substitutions we obtain power series $\bar{f},\bar{g}$ in two variables with equal Newton polygons, then the Newton polytopes of $f$ and $g$ are also equal (see Corollary 7.3). In Section 8 we extend the notion of Eggers tree introduced in [Eg], to quasi-ordinary settings. Remark that the tree we use here is not exactly the Eggers-Wall tree introduced in [Po] for the quasi-ordinary situation. The main result of Section 9 is Theorem 9.1, where we characterize the irreducible factors of higher derivatives of quasi-ordinary Weierstrass polynomials. Theorem 9.1 allows us to give factorizations of higher derivatives, in terms of the Eggers tree, in Section 10. Theorem 10.4 generalizes the factorization from [Ca] on higher order polars to quasi-ordinary singularities (not necessary irreducible) and also the factorization from [GB-GP] to higher order polars. Theorem 10.11 and Proposition 10.12 extend the statements of [GB-Gw4, Theorem 6.2] to irreducible quasi-ordinary Weierstrass polynomials. Finally in Section 11 we establish that our results also hold for quasi-ordinary power series.

2 Newton polytopes

Let $\alpha=\sum\alpha_{\bf i}\underline{x}^{\bf i}\in S[[\underline{x}]]$ be a non zero formal power series with coefficients in a ring $S$, where $\underline{x}=(x_{1},\ldots,x_{d})$ and $\underline{x}^{\bf i}=x_{1}^{i_{1}}\cdots x_{d}^{i_{d}}$, with ${\bf i}=(i_{1},\ldots,i_{d})$. The Newton polytope $\Delta(\alpha)\subset\mathbf{R}^{d}$ of $\alpha$ is the convex hull of the set $\bigcup_{\alpha_{\bf i}\neq 0}{\bf i}+\mathbf{R}^{d}_{\geq 0}$. By convention the Newton polytope of the zero power series is the empty set.

The Newton polytope of a polynomial $f=\sum_{{\bf i},j}a_{{\bf i},j}\underline{x}^{\bf i}y^{j}\in S[[\underline{x}]][y]$ is the polytope $\Delta(f)\subset\mathbf{R}^{d}\times\mathbf{R}$ of $f$ viewed as a power series in $x_{1},\ldots,x_{d},y$. If $\Gamma$ is a compact face of $\Delta(f)$ then $f|_{\Gamma}:=\sum_{({\bf i},j)\in{\Gamma}}a_{{\bf i},j}\underline{x}^{\bf i}y^{j}\in S[\underline{x}][y]$ is called the symbolic restriction of $f$ to $\Gamma$.

We say that a subset of $\mathbf{R}^{d+1}$ is a Newton polytope if it is the Newton polytope of some polynomial in $S[[\underline{x}]][y]$.

Let ${\bf q}=(q_{1},\dots,q_{d})\in\mathbf{Q}_{\geq 0}^{d}$ and let $k$ be a positive integer. We define the elementary Newton polytope

[TABLE]

Its inclination is, by definiton, $\frac{1}{k}{\bf q}$.

We denote by \Bigl{\{}\begin{picture}(3.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){3.0}} \put(0.0,0.85){\line(1,0){3.0}} \put(1.5,1.3){\makebox(0.0,0.0)[b]{\infty}} \put(1.5,0.7){\makebox(0.0,0.0)[t]{k}} \end{picture}\Bigr{\}} the Newton polytope $\Delta(y^{k})$, which is the first orthant translated by $(0,\dots,0,k)$. By convention we consider it as an elementary polytope.

Example 2.1

The elementary Newton polytope \left\{\begin{picture}(5.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){5.0}} \put(0.0,0.85){\line(1,0){5.0}} \put(2.5,1.3){\makebox(0.0,0.0)[b]{(4,2)}} \put(2.5,0.7){\makebox(0.0,0.0)[t]{8}} \end{picture}\right\} is

(0,\!0,\!8)$$(4,\!2,\!0)

A Newton polytope is polygonal if the maximal dimension of its compact faces is one.

Remember that the Minkowski sum of $A,B\subset\mathbf{R}^{d+1}$ is the set $A+B:=\{a+b\;:\;a\in A,b\in B\}$. If a Newton polytope $\Delta$ has a representation of the type

[TABLE]

then summing all the elementary Newton polytopes of the same inclination in (1) we obtain a unique representation, up to the order of the terms, called canonical representation of $\Delta$. If the inclinations can be well-ordered then $\Delta$ is polygonal.

2.1 Newton polytopes and factorizations

Let $\mathbf{K}$ be a field of characteristic zero. We denote by $\mathbf{K}[[x_{1}^{1/k},\dots,x_{d}^{1/k}]]$ the ring of fractional power series in $d$ variables where all the exponents are nonnegative rational numbers with denominator $k\in\mathbf{N}\backslash\{0\}$. Put $\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]:=\bigcup_{k\in\mathbf{N}\backslash\{0\}}\mathbf{K}[[x_{1}^{1/k},\dots,x_{d}^{1/k}]]$. We will denote by

[TABLE]

the ideal of $\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ generated by $\alpha\in\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$.

A Weierstrass polynomial is a monic polynomial where the coefficients different from the leading coefficient are non-units of the formal power series. Notice that, according to this definition, the constant polynomial 1 is a Weierstrass polynomial.

The next lemma gives sufficient conditions for reducibility of Weierstrass polynomials. One of the consequences of this lemma is that a Weierstrass polynomial with a polygonal Newton polytope admits a decomposition into coprime factors such that the Newton polytope of each factor is elementary (see Theorem 2.4, see also [GB-GP, Theorem 3]).

Lemma 2.2

Let $g=y^{m}+c_{1}y^{m-1}+\cdots+c_{m}\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial. Assume that there exists ${\bf q}\in\mathbf{Q}^{d}$ such that $c_{i}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]\subseteq\underline{x}^{i{\bf q}}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ for all $1\leq i\leq m$ with equality for some $i=i_{0}$, $1\leq i_{0}<m$ and strict inclusion for $i=m$. Then $g$ has at least two coprime factors.

Proof. We will apply [R-S, Theorem 2.4]. Without lost of generality we may assume that $i_{0}$ is the maximal index $i\in\{1,\ldots,m-1\}$ such that $c_{i}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]=\underline{x}^{i{\bf q}}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$. Then the segment $\Gamma$ with endpoints $(0,\dots,0,m)$ and $(i_{0}{\bf q},m-i_{0})$ is an edge of $\Delta(g)$. The symbolic restriction of $g$ to $\Gamma$ is the product $g|_{\Gamma}=y^{m-i_{0}}\cdot\tilde{g}$, where $\tilde{g}\in\mathbf{K}[\underline{x}][y]$ is coprime with $y$. The associated polyhedron of $g$, in the sense of Rond-Schober (see [R-S, page 4732] is $m{\bf q}+\mathbf{R}_{\geq 0}^{d}$. Hence the polynomial $g$ verifies the hypothesis of [R-S, Theorem 2.4] and the lemma follows.

Remark 2.3

The assumptions of Lemma 2.2 mean geometrically that the Newton polytope $\Delta(g)$ is included in the elementary polytope \left\{\begin{picture}(4.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){4.0}} \put(0.0,0.85){\line(1,0){4.0}} \put(2.0,1.3){\makebox(0.0,0.0)[b]{m{\bf q}}} \put(2.0,0.7){\makebox(0.0,0.0)[t]{m}} \end{picture}\right\}, and $\Delta(g)$ has an edge $\Gamma$, which endpoints $(0,\dots,0,m)$ and $(i_{0}{\bf q},m-i_{0})$, for some $1\leq i_{0}<m$. The next picture illustrates the situation:

({\bf 0},m)$$(i_{0}{\bf q},m-i_{0})$$(m{\bf q},0)

Theorem 2.4

Let $f\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial. Assume that $\Delta(f)$ is a polygonal Newton polytope with canonical representation \sum_{i=1}^{r}\left\{\begin{picture}(2.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){2.0}} \put(0.0,0.85){\line(1,0){2.0}} \put(1.0,1.3){\makebox(0.0,0.0)[b]{{\bf q}{i}}} \put(1.0,0.7){\makebox(0.0,0.0)[t]{k{i}}} \end{picture}\right\}. Then $f$ admits a factorization $f_{1}\cdots f_{r}$, where $f_{i}\in\mathbf{K}[[\underline{x}]][y]$ are Weierstrass polynomials, not necessarily irreducible, such that \Delta(f_{i})=\left\{\begin{picture}(2.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){2.0}} \put(0.0,0.85){\line(1,0){2.0}} \put(1.0,1.3){\makebox(0.0,0.0)[b]{{\bf q}{i}}} \put(1.0,0.7){\makebox(0.0,0.0)[t]{k{i}}} \end{picture}\right\} for $i=1,\dots,r$.

Proof. Let $f=g_{1}\cdots g_{s}$ be the factorization of $f$ into irreducible Weierstrass polynomials. Since the Newton polytope of a product is the Minkowski sum of the Newton polytopes of the factors, by hypothesis we get \Delta(g_{j})=\sum_{i=1}^{r}b_{ij}\left\{\begin{picture}(2.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){2.0}} \put(0.0,0.85){\line(1,0){2.0}} \put(1.0,1.3){\makebox(0.0,0.0)[b]{{\bf q}{i}}} \put(1.0,0.7){\makebox(0.0,0.0)[t]{k{i}}} \end{picture}\right\} for some $b_{ij}\in\mathbf{Q}_{\geq 0}$. By Remark 2.3 $\Delta(g_{j})$ is elementary, hence for fixed $j$ only one term of the previous sum is nonzero. On the other hand, for fixed $i$, we get $\sum_{j}b_{ij}=1$. Put $f_{i}:=\prod g_{j}$, where the product runs over all $g_{j}$ such that $b_{ij}\neq 0$. Then $f=f_{1}\cdots f_{r},$ where \Delta(f_{i})=\left\{\begin{picture}(2.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){2.0}} \put(0.0,0.85){\line(1,0){2.0}} \put(1.0,1.3){\makebox(0.0,0.0)[b]{{\bf q}{i}}} \put(1.0,0.7){\makebox(0.0,0.0)[t]{k{i}}} \end{picture}\right\} for $i=1,\dots,r$.

Theorem 2.5

Let $f(y),g(y)\in\mathbf{L}[y]$ be monic polynomials, where $\mathbf{L}$ is a field of characteristic zero. If $g(y)$ is irreducible in the ring $\mathbf{L}[y]$ then the polynomial $R(T)=\mathrm{Res}\,_{y}(T-f(y),g(y))$, where $\mathrm{Res}\,_{y}(-,-)$ denotes the resultant, is either irreducible in $\mathbf{L}[T]$ or is a power of an irreducible polynomial.

Proof. Let $y_{1},\dots,y_{m}$ be the roots of $g(y)$ in the algebraic closure of the field $\mathbf{L}$. Then $R(T)=\prod_{i=1}^{m}(T-f(y_{i}))$. Since $\mathbf{L}$ is a field of characteristic zero and $g(y)$ is irreducible, the Galois group of the field extension $\mathbf{L}\hookrightarrow\mathbf{L}(y_{1},\dots,y_{m})$ acts transitively on the set $\{y_{1},\dots,y_{m}\}$. It follows that this group acts transitively on the set $\{f(y_{1}),\dots,f(y_{m})\}$. Hence if $R=R_{1}\cdots R_{s}$ is a factorization of $R=R(T)$ into irreducible monic polynomials in the ring $\mathbf{L}[T]$ then $R_{i}=R_{j}$ for $i\neq j$.

Next corollary will be used in the proof of the main result of the decompositions of higher polars, which is Theorem 9.1.

Corollary 2.6

Let $f(y)$, $g(y)\in\mathbf{K}[[\underline{x}]][y]$ be Weierstrass polynomials. If the resultant $\mathrm{Res}\,_{y}(g(y),f(y)-T)\in\mathbf{K}[[\underline{x}]][T]$ satisfies the assumptions of Lemma 2.2, then $g(y)$ is not irreducible in the ring $\mathbf{K}[[\underline{x}]][y]$.

Proof. By Lemma 2.2 the polynomial $R(T)$ has at least two coprime factors. By Theorem 2.5, $g(y)$, considered as a polynomial in $\mathbf{K}((\underline{x}))[y]$, is not irreducible, thus by Gauss Lemma it is not irreducible as a polynomial in $\mathbf{K}[[\underline{x}]][y]$.

Remark 2.7

Beata Hejmej in [He] generalizes Theorem 2.5 to polynomials with coefficients in a field of any characteristic. Hence the results of this section hold for fields of arbitrary characteristic.

