Numerical Semigroups and Applications (RSME Springer Series, 3) 3030549429, 9783030549428

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RSME Springer Series 3

Abdallah Assi Marco D’Anna Pedro A. García-Sánchez

Numerical Semigroups and Applications Second Edition

RSME Springer Series Volume 3

Editor-in-Chief José Bonet, Instituto Universitario de Matemática Pura y Aplicada (IUMPA), Universitat Politècnica de València, Valencia, Spain Series Editors Nicolas Andruskiewitsch, FaMAF - CIEM (CONICET), Universidad Nacional de Córdoba, Córdoba, Argentina María Emilia Caballero, Instituto de Matemáticas, Universidad Nacional Autónoma de México, México, Mexico Pablo Mira, Departamento de Matematica Aplicada y Estadistica, Universidad Politécnica de Cartagena, Cartagena, Spain Timothy G. Myers, Centre de Recerca Matemàtica, Barcelona, Spain Marta Sanz-Solé, Department of Mathematics and Informatics, Barcelona Graduate School of Mathematics (BGSMath), Universitat de Barcelona, Barcelona, Spain Karl Schwede, Department of Mathematics, University of Utah, Salt Lake City, UT, USA

As of 2015, RSME - Real Sociedad Matemática Española - and Springer cooperate in order to publish works by authors and volume editors under the auspices of a co-branded series of publications including advanced textbooks, Lecture Notes, collections of surveys resulting from international workshops and Summer Schools, SpringerBriefs, monographs as well as contributed volumes and conference proceedings. The works in the series are written in English only, aiming to offer high level research results in the fields of pure and applied mathematics to a global readership of students, researchers, professionals, and policymakers.

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Abdallah Assi Marco D’Anna Pedro A. García-Sánchez •

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Numerical Semigroups and Applications Second Edition

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Abdallah Assi Département de Mathématiques Université d’Angers Angers, France

Marco D’Anna Dipartimento di Matematica e Informatica Università di Catania Catania, Italy

Pedro A. García-Sánchez Departamento de Álgebra Universidad de Granada Granada, Spain

ISSN 2509-8888 ISSN 2509-8896 (electronic) RSME Springer Series ISBN 978-3-030-54942-8 ISBN 978-3-030-54943-5 (eBook) https://doi.org/10.1007/978-3-030-54943-5 1st edition: © Springer International Publishing Switzerland 2016 2nd edition: © Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

To our families

Preface

Early versions of this manuscript were developed for a course on numerical semigroups and their application to the study of planar curves, which was taught at the Lebanese University. Since the first edition, we have enriched the text with more applications that relate numerical semigroups to ongoing research in a number of fields. Selected topics from the first three chapters and from the last one were the core of a course on numerical semigroups that was taught at the Scuola Superiore di Catania. Nevertheless, the text is intended to be self contained and should be accessible to beginning graduate student in mathematics. We have included numerous examples and computational experiments to insure that the reader develops a solid understanding of the fundamentals before moving forward. In each case, it should be possible to check the examples by hand, or by plotting the code into a computer. Some of the more complicated examples can be performed with the aid of the numerical semigroups package in GAP, which is software tool for mathematical computation available free online. We begin with the basic notions and terminology related to numerical semigroups. Next we focus on the study of irreducible numerical semigroups, and in particular, free numerical semigroups, which arise in the study of planar irreducible curves. Afterwards we discuss the computation of minimal presentations and how they are used to calculate nonunique factorization invariants. Factorization and division are closely related, which will become apparent in studying the Feng–Rao distance and its connection to Coding Theory. Successively, we develop some ideal theory, with particular attention to the canonical ideal and the associated duality, which plays a prominent part in understanding several properties of the semigroup itself. We also show how to determine the multiplicity sequence of a numerical semigroup, and using Arf numerical semigroups, we present the characterization of all the numerical sequences that can be realized as multiplicity sequences. Numerical semigroups naturally arose as the set of values of b which have nonnegative integer solutions to Diophantine equations of the form a1 x1 þ . . . þ an xn ¼ b, where a1 ; . . .; an ; b 2 N (here N denotes the nonnegative integers). We reduce to the case gcd ða1 ; . . .; an Þ ¼ 1. In his lectures Frobenius asked what is the largest integer b such that a given equation has no solutions over vii

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Preface

the nonnegative integers. Sylvester and others solved the n = 2 case, and since then finding the largest such b has been known as the Frobenius Problem. A thorough introduction to the Frobenius Problem and related topics is given in [65]. An active area of study where numerical semigroups continue to play a role is within commutative algebra and algebraic geometry. Let K be a field, and let A = K½ta1 ; . . .; tan  be the K-algebra of polynomials in ta1 ; . . .; tan . The ring A is the coordinate ring of the curve parametrized by ta1 ; . . .; tan , and information from A can be derived from the properties of the numerical semigroup generated by the exponents a1 ; . . .; an . As a result it is often the case that names of invariants in numerical semigroup theory are inherited from Algebraic Geometry. Similarly, Bertin and Carbonne [17], Delorme [36], Watanabe [76], and others have successfully identified properties of numerical semigroups which equate to its associated numerical semigroup ring fitting within various standard classifications in ring theory. In the monograph [15], one can find a dictionary relating much of the overlapping terminology between commutative algebra and numerical semigroup theory. Numerical semigroups are also useful in the study of singularities over planar algebraic curves. Let K be an algebraically closed field of characteristic zero, and let f(x, y) be an element of K[[x, y]]. Given another element g 2 K[[x, y]], we define the local intersection multiplicity of f with g to be the rank of the K-vector space K[[x, y]]/(f, g). When g runs over the set of elements of K[[x, y]]\(f). These numbers define a semigroup. If in addition f is irreducible, then the semigroup is a numerical semigroup. This leads to a classification of irreducible formal power series in terms of their associated numerical semigroups. This classification can be generalized to polynomials with one place at infinity. With regard to this topic, arithmetic properties of numerical semigroups played an essential role in the proof of the Abhyankar–Moh Lemma, which says that a coordinate has a unique embedding in the plane. Numerical semigroups associated to planar curves are free, and thus irreducible. This is why we spend some time explaining irreducible numerical semigroups and their two big subfamilies: symmetric and pseudo-symmetric numerical semigroups. Recently, due to the use of algebraic codes and Weierstrass numerical semigroups, some applications to coding theory and cryptography have arisen. The idea is to find properties of codes in terms of an associated numerical semigroup. See for instance [34], and the references therein. With this in mind we discuss Feng–Rao distances and their generalization to higher orders. Another focus of recent interest has been the study of factorizations in monoids. Considering the equation a1 x1 þ . . . þ an xn ¼ b, we can think of the set of nonnegative integer solutions as the set of factorizations of b in terms of a1 ; . . .; an . It can be easily shown that no numerical semigroup other than N is half-factorial, or in other words, there are always elements with factorizations of different lengths. We will discuss some of the invariants which measure how far monoids are from being half-factorial, and how wild the sets of factorizations are. Over the last decade many algorithms for computing such invariants over numerical semigroups have been developed. As a result studying these invariants over numerical semigroups