3 Kuo-Lu tree of a quasi-ordinary polynomial

From now on $\mathbf{K}$ will be an algebraically closed field of characteristic zero. Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial of degree $n$. Such a polynomial is quasi-ordinary if its $y$-discriminant equals $\underline{x}^{\bf i}u(\underline{x})$, where $u(\underline{x})$ is a unit in $\mathbf{K}[[\underline{x}]]$ and ${\mathbf{i}}\in\mathbf{N}^{d}$. After Jung-Abhyankar theorem (see [Pa-R, Theorem 1.3]) the roots of $f$ are in the ring $\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ of fractional power series and we may factorize $f(y)$ as $\prod_{i=1}^{n}(y-\alpha_{i})$, where $\alpha_{i}$ is zero or a fractional power series of nonnegative order. Put $\mathop{\mathrm{missing}}{Zer}f:=\{\alpha_{i}\,:\;1\leq i\leq n\}$. Since the differences of roots divide the discriminant, for $i\neq j$ we have

[TABLE]

The contact of $\alpha_{i}$ and $\alpha_{j}$ is by definition $O(\alpha_{i},\alpha_{j}):={\bf q}_{ij}$. By convention $O(\alpha_{i},\alpha_{i})=+\infty$.

We introduce in $\mathbf{Q}_{\geq 0}^{d}$ the partial order: ${\bf q}\leq{\bf q}^{\prime}$ if ${\bf q}^{\prime}-{\bf q}\in\mathbf{Q}_{\geq 0}^{d}$. By convention $+\infty$ is bigger than any element of $\mathbf{Q}_{\geq 0}^{d}$.

After [B-M, Lemma 4.7], for every $\alpha_{i},\alpha_{j},\alpha_{k}\in\mathop{\mathrm{missing}}{Zer}f$ one has $O(\alpha_{i},\alpha_{k})\leq O(\alpha_{j},\alpha_{k})$ or $O(\alpha_{i},\alpha_{k})\geq O(\alpha_{j},\alpha_{k})$.

Moreover, we have the strong triangle inequality:

[TABLE]

In general, we say that the contact between the fractional power series $\alpha$ and $\beta$ is well-defined if and only if $\alpha-\beta=\underline{x}^{{\bf q}}w(\underline{x})$, for some ${\bf q}\in\mathbf{Q}^{d}$ and $w\in\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ such that $w(0)\neq 0$. In such a case we put $O(\alpha,\beta)={\bf q}$.

Now we construct the Kuo-Lu tree of a quasi-ordinary Weierstrass polynomial $f$. Given ${\bf q}\in\mathbf{Q}^{d}_{\geq 0}$ we put $\alpha_{i}\equiv\alpha_{j}$ mod ${\bf q}^{+}$ if $O(\alpha_{i},\alpha_{j})>{\bf q},$ for $\alpha_{i},\alpha_{j}\in\mathop{\mathrm{missing}}{Zer}f$. Let ${\bf h}_{0}$ be the minimal contact between the elements of $\mathop{\mathrm{missing}}{Zer}f$. We represent $\mathop{\mathrm{missing}}{Zer}f$ as a horizontal bar $B_{0}$ and call $h(B_{0})$ the height of $B_{0}$. The equivalence relation $\equiv$ mod $h(B_{0})^{+}$ divides $B_{0}=\mathop{\mathrm{missing}}{Zer}f$ into cosets $B_{1},\ldots,B_{r}$. We draw $r$ vertical segments from the bar $B_{0}$ and at the end of the $j$th vertical segment we draw a horizontal bar which represents $B_{j}$. The bar $B_{j}$ is called a postbar of $B_{0}$ and in such a situation we write $B_{0}\perp B_{j}$. We repeat this construction recursively for every $B_{j}$ with at least two elements. The set of bars ordered by the inclusion relation is a tree. Following [K-Lu] we call this tree the Kuo-Lu tree of $f$ and denote it $T(f)$. The bar $B_{0}$ of minimal height is called the root of $T(f)$. For every bar $B$ of $T(f)$ there exists a unique sequence $B_{0}\perp B^{\prime}\perp B^{\prime\prime}\perp\cdots\perp B$, starting in $B_{0}$ and ending in $B$.

In the above construction, we do not draw the bars $\{\alpha_{i}\}\subset\mathop{\mathrm{missing}}{Zer}f$. These bars are the leaves of $T(f)$ and they are the only bars of infinite height.

Let $B,B^{\prime}\in T(f)$ be such that $B\perp B^{\prime}$. All fractional power series belonging to $B^{\prime}$ have the same term with the exponent $h(B)$. Let $c$ be the coefficient of such term. We say that $B^{\prime}$ is supported at the point $c$ on $B$ and we denote it by $B\perp_{c}B^{\prime}$. Observe that different postbars of $B$ are supported at different points.

This construction is adapted from [K-Lu] to quasi-ordinary case.

Example 3.1

Let $f=f_{1}f_{2}\in\mathbf{C}[[x_{1},x_{2}]][y]$, where $f_{1}=y^{2}-x_{1}^{3}x_{2}^{2}$ and $f_{2}=y-x_{1}^{5}x_{2}^{2}$. Observe that $f$ is quasi-ordinary since its $y$-discriminant equals $4x_{1}^{9}x_{2}^{6}(-1+x_{1}^{7}x_{2}^{2})^{2}$. The roots of $f$ are $\alpha=x_{1}^{3/2}x_{2}$, $\beta=-x_{1}^{3/2}x_{2}$ and $\gamma=x_{1}^{5}x_{2}^{2}$. The Kuo-Lu tree of $f$ is:

\left(\frac{3}{2},1\right)$$\alpha$$\beta$$\gamma$$T(f)

In the above picture we draw also a vertical segment supporting $T(f)$ called by Kuo and Lu in [K-Lu] the main trunk of the tree.

4 Compatibility with pseudo-balls

Let $\alpha\in\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ be a fractional power series and ${\bf h}\in\mathbf{Q}^{d}_{\geq 0}$. The pseudo-ball centered in $\alpha$ and of height ${\bf h}$ is the set $\alpha+\underline{x}^{\bf h}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$. The pseudo-ball centered in $\alpha$ of infinite height is the set $\{\alpha\}$.

Let $f$ be a quasi-ordinary polynomial $f$. Consider the bar $B=\{\alpha_{i_{1}},\dots,\alpha_{i_{s}}\}$ with finite height ${\bf h}$ of the Kuo-Lu tree $T(f)$. Set $\tilde{B}:=\alpha+\underline{x}^{\bf h}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$, where $\alpha\in B$. As $\alpha_{i_{k}}-\alpha_{i_{l}}\in\underline{x}^{\bf h}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ for $1\leq k\leq l\leq s$ the pseudo-ball $\tilde{B}$ is independent of the choice of $\alpha$. If $B=\{\alpha_{i}\}$ is a bar of infinite height then we put $\tilde{B}=B$. The mapping $B\to\tilde{B}$ is a one-to-one correspondence between $T(f)$ and the set of pseudo-balls $\tilde{T}(f):=\{\alpha_{i}+(\alpha_{i}-\alpha_{j})\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]:\alpha_{i},\alpha_{j}\in\mathop{\mathrm{missing}}{Zer}f\}$. For the purposes of this article it is easier to deal with pseudo-balls, hence from now on, we shall identify the elements of $T(f)$ with corresponding pseudo-balls. Such pseudo-balls will be called quasi-ordinary pseudo-balls.

Let $B=\alpha+\underline{x}^{h(B)}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ be a quasi-ordinary pseudo-ball of finite height. Every $\gamma\in B$ has a form $\gamma=\lambda_{B}(\underline{x})+c_{\gamma}\underline{x}^{h(B)}+\cdots$, where $\lambda_{B}(\underline{x})$ is obtained from any $\beta\in B$ by omitting all the terms of order bigger than or equal to $h(B)$ and ellipsis means terms of higher order. We call the number $c_{\gamma}$ the *leading coefficient of * $\gamma$ with respect to $B$ and denote it $\mbox{\rm lc}_{B}(\gamma)$. Remark that $c_{\gamma}$ can be zero.

Let $\mathbf{L}$ be the field of fractions of $\mathbf{K}[[\underline{x}]]$. It follows from [GP, Remark 2.3] that any truncation of a root of a quasi-ordinary polynomial is a root of a quasi-ordinary polynomial. Hence the field extensions $\mathbf{L}\hookrightarrow\mathbf{L}(\lambda_{B}(\underline{x}))\hookrightarrow\mathbf{L}(\lambda_{B}(\underline{x}),\underline{x}^{h(B)})$ are algebraic and we can associate with $B$ two numbers:

• •

the degree of the field extension $\mathbf{L}\hookrightarrow\mathbf{L}(\lambda_{B}(\underline{x}))$ that we will denote $N(B)$,

• •

the degree of the field extension $\mathbf{L}(\lambda_{B}(\underline{x}))\hookrightarrow\mathbf{L}(\lambda_{B}(\underline{x}),\underline{x}^{h(B)})$ that we will denote $n(B)$.

In this section we introduce the notion of compatibility of a Weierstass polynomial $g$ with a pseudo-ball $B$. We define a polynomial $G_{B}(z)$ which will play an important role in the sequel.

Definition 4.1

Let $g(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial and $B$ be a pseudo-ball of finite height. If

[TABLE]

for some $G_{B}(z)\in\mathbf{K}[z]\setminus\{0\}$ and some exponent $q(g,B)\in(\mathbf{Q}_{\geq 0})^{d}$ then we will say that $g$ is compatible with $B$. In (3) $\cdots$ means terms of higher order. The polynomial $G_{B}(z)$ will be called the $B$-characteristic polynomial of $g$.

Example 4.2

Return to Example 3.1. Let $B=\alpha+x_{1}^{3/2}x_{2}\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ be a pseudo-ball of $T(f)$ of height $h(B)=\left(\frac{3}{2},1\right)$. Observe that

[TABLE]

Hence the polynomial $f$ is compatible with the pseudo-ball $B$ and its $B$-characteristic polynomial is $F_{B}(z)=z(z^{2}-1)$, but for example the polynomial $g(y)=y-x_{1}-x_{2}$ is not compatible with $B$.

Our next goal is to prove in Corollary 4.5 that if a Weierstrass polynomial is compatible with a pseudo-ball then any factor of it is also compatible.

Lemma 4.3

Let $g(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial and let $B$ be a pseudo-ball of finite height. Consider $g(\lambda_{B}(\underline{x})+z\underline{x}^{h(B)})$ as a fractional power series $\tilde{g}(\underline{x})$ with coefficients in $\mathbf{K}[z]$. Then $g(y)$ is compatible with $B$ if and only if the Newton polytope of $\tilde{g}(\underline{x})$ equals the Newton polytope of a monomial.

Proof. If $g$ is compatible with $B$ then by (3) we get \Delta\bigl{(}\tilde{g}(\underline{x})\bigr{)}=\Delta\bigl{(}\underline{x}^{q(g,B)}\bigr{)}. Conversely, suppose that the Newton polytope of $\tilde{g}(\underline{x})$ equals the Newton polytope of the monomial $\underline{x}^{\bf q}$. Then $\tilde{g}(\underline{x})$ has a form $\underline{x}^{\bf q}\sum_{i=0}^{n}a_{i}(\underline{x})z^{n-i}$, where at least one of the values $a_{i}(0)$ is nonzero. Hence the $B$-characteristic polynomial of $g$ is $G_{B}(z)=\sum_{i=0}^{n}a_{i}(0)z^{n-i}$.

Remark 4.4

From the proof of Lemma 4.3 we get that $\tilde{g}(\underline{x})$ has the form $G_{B}(z)\underline{x}^{q(g,B)}+\sum_{h>q(g,B)}a_{h}(z)\underline{x}^{h}$, where $a_{h}(z)\in\mathbf{K}[z]$.

Corollary 4.5

Let $g\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial compatible with a pseudo-ball $B$. Then any factor of $g$ is compatible with $B$.

Proof. The Newton polytope of the product is the Minkowki sum of Newton polytopes of the factors. Hence, if $\Delta(\tilde{g})=\Delta(\underline{x}^{\bf q})$ and $\tilde{g}=\tilde{g}_{1}\tilde{g}_{2}$ then $\Delta(\tilde{g}_{i})$ have the form $\Delta(\underline{x}^{{\bf q}_{i}})$ for some ${\bf q}_{1},{\bf q}_{2}$ such that ${\bf q}={\bf q}_{1}+{\bf q}_{2}.$

Next lemma generalizes to $d$ variables [GB-Gw4, Lemma 3.1].

Lemma 4.6

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial of degree $n$ compatible with the pseudo-ball $B$. Then for every $k\in\{1,\dots,\deg F_{B}(z)\}$ the Weierstrass polynomial $g(y)=\frac{(n-k)!}{n!}\frac{d^{k}}{dy^{k}}f(y)$ is also compatible with $B$ and its $B$-characteristic polynomial is $G_{B}(z)=\frac{(n-k)!}{n!}\frac{d^{k}}{dz^{k}}F_{B}(z)$.

Proof. Differentiating identity $f(\lambda_{B}(\underline{x})+z\underline{x}^{h(B)})=F_{B}(z)\underline{x}^{q(f,B)}+\cdots$ with respect to $z$ we get $f^{\prime}(\lambda_{B}(\underline{x})+z^{h(B)})\underline{x}^{h(B)}=F^{\prime}_{B}(z)\underline{x}^{q(f,B)}+\cdots$. Hence $f^{\prime}(\lambda_{B}(\underline{x})+z\underline{x}^{h(B)})=F^{\prime}_{B}(z)\underline{x}^{q(f,B)-h(B)}+\cdots$, which proves the lemma for $k=1$. The proof for higher derivatives runs by induction on $k$.

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial of degree $n$. The Weierstrass polynomial $\frac{(n-k)!}{n!}\frac{d^{k}}{dy^{k}}f(y)$ of Lemma 4.6 will be called the normalized $k$th derivative of the Weierstrass polynomial $f(y)\in\mathbf{K}[[\underline{x}]][y]$ and we will denote it by $f^{(k)}(y)$. The variety of equation $f^{(k)}=0$ is called the $k$th polar of $f=0$. Since the normalized $n$th derivative of $f$ is constant, in the rest of the paper we consider normalized $k$th derivatives of $f$ for $1\leq k<\deg f$.

Lemma 4.7

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial and let $B$ be a pseudo-ball of finite height.

If $f$ is compatible with $B$, then for any $\gamma\in B$ we have

[TABLE] 2. 2.

If $f(y)=\prod_{i=1}^{n}(y-\alpha_{i})$ and we assume that one of the following holds: $\underline{x}$ is a single variable and $B$ is arbitrary or $f$ is quasi-ordinary and $B\in\tilde{T}(f)$ then $f$ is compatible with $B$ and we have

[TABLE]

and

[TABLE]

Proof. Since $\gamma\in B$ we can write $\gamma=\lambda_{B}(\underline{x})+z\underline{x}^{h(B)}$, where $z=\mbox{\rm lc}_{B}(\gamma)+\cdots$. By Remark 4.4 we have $f(\gamma)=f(\lambda_{B}(\underline{x})+z\underline{x}^{h(B)})=F_{B}(z)\underline{x}^{q(f,B)}+\cdots=F_{B}(\mbox{\rm lc}_{B}\gamma)\underline{x}^{q(f,B)}+\cdots$. This proves (4).

Suppose $\gamma=\lambda_{B}(x)+zx^{h},$ where $z$ is a constant. We have $f(\gamma)=\prod_{i=1}^{n}(\gamma-\alpha_{i})$. In order to prove (5) and (6), it is enough to compute the initial term of every factor $\gamma-\alpha_{i}$. If $\alpha_{i}\in B$ then the initial term of $\gamma-\alpha_{i}$ equals $(\mbox{\rm lc}_{B}\gamma-\mbox{\rm lc}_{B}\alpha_{i})\underline{x}^{h(B)}$. Otherwise the initial terms of $\gamma-\alpha_{i}$ and $\lambda_{B}-\alpha_{i}$ are equal. We finish the proof multiplying the initial terms.

Corollary 4.8

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be a quasi-ordinary Weierstrass polynomial. Then every factor of $f(y)$ is compatible with all bars $B\in T(f)$ of finite height.

Lemma 4.9

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be a quasi-ordinary Weierstrass polynomial, $p(y)$ be a factor of $f(y)$ and $B,B^{\prime}$ be bars of finite heights in $T(f)$ such that $B\perp B^{\prime}$. Then

[TABLE]

Proof. Put $p(y)=\prod_{\alpha\in\mathop{\mathrm{missing}}{Zer}p}(y-\alpha)$. Let $\gamma\in B,\gamma^{\prime}\in B^{\prime}$ be such that $O(\gamma,\alpha)=h(B)$ for all $\alpha\in B\cap\mathop{\mathrm{missing}}{Zer}p$ and $\mathrm{cont}(\gamma^{\prime},\alpha)=h(B^{\prime})$ for all $\alpha\in B^{\prime}\cap\mathop{\mathrm{missing}}{Zer}p$. By the STI we get $O(\gamma^{\prime},\alpha)=O(\gamma,\alpha)$ for any $\alpha\in\mathop{\mathrm{missing}}{Zer}p\backslash B^{\prime}$. If $\alpha\in\mathop{\mathrm{missing}}{Zer}p\cap B^{\prime}$ then $O(\gamma,\alpha)=h(B)$ and $O(\gamma^{\prime},\alpha)=h(B^{\prime})$. Hence

[TABLE]

Lemma 4.9 is similar in spirit to [GB-Gw-L, Lemma 2.7].

Lemma 4.10

Let $B$ be a quasi-ordinary pseudo-ball and let $g(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial compatible with $B$. Then

$G_{B}(z)=z^{k}\cdot H(z^{n(B)})$, for some $k\in\mathbf{N}$ and $H(z)\in\mathbf{K}[z]$. 2. 2.

If $g$ is irreducible and quasi-ordinary then $G_{B}(z)=az^{k}$ or $G_{B}(z)=a(z^{n(B)}-c)^{l}$, for some non-zero $a,c\in\mathbf{K}$ and some $l\in\mathbf{N}$.

Proof. Let $\mathbf{L}$ be the field of quotients of $\mathbf{K}[[\underline{x}]]$. By [Li, Lemma 5.7] and [GP, Remark 2.7] the algebraic extension $\mathbf{L}(\lambda_{B}(\underline{x}))\hookrightarrow\mathbf{L}(\lambda_{B}(\underline{x}),\underline{x}^{h(B)})$ is cyclic. Hence the generator $\varphi$ of the group $Gal(\mathbf{L}(\lambda_{B}(\underline{x})\hookrightarrow\mathbf{L}(\lambda_{B}(\underline{x}),\underline{x}^{h(B)})$ acts as follows: $\varphi(\lambda_{B}(\underline{x}))=\lambda_{B}(\underline{x})$ and $\varphi(\underline{x}^{h(B)})=\omega\underline{x}^{h(B)},$ where $\omega$ is a primitive $n(B)$th root of the unity. Applying $\varphi$ to (3) we get

[TABLE]

for some $0\leq k<n(B)$. Substituting $\omega z$ for $z$ in (3) and comparing with (7) we get $G_{B}(z)\omega^{k}=G_{B}(\omega z)$. Multiplying this equality by $(\omega z)^{n(B)-k}$ and putting $W(z):=z^{n(B)-k}G_{B}(z)$ we obtain $W(z)=W(\omega z).$ This implies that $W(z)=\overline{W}(z^{n(B)})$, for some $\overline{W}(z)\in\mathbf{K}[z]$. We finish the proof putting $H(z^{n(B)})=z^{-n(B)}\overline{W}(z^{n(B)})$. This proves the first part of the lemma.

Suppose now that $g$ is irreducible and quasi-ordinary. Let $\gamma=\lambda_{B}(\underline{x})+cx^{h(B)}+\cdots\in B\cap\mathop{\mathrm{missing}}{Zer}g$. Since the extension $\mathbf{L}(\lambda_{B}(\underline{x}))\hookrightarrow\mathbf{L}(\lambda_{B}(\underline{x}),\underline{x}^{h(B)})\hookrightarrow\mathbf{L}(\gamma)$ is Galois, any other root of $g$ belonging to $B$ has the form $\lambda_{B}(\underline{x})+\omega^{i}cx^{h(B)}+\cdots$, for some $0\leq i<n(B)$. Using the first part of the lemma and the equality (5) we complete the proof.

5 Conjugate pseudo-balls

In this section we define an equivalence relation between pseudo-balls called conjugacy relation. This will allow us to introduce, in Section 8, the notion of the Eggers tree of a quasi-ordinary Weierstrass polynomial.

Let $\mathbf{L}$ be the field of fractions of $\mathbf{K}[[\underline{x}]]$ and $\mathbf{M}$ be the field of fractions of $\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$.

Lemma 5.1

Let $\varphi$ be an $\mathbf{L}$-automorphism of $\mathbf{M}$. Then

For any ${\bf q}\in\mathbf{Q}^{d}$ there exists a root $\omega$ of the unity such that $\varphi(\underline{x}^{\bf q})=\omega\cdot\underline{x}^{\bf q}$, 2. 2.

$\varphi(\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]])=\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$, 3. 3.

If $u$ is a unit of the ring $\in\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ and ${\bf q}\in(\mathbf{Q}_{\geq 0})^{d}$ then $\varphi(u\cdot\underline{x}^{\bf q})=\tilde{u}\cdot\underline{x}^{\bf q}$ for some unit $\tilde{u}\in\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$.

Proof. Let $k$ be a positive integer. Observe that x_{i}=\varphi(x_{i})=\varphi\bigl{(}(x_{i}^{1/k})^{k}\bigr{)}=\varphi(x_{i}^{1/k})^{k}. Hence $\varphi(x_{i}^{1/k})=c\cdot x_{i}^{1/k}$ for some $c\in\mathbf{K}\backslash\{0\}$ such that $c^{k}=1$. It follows that for any ${\bf q}\in\mathbf{Q}^{d}$ there exists $\omega\in\mathbf{K}$ such that $\varphi(\underline{x}^{{\bf q}})=\omega\underline{x}^{\bf q}$ and $\omega^{m}=1$ for some positive integer $m$.

Every element of the ring $\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$ can be represented as a finite sum $\sum_{\bf q}a_{\bf q}\underline{x}^{\bf q}$ where ${\bf q}=(q_{1},\dots,q_{d})\in(\mathbf{Q}_{\geq 0})^{d}$ ($0\leq q_{i}<1$) and $a_{\bf q}\in\mathbf{K}[[\underline{x}]]$. This together with 1. proves items 2. and 3. of the lemma.

Let $B$, $B^{\prime}$ be pseudo-balls. We say that $B$ and $B^{\prime}$ are conjugate if there exists an $\mathbf{L}$-automorphism $\varphi$ of $\mathbf{M}$ such that $B^{\prime}=\varphi(B)$. The conjugacy of pseudo-balls is an equivalence relation. It follows from Lemma 5.1 that conjugate pseudo-balls have the same height. Moreover two quasi-ordinary pseudo-balls $B$ and $B^{\prime}$ of the same height are conjugate if any irreducible quasi-ordinary polynomial which has one of its roots in $B$ has another root in $B^{\prime}$ (in this way conjugate bars were defined in [K-Pa, Definition 6.1]). If $B^{\prime}=\varphi(B)$ then $\lambda_{B^{\prime}}=\varphi(\lambda_{B})$. The converse is also true; if $h(B)=h(B^{\prime})$ and there exists an $\mathbf{L}$-automorphism $\varphi$ of $\mathbf{M}$ such that $\lambda_{B^{\prime}}=\varphi(\lambda_{B})$ then $B$ and $B^{\prime}$ are conjugate. It follows from the above that the number of pseudo-balls conjugate with $B$ is equal to the degree of the minimal polynomial of $\lambda_{B}$, which is the degree $N(B)$ of the field extension $\mathbf{L}\hookrightarrow\mathbf{L}(\lambda_{B}(\underline{x}))$.

Lemma 5.2

Let $B,B^{\prime}$ be quasi-ordinary conjugate pseudo-balls. If $p(y)\in\mathbf{K}[[\underline{x}]][y]$ is a Weierstrass polynomial compatible with $B$ then

$p(y)$* is compatible with $B^{\prime}$.* 2. 2.

$q(p,B)=q(p,B^{\prime})$. 3. 3.

The characteristic polynomials $P_{B^{\prime}}(z)$ and $P_{B}(z)$ of $p(y)$ verify the equality $P_{B^{\prime}}(z)=\theta P_{B}(\omega z)$, for some roots of the unity $\theta$ and $\omega$.

Proof. Let $\mathbf{L}$ be the field of quotients of $\mathbf{K}[[\underline{x}]]$ and let $\varphi$ be a $\mathbf{L}$-automorphism of $\mathbf{M}$ such that $\varphi(B)=B^{\prime}$. Then $\varphi(\lambda_{B})=\lambda_{B^{\prime}}$. By Lemma 5.1 we have $\varphi(\underline{x}^{h(B)})=\omega^{-1}\underline{x}^{h(B)}$ and $\varphi(\underline{x}^{q(p,B)})=\theta\underline{x}^{q(p,B)}$ for some roots of the unity $\theta$ and $\omega$. Applying $\varphi$ to (3), with $g$ replaced by $p$, we get

[TABLE]

This gives $q(p,B)=q(p,B^{\prime})$ and $P_{B^{\prime}}(\omega^{-1}z)=\theta P_{B}(z)$.

6 Kuo-Lu Lemma for higher derivatives

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be a quasi-ordinary Weierstrass polynomial. We begin with combinatorial results concerning the Kuo-Lu tree $T(f)$. Remember that we identify any bar of $T(f)$ with the corresponding quasi-ordinary pseudo-ball. At the end of the section we apply these results to Newton-Puiseux roots of higher derivatives of $f(y)$.

Take an integer $k$ such that $1\leq k\leq\deg f$. With every bar $B$ of $T(f)$ we associate the numbers:

• •

$m(B)$ which is the number of roots of $f(y)$ which belong to $B$,

• •

$n_{k}(B)=\max\{m(B)-k,0\}$, and

• •

$t_{k}(B)=n_{k}(B)-\sum_{B\perp B^{\prime}}n_{k}(B^{\prime})$.