Preface

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offers a good chance to obtain families of examples, which can be used to test conjectures. Two factorizations are expressions of the same element in terms of atoms, and one can go from one factorization to another by using a minimal presentation. Hence, minimal presentations are an important tool in the study of nonunique factorization invariants. We will show how to compute a minimal presentation of a numerical semigroup both by using graphs and combinatorics, and through elimination theory. The graphs used to compute minimal presentations can be generalized to simplicial complexes. Those having nonzero Euler characteristic are important in the expression of the generating function (Hilbert series) of the semigroup as a quotient of two polynomials. The aim of this manuscript is to give some basic notions related to numerical semigroups, and from these, on the one hand, describe a classical application to the study of singularities of plane algebraic curves, and on the other, show how numerical semigroups can be used to obtain handy examples of nonunique factorization invariants. Supplementary material, examples, jupyter notebooks and tutorials are available at: https://github.com/numerical-semigroups. Angers, France Catania, Italy Granada, Spain May 2020

Abdallah Assi Marco D’Anna Pedro A. García-Sánchez

Acknowledgments

The first author is partially supported by the project GDR CNRS 2945 and a GENILSSV 2014 grant. The second author is supported by the project “Proprietà algebriche locali eliglobali di anelli associati a curve e ipersuperfici” PTR 2016-18–Dipartimento di Matematica e Informatica—Università di Catania. The third author is supported by the projects MTM2010-15595, MTM201455367-P, MTM2017-84890-P, PGC2018-096446-B-C21, FQM-343, FQM-5849, Géanpyl (FR n° 2963 du CNRS), and FEDER funds. The authors would like to thank B. Alarcón Heredia, M. Delgado, J. I. Farrán, A. M. Jiménez Macías, M. J. Leamer, D. Llena, V. Micale, and H. Martín Cruz for their comments and corrections. We would also thank the referees and the editor for their patience.

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Contents

1 Numerical Semigroups, the Basics . . . . . . . . . . . . . . . . 1.1 Notable Elements . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Numerical Semigroups with Maximal Embedding Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Special Gaps and Unitary Extensions of a Numerical Semigroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2 Irreducible Numerical Semigroups . . . . . . . . . . . . . . . . . . . . 2.1 Characterizations of Irreducible Numerical Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Decomposition of a Numerical Semigroup into Irreducible Semigroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Free Numerical Semigroups . . . . . . . . . . . . . . . . . . . . . . 3 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Basic Definitions and Relevant Invariants . . 3.2 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Irreducibility . . . . . . . . . . . . . . . . . . . . . . . 3.4 Reduction Number, Blowup and Multiplicity Semigroups . . . . . . . . . . . . . . . . . . . . . . . .

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4 Semigroup of an Irreducible Meromorphic Series . . . . . . . . . . 4.1 Some Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Characteristic Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 The Local Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4.1 Module of Kähler Differentials . . . . . . . . . . . . . . . . . 4.5 The Case of Curves with One Place at Infinity . . . . . . . . . . . 4.5.1 Module of Kähler Differentials on Polynomial Curves

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5 Minimal Presentations . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Generators and Relations . . . . . . . . . . . . . . . . . . . . 5.2 Free Numerical Semigroups . . . . . . . . . . . . . . . . . . 5.3 Graphs of Factorizations and Minimal Presentations . 5.4 Presentations and Binomial Ideals . . . . . . . . . . . . . . 5.5 Shaded Sets and Generating Functions . . . . . . . . . . .

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6 Factorizations and Divisibility . . . . . . . . . 6.1 Length Based Invariants . . . . . . . . . . 6.2 Distance Based Invariants . . . . . . . . . 6.3 How Far Is an Irreducible from Being 6.4 Divisors and Feng–Rao Distances . . .

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References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

Chapter 1

Numerical Semigroups, the Basics

In this chapter we introduce the basic notions related to numerical semigroups. Numerical semigroups have not been always been referred to as such. In the past some authors called them semimodules, or demimodules and recently many authors (mainly those concerned with factorization properties) are starting to refer to them as numerical monoids. Numerical semigroups are monoids, and thus it makes sense to use this latter notation, but historically the term numerical semigroup has been preferred, and this is the one we will use in this monograph.

1.1 Notable Elements Numerical semigroups arise naturally in the study of several seemingly unrelated problems in various fields of mathematics. Since many early developments in numerical semigroups were aimed at solving problems in already established fields of mathematics, many numerical semigroups invariants were named after the related concept in the problem treated. This is particularly truly with its relation to the study of curves. For instance [15] illustrates a poignant example of this phenomena. We say that S is a submonoid of N if the following conditions hold: (i) S ⊆ N; (ii) 0 ∈ S; (iii) if a, b ∈ S then a + b ∈ S. Clearly, {0} and N are submonoids of N. Also, if S contains a nonzero element a, then da ∈ S for all d ∈ N; hence S is an infinite set whenever it is nonzero.  Let t be an indeterminant, and K be a field. Then K[t s | s ∈ S] = s∈S Kt s is a subring of the polynomial ring K[t]. Note that the set of polynomials with exponents in S is closed under multiplication, because S is closed under addition. This ring is usually denoted K[S] and is called the semigroup ring of S over K. Let S be a submonoid of N and let G be the subgroup of Z generated by S (that is, G = {x − y | x, y ∈ S}). If 1 ∈ G, then we say that S is a numerical semigroup. © Springer Nature Switzerland AG 2020 A. Assi et al., Numerical Semigroups and Applications, RSME Springer Series 3, https://doi.org/10.1007/978-3-030-54943-5_1

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Example 1 Let S = 2N, the set of even nonnegative integers. Then clearly, S is a submonoid of N. The group spanned by S is 2Z; hence S is not a numerical semigroup. Observe that there are infinitely many nonnegative integers that do not belong to S. As seen in the next result, this is not the case with numerical semigroups. Proposition 1 Let S be a submonoid of N. Then S is a numerical semigroup if and only if N \ S is a finite set. Proof Let S be a numerical semigroup and let G be the group generated by S in Z. Since 1 ∈ G, there exists s ∈ S such that s + 1 ∈ S. We claim that n ∈ S whenever n ≥ (s − 1)(s + 1). Let n ≥ (s − 1)(s + 1), and write n = qs + r with 0 ≤ r < s. Since n = qs + r ≥ (s − 1)s + (s − 1), we have q ≥ s − 1 ≥ r . Thus n = qs + r = (q − r )s + r (s + 1) ∈ S. Conversely, assume that N \ S has finitely many elements. Then there exist s ∈ S such that s + 1 ∈ S. Hence 1 = (s + 1) − s ∈ G. We set G(S) = N \ S and we call it the set of gaps of S. We denote by g(S) the cardinality of G(S), the genus of S. The genus of a semigroup associated with a planar curve with one place at infinity coincides with the geometric genus of the curve (see Remark 10). Also for Weierstrass numerical semigroups, the genus of the semigroup is connected with the genus of the curve associated with it. Let C be a smooth algebraic projective curve over the complex numbers of genus g. Weirstrass Lückensatz states that for every P ∈ C, there are exactly g integers α1 (P), . . . , αg (P) with 1 = α1 (P) < · · · < αg (P) = 2g − 1, such that for all i there is no meromorphic function on X with a pole at P of multiplicity αi (P) as its only singularity. The set S = N \ {α1 (P), . . . , αg (P)} is a numerical semigroup, known as the Weirstrass semigroup at P (see [32] for a review of the problem). The integers α1 (P), . . . , αg (P) are the gaps of X at P. Not every numerical semigroup can be realized as a Weierstrass semigroup (the question whether this would be the case was initially proposed by Hurwitz, since then many authors studied the problem; one can get an idea of the proportion of these semigroups in [53]). Let again t be an unknown and K be a field. As in the polynomial case, K[[t s | s ∈ S]], denoted by K[[S]], is a subring of the ring of power series K[[t]]. The ring K[[S]] is a local ring with maximal ideal m = K[[t s | s ∈ S ∗ ]], where S ∗ = S \ {0}. Some authors use degree of singularity to refer to the genus of a numerical semigroup. This is because it coincides with the length of the K[[S]]-module K[[t]]/K[[S]], which is known as the degree of singularity K[[S]] (see for instance [15, 58]). The idea of focusing on numerical semigroups instead of submonoids of N is because in each isomorphy class of submonoids of N, one can always find a numerical semigroup (and only one). Proposition 2 Let S be a submonoid of N. Then S is isomorphic to a numerical semigroup. Proof Let d be gcd(S), that is, d is the generator of the group generated by S in Z. Let S1 = {s/d | s ∈ S}, which is a numerical semigroup. The map φ : S → S1 , φ(s) = s/d is a homomorphism of monoids that is clearly bijective.