Remark 6.1

For $1\leq k<m(B)$ we have $n_{k}(B)>0$, $t_{k}(B)>0$ and for $m(B)\leq k\leq\deg f$ we have $n_{k}(B)=t_{k}(B)=0$.

We denote by $T_{k}(f)$ the sub-tree of $T(f)$ consisting of the bars $B\in T(f)$ such that $m(B)\geq k$.

Let $F\in\mathbf{K}[z]$ be a non constant polynomial. Let $F^{(k)}$ denotes the $k$th derivative of $F$.

Definition 6.2

We will say that $F$ is $k$-regular if one of the following conditions holds:

$F^{(k)}$* is zero or* 2. 2.

$F^{(k)}$* is nonzero and there is not a root of $F$ of multiplicity $\leq k$ which is a root of $F^{(k)}$.*

Recall that common roots of a polynomial $F$ and its first derivative are multiple roots of $F$. Hence any polynomial is 1-regular.

In general it is not easy to verify the $k$-regular property. In this papers polynomials of the form

[TABLE]

play an important role. Their $k$-regularity, for any $k$, is a consequence of Lemma 10.9.

Remark 6.3

Let $F(z)=\mbox{const}\prod_{i=1}^{r}(z-z_{i})^{m_{i}}$, where $z_{i}$ are pairwise different, $m_{i}\geq k$ for $1\leq i\leq s$ and $m_{i}<k$ for $s<i\leq r$. After differentiating, the multiplicity of any root drops by one. Hence putting ${\cal F}^{\oplus}(z)=\prod_{i=1}^{s}(z-z_{i})^{m_{i}-k}$ we obtain the decomposition

[TABLE]

into two coprime polynomials. A polynomial $F$ is $k$-regular if and only if $F$ and ${\cal F}^{\ominus}$ do not have common roots.

Definition 6.4

Let $f\in\mathbf{K}[[\underline{x}]][y]$ be a quasi-ordinary Weierstrass polynomial. We say that $f$ is Kuo-Lu $k$-regular if for every $B\in T(f)$ of finite height the polynomial $F_{B}(z)$ is $k$-regular.

We finish this subsection with some results for Weierstrass polynomials with coefficients in the ring of the formal power series in one variable.

Let $f(y)\in\mathbf{K}[[x]][y]$ be a square-free Weierstrass polynomial. Fix $B\in T_{k}(f)$ and assume that $\{B_{1},\dots,B_{s}\}$ is the set of post-bars of $B$ in $T_{k}(f)$. Denote $B^{\circ}=B\setminus(B_{1}\cup\dots\cup B_{s})$.

Theorem 6.5

Let $f(y)\in\mathbf{K}[[x]][y]$ be a square-free Weierstrass polynomial over the ring of formal power series in one variable. Let $f(y)=\prod_{i=1}^{n}(y-\alpha_{i})$ and $f^{(k)}(y)=\prod_{j=1}^{n-k}(y-\beta_{j})$ be the Newton-Puiseux factorizations of $f$ and $f^{(k)}$. Then

• (i)

for every $B\in T_{k}(f)$ the set $\{j:\beta_{j}\in B\}$ has $n_{k}(B)$ elements,

• (ii)

for every $B\in T_{k}(f)$ the set $\{j:\beta_{j}\in B^{\circ}\}$ has $t_{k}(B)$ elements,

• (iii)

for every $\beta_{j}$ there exists a unique $B\in T_{k}(f)$ such that $\beta_{j}\in B^{\circ}$.

• (iv)

Let $B\in T_{k}(f)$. If the polynomial $F_{B}(z)$ is $k$-regular then for every $\alpha_{i}\in B$, $\beta_{j}\in B^{\circ}$ one has $O(\alpha_{i},\beta_{j})=h(B)$. Otherwise there exist $\alpha_{i}\in B$, $\beta_{j}\in B^{\circ}$ such that $O(\alpha_{i},\beta_{j})>h(B)$.

Proof. Proof of (i). Suppose first, that $B\in T_{k}(B)$ has finite height. Then by Lemma 4.7 $F_{B}(z)=\mbox{const}\prod_{i:\alpha_{i}\in B}(z-\mbox{\rm lc}_{B}(\alpha_{i}))$. By equality $(4)$ of this lemma and Lemma 4.6 we get $F_{B}^{(k)}(z)=\mbox{const}\prod_{j:\beta_{j}\in B}(z-\mbox{\rm lc}_{B}(\beta_{j}))$. Hence, the set $\{j:\beta_{j}\in B\}$ has $\deg F_{B}-k=n_{k}(B)$ elements.

If the height of $B$ is infinite then $B=\{\alpha_{i}\}$ for exactly one Newton-Puiseux root $\alpha_{i}$ of $f(y)$. Hence for $k=1$ $n_{1}(B)=0$ and $f^{\prime}(y)$ does not have roots in $B$ and for $k>1$ $B\notin T_{k}(f)$.

Proof of (ii). It is enough to count the elements of the set $\{j:\beta_{j}\in B^{\circ}\}$ using (i).

Proof of (iii). Let $B_{0}$ be the root of the tree $T(f)$. By (i), $\{\beta_{1},\dots,\beta_{n-k}\}$ is a subset of $B_{0}$. It is clear that the sets $B^{\circ}$ for $B\in T_{k}(f)$ are pairwise disjoint and their union is equal to $B_{0}$. This proves (iii).

Proof of (iv). Assume that $B_{1}$, …, $B_{r}$ are the post-bars of $B$ supported at points $z_{1}$, …, $z_{r}$ respectively, and that $m(B_{i})\geq k$ for $i\in\{1,\dots s\}$, $m(B_{i})<k$ for $i\in\{s+1,\dots r\}$. Then by Lemma 4.7 $F_{B}(z)=\prod_{i=1}^{r}(z-z_{i})^{m(B_{i})}$.

After Remark 6.3 the $k$th derivative of $F_{B}(z)$ is the product of two coprime polynomials

[TABLE]

where ${\cal F}^{\oplus}_{B}(z):=\prod_{i=1}^{s}(z-z_{i})^{n_{k}(B_{i})}$.

We get $\deg{\cal F}^{\ominus}_{B}(z)=t_{k}(B)$. Hence it follows from (ii) and (iii) that all roots of ${\cal F}^{\ominus}_{B}(z)$ correspond to those Newton-Puiseux roots of $f^{(k)}(y)$ that belong to $B^{\circ}$. For $\alpha_{i}\in B$, $\beta_{j}\in B^{\circ}$ one has $O(\alpha_{i},\beta_{j})>h(B)$ if and only if $\mbox{\rm lc}_{B}(\alpha_{i})=\mbox{\rm lc}_{B}(\beta_{j})$, which means that the polynomials $F_{B}(z)$ and ${\cal F}^{\ominus}_{B}(z)$ have a common root. Since $F_{B}(z)$ is $k$-regular if and only if ${\cal F}^{\ominus}_{B}(z)$ and $F_{B}(z)$ do not have common roots we get (iv).

Remark 6.6

Let $f(y)\in\mathbf{K}[[x]][y]$ be a square-free Weierstrass polynomial over the ring of formal power series in one variable. Let $B\in T_{k}(f)$, $\beta_{i}\in B\cap\mathop{\mathrm{missing}}{Zer}f^{(k)}$ and put $c=\mbox{\rm lc}_{B}\beta_{i}$. Then ${\cal F}^{\oplus}_{B}(c)\neq 0$ if and only if $\beta_{i}\in B^{\circ}$. If ${\cal F}^{\oplus}_{B}(c)=0$ then there exists a sequence of postbars $B\perp_{c}B_{1}\perp\cdots\perp B_{l}$ such that $\beta_{i}\in B_{l}^{\circ}$ and $B_{l}\in T_{k}(f)$.

For quasi-ordinary Weierstrass polynomials which are Kuo-Lu $k$-regular, the counterpart of [K-Lu, Lemma 3.3] is true:

Corollary 6.7

Let $f(y)\in\mathbf{K}[[x]][y]$ be a square-free Weierstrass polynomial over the ring of formal power series in one variable. Assume that $f$ is Kuo-Lu $k$-regular. Then under assumptions and notations of Theorem 6.5, for every $\alpha_{i}\in\mathop{\mathrm{missing}}{Zer}f$, $\beta_{s}\in\mathop{\mathrm{missing}}{Zer}f^{(k)}$ there exists $\alpha_{j}\in\mathop{\mathrm{missing}}{Zer}f$ such that $O(\alpha_{i},\beta_{s})=O(\alpha_{i},\alpha_{j})$.

7 Newton polytopes of resultants

In this section we give a formula for the Newton polytope of the resultant $\mathrm{Res}\,_{y}(f^{(k)}(y),p(y)-T)$, where $f(y)$ is a Kuo-Lu $k$-regular quasi-ordinary Weierstrass polynomial, $p(y)$ is a factor of $f(y)$ and $T$ is a new variable. We prove that for irreducible $p(y)$, the Newton polytope of the resultant is polygonal.

7.1 Monomial substitutions

Let $g(\underline{x},y)\in\mathbf{K}[[\underline{x},y]]$. For any monomial substitution $x_{1}=u^{r_{1}}$,…, $x_{d}=u^{r_{d}}$, where $r_{i}$ are positive integers, we put

[TABLE]

We will write simply $\bar{g}(u,y)$ when no confusion can arise.

Observe that for $g=\underline{x}^{\bf s}$ we get $\bar{g}^{[{\bf r}]}=u^{\langle\bf{r},\bf{s}\rangle}$, where $\langle\cdot,\cdot\rangle$ denotes the scalar product.

Lemma 7.1

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be a quasi-ordinary Weierstrass polynomial. There is a one-to-one correspondence between the bars of $T(f)$ and the bars of $T(\bar{f}^{[{\bf r}]})$. If $B$ and $\bar{B}$ are the corresponding bars of $T(f)$ and $T(\bar{f}^{[{\bf r}]})$ respectively then

$h(\bar{B})=\langle{\bf r},h(B)\rangle$* and $t_{k}(\bar{B})=t_{k}(B)$.* 2. 2.

For any factor $g$ of $f$, the $B$-characteristic polynomial of $g$ and the $\bar{B}$-characteristic polynomial of $\bar{g}^{[{\bf r}]}$ are equal and $q(\bar{g}^{[{\bf r}]},\bar{B})=\langle{\bf r},q(g,B)\rangle$.

Proof. Set $u^{\bf r}=(u^{r_{1}},\dots,u^{r_{d}})$. If $\mathop{\mathrm{missing}}{Zer}f=\{\alpha_{i}(\underline{x})\}_{i=1}^{n}$ then $\mathop{\mathrm{missing}}{Zer}\bar{f}^{[{\bf r}]}=\{\alpha_{i}(u^{\bf r})\}_{i=1}^{n}$ and $O(\alpha_{i}(u^{\bf r}),\alpha_{j}(u^{\bf r}))=\langle{\bf r},O(\alpha_{i}(\underline{x}),\alpha_{j}(\underline{x}))\rangle$ for $i\neq j$.

Hence every bar $B=\{\alpha_{i_{j}}(\underline{x})\}_{j=1}^{k}$ of $T(f)$ yields the bar $\bar{B}=\{\alpha_{i_{j}}(u^{{\bf r}})\}_{j=1}^{k}$ of $T(\bar{f}^{[{\bf r}]})$ of height $\langle{\bf r},h(B)\rangle$.

Substituting $u^{r_{i}}$ for $x_{i}$ in the equation (3) appearing in Definition 4.1, we get

[TABLE]

hence the second part of the lemma follows.

The proof of the next lemma is similar in spirit to the proof of [GB-Gw3, Theorem 4.1] and the proof of [GB-Gw3, Theorem 9.2]. The same arguments were used there in special situation. Here we repeat the proof for the convenience of the reader.

Lemma 7.2

Let $g(\underline{x},y)\in\mathbf{K}[[\underline{x},y]]$ and $\Delta\subseteq\mathbf{R}^{d+1}$ be a Newton polytope. For any ${\bf r}\in(\mathbf{R}_{>0})^{d}$ let $\bar{\Delta}^{[{\bf r}]}$ be the image of $\Delta$ by the linear mapping $\pi_{\bf r}:\mathbf{R}^{d}\times\mathbf{R}\longrightarrow\mathbf{R}^{2}$ given by $({\bf a},b)\mapsto(\langle{\bf r},{\bf a}\rangle,b)$. If $\Delta(\bar{g}^{[{\bf r}]})=\bar{\Delta}^{[{\bf r}]}$ for every ${\bf r}\in(\mathbf{N}\backslash\{0\})^{d}$, then $\Delta(g)=\Delta$.

Proof. For every Newton polytope $\Delta\subseteq(\mathbf{R}_{\geq 0})^{d+1}$ and every $v\in(\mathbf{R}_{\geq 0})^{d+1}$ we define the support function $l(v,\Delta)=\min\{\langle v,\alpha\rangle\;:\;\alpha\in\Delta\}$. To prove the lemma it is enough to show that the support functions $l(\cdot,\Delta(g))$ and $l(\cdot,\Delta)$ are equal. As these functions are continuous it suffices to show the equality on a dense subset of $\mathbf{R}_{\geq 0}^{d+1}$.

Let $\vec{r}=(r_{1},\dots,r_{d+1})=({\bf r},r_{d+1})\in\mathbf{R}_{\geq 0}^{d+1}$, where ${\bf r}=(r_{1},\dots,r_{d})$.

Perturbing $\vec{r}$ a little we may assume that the hyperplane $\{\,\alpha\in\mathbf{R}^{d+1}:\langle\vec{r},\alpha\rangle=l(\vec{r},\Delta(g)\,\}$ supports $\Delta(g)$ at exactly one point $\check{\alpha}=(\underline{\check{\alpha}},\check{\alpha}_{d+1})$. Since after a small change of $\vec{r}$ the support point remains the same, we can assume, perturbing $\vec{r}$ again if necessary, that all $r_{i}$ are positive rational numbers.

We will show that

[TABLE]

Multiplying $\vec{r}$ by the common denominator of $r_{1}$, …, $r_{d+1}$ we may assume that all $r_{i}$ are positive integers. At this point of the proof we fix $\vec{r}$. We claim that l(\vec{r},\Delta)=l\bigl{(}(1,r_{d+1}),\bar{\Delta}^{[{\bf r}]}\bigr{)} and l(\vec{r},\Delta(g))=l\bigl{(}(1,r_{d+1}),\Delta\left(\bar{g}^{[{\bf r}]}\right)\bigr{)}.

First equality follows from the definition of $\pi_{\bf r}$ and the identity

[TABLE]

for $\alpha\in\mathbf{R}^{d+1}.$

Write $\alpha=(\underline{\alpha},\alpha_{d+1})\in\mathbf{R}^{d+1}$ and $g(\underline{x},y)=\sum_{\alpha}d_{\alpha}\underline{x}^{\underline{\alpha}}y^{\alpha_{d+1}}\in\mathbf{K}[[\underline{x},y]]$. Since the hyperplane $\{\,\alpha\in\mathbf{R}^{d+1}:\langle\vec{r},\alpha\rangle=l(\vec{r},\Delta(g)\,\}$ supports $\Delta(g)$ at $\check{\alpha}$, the term $d_{\check{\alpha}}u^{\langle{\bf r},\underline{\check{\alpha}}\rangle}y^{\check{\alpha}_{d+1}}$ of $\bar{g}^{[{\bf r}]}$, satisfies the equality $\langle{\bf r},\underline{\check{\alpha}}\rangle+r_{d+1}\check{\alpha}_{d+1}=l(\vec{r},\Delta(g))$, while for all other terms $d_{\alpha}u^{\langle{\bf r},\underline{\alpha}\rangle}y^{\alpha_{d+1}}$ with $d_{\alpha}\neq 0$ appearing in $\bar{g}^{[{\bf r}]}$, we have $\langle{\bf r},\underline{\alpha}\rangle+r_{d+1}\alpha_{d+1}>l(\vec{r},\Delta(g))$.

Hence l\bigl{(}(1,r_{d+1}),\Delta(g)\bigr{)}=\langle{\bf r},\underline{\check{\alpha}}\rangle+r_{d+1}\check{\alpha}_{d+1}=l(\vec{r},\Delta(g)), so we get (11).

Corollary 7.3

Let $g_{1}(\underline{x},y)$, $g_{2}(\underline{x},y)\in\mathbf{K}[[\underline{x},y]]$. Suppose that $\Delta(\bar{g}_{1}^{[{\bf r}]})=\Delta(\bar{g}_{2}^{[{\bf r}]})$ for every ${\bf r}\in(\mathbf{N}\backslash\{0\})^{d}$. Then $\Delta(g_{1})=\Delta(g_{2})$.

Theorem 7.4

Assume that $f\in\mathbf{K}[[\underline{x}]][y]$ is a Kuo-Lu $k$-regular quasi-ordinary Weierstrass polynomial and $p$ is a Weierstrass polynomial which is a factor of $f$ in $\mathbf{K}[[\underline{x}]][y]$. Then the Newton polytope of $R(T):=\mathrm{Res}\,_{y}(f^{(k)}(y),p(y)-T)\in\mathbf{K}[[\underline{x}]][T]$ is equal to

[TABLE]

Proof. First we will prove the theorem for $d=1$. We use the notation of Theorem 6.5. Let $\prod_{j=1}^{n-k}(y-\beta_{j})$ be the Newton-Puiseux factorization of $f^{(k)}(y)$. By the well-known properties of the resultants we have

[TABLE]

By Theorem 6.5, for every $\beta_{j}$ there exists a unique bar $B\in T(f)$ such that $\beta_{j}\in B^{\circ}.$ For such a bar, $h(B)$ is finite and $t_{k}(B)\neq 0$. By Corollary 4.5 the polynomial $p$ is compatible with $B$ and by (5) of Lemma 4.7 ${P}_{B}(z)$ is a factor of ${F}_{B}(z)$. By Theorem 6.5 $(iv)$ we get that $O(\alpha_{i},\beta_{j})=h(B)$ for any $\alpha_{i}\in B$. Hence $\mbox{\rm lc}_{B}\beta_{j}$ does not belong to the set $\{\mbox{\rm lc}_{B}\alpha_{i}\;:\;\alpha_{i}\in B\}$. So by the equality (5) in Lemma 4.7 we have ${F}_{B}(\mbox{\rm lc}_{B}\beta_{j})\neq 0$ and consequently ${P}_{B}(\mbox{\rm lc}_{B}\beta_{j})\neq 0$. Now, using equality (4) of Lemma 4.7 we conclude that the Newton polytope of $p(\beta_{j})-T$ is equal to \left\{\begin{picture}(7.0,3.0)(0.0,0.4)\put(0.0,1.15){\line(1,0){7.0}} \put(0.0,0.85){\line(1,0){7.0}} \put(3.5,1.3){\makebox(0.0,0.0)[b]{q(p,B)}} \put(3.5,0.7){\makebox(0.0,0.0)[t]{1}} \end{picture}\right\}. Using the property that the Newton polytope of a product is the Minkowski sum of the Newton polytopes of its factors, and $(ii)$ of Theorem 6.5 we finish the proof for $d=1$.

Assume now that $d>1$.

Let $x_{1}=u^{r_{1}}$,…, $x_{d}=u^{r_{d}}$ be a monomial substitution, where $r_{i}$ are positive integers. By Lemma  7.1 $f^{[{\bf r}]}$ is Kuo-Lu $k$-regular, hence by the first part of the proof ($d=1$)

[TABLE]

For any elementary polytope of the above sum, Lemma 7.1 gives

[TABLE]

Since the image of the Minkowski sum of Newton polytopes is the Minkowski sum of the images, we get $\Delta(\bar{R}^{[{\bf r}]})=\pi_{\bf r}(\Delta)$, where $\Delta$ denotes the Newton polytope given in (12). By Lemma 7.2 we get $\Delta(R)=\Delta$.

8 Eggers tree of a quasi-ordinary Weierstrass polynomial

In this section we introduce the Eggers tree of a quasi-ordinary Weierstrass polynomial $f$, after the conjugacy relation defined in Section 5. Denote by $[B]$ the conjugacy class of the pseudo-ball $B$ of the Kuo-Lu $T(f)$. By definition, the Eggers tree of $f$, denoted by $E(f)$, is the set of conjugacy classes with the natural order induced by the Kuo-Lu tree. This is the natural generalization of the Eggers tree associated with plane curves in [Eg]. The notion of Eggers tree, for quasi-ordinary singularities, was introduced by Popescu-Pampu in [Po]. He defined a slightly different notion of the Eggers tree, since he generalized to quasi-ordinary singularities the version of Eggers tree defined for curves in [W].

The leaves of $E(f)$ correspond with irreducible factors of $f$. Following Eggers we draw them in white color. By definition, the root of $E(f)$ is its vertex of minimum height. The branches of $E(f)$ are the smallest sub-trees of $E(f)$ containing the root and one of its leaves. Let $[B]$ be a vertex in the branch of $E(f)$ corresponding with the irreducible componente $f_{i}$ of $f$. Eggers draws in a dashed way the edge leaving from the vertex $[B]$ in this branch if there are not two roots of $f_{i}$ with contact $h(B)$.

Recall that the number of pseudo-balls conjugate with a quasi-ordinary pseudo-ball $B$ is $N(B)$ (see page 5.1).

Let $[B]$ be a vertex of the Eggers tree of a quasi-ordinary polynomial $f$. By Lemma 5.2, for any $k\in\{1,\ldots,\deg f\}$, the numbers $n_{k}(B)$ and $t_{k}(B)$ do not depend on the representative of $[B]$. Moreover, if $p(y)\in\mathbf{K}[[\underline{x}]][y]$ is a Weierstrass polynomial compatible with $B$ then the number $q(p,B)$ and the degree of its $B$-characteristic polynomial are also independent of the representative of $[B]$.

The Eggers tree of the quasi-ordinary polynomial $f=f_{1}f_{2}$ from Example 3.1 is

[B]$$f_{1}$$f_{2}

Remark 8.1

If $p$ is an irreducible factor of $f$ then, following Lemma 4.9, the sequence $\{q(p,B)\}_{[B]}$ is increasing along the branch ${P}$ of the Eggers tree of $f$ containing the leave representing $p$. Moreover, if $[B]$ does not belong to $P$ then $q(p,B)=q(p,B_{0})$, where $[B_{0}]$ is the last common vertex of $P$ and the branches of the Eggers tree containing $[B]$. Hence, the set $\{q(p,B)\}_{[B]}$ is well-ordered.

After Remark 8.1 we get

Corollary 8.2

Let $f\in\mathbf{K}[[\underline{x}]][y]$ be a Kuo-Lu $k$-regular quasi-ordinary Weierstrass polynomial and $p$ a Weierstrass polynomial which is an irreducible factor of $f$ in $\mathbf{K}[[\underline{x}]][y]$. Then the Newton polytope in (12) is polygonal.

9 Irreducible factors of higher derivatives

Let $f$ be a quasi-ordinary Weierstrass polynomial. In this section we study irreducible factors of normalized higher derivatives $f^{(k)}$. We show that every such an irreducible factor can be associated with a certain vertex $[B]$ of the Eggers tree of $f$. By definition an Eggers factor will be the product of all irreducible factors associated with the same vertex of $E(f)$. The Eggers factorization of a higher derivative is the product of all its Eggers factors. It generalizes to higher derivatives the factorization of the first polar given in [Eg] and [GB] for plane curves and in [GB-GP] for quasi-ordinary polynomials.

Let $F_{B}(z)$ be the $B$-characteristic polynomial of $f$. After Remark 6.3, the polynomial $F_{B}^{(k)}(z)$ is the product of two coprime polynomials ${\cal F}^{\oplus}_{B}(z)$ and ${\cal F}^{\ominus}_{B}(z),$ where

[TABLE]

Theorem 9.1

Let $f(y)$ be a quasi-ordinary Weierstrass polynomial and let $g(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial which is an irreducible factor of $f^{(k)}(y)$. Then there exists $[B]\in E(f)$, with $B\in T_{k}(f)$, such that:

If $B^{\prime}\in T_{k}(f)\backslash[B]$ then every root of $G_{B^{\prime}}(z)$ is a root of ${\cal F}^{\oplus}_{B^{\prime}}(z)$. 2. 2.

If $B^{\prime}\in T_{k}(f)\cap[B]$ then $G_{B^{\prime}}(z)$ and ${\cal F}^{\oplus}_{B^{\prime}}(z)$ do not have common roots. Moreover

[TABLE]

for some $l\geq 1$ and $a,c\in\mathbf{K}\backslash\{0\}$. If $l=1$ then $g(y)$ is quasi-ordinary.

Proof. Let ${\cal T}=\{\,B\in T_{k}(f):G_{B}(z)\mbox{ has a root which is not a root of }{\cal F}^{\oplus}_{B}(z)\,\}$. By Remark 6.6, $B\in{\cal T}$ if and only if for any monomial substitution $\bar{g}$ has a Newton-Puiseux root that belongs to $\bar{B}^{\circ}$.

Let ${\cal E}=\{\,[B]\in E(f):B\in{\cal T}\,\}$. We will show that ${\cal E}$ has only one element. Suppose that this is not the case, and let $[B_{0}]$ be the infimum of ${\cal E}$ in the ordered set $E(f)$ (the infimum exists because $E(f)$ has the structure of a tree).

Let $[B^{\prime}]$ be an element of ${\cal E}$ different from $[B_{0}]$ and let $p$ be any irreducible factor of $f$ such that one of its roots belongs to $B^{\prime}$. By definition of the Eggers tree there exists $B_{1}\in[B_{0}]$ such that $B^{\prime}\subsetneq B_{1}$. Since $B^{\prime}\in T_{k}(f)$, one of the roots of $P_{B_{1}}(z)$ has multiplicity bigger or equal than $k$. Hence, by the second statement of Lemma 4.10 all the roots of $P_{B_{1}}(z)$ have this property. By (9), the polynomial $P_{B_{1}}(z)$ could only share roots with ${\cal F}^{\oplus}_{B_{1}}(z)$. Hence, by Remark 6.3 the polynomials $P_{B_{1}}(z)$ and ${\cal F}^{\ominus}_{B_{1}}(z)$ are coprime.