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Even though any numerical semigroup has infinitely many elements, it can be described by means of finitely many of them. The rest can be obtained as linear combinations with nonnegative integer coefficients from these finitely many. Let S be a numerical semigroup and let A ⊆ S. We say that S is generated by A and we write S = A if for all s ∈ S, there exist a1 , . . . , ar ∈ A and λ1 , . . . , λr ∈ N such that a = ri=1 λi ai . Every numerical semigroup S is finitely generated, that is, S = A with A ⊆ S and A is a finite set. The smallest nonzero element of S is called the multiplicity of S, m(S) = min(S ∗ ). The multiplicity of S is precisely the multiplicity of m in K[[S]], which in this setting corresponds with leading coefficient of the Hilbert–Samuel polynomial of K[[S]] ([38, Chap. 12]). Proposition 3 Every numerical semigroup is finitely generated. Proof Let A be a system of generators of S (S itself is a system of generators). Let m be the multiplicity of S. Clearly m ∈ A. Assume that a < a are two elements in A such that a ≡ a mod m. Then a = km + a for some positive integer k. So we can remove a from A and we still have a generating system for S. Observe that at the end of this process we have at most one element in A in each congruence class modulo m, and we conclude that we can choose A to have finitely many elements. The underlying idea in the last proof motivates the following definition. Let n ∈ S ∗ . We define the Apéry set of S with respect to n, denoted Ap(S, n), to be the set Ap(S, n) = {s ∈ S | s − n ∈ / S}. This is why some authors call {n} ∪ (Ap(S, n) \ {0}) a standard basis of S, when n is chosen to be the multiplicity of S. Apéry sets were introduced by Apéry in [7], where he proved some of its basic properties, revealing the relevance of these sets later (and we would even say ubiquity when studying numerical semigroups). The reader will see examples of the importance of this concept throughout this manuscript; we will see that they can be used for the calculation of many properties and invariants of numerical semigroups. As we see next, Ap(S, n) has precisely n elements, as expected from the proof of Proposition 3. Lemma 1 Let S be a numerical semigroup and let n ∈ S ∗ . For all i ∈ {1, . . . , n}, let w(i) be the smallest element of S such that w(i) ≡ i mod n. Then Ap(S, n) = {0, w(1), . . . , w(n − 1)}. Proof Let i ∈ {0, . . . , n − 1}. By definition, w(i) ∈ S and clearly w(i) − n ≡ i mod n, whence w(i) − n ∈ / S. In particular, w(i) ∈ Ap(S, n). This proves one inclusion. Observe that there are no elements a, b ∈ Ap(S, n) such that a ≡ b mod n. Hence we get an equality, because we are ranging all possible congruence classes modulo n.

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Example 2 Let S be the numerical semigroup generated by {4, 7, 10}. In order to compute the Apéry set of 4 in S, we need to now the least element in S of every congruence class modulo 4. Clearly 0 is the least element in S congruent with 0 modulo 4. As 1, 5, 9, and 13 are not in S, the least element in S congruent with 1 modulo 4 is 17. The classes of 2 and 3 are easier, since here 10 and 7 are minimal generators congruent with 2 and 3 modulo 4, respectively. Hence Ap(S, 4) = {0, 7, 10, 17}. Next we give an example using the numericalsgps GAP package ([35, 44], respectively). We will do this several times along the manuscript, since it is also our intention to show how calculations can easily be accomplished with the help of this package. GAP Example 1 Let us start defining a numerical semigroup. gap> s:=NumericalSemigroup(5,9,21);; gap> SmallElements(s); [ 0, 5, 9, 10, 14, 15, 18, 19, 20, 21, 23 ]

This means that our semigroup is {0, 5, 9, 10, 14, 15, 18, 19, 20, 21, 23, →}, where the arrow means that every integer larger than 23 is in the set. If we take a nonzero element n in the semigroup, its Apéry set has exactly n elements. gap> AperyList(s,5); [ 0, 21, 27, 18, 9 ]

We can define the Apéry set for other integers as well, but the above feature no longer holds. gap> AperyList(s,6); [ 0, 5, 9, 10, 14, 18, 19, 23, 28 ]

Apéry sets are one of the most important tools when dealing with numerical semigroups. Next we see that they can be used to represent elements in a numerical semigroup in a unique way. Proposition 4 Let S be a numerical semigroup and let n ∈ S ∗ . For all s ∈ S, there exists a unique (k, w) ∈ N × Ap(S, n) such that s = kn + w. Proof Let s ∈ S. If s ∈ Ap(S, n), then we set k = 0, w = s. If s ∈ / Ap(S, n), then s1 = s − n ∈ S. We restart with s1 . Clearly there exists k such that sk = s − kn ∈ Ap(S, n). Let s = k1 n + w1 with k1 ∈ N, w1 ∈ Ap(S, n). Suppose that k1 = k. Hence 0 = (k1 − k)n = w − w1 . In particular w ≡ w1 mod n. This is a contradiction. Remark 1 Proposition 4 is also true for any s ∈ Z. Indeed, it is not hard to prove that if s = kn + w, then s ∈ S if and only if k ≥ 0.