Let $\bar{g}=\prod_{i=1}^{m}(y-\bar{\beta}_{i})$ be the Newton-Puiseux factorization of $g$ after some monomial substitution. Fix $B\in[B_{0}]$. By Lemmas 5.2 and 7.1 we get $q(\bar{p},\bar{B})=q(\bar{p},\bar{B}_{0})$. Let us define two sets of indexes associated with $\bar{B}$:

[TABLE]

[TABLE]

Directly from the definition of $P_{B}$ we have: if $i\in I_{\bar{B}}$ then $\mathrm{ord}\,\bar{p}(\bar{\beta}_{i})=q(\bar{p},\bar{B}_{0})$, and if $i\in J_{\bar{B}}$ then $\mathrm{ord}\,\bar{p}(\bar{\beta}_{i})>q(\bar{p},\bar{B}_{0})$.

The cardinality of $I_{\bar{B}}$ is equal to the number of roots of $G_{B}(z)$ counted with multiplicities which are not the roots of $P_{B}(z)$. Similarly the cardinality of $J_{\bar{B}}$ is equal to the number of roots of $G_{B}(z)$ counted with multiplicities which are the roots of $P_{B}(z)$. Hence the cardinality of these sets does not depend on the choice of the monomial substitution. Let $I:=\bigcup_{B\in[B_{0}]}I_{\bar{B}}$ and $J:=\bigcup_{B\in[B_{0}]}J_{\bar{B}}$. Observe that $\mathrm{ord}\,\bar{p}(\bar{\beta}_{i})=q(\bar{p},\bar{B}_{0})$ for $i\in I$ and $\mathrm{ord}\,\bar{p}(\bar{\beta}_{i})>q(\bar{p},\bar{B}_{0})$ for $i\in J$.

The sets $I$ and $J$ depend on the choice of the monomial substitution but their cardinality does not. We will show that the set $J$ is nonempty. Since $B^{\prime}\in{\cal T}$, there exists $\bar{\beta}_{i}\in\bar{B}^{\prime}$. Any root of $\bar{p}$ that belongs to $\bar{B}^{\prime}$ has the same leading coefficient with respect to $\bar{B}_{1}$ as $\bar{\beta}_{i}$. Hence $P_{B_{1}}(\mbox{\rm lc}_{\bar{B}_{1}}\bar{\beta}_{i})=0$, which gives $i\in J_{\bar{B}_{1}}\subset J$.

Now we will prove that the set $I$ is empty. Suppose that it is not the case. Put $R(T):=\mathrm{Res}\,_{y}(g,p-T)$ and $\bar{R}(T):=\mathrm{Res}\,_{y}(\bar{g},\bar{p}-T)$. We can write

[TABLE]

for some $c_{i}\in\mathbf{K}[[\underline{x}]]$. By a well-known formula for the resultant we have $\bar{R}(T)=\pm\prod_{i=1}^{m}(\bar{p}(\bar{\beta}_{i})-T)$. Since the Newton polygon of a product is the Minkowski sum of the Newton polygons of its factors, $\Delta(\bar{R}(T))$ has an edge of inclination $q(\bar{p},\bar{B}_{0})$ starting in the point $(0,m)$. The projection of this edge to the vertical axis has length $\sharp I$. This gives

[TABLE]

Since the monomial substitution was arbitrary, we have

[TABLE]

By Corollary 2.6 $g$ is not irreducible and we get a contradiction.

We conclude that $I=\emptyset$. This means that for every $\bar{\beta}_{i}$ there exists $B\in[B_{0}]$ such that $\bar{\beta}_{i}\in\bar{B}$ and $\mathrm{ord}\,\bar{p}(\bar{\beta}_{i})>q(\bar{p},\bar{B})$. By Remark 6.6, $\bar{\beta}_{i}$ belongs to a post-bar of $\bar{B}$, which has a nonempty intersection with $\mathop{\mathrm{missing}}{Zer}\bar{p}$. All post-bars of $B\in[B_{0}]$ that have nonempty intersection with $\mathop{\mathrm{missing}}{Zer}p$ conjugate. They form the vertex of $E(f)$, bigger than $[B_{0}]$, which is smaller or equal (with the natural order in $E(f)$) than any element of $\cal E$. Hence $[B_{0}]$ cannot be the infimum of ${\cal E}$ and we arrive again at a contradiction.

We have shown that ${\cal E}$ has only one element. Denote it by $[B_{0}]$. Hence for any monomial substitution we have $\mathop{\mathrm{missing}}{Zer}\bar{g}\subset\bigcup_{B\in[B_{0}]}\bar{B}^{\circ}$. By Remark 6.6 we get 1.  and the first part of 2.

Now we will find the form of $G_{B}(z)$, for any $B\in[B_{0}]$. If for every $\bar{\beta}_{i}\in B$ the leading coefficient $\mbox{\rm lc}_{\bar{B}}\bar{\beta}_{i}$ is [math], then obviously $G_{B}(z)=az^{l}$. Otherwise by Lemma 4.10 there exist $c\neq 0$ and a polynomial $G_{1}(z)$ coprime with $z^{n(B)}-c^{n(B)}$ such that $G_{B}(z)=G_{1}(z)(z^{n(B)}-c^{n(B)})^{l}$. Let $p(y)$ be the minimal Weierstrass polynomial of $\lambda_{B}(\underline{x})+c\underline{x}^{h(B)}$. Then $P_{B}(z)=\mbox{const}\cdot(z^{n(B)}-c^{n(B)})$.

Proceeding as in the first part of the proof we define again the sets $I$, $J$ of indexes. By the choice of $p(y)$ the set $J$ is nonempty. If the polynomial $G_{1}(z)$ has positive degree then the set $I$ is nonempty and we arrive at a contradiction. Hence $G_{1}(z)$ is a constant which proves the second part of the theorem.

Now we prove that if $l=1$ in (14) then $g(y)$ is quasi-ordinary. Let $p(y)\in\mathbf{K}[[\underline{x}]][y]$ be the minimal polynomial of $\lambda_{B}(\underline{x})$ if $G_{B}(z)=az$ or the minimal polynomial of $\lambda_{B}(\underline{x})+c{\underline{x}}^{h(B)}$ if $G_{B}(z)=a(z^{n(B)}-c^{n(B)})$. Then $G_{B}(z)$ is equal to $P_{B}(z)$ up to multiplication by a constant. By Lemma 5.2 for any $B^{\prime}\in[B_{0}]=[B]$ the characteristic polynomials $G_{B^{\prime}}(z)$ and $P_{B^{\prime}}(z)$ have the same form, in particular have the same number of roots and all their roots are simple. Take any monomial substitution and let $\bar{\beta}^{\prime}$, $\bar{\beta}^{\prime\prime}$ be different roots of $\bar{g}(y)$. Since $\mathop{\mathrm{missing}}{Zer}\bar{g}\subset\bigcup_{B\in[B_{0}]}\bar{B}^{\circ}$ there exist $B^{\prime},B^{\prime\prime}\in[B_{0}]$ such that $\bar{\beta}^{\prime}\in\bar{B}^{\prime}$ and $\bar{\beta}^{\prime\prime}\in\bar{B}^{\prime\prime}$. If $B^{\prime}=B^{\prime\prime}$ then $O(\bar{\beta}^{\prime},\bar{\beta}^{\prime\prime})=h(\bar{B}^{\prime})$ because $\bar{\beta}^{\prime}$ and $\bar{\beta}^{\prime\prime}$ have different leading coefficients with respect to $\bar{B}^{\prime}$. If $B^{\prime}\neq B^{\prime\prime}$ then $O(\bar{\beta}^{\prime},\bar{\beta}^{\prime\prime})=O(\lambda_{\bar{B}^{\prime}},\lambda_{\bar{B}^{\prime\prime}})$. In both cases the contact $O(\bar{\beta}^{\prime},\bar{\beta}^{\prime\prime})$ depends only on $B^{\prime}$ and $B^{\prime\prime}$. The same argument applies to the roots of $\bar{p}(y)$. As a consequence any bijection $\Phi:\mathop{\mathrm{missing}}{Zer}\bar{g}\to\mathop{\mathrm{missing}}{Zer}\bar{p}$ such that $\Phi(\bar{B}^{\prime}\cap\mathop{\mathrm{missing}}{Zer}\bar{g})=\bar{B}^{\prime}\cap\mathop{\mathrm{missing}}{Zer}\bar{p}$ for $B^{\prime}\in[B_{0}]$ preserves contacts.

Since the discriminant of a monic polynomial is the product of differences of its roots, the discriminant of $\bar{g}$ and the discriminant of $\bar{p}$ have the same order. Then by Corollary 7.3 the Newton polytopes of the discriminants of $g(y)$ and $p(y)$ are equal and we conclude that $g(y)$ is quasi-ordinary.

For $k$-regular quasi-ordinary Weierstrass polynomials we can say more.

Corollary 9.2

Let $f(y)$ be a $k$-regular Kuo-Lu quasi-ordinary Weierstrass polynomial and let $g(y)\in\mathbf{K}[[\underline{x}]][y]$ be a Weierstrass polynomial which is an irreducible factor of $f^{(k)}(y)$. Then there exists $[B]\in E(f)$ with $B\in T_{k}(f)$ such that:

If $B^{\prime}\in T(f)\cap[B]$ then $G_{B^{\prime}}(z)$ and $F_{B^{\prime}}(z)$ do not have common roots. 2. 2.

If $B^{\prime}\in T_{k}(f)\backslash[B]$ then every root of $G_{B^{\prime}}(z)$ is a root of ${\cal F}^{\oplus}_{B^{\prime}}(z)$. 3. 3.

If $B^{\prime}\in T(f)\setminus T_{k}(f)$ then $G_{B^{\prime}}(z)$ is a non-zero constant polynomial.

Proof. Take $B^{\prime}\in T_{k}(f)$. Then by Lemma 4.6 and the definition of $k$-regularity $G_{B^{\prime}}(z)$ and ${\cal F}_{B^{\prime}}^{\ominus}(z)$ do not have common roots. Hence for $B^{\prime}\in T_{k}(f)$ it is enough to use Theorem 9.1. This proves 1. The second statement is the first item of Theorem 9.1.

Now let $B^{\prime}\in T(f)\setminus T_{k}(f)$. Consider the chain of bars $B_{0}\perp_{c}B_{1}\perp\cdots\perp B_{s}=B^{\prime}$ of $T(f)$ such that $B_{0}\in T_{k}(f)$ and $B_{i}\notin T_{k}(f)$ for $1\leq i\leq s$. By the $k$-regularity of $F_{B_{0}}(z)$, we get $G_{B_{0}}(c)\neq 0$. Since $g$ is compatible with $B_{0}$, after (4) of Lemma 4.7, we have

[TABLE]

which shows that $g$ is also compatible with $B^{\prime}$ and its $B^{\prime}$-characteristic polynomial $G_{B^{\prime}}(z)$ equals $G_{B_{0}}(c)$.

10 Eggers factorizations of higher derivatives

Let $f$ be a quasi-ordinary Weierstrass polynomial. In this section we propose a factorization of the normalized derivative $f^{(k)}$ into factors associated with points of Eggers tree $E(f)$.

Definition 10.1

Let $g,p\in\mathbf{K}[[\underline{x}]][y]$ be Weierstrass polynomials. The P-contact between $g$ and $p$ is

[TABLE]

The notion of P-contact has its counterpart in the theory of plane analytic curves: for $y$-regular plane branches it is related with the logarithmic distance studied by Płoski in [Pł], since in such case $\Delta(\mathrm{Res}\,_{y}(g,p))$ equals the Newton polygon of a monomial $x^{m}$, where $m$ is the intersection multiplicity of the branches $g=0$ and $p=0$.

If $g$ is compatible with a pseudo-ball $B$ then we put

[TABLE]

Proposition 10.2

Let $B$ be a quasi-ordinary pseudo-ball of finite height and let $f$ be an irreducible quasi-ordinary Weierstrass polynomial compatible with $B$ such that $\mathop{\mathrm{missing}}{Zer}f\cap B\neq\emptyset$ or equivalently such that $F_{B}(z)$ has positive degree. Then $\mathrm{cont}_{\rm P}(f,B)$ does not depend on $f$.

Proof. Take any $f_{1}$, $f_{2}$ satisfying the assumptions of the proposition and let $\alpha_{1}\in\mathop{\mathrm{missing}}{Zer}f_{1}\cap B$, $\alpha_{2}\in\mathop{\mathrm{missing}}{Zer}f_{2}\cap B$. Choose a constant $c\in\mathbf{K}$ such that $(F_{i})_{B}(c)\neq 0$, for $i=1,2$ and let $\gamma=\lambda_{B}+c\underline{x}^{h(B)}$. Then $O(\gamma,\alpha_{1})=O(\gamma,\alpha_{2})=h(B)$ and for any $\xi\in\mathop{\mathrm{missing}}{Zer}f_{1}\cup\mathop{\mathrm{missing}}{Zer}f_{2}$ we have $O(\xi,\gamma)\leq h(B)$.

Let $G$ be a finite subgroup of $\mathbf{L}$-automorphisms of $\mathbf{M}$ that acts transitively on the sets $\mathop{\mathrm{missing}}{Zer}f_{1}$ and $\mathop{\mathrm{missing}}{Zer}f_{2}$. By the orbit stabilizer theorem, for $i\in\{1,2\}$ we get

[TABLE]

By STI we have $O(\gamma,\sigma(\alpha_{1}))=O(\gamma,\sigma(\alpha_{2}))$ for all $\sigma\in G$. Thus $\frac{1}{\deg f_{1}}q(f_{1},B)=\frac{1}{\deg f_{2}}q(f_{2},B)$.

After Proposition 10.2 we define the self-contact of a pseudo-ball $B$ of finite height as

[TABLE]

for any $f$ satisfying the assumptions of this proposition.

By Lemma 5.2 conjugate pseudo-balls have the same self-contact, hence the self-contact of $[B]$ is well-defined for any vertex $[B]$ of ${E}(f)$, where $B$ is of finite height.

In the set of Newton polytopes we define the next partial order: $\Delta_{1}\succeq\Delta_{2}$ if and only if $\Delta_{1}\subseteq\Delta_{2}$. Observe that $\Delta(\underline{x}^{{\bf q}_{1}})\succeq\Delta(\underline{x}^{{\bf q}_{2}})$ if and only if ${\bf q}_{1}\geq{\bf q}_{2}$. Now we show how the self-contacts of $[B]\in E(f)$ determine the P-contacts between irreducible factors of $f$.

Proposition 10.3

Let $f$ be a quasi-ordinary Weierstrass polynomial. Then the self contacts of vertices of finite height increase along the branches of $E(f)$. Moreover for any different irreducible factors $f_{1}$, $f_{2}$ of $f$

[TABLE]

where the maximum is taken over all $[B]\in E(f)$ such that $\mathop{\mathrm{missing}}{Zer}f_{i}\cap B\neq\emptyset$ for $i=1,2$.

Proof. Let $B$, $B^{\prime}$ be pseudo-balls of $T(f)$ of finite height such that $B^{\prime}\subsetneq B$. Choose an irreducible factor $f_{i}$ of $f$ such that $\mathop{\mathrm{missing}}{Zer}f_{i}\cap B^{\prime}\neq\emptyset$. By Lemma 4.9 we get $q(f_{i},B)<q(f_{i},B^{\prime})$, hence $\mbox{self-contact}(B)\prec\mbox{self-contact}(B^{\prime})$.

Let $[B]\in E(f)$ be the maximum (with the order defined in $E(f)$) of the set of all vertices $[B^{\prime}]\in E(f)$ such that $\mathop{\mathrm{missing}}{Zer}f_{i}\cap B^{\prime}\neq\emptyset$ for $i=1,2$. The pseudo-ball $B$ has the form $\gamma+(\gamma-\delta)\mathbf{K}[[\underline{x}^{1/\mathbf{N}}]]$, for some $\gamma\in\mathop{\mathrm{missing}}{Zer}f_{1}$ and $\delta\in\mathop{\mathrm{missing}}{Zer}f_{2}$ with maximal possible contact. By the choice of $\gamma$ and $\delta$, we have $O(\gamma,\delta^{\prime})\leq h(B)$ for all $\delta^{\prime}\in\mathop{\mathrm{missing}}{Zer}f_{2}\cap B$, consequently $(F_{2})_{B}(\mbox{\rm lc}_{B}\gamma)\neq 0$. Then $f_{2}(\gamma)=(F_{2})_{B}(\mbox{\rm lc}_{B}\gamma)\underline{x}^{q(f_{2},B)}+\cdots$.

Applying the Galois action associated with the irreducible polynomial $f_{2}$ we get $\Delta({f_{2}(\gamma)})=\Delta({f_{2}(\gamma^{\prime})})$, for any $\gamma,\gamma^{\prime}\in\mathop{\mathrm{missing}}{Zer}f_{1}$. Hence by the definition of the self-contact and the identity $\Delta(\mathrm{Res}\,_{y}(f_{1},f_{2}))=\sum_{\gamma\in\mathop{\mathrm{missing}}{Zer}f_{1}}\Delta({f_{2}(\gamma)})$ we have

[TABLE]

Theorem 10.4

Let $f\in\mathbf{K}[[\underline{x}]][y]$ be a quasi-ordinary Weierstrass polynomial. Then

[TABLE]

where $p_{[B]}$ are Weierstrass polynomials such that

The $B$-characteristic polynomial of $p_{[B]}$ equals $F_{B}^{\ominus}$ up to multiplication by constants and $\deg p_{[B]}=N(B)t_{k}(B)$. 2. 2.

For every irreducible factor $g$ of $p_{[B]}$ and every irreducible factor $f_{i}$ of $f$ we get

1. (a)

$\mathrm{cont}_{\rm P}(g,B)=\mbox{\rm self-contact}(B).$** 2. (b)

If $\mathrm{cont}_{\rm P}(f_{i},B)\prec\mbox{\rm self-contact}(B)$ then $\mathrm{cont}_{\rm P}(f_{i},g)=\mathrm{cont}_{\rm P}(f_{i},B)$. 3. (c)

If $\mathrm{cont}_{\rm P}(f_{i},B)=\mbox{\rm self-contact}(B)$ then $\mathrm{cont}_{\rm P}(f_{i},g)\succeq\mathrm{cont}_{\rm P}(f_{i},B)$. 3. 3.

If $f$ is $k$-regular then the inequalities $\succeq$ in (c) become equalities. 4. 4.

For every irreducible factor $g$ of $p_{[B]}$ there is an irreducible factor $f_{i}$ of $f$ such that $\mathrm{cont}_{\rm P}(f_{i},g)=\mathrm{cont}_{\rm P}(f_{i},B)=\mbox{\rm self-contact}(B).$

Proof. We define $p_{[B]}$ as the product of all irreducible factors of $f^{(k)}$ having the same $[B]$ in Theorem 9.1 (by convention the product of an empty family is $1$). After some monomial substitution $\bar{p}_{[B]}$ has $N(B)t_{k}(B)$ roots and all of them are in $\bigcup_{B^{\prime}\in[B]}\bar{B^{\prime}}^{0}$. Consequently $\deg p_{[B]}=N(B)t_{k}(B)$.

Now we will prove the second statement. Since $p_{[B]}$ has positive degree we may assume that $B\in T_{k}(f)$. Let $f_{i}$ be an irreducible factor of $f$. If $\mathrm{cont}_{\rm P}(f_{i},B)\prec\mbox{ self-contact}(B)$ then by Proposition 10.2 $(F_{i})_{B}(z)$ is a non-zero constant polynomial. Hence, for any $\bar{\gamma}\in\mathop{\mathrm{missing}}{Zer}\bar{p}_{[B]}$ we have $\mathrm{ord}\,\bar{f_{i}}(\bar{\gamma})=q(\bar{f}_{i},\bar{B})$, which proves 2(b). Suppose now that $\mathrm{cont}_{\rm P}(f_{i},B)=\mbox{ self-contact}(B)$. For every root $\bar{\gamma}\in\mathop{\mathrm{missing}}{Zer}\bar{p}_{[B]}$ we have $\mathrm{ord}\,\bar{f_{i}}(\bar{\gamma})\geq q(\bar{f}_{i},\bar{B})$ with equality in the $k$-regular case. Hence if $g$ is an irreducible factor of $p_{[B]}$ then $\mathrm{ord}\,\mathrm{Res}\,_{y}(\bar{f}_{i},\bar{g})\geq(\deg g)\cdot q(\bar{f}_{i},\bar{B})$ with equality in the $k$-regular case. This gives 2(c) and 3.

If the polynomial $F_{B}(z)$ is as in (8) then it is $k$-regular. In this case for any irreducible factor $f_{i}$ of $f$, with $(F_{i})_{B}(z)$ of positive degree, the polynomials $(F_{i})_{B}(z)$ and $G_{B}(z)$ do not have common factors.

If $F_{B}(z)$ is not as in (8), then by Lemma 4.10 there is an irreducible factor $f_{i}$ of $f$ such that the polynomials $(F_{i})_{B}(z)$ and $G_{B}(z)$ do not have common factors and $(F_{i})_{B}(z)$ has positive degree. After any monomial substitution, we have $\mathrm{ord}\,\bar{f}_{i}(\bar{\gamma})=q(\bar{f}_{i},\bar{B})$, for every $\bar{\gamma}\in\mathop{\mathrm{missing}}{Zer}\bar{g}$. This gives $\mathrm{ord}\,\mathrm{Res}\,(\bar{f}_{i},\bar{g})=\deg g\cdot q(\bar{f}_{i},\bar{B})$. Since the monomial substitution was arbitrary, the fourth statement of the theorem holds true in all cases.

It rests to prove 2(a). Choose $f_{i}$ as in the proof of the fourth statament. Then $\Delta(g(\alpha))=\Delta(\underline{x}^{q(g,B)})$ for any $\alpha\in B\cap\mathop{\mathrm{missing}}{Zer}f_{i}$.

Applying the same argument as in the end of the proof of Proposition 10.3, we get $\Delta(\mathrm{Res}\,_{y}(f_{i},g))=\deg f_{i}\Delta(g(\alpha))=\deg f_{i}\cdot\Delta(\underline{x}^{q(g,B)})$. After the fourth statement and the definition of the P-contact: $\mbox{self-contact}(B)=\mathrm{cont}_{\rm P}(f_{i},g)=\frac{1}{\deg g\deg f_{i}}\Delta(\mathrm{Res}\,_{y}(f_{i},g))=\frac{1}{\deg g}\Delta(\underline{x}^{q(g,B)})=\mathrm{cont}_{\rm P}(g,B).$

Example 10.5

We consider the example in [GB-GP, Section 10]: let $f=f_{1,1}f_{1,2}f_{2,1}f_{2,2}$, where $f_{i,j}=(y^{2}-ix_{1}^{3}x_{2}^{2})^{2}-jx_{1}^{5}x_{2}^{4}y$ are irreducible quasi-ordinary polynomials for $i,j\in\{1,2\}$. The Kuo-Lu and the Eggers tree of $f$ are

[B_{1}]$$[B_{2}]$$[B_{3}]$$f_{1,1}$$f_{1,2}$$f_{2,1}$$f_{2,2}

The heights of the vertices of the Eggers tree are: $h[B_{1}]=\left(\frac{3}{2},1\right)$, $h([B_{2}])=h([B_{3}])=\left(\frac{7}{4},\frac{3}{2}\right)$; the self-contacts are $\hbox{\rm self-contact}([B_{1}])=\frac{1}{4}\Delta(\underline{x}^{(6,4)})$ and $\hbox{\rm self-contact}([B_{2}])=\hbox{\rm self-contact}([B_{3}])=\frac{1}{4}\Delta\left(\underline{x}^{(13,10)}\right)$.

For any $1\leq k\leq 16$, the degrees of polynomials $p_{[B_{i}]}$ are

[TABLE]

The characteristic polynomials are $F_{B_{1}}(z)=(z^{2}-1)^{4}(z^{2}-2)^{4}$, $F_{B_{2}}(z)=(4z^{2}-1)(4z^{2}-2)$ and $F_{B_{3}}(Z)=(8z^{2}-\sqrt{2})(8z^{2}-2\sqrt{2})$. We can verify that these polynomials are $k$-regular for any $k$.

Theorem 10.4 allows us to compute the P-contact between the irreducible factors of $f$ and the irreducible factors of its higher order polars. For any $k$ and any irreducible factor $g$ of $p_{[B_{1}]}$, we have $\mathrm{cont}_{\rm P}(f_{i,j},g)=\hbox{\rm self-contact}([B_{1}])$. For any $k$ and any irreducible factor $g$ of $p_{[B_{2}]}$, we have $\mathrm{cont}_{\rm P}(f_{1,j},g)=\hbox{\rm self-contact}([B_{2}])$ and $\mathrm{cont}_{\rm P}(f_{2,j},g)=\hbox{\rm self-contact}([B_{1}])$, for any $j=1,2$. We have the symmetric situation for the irreducible factors of $p_{[B_{3}]}$.

Example 10.6

The second polar of the quasi-ordinary polynomial $f$ from the Example 3.1 (see page 8 for its Eggers tree) has only one Eggers factor $p_{[B]}=y$, with $\mathrm{cont}_{\rm P}(f_{2},y)=\Delta(\underline{x}^{(5,2)})\succ\mathrm{cont}_{\rm P}(f_{1},y)=\Delta(\underline{x}^{(3/2,1)})=\mbox{\rm self-contact}(B),$ hence in item 2. (c) of Theorem 10.4 we have equality for $f_{1}$ and strict inequality for $f_{2}$.