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This gives an alternative proof that S is finitely generated. Corollary 1 Let S be a numerical semigroup. Then S is finitely generated. Proof Let n ∈ S ∗ . By the proposition above, S = {n} ∪ Ap(S, n) \ {0}} . The cardinality of Ap(S, n) = n. This proves the result. Example 3 Let S = 3, 5 . Then S = {0, 3, 5, 6, 8, →}. The smallest element in S congruent with 0 modulo 3 is 0; congruent with 1 modulo 3 we have 10; and element in S with 2 modulo 3 is 5. This means that every s ∈ S is either in (3N) or (10 + 3N) or (5 + 3N). Let S be a numerical semigroup and let A ⊆ S. We say that A is a minimal set of generators of S if S = A and no proper subset of A has this property. Corollary 2 Let S be a numerical semigroup. Then S has a minimal set of generators. This set is finite and unique: it is actually S ∗ \ (S ∗ + S ∗ ). Proof Notice that by using the argument in the proof of Proposition 3, every generating set can be refined to a minimal generating set. Let A = S ∗ \ (S ∗ + S ∗ ) and let B be another minimal generating set. If B is not included in A, there exists a ∈ B and b, c ∈ S ∗ such that a = b + c. But this contradicts the minimality of B, since in this setting B \ {a} is a generating system for S. This proves B ⊆ A.  Now take a ∈ A ⊆ S = B . Then a = b∈B λb b. But a ∈ S ∗ \ (S ∗ + S ∗ ), and  so b∈B λb = 1. This means that there exists b ∈ B with λb = 1 and λb = 0 for the rest of b ∈ B. We conclude that a = b ∈ B, which proves the other inclusion. Let S be a numerical semigroup. The cardinality of a minimal set of generators of S is called the embedding dimension of S. We denote it by e(S). Clearly, the embedding dimension of S is the cardinality of the set of minimal generators of the maximal ideal of K[[S]], and thus it is the embedding dimension of the local ring (K[[S]], m). Lemma 2 Let S be a numerical semigroup. We have e(S) ≤ m(S). Proof The proof easily follows from the proof of Corollary 1 or from that of Proposition 3. Example 4 Next we give some examples, including extremal ones. (i) S = N if and only if e(S) = 1. (ii) Let m ∈ N∗ and let S = m, m + 1, . . . , 2m − 1 . We have Ap(S, m) = {0, m + 1, . . . , 2m − 1} and {m, m + 1, . . . , 2m − 1} is a minimal set of generators of S. In particular e(S) = m(S) = m. (iii) Let S = {0, 4, 6, 8, 9, 10, . . . }. We have Ap(S, 4) = {0, 9, 6, 15}. In particular m(S) = 4 and S = 4, 6, 9, 15 = 4, 6, 9 . Hence e(S) = 3.

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Let S be a numerical semigroup. We set F(S) = max(Z \ S) and we call it the Frobenius number of S. This invariant is called after Frobenius because, as we mentioned in the Introduction, he proposed in one of his lectures the problem of finding the largest integer f such that for fixed coprime positive integers a and b, the Diophantine linear equation ax + by = f has no solution over N. Frobenius was asking precisely for a formula for the maximum of Z \ a, b . Sylvester (and others) solved this problem at the end of the nineteenth century, and many mathematicians started to work in the generalization of this question (see [65] for a detailed overview). We set C(S) = F(S) + 1 and we call it the conductor of S. Again this name is inherited from another invariant. The conductor of K[[S]] in K[[t]], (K[[S]] :   classical K[[t]]), is precisely t C(S) , see [15]. Recall that we denoted G(S) = N \ S and we called it the set of gaps of S. Also we used g(S) to denote the cardinality of G(S) and we call g(S) the genus of S. We denote by n(S) the cardinality of {s ∈ S | s ≤ F(S)}. Selmer in [73] proved that the machinery developed by Apéry can be used to calculate the genus and Frobenius number of a numerical semigroup. Proposition 5 Selmer’s formulas Let S be a numerical semigroup and let n ∈ S ∗ . (i) F(S) = max(Ap(S, n)) − n.  (ii) g(S) = n1 w∈Ap(S,n) w −

n−1 . 2

Proof (i) Clearly max(Ap(S, n)) − n ∈ / S. If x > max(Ap(S, n)) − n, then x + n > max(Ap(S, n)). Write x + n = qn + i, q ∈ N, i ∈ {0, . . . , n − 1} and let w(i) ∈ Ap(S, n) be the smallest element of S that is congruent to i modulo n. Since x + n > w(i), we have x + n = kn + w(i) with k > 0. Hence x = (k − 1)n + w(i) ∈ S. (ii) For all w ∈ Ap(S, n), write w = ki n + i, ki ∈ N, i ∈ {0, . . . , n − 1}. We have: Ap(S, n) = {0, k1 n + 1, . . . , kn−1 n + n − 1}. Let x ∈ N and suppose that x ≡ i mod n. We claim that x ∈ S if and only if w(i) ≤ x. In fact, if x = qi n + i, then x − w(i) = (qi − ki )n. Hence w(i) ≤ x if and only if ki ≤ qi if and only if x = (qi − ki )n + w(i) ∈ S. It follows that x∈ / S if and only if x = qi n + i, qi < ki . Consequently

g(S) =

n−1  i=1

⎞ ⎛ ⎛ n−1 1 1 ⎝ n − 1 = ⎝ ki = (ki n + i)⎠ − n 2 n i=1



w∈Ap(S,n)

Example 5 Let S = a, b be a numerical semigroup. We have Ap(S, a) = {0, b, 2b, . . . , (a − 1)b}. Hence

⎞ w⎠ −

n−1 . 2

1.1 Notable Elements

(i) F(S) = (a − 1)b − a = ab − a − b. (ii) g(S) = a1 (b + 2b + · · · + (a − 1)b) −

7

a−1 2

=

(a−1)(b−1) 2

=

F(S)+1 . 2

Example 6 We revisit S = 3, 5 . We know that S = {0, 3, 5, 6, 8, →} and Ap(S, 3) = {0, 5, 10}. Then g(S) = 4 = 13 (5 + 10) − 22 and F(S) = 7 = 10 − 3. Lemma 3 Let S be a numerical semigroup. We have g(S) ≥

F(S)+1 . 2

Proof Let s ∈ N. If s ∈ S, then F(S) − s ∈ / S. Thus g(S) is greater than or equal to n(S) (the cardinality of N(S) = {s ∈ S | s < F(S)}). But n(S) + g(S) = F(S) + 1. This proves the result. GAP Example 2 Let S = 5, 7, 9 . gap> s:=NumericalSemigroup(5,7,9);

gap> FrobeniusNumber(s); 13 gap> Conductor(s); 14 gap> ap:=AperyList(s,5); [ 0, 16, 7, 18, 9 ] gap> Maximum(ap)-5; 13 gap> Sum(ap)/5 -2; 8 gap> Genus(s); 8 gap> Gaps(s); [ 1, 2, 3, 4, 6, 8, 11, 13 ]

There have been many attempts to see how many numerical semigroups there are with given genus. As of today, it is known that asymptotically this number increases like the Fibonacci sequence when we increase the genus. However for an arbitrary genus g, we still do not know if there are more numerical semigroups with genus g + 1 than numerical semigroups with genus g. Conjecture 1 Let g be positive integer and let n g be the number of numerical semigroups S with g(S) = g. Is n g ≤ n g+1 ? This conjecture is known to be true for g ≤ 67 but it is still open in general (J. Fromentin, personal communication; see also Manuel Delgado’s web page). GAP Example 3 We enumerate numerical semigroups with genus i for i ∈ {1, . . . , 20}. gap> List([1..20],i->Length(NumericalSemigroupsWithGenus(i))); [ 1, 2, 4, 7, 12, 23, 39, 67, 118, 204, 343, 592, 1001, 1693, 2857, 4806, 8045, 13467, 22464, 37396 ]

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The following result allows to prove Johnson’s formulas (see Corollary 3 below). These formulas were introduced in an attempt to simplify the calculation of the Frobenius number by reducing the problem to a numerical semigroup with smaller embedding dimension. Unfortunately these formulas can only be applied in a quite particular case. Proposition 6 Let S be a numerical semigroup minimally generated by n 1 , . . . , n p . Let d = gcd(n 1 , . . . , n p−1 ) and let T = n 1 /d, . . . , n p−1 /d, n p . We have Ap(S, n p ) = dAp(T, n p ). Proof Let w n p ). Since w − n p ∈ / S, we deduce that w ∈ n 1 , . . . , n p−1 .

∈ Ap(S, n . If wd − n p ∈ T , then w − dn p ∈ S, which is a contradicThus, wd ∈ nd1 , . . . , p−1 d tion. Hence wd ∈ Ap(T, n p ), in particular w ∈ dAp(T, n p ). 

n Conversely, if w ∈ Ap(T, n p ), then w ∈ nd1 , . . . , p−1 . Consequently, d  p dw ∈ n 1 , . . . , n p−1 ⊆ S. Suppose that dw − n p ∈ S. We have then dw − n p = i=1 λi n i  p−1 for some nonnegative integers λ1 , . . . , λ p . This implies that dw = i=1 λin i +  p−1 λ +1 n p, (λ p + 1)n p . In particular, d divides λ p + 1. Write w = i=1 λi ndi + pd whence w − n p ∈ T , which is a contradiction. This proves that dw ∈ Ap(S, n p ), and we have both inclusions. Corollary 3 Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p }. n , n p . Let d = gcd(n 1 , . . . , n p−1 ) and let T =  nd1 , . . . , p−1 d (i) F(S) = dF(T ) + (d − 1)n p . (d−1)(n p −1) (ii) g(S) = dg(T ) + . 2 Proof (i) F(S)= max Ap(S, n p ) − n p =d max Ap(T, n p ) − n p =d(max Ap(T, n p) − n p)  + (d − 1)n p =dF(T ) + (d − 1)n   p.  n p −1 n p −1 1  d (ii) g(S) = n p w∈Ap(S,n p ) w − 2 = n p w∈Ap(T,n p ) w − 2    n −1 (d−1)(n p −1) + = d n1p w∈Ap(T,n p ) w − p2 . 2 Example 7 Let S = 20, 30, 17 , T = 2, 3, 17 = 2, 3 ; F(S) = 10F(T ) + 9 × 17 = 163 and g(S) = 10 + (9 × 16)/2 = 82. Let S be a numerical semigroup. We say that x ∈ Z is a pseudo-Frobenius number if x ∈ / S and x + s ∈ S for all s ∈ S ∗ . We denote by PF(S) the set of pseudo-Frobenius numbers. The cardinality of PF(S) is denoted by t(S) and we call it the type of S. Observe that F(S) = max(PF(S)). The ring K[[S]] is Cohen–Macaulay, and t(S) is precisely the Cohen–Macaulay type of K[[S]] (see for instance [58]). We show next several ways to characterize the set of pseudo-Frobenius numbers of a numerical semigroup. Let a, b ∈ Z. We define ≤ S as follows: a ≤ S b if b − a ∈ S. Clearly ≤ S is a (partial) order relation. With this order relation Z becomes a poset. The following result states that PF(S) are precisely the maximal gaps of S with respect to ≤ S .

1.1 Notable Elements

9

Proposition 7 Let S be a numerical semigroup. We have PF(S) = Maximals≤S (Z \ S). Proof Let x ∈ PF(S): x ∈ / S and x + S ∗ ⊆ S. Let y ∈ N \ S and assume that x ≤ S y. If x = y, then y − x = s ∈ S ∗ , hence y = x + s ∈ x + S ∗ ⊆ S. This is a contradiction. / S for some s ∈ S ∗ , then x ≤ S Conversely, let x ∈ Maximals≤S (Z \ S). If x + s ∈ x + s; a contradiction. We can also recover the pseudo-Frobenius elements tin terms of the Apéry sets. Proposition 8 Let S be a numerical semigroup and let n ∈ S ∗ . Then   PF(S) = w − n | w ∈ Maximals≤S Ap(S, n) . Proof Let x ∈ PF(S). By definition, x + n ∈ S and x ∈ / S. Hence x + n ∈ Ap(S, n). Let us prove that x + n is maximal with respect to ≤ S . Let w ∈ Ap(S, n) such that x + n ≤ S w. Then there exists s ∈ S such that w − x − n = s. We have w − n = / S, a contradiction. x + s. If s ∈ S ∗ , then x + s ∈ S. But w − n ∈ For the other inclusion, let w ∈ Maximals≤S Ap(S, n) and let s ∈ S ∗ . If w − n + s∈ / S, then w + s ∈ Ap(S, n). This contradicts the maximality of w. Example 8 (i) Let S = 5, 6, 8 ; Ap(S, 5) = {0, 6, 12, 8, 14}. Hence PF(S) = {12 − 5, 14 − 5} = {7, 9}. In particular, t(S) = 2. Figure 1.1 depicts the Apéry set of 5 in S as a Hasse diagram with respect to the order ≤ S . (ii) Let S = a, b where a, b ∈ N \ {0, 1} and gcd(a, b) = 1. We have Ap(S, a) = {0, b, 2b, . . . , (a − 1)b}. Thus PF(S) = {F(S) = (a − 1)b − a} and t(S) = 1. As a consequence of Lemma 1 and Proposition 8, we obtain an upper bound for the type of a numerical semigroup. Corollary 4 Let S be a numerical semigroup other than N. We have t(S) ≤ m(S) − 1.

Fig. 1.1 Ap(5, 6, 8 , 5) as a Hasse diagram

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Proof The type S is nothing but the cardinality of the set of maximal elements of Ap(S, m(S)) with respect to ≤ S . Since 0 is not a maximal element, the result follows. Remark 2 Let S be a numerical semigroup. In the above inequality, one cannot replace m(S) − 1 by e(S) as it is shown in the following example. Take for instance S = 43, 97, 128, 165 . Then t(S) = 5 > 4 (this example was found using the package numericalsgps). Moreover, in [41] there is a family of numerical semigroups with embedding dimension four and type arbitrarily large. The type, the number of sporadic elements (elements below the Frobenius number) and the genus of a numerical semigroup are related in the following way. Proposition 9 Let S be a numerical semigroup and recall that n(S) is the cardinality of N(S) = {s ∈ S | s < F(S)}. With these notations we have g(S) ≤ t(S)n(S). Proof Let x ∈ S \ N and let f x = min{ f ∈ PF(S) | f − x ∈ S}. Let φ : G(S) → PF(S) × N(S), φ(x) = ( f x , f x − x). The map φ is clearly injective. In particular g(S) is less than or equal than the cardinality of PF(S) × N(S), which is t(S)n(S). If we use the fact that g(S) + n(S) = F(S) + 1 = C(S), then we obtain the following consequence. Corollary 5 Fröberg–Gottlieg–Haggvist Let S be a numerical semigroup. We have C(S) ≤ (t(S) + 1)n(S). GAP Example 4 We go back to S = 5, 7, 9 . gap> PseudoFrobenius(s); [ 11, 13 ] gap> Type(s); 2 gap> Multiplicity(s); 5 gap> SmallElements(s); [ 0, 5, 7, 9, 10, 12, 14 ] gap> Length(last-1); 7

Conjecture 2 Wilf C(S) ≤ e(S)n(S).

1.1 Notable Elements

11

Recently many authors have done some progress in this conjecture. Dobbs and Matthews gathered some folklore and proved new cases in [37]: for a numerical semigroup S if either • • • • • • •

e(S) ≤ 3 (this was also proved in [41]) or e(S) = m(S) (these semigroups are studied in the next section) or F(S) − 1 + g(S) ≤ 4 or 4g(S) ≤ 3C(S) or n(S) ≤ 4 or 4n(S) ≥ C(S) or F(S) ≤ 20 (one can nowadays check this by computer even for higher Frobenius number) or • S is symmetric or pseudo-symmetric (semigroups studied in the next chapter).

Kaplan in [52] was able to show that the conjecture holds true for numerical semigroups such that 2g(S) < 3m(S), and also for those fulfilling F(S) < 2m(S). Eliahou relaxed in [39] the latter condition to C(S) ≤ 3m(S) (he also announced that some other cases will appear in a separate paper). The average of semigroups fulfilling this inequality tends to one when the genus goes to infinity (this interesting fact was shown by Zhai in [79]). An exhaustive search up to genus 60 done by Fromentin and Hivert in [42] shows that for numerical semigroups with genus less than 60, the conjecture holds. Sammartano in [72] showed that the Wilf’s conjecture is also true for numerical semigroups generated by almost arithmetic sequences and also for numerical semigroups S such that m(S) ≤ 2e(S) (and as a corollary for all semigroups with multiplicity less than or equal to 8). Moscariello and Sammartano proved that for fixed m(S)/e(S), the result is also true for all values of m(S) large enough and not divisible by a finite set of primes [62]. There is another way to relate the type, the Frobenius number and genus of a numerical semigroup by using Nari’s correspondence [63]. Let S be a numerical semigroup. For every n in N(S) ∪ PF(S) \ {F(S)}, the integer F(S) − n is a gap of S. Thus n(S) + t(S) − 1 ≤ g(S). As n(S) = F(S) + 1 − g(S), we deduce that F(S + t(S) ≤ 2 g(S), whence we obtain the following result. Proposition 10 Let S be a numerical semigroup. Then g(S) ≥

F(S) + t(S) . 2

Numerical semigroups attaining the equality in Proposition 10 are known as almost symmetric numerical semigroups . Example 9 Let S = 6, 7, 9, 10 = {0, 6, 7, 9, 10, 12, →}. Then t(S) = 3, F(S) = 11, and g(S) = 7 = (11 + 3)/2.

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1.2 Numerical Semigroups with Maximal Embedding Dimension Let S be a numerical semigroup and recall that e(S) ≤ m(S). In the following we shall consider the case where e(S) = m(S). We are going to see that if this is the case, then the type is also maximal. Let S be a numerical semigroup. We say that S has maximal embedding dimension if e(S) = m(S). Maximal embedding dimension numerical semigroups have been widely treated in the literature. This is due to several reasons, their minimal presentations have special form and are relatively easy to compute, and also because some families of numerical semigroups related to the resolution of singularities have maximal embedding dimension. This is the case of saturated and Arf numerical semigroups. Example 10 Let S = 3, 5 . We already know that Ap(S, 3) = {0, 5, 10}. The semigroup S is not of maximal embedding dimension, but S = 3, 8, 13 = 3 + Ap(S, 3) is. This construction always yields maximal embedding dimension numerical semigroups ([69, Corollary 3.7]). Trivially, any minimal generator is in the Apéry set of any nonzero element different from it. We write the short proof for this. / Lemma 4 Let x be a minimal generator of S and let n ∈ S ∗ , n = x. We have x − n ∈ S. In particular x ∈ Ap(S, n). Proof If x − n ∈ S, since x = n + (x − n), this contradicts the the fact that x is a minimal generator. As a consequence, we get that the Apéry set of the multiplicity in a maximal embedding dimension numerical semigroup consists of 0 plus the rest of minimal generators. Proposition 11 Let {n 1 , . . . , n e } be a minimal set of generators of S with n 1 < n 2 < · · · < n e . Then S has maximal embedding dimension if and only if Ap(S, n 1 ) = {0, n 2 , . . . , n e }. Proof Assume that S has maximal embedding dimension. By Lemma 4, n 2 , . . . , n e ∈ Ap(S, n 1 ). But n 1 = m(S) = e, whence Ap(S, n 1 ) = {0, n 2 , . . . , n e }. Conversely, by Lemma 1, the cardinality of Ap(S, n 1 ) is n 1 . Thus e = e(S) = n 1 = m(S). As we already mentioned above, the type is also maximal in this kind of numerical semigroup. Actually this also characterizes maximal embedding dimension. Proposition 12 Let {n 1 , n 2 , . . . , n e } be a minimal set of generators of S with n 1 < n 2 < · · · < n e . The following are equivalent: (i) S has maximal embedding dimension,

1.2 Numerical Semigroups with Maximal Embedding Dimension

13

e (ii) g(S) = n11 i=2 n i − n 12−1 , (iii) t(S) = n 1 − 1 = m(S) − 1. . . , n e }. Proof If S has maximal embedding dimension, then Ap(S, n 1 ) = {0, n 2 , . e By Selmer’s formulas (Proposition 5), g(S)= n11 w∈Ap(S,n 1 ) w − n 12−1 = n11 i=2 ni  n 1 −1 1 − 2 . Conversely, we have {n 2 , . . . , n e } ⊆ Ap(S, n 1 ) and also n 1 w∈Ap(S,n 1 ) w = 1 e i=2 n i . Hence Ap(S, n 1 ) = {0, n 2 , . . . , n e }. In particular, S has maximal embedn1 ding dimension. This proves that (i) and (ii) are equivalent. Finally we prove that (i) is equivalent to (iii). If S has maximal embedding dimension, then Ap(S, n 1 ) = {0, n 2 , . . . , n e }. It easily follows that Maximals≤S Ap(S, n 1 ) = {n 2 , . . . , n e }, whence t(S) = n 1 − 1 = m(S) − 1. Now assume that t(S) = n 1 − 1. Then the cardinality of PF(S) is n 1 − 1 = m(S) − 1. According to Proposition 8, this means that all the elements in Ap(S, n 1 ) with the exception of 0 are maximal with respect to ≤ S . We also know that {n 2 , . . . , n e } ⊆ Ap(S, n 1 ) (Lemma 4). Hence all minimal generators (other than n 1 ) are maximal in Ap(S, n 1 ) with  respect to ≤ S . Assume that there exists x ∈ Ap(S, n 1 ) \ {0, n 2 , . . . , n e }. Then e λi n i , λi ≥ 0, and since x − n 1 ∈ / S, we deduce that λ1 = 0. Since x = 0, x = i=1 λk = 0 for some k. Thus x − n k ∈ S, and consequently n k is not maximal with respect to ≤ S . This is a contradiction. Hence Ap(S, n 1 ) = {0, n 2 , . . . , n e }, and this yields n 1 = m(S) = e(S). Observe that, as a corollary of the last result, for any maximal embedding dimension numerical semigroup t(S) + 1 = m(S) = e(S). Hence by applying Corollary 5, we have that Wilf’s conjecture holds for maximal embedding dimension numerical semigroups, as announced in the last section. As another consequence of Selmer’s formulas, we get an easy expression of the Frobenius number of a maximal embedding dimension numerical semigroup. Proposition 13 Let {n 1 , n 2 , . . . , n e } be a minimal set of generators of S with n 1 < n 2 < · · · < n e and e = n 1 . Then F(S) = n e − n 1 . Proof This follows from the fact that F(S) = max Ap(S, n 1 ) − n 1 (Proposition 5). Example 11 The converse to this proposition is not true. Just take S = 4, 5, 11 . GAP Example 5 As we already mentioned in Example 10, one can always construct maximal embedding dimension numerical semigroups from any numerical semigroup in the following way (see [69, Chap. 2]). gap> s:=NumericalSemigroup(4,7,9);

gap> AperyList(s); [ 0, 9, 14, 7 ] gap> t:=NumericalSemigroup(4+last);

gap> MinimalGenerators(t); [ 4, 11, 13, 18 ]

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Remark 3 Semple proposed a classification of the singularities of a curve in the three dimensional space based on the analysis of the successive blow-ups of the curve. Du Val extended the argument for arbitrary dimensions and wondered if his geometric procedure could be developed in an algebraic way. Arf, who was listening to his lectures gave an answer a week later, and his solution makes use of numerical semigroups [8]. He starts with the local ring of the curve localized at the given point. And then computes what was called later by Lipman the Arf closure of this ring [57]. Finally he determines the semigroup of values of the ring obtained in the preceding step. From this semigroup one can recover the multiplicity sequence of the successive blow-ups, and thus one can do the classification of the singularity by means of this sequence (or its successive sums). See [74] for a description of the process and several historical interesting remarks. Let us give the definition of an Arf numerical semigroup, which are the value semigroups appearing in the above paragraph. A numerical semigroup  has the Arf property if for every x, y, z ∈ , with x ≥ y ≥ z, then x + y − z ∈ . In particular, if we take z = m(), the multiplicity of , then for every x, y ∈  ∗ , x + y − m() ∈ . So for every x and y minimal generators of , x + y is no longer in the Apéry set of m(), and by Proposition 11, we have that  has maximal embedding dimension. Hence numerical semigroups appearing in Arf construction are numerical semigroups with maximal embedding dimension. See [15, Theorem I.3.4] for other fifteen characterizations of Arf numerical semigroups (some of them use the concept of blow-up of a numerical semigroup or Lipman semigroup).

1.3 Special Gaps and Unitary Extensions of a Numerical Semigroup We introduce in this section another set of notable elements of numerical semigroups, that is, in some sense dual to the concept of minimal generating system. Let S be a numerical semigroup. Notice that an element s ∈ S is a minimal generator if and only if S\{s} is a numerical semigroup. We define the set of special gaps of S as SG(S) = {x ∈ PF(S) | 2x ∈ S}. The duality we mentioned above comes in terms of the following result. Lemma 5 Let x ∈ Z \ S. Then x ∈ SG(S) if and only if S ∪ {x} is a numerical semigroup. Proof Easy exercise. Example 12 Let S = 3, 7, 8 . Then the set of gaps of S is {1, 2, 4, 5}. We see that 1 ≤ S 4 since 4 − 1 = 3 ∈ S; and for the same reason 2 ≤ S 5. Hence, PF(S) ⊆ {4, 5}. As 4 and 5 are incomparable with respect to ≤ S , we get an equality. Notice that 2 × 4, 2 × 5 ∈ S. Consequently SG(S) = {4, 5}.

1.3 Special Gaps and Unitary Extensions of a Numerical Semigroup

15

Clearly, S ∪ {5} is again a numerical semigroup. Actually this always holds for any numerical semigroup by taking as gap its Frobenius number. If S and T are numerical semigroups, with S ⊂ T , we can construct a chain of numerical semigroups S = S1 ⊂ S2 ⊂ · · · ⊂ Sk = T such that for every i, Si+1 is obtained from Si by adjoining a special gap. This can be done thanks to the following result. Lemma 6 Let T be a numerical semigroup and assume that S ⊂ T . Then max(T \ S) ∈ SG(S). In particular, S ∪ {max(T \ S)} is a numerical semigroup. Proof Let x = max(T \ S). Clearly 2x ∈ S. Take s ∈ S ∗ . Then x + s ∈ T and x < x + s. Hence x + s ∈ S. Remark 4 Let O(S) be the set of oversemigroups of S, that is, the set of numerical semigroups T such that S ⊆ T . Since N \ S is a finite set, we deduce that O(S) is a finite set. Example 13 The graph in Fig. 1.2 represents the set of oversemigroups of 3, 7, 11 . We have drawn a node for each of the oversemigroups. If an edge connects two semigroups, it is labeled with the special gap added to the semigroup below to obtain the other semigroup containing it (or if we look downwards, the minimal generator we remove to obtain the semigroup below). GAP Example 6 We show how to compute oversemigroups using GAP.

Fig. 1.2 The Hasse diagram of oversemigroups of 3, 7, 11

16

1 Numerical Semigroups, the Basics gap> s:=NumericalSemigroup(7,9,11,17);; gap> Genus(s); 12 gap> o:=OverSemigroups(s);; gap> Length(o) 51 gap> s:=NumericalSemigroup(3,5,7);; gap> o:=OverSemigroups(s);; gap> List(last,MinimalGenerators); [ [ 1 ], [ 2, 3 ], [ 3 .. 5 ], [ 3, 5, 7 ] ]

Chapter 2

Irreducible Numerical Semigroups

A numerical semigroup S is irreducible if it cannot be expressed as the intersection of two proper oversemigroups. The motivation of the study of these semigroups was initially to express any numerical semigroup as a finite intersection of irreducible numerical semigroups, and then derive properties of the original semigroup in terms of the irreducibles that appear in this decomposition. Historically this was not the reason to study these semigroups. It turns out that irreducible numerical semigroups are either symmetric (when their Frobenius number is odd) or pseudo-symmetric (even Frobenius number); and every symmetric or pseudo-symmetric numerical semigroup is irreducible. Kunz in [55] proved that K [[S]] is a Gorenstein ring if and only if S is symmetric. Consequently obtaining examples of symmetric numerical semigroups would yield examples of one dimensional Gorenstein rings. This motivated a series of tools and machinery to produce families of symmetric numerical semigroups. The name symmetric comes from the symmetry in the set of nonnegative integers less than the Frobenius number of the semigroup: there are as many gaps as elements in this interval. The closest we can get to this symmetry when the Frobenius number is even, taking into account that its half is forced to be a gap, is precisely when the semigroup is pseudo-symmetric. Semigroups appearing in Chap. 4 are symmetric, but this is not the only reason to study them. Symmetric numerical semigroups have attracted the attention of many algebraists due to their connections to curves and their coordinate rings. This resulted in the development of a new theory and machinery for calculating examples and properties of algebraic curves.

2.1 Characterizations of Irreducible Numerical Semigroups Fröberg, Gottlieb and Häggkvist proved in [41] that maximal (with respect to set inclusion) numerical semigroups in the set of numerical semigroups with fixed Frobenius number correspond to symmetric (if this Frobenius number is odd) or © Springer Nature Switzerland AG 2020 A. Assi et al., Numerical Semigroups and Applications, RSME Springer Series 3, https://doi.org/10.1007/978-3-030-54943-5_2

17

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2 Irreducible Numerical Semigroups

pseudo-symmetric (Frobenius number even) numerical semigroups. We are going to see this from the unified point of view of irreducible numerical semigroups, which gather both families. The following lemma is just a particular case of Lemma 6, taking T = N; this fact was already pointed out in Example 12. Lemma 7 Let S be a numerical semigroup other than N. Then S ∪ {F(S)} is a numerical semigroup. Example 14 The above construction allows to construct a path connecting the semigroup S and N in the graph of all oversemigroups of S. In Example 13, this path is 3, 7, 11, 3, 7, 8, 3, 5, 7, 3, 4, 5, 2, 3, N. Theorem 1 Let S be a numerical semigroup. The following are equivalent. (i) S is irreducible. (ii) S is maximal (with respect to set inclusion) in the set of numerical semigroups T such that F(S) = F(T ). (iii) S is maximal (with respect to set inclusion) in the set of numerical semigroups T such that F(S) ∈ / T. Proof (i) implies (ii) Let T be a numerical semigroup such that F(S) = F(T ). If S ⊆ T , then S = T ∩ (S ∪ {F(S)}). Since S = S ∪ {F(S)}, we deduce S = T . (ii) implies (iii) Let T be a numerical semigroup such that F(S) ∈ / T and assume that S ⊆ T . The set T1 = T ∪ {F(S) + 1, F(S) + 2, . . . } is a numerical semigroup with F(T1 ) = F(S). But S ⊆ T1 , whence S = T1 . Since F(S) + k ∈ S for all k ≥ 1, it follows that S = T . (iii) implies (i) Let S1 , S2 be two numerical semigroups such that S ⊆ S1 , S ⊆ S2 and S = S1 ∩ S2 . Since F(S) ∈ / S, F(S) ∈ / Si for some i ∈ {1, 2}. By (iii), Si = S. Example 15 In Fig. 2.1 we have drawn the Hasse diagram (ordered with respect to set inclusion) of the set of numerical semigroups with Frobenius number seven. We see that we have three maximal elements: 3, 5, 2, 9, and 4, 5, 6. Thus these are the only irreducible numerical semigroups with Frobenius number seven. Let S be a numerical semigroup. We say that S is symmetric if (i) S is irreducible and (ii) F(S) is odd. We say that S is pseudo-symmetric if (i) S is irreducible and (ii) F(S) is even. Next we collect some classical characterizations of symmetric and pseudosymmetric numerical semigroups. We first prove the following. Sometimes the set H appearing in the next proposition is known as the set of holes of the semigroup.

2.1 Characterizations of Irreducible Numerical Semigroups

19

Fig. 2.1 Numerical semigroups with Frobenius number seven

Proposition 14 Let S be a numerical semigroup and suppose that    / S, x = H = x ∈ Z \ S  F(S) − x ∈

F(S) 2



is not empty. If h = max H , then S ∪ {h} is a numerical semigroup. Proof Since S ⊆ S ∪ {h}, the set N \ (S ∪ {h}) has finitely many elements. Let a, b ∈ S ∪ {h}. • If a, b ∈ S, then a + b ∈ S. • Let a ∈ S and b = h. If a = 0, then a + h = h ∈ S ∪ {h}. So assume that a = 0 and a + h ∈ / S. By the maximality of h, we deduce F(S) − a − h = F(S) − (a + h) ∈ S. Hence F(s) − h = a + F(S) − a − h ∈ S. This contradicts the definition of h. • Finally assume that a = b = h. If 2h ∈ / S, then the maximality of h implies that F(S) − 2h = s ∈ S ∗ . This implies that F(S) − h = h + s, which by the preceding paragraph is in S, contradicting the definition of h. GAP Example 7 In light of Proposition 14 and Lemma 5, if for a numerical semigroup, there exists a maximum of {x ∈ Z \ (S ∪ {F(S)/2}) | F(S) − x ∈ / S}, then it is a special gap. gap> s:=NumericalSemigroup(7,9,11,17);

gap> g:=Gap(s); [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 13, 15, 19 ] gap> Filtered(g, x-> (x19/2) and not(19-x in s)); [ 4, 6, 13, 15 ] gap> SpecialGaps(s); [ 13, 15, 19 ]

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2 Irreducible Numerical Semigroups

We have introduced the concepts of symmetric and pseudo-symmetric numerical semigroups as subclasses of the set of irreducible numerical semigroups. However as we said at the beginning of this chapter, these two concepts existed before that of irreducible numerical semigroup, and thus the definitions were different than the ones we have given above. Needless to say that as in the case of irreducible numerical semigroups, there are many different characterizations of these properties. In the literature, sometimes these are chosen to be the definition of symmetric and pseudosymmetric numerical semigroups. Proposition 15 Let S be a numerical semigroup. (i) S is symmetric if and only if for all x ∈ Z \ S, we have F(S) − x ∈ S. (ii) S is pseudo-symmetric if and only if F(S) is even and for all x ∈ Z \ S, F(S) − . x ∈ S or x = F(S) 2 Proof (i) Assume that S is symmetric. Then F(S) is odd, and thus H = {x ∈ Z \ S | F(S) − x ∈ / S} = {x ∈ Z \ S | F(S) − x ∈ / S, x = F(S)/2}. If H is not the emptyset, then T = S ∪ {h}, with h = max H , is a numerical semigroup with Frobenius number F(S) containing properly S, which is impossible in light of Theorem 1. For the converse note that F(S) cannot be even, since otherwise as F(S)/2 ∈ / S, we would have F(S) − F(S)/2 = F(S)/2 ∈ S; a contradiction. So, we only need to prove that S is irreducible. Let to this end T be a numerical semigroup such that F(S) ∈ / T and suppose that S ⊂ T . Let x ∈ T \ S. By hypothesis F(S) − x ∈ S. This implies that F(S) = (F(S) − x) + x ∈ T . This is a contradiction (we are using here Theorem 1 once more). (ii) The proof is the same as the proof of (i). The maximality of irreducible numerical semigroups in the set of numerical semigroups with the same Frobenius number, translates to minimality in terms of gaps. This is highlighted in the next result. Observe that for any numerical semigroup S, if x ∈ S, then F(S) − x ∈ / S. In particular, n(S) = #(S ∩ [0, F(S)]) ≥ g(S). As F(S) + 1 = n(S) + g(S), we deduce that g(S) ≥ (F(S) + 1)/2. Corollary 6 Let S be a numerical semigroup. . (i) S is symmetric if and only if g(S) = F(S)+1 2 (ii