Now we study the examples of [Ca]:

Example 10.7

([Ca, Example 5.1]) Let $f=y^{3}+x^{2}y$. The Eggers tree of $f$ has only one vertex $[B]$ of finite height, where $B=x\mathbf{K}[[x^{1/\mathbf{N}}]]$. The $B$-characteristic polynomial of $f$ is as in the previous example, so it is not $2$-regular. We get $f^{(2)}=p_{[B]}=y$. If $f_{1}=y-x$, $f_{2}=y+x$ and $f_{3}=y$ then $\emptyset=\mathrm{cont}_{\rm P}(f_{3},y)\succeq\mathrm{cont}_{\rm P}(f_{i},y)=\mathrm{cont}_{\rm P}(f_{i},B)=\Delta(x)=\mbox{\rm self-contact}(B)$ for $i=1,2$. This illustrates the fourth statement of Theorem 10.4.

Example 10.8

([Ca, Example 5.2]) Let $f_{a}=y^{4}+ax^{2}y^{2}+x^{2}y+x^{10}$. We get $f_{a}=f_{a1}f_{a2}$, where $f_{a1}$ is irreducible and the contact of any two different roots of it is $\frac{2}{3}$, and $f_{a2}=0$ is a smooth curve tangent to $y=0$. The Eggers tree of $f_{a}$ is:

[B]$$f_{a1}$$f_{a2}

The characteristic polynomial $F_{B}(z)$ equals $z^{4}+z$. Hence $f_{a}$ is not $2$-regular. For any irreducible factor $g$ of $f^{(2)}_{a}$ we get $\mathrm{cont}_{\rm P}(f_{a1},g)=\Delta(x^{2/3})=\mbox{ self-contact}(B)$. For $f_{a2}$ the P-contact depends on $a$:

[TABLE]

10.1 Irreducible case

Assume that $f(y)\in\mathbf{K}[[\underline{x}]][y]$ is an irreducible quasi-ordinary Weierstrass polynomial of degree $n>1$ and $\mathop{\mathrm{missing}}{Zer}f=\{\alpha_{i}\}_{i=1}^{n}$. By [Li] the set $\{O(\alpha_{i},\alpha_{j})\,:\,i\neq j\}:=\{{\bf h}_{1},\ldots,{\bf h}_{s}\}$ is well-ordered, so we may assume that ${\bf h}_{1}\leq{\bf h}_{2}\leq\cdots\leq{\bf h}_{s}$. These values are the finite heights of the bars of $T(f)$. The sequence ${\bf h}_{1},\ldots,{\bf h}_{s}$ is called the sequence of characteristic exponents of $f(y)$. Let $B_{i}$ be any bar in $T(f)$ of height ${\bf h}_{i}$. By [GP, Remark 2.7] the degree $n(B_{i})$ of the field extension $\mathbf{L}(\lambda_{B_{i}}(\underline{x}))\hookrightarrow\mathbf{L}(\lambda_{B_{i}}(\underline{x}),\underline{x}^{{\bf h}_{i}})$ does not depend on the choice of $B_{i}$ and will be denoted by $n_{i}$. Put $e_{i}:=n_{i+1}\cdots n_{s}$ for $0\leq i\leq s$ (by convention the empty product is one). Observe that $T(f)$ has a special structure: all bars of the same height are conjugate and there are $n_{1}\cdots n_{i-1}$ conjugate bars of height ${\bf h}_{i}$ (see [GB-Gw3, Theorem 6.2]).

By (12) we get

[TABLE]

where $B_{i}$ is any ball of $T(f)$ of height ${\bf h}_{i}$ and

[TABLE]

Let $i_{k}\in\{1,\ldots,s\}$ be such that $e_{i_{k}}\leq k<e_{i_{k}-1}$. Then $t_{k}(B_{i})$ is positive if and only if $1\leq i\leq i_{k}$.

The Newton polytope of (16) is polygonal (see Corollary 8.2) and has $i_{k}$ edges of different inclinations. After Theorem 2.4 we decompose $\mathrm{Res}\,_{y}(f^{(k)},f-T)=\prod_{i=1}^{i_{k}}R_{i}$, where $\deg_{T}R_{i}=(n_{1}\cdots n_{i-1})t_{k}(B_{i})$ and any $R_{i}$ has an elementary Newton polytope of inclination $q(f,B_{i})$.

Such a decomposition of the resultant can be also obtained from Eggers factorization of $f^{(k)}$. By Lemma 4.10 the $B_{i}$-characteristic polynomial of $f$ has the form

[TABLE]

for some $c_{B_{i}}\in\mathbf{K}\setminus\{0\}$. The properties of such polynomials are described in the following lemma, which was proved in [GB-Gw4, Lemma 5.3] for complex polynomials but by Lefschetz Principle it holds true for polynomials over any algebraically closed field of characteristic zero.

Lemma 10.9

Let $\mathbf{K}$ be an algebraically closed field of charactersitic zero. If $F(z)=(z^{n}-c)^{e}\in\mathbf{K}[z]$ with $c\neq 0$ then for $1\leq k<\deg F(z)$ one has $\frac{d^{k}}{dz^{k}}F(z)=Cz^{a}(z^{n}-c)^{b}\prod_{i=1}^{d}(z^{n}-c_{i})$, where $C\neq 0$ and

1. (1)

$0\leq a<n$* and $a+k\equiv 0\pmod{n}$,* 2. (2)

$b=\max\{e-k,0\}$, 3. (3)

$d=\min\{e,k\}-\lceil\frac{k}{n}\rceil$, where $\lceil x\rceil$ denotes the smallest integer bigger than or equal to $x$, 4. (4)

$c_{i}\neq c_{j}$* for $1\leq i<j\leq d$ and $0\neq c_{i}\neq c$ for $1\leq i\leq d$.*

Corollary 10.10

Every irreducible quasi-ordinary Weierstrass polynomial is Kuo-Lu $k$-regular for any positive integer $k$.

Theorem 10.11

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be an irreducible quasi-ordinary Weierstrass polynomial of degree $n>1$ and characteristic exponents ${\bf h}_{1},\ldots,{\bf h}_{s}$. Then

[TABLE]

where

$p_{i}$* is a Weierstrass polynomial in $\mathbf{K}[[\underline{x}]][y]$ of degree $n_{1}\cdots n_{i-1}t_{k}(B_{i})$.* 2. 2.

Any irreducible factor $g$ of $p_{i}$ verifies

[TABLE] 3. 3.

The $B_{i}$-characteristic polynomial of $p_{i}$ is $(P_{i})_{B_{i}}=constF_{B_{i}}^{\ominus}.$

Proof. The theorem follows from Corollary 10.10 and the first, second and third part of Theorem 10.4.

Proposition 10.12

Let $f(y)\in\mathbf{K}[[\underline{x}]][y]$ be an irreducible quasi-ordinary Weierstrass polynomial with characteristic exponents ${\bf h}_{1},\ldots,{\bf h}_{s}$. Let $a$, $d$ be integers such that $0\leq a<n_{i}$, $a+k\equiv 0\pmod{n_{i}}$ and $d=\min\{e_{i},k\}-\lceil\frac{k}{n_{i}}\rceil$. Then every $p_{i}$ of (18) admits a factorization of the form $p_{i}=p_{i0}p_{i1}\cdots p_{id},$ where

the corresponding $B_{i}$-characteristic polynomials are $P_{i0}(z)=\mbox{const}\cdot z^{a}$, $P_{ij}(z)=\mbox{const}\cdot(z^{n_{i}}-c_{j})$ with $c_{j}\neq c_{l}$ for $1\leq j<l\leq d$ and $c_{j}\neq 0$. 2. 2.

$p_{i0}$* is a Weierstrass polynomial of degree $a\cdot n_{1}\cdots n_{i-1}$ not necessarily quasi-ordinary.* 3. 3.

Every $p_{ij}$ for $1\leq j\leq d$ is a quasi-ordinary irreducible Weierstrass polynomial of degree $n_{1}\cdots n_{i}$ and characteristic exponents ${\bf h}_{1},\ldots,{\bf h}_{i}$.

Proof. After (17) $F_{B_{i}}(z)$ has the form $a(z^{n_{i}}-c)^{e_{i}}$ for some nonzero $a$ and $c$.

By the first part of Theorem 10.4 and Lemma 10.9 the polynomial $P_{i,B_{i}}(z)=\hbox{\rm const}\cdot z^{a}\prod_{j=1}^{d}(z^{n_{i}}-c_{j})$. This polynomial is the product of the $B_{i}$-characteristic polynomials of the irreducible factors of $p_{i}$. From the second part of Theorem 9.1, we know that $p_{i}$ has $d$ irreducible factors $\{p_{ij}\}_{j=1}^{d}$ such that $P_{ij}(z)=\mbox{const}\cdot(z^{n_{i}}-c_{j})$. If $p_{i}$ has other irreducible factors, then $p_{i0}$ is their product. It also follows from Theorem 9.1 that $p_{ij}$ are quasi-ordinary for $1\leq j\leq d$.

By a similar argument as in the first part of the proof of Theorem 10.4 we get $\deg p_{ij}=N(B_{i})\deg P_{i,jB_{i}}(z)$. Since $N(B_{i})=n_{1}\cdots n_{i-1}$, we obtain the statements about the degrees of $p_{ij}$.

Fix $p_{ij}$ for $j\in\{1,\ldots,d\}$. The pseudo-ball $B_{i}$ has $n_{1}\cdots n_{i-1}$ conjugate pseudo-balls. Each of these pseudo-balls contains $n_{i}$ roots of $p_{ij}$. Since the roots of $P_{i,j}(z)$ are simple, any two roots of $p_{ij}$ belonging to the same pseudo-ball have different leading coefficients with respect to $B_{i}$, so their contact equals ${\bf h}_{i}$. Now, if we consider two roots of $p_{ij}$ belonging to different conjugate pseudo-balls, then their contact depends only on these two pseudo-balls, hence it is equal to ${\bf h}_{l}$ for some $l\in\{1,\ldots,i-1\}$. We conclude that the characteristic exponents of $p_{ij}$ are ${\bf h}_{1},\ldots,{\bf h}_{i}$.

In Proposition 10.12 the integer $a$ can be [math], in such a case $p_{i0}=1$. If $a=1$ then $p_{i0}$ is quasi-ordinary with characteristic exponents ${\bf h}_{1},\ldots,{\bf h}_{i-1}$. Moreover $d$ can be zero and in such a case $p_{i}=p_{i0}$.

11 Eggers decomposition for power series

In this section we deal with power series in variables $\underline{x}$ and $y$. A power series will be called quasi-ordinary if it is a product of a unity and a quasi-ordinary Weierstrass polynomial. We outline how to generalize the results of previous sections to quasi-ordinary power series. For that we need the next generalization of Lemma 4.6:

Lemma 11.1

Let $f=uf^{*}$ and $\frac{\partial^{k}}{\partial y^{k}}f=wg^{*},$ where $u$, $w\in\mathbf{K}[[\underline{x},y]]$ are unities, $f^{*}$, $g^{*}\in\mathbf{K}[[\underline{x}]][y]$ are Weierstrass polynomials and $1\leq k\leq n=\deg f^{*}$. Assume that $f^{*}$ is compatible with a pseudo-ball $B$. Then $g^{*}$ is compatible with $B$ and $G^{*}_{B}(z)=\frac{(n-k)!}{n!}\frac{d^{k}}{dz^{k}}F_{B}^{*}(z)$.

Proof. Substituting $\underline{x}=0$ we get $f(0,y)=u(0,0)y^{n}+\cdots$. Hence $\frac{\partial^{k}f}{\partial y^{k}}(0,y)=\frac{n!}{(n-k)!}u(0,0)y^{n-k}+\cdots$. On the other hand $\frac{\partial^{k}f}{\partial y^{k}}(0,y)=w(0,y)g^{*}(0,y)$ which implies that

[TABLE]

By the assumption of compatibility of $f^{*}$ we have

[TABLE]

Hence $f_{1}(\underline{x},z):=\underline{x}^{-q(f^{*},B)}f(\underline{x},\lambda_{B}(\underline{x})+z\underline{x}^{h(B)})$ is a fractional power series such that

[TABLE]

By the chain rule of differentiation

[TABLE]

Differentiating (20) yields $\frac{\partial^{k}f_{1}}{\partial z^{k}}(0,z)=u(0,0)\frac{d^{k}}{dz^{k}}F_{B}^{*}(z)$. Thus

[TABLE]

Comparing (21) and (22) we get

[TABLE]

By the definition of $g^{*}$, the left hand side of the above equality can be written as

[TABLE]

which gives, after (19)

[TABLE]

and finishes the proof.

Theorem 6.5, Corollary 6.7, Theorem 7.4, Theorem 9.1, Corollary 9.2, Theorem 10.4, Theorem  10.11 and Proposition 10.12, where $f^{(k)}$ stands for the Weierstrass polynomial of $k$th derivative, remain true for quasi-ordinary power series. For the proofs it is enough to replace the power series by their Weierstrass polynomials and use Lemma 11.1 instead of Lemma 4.6 when required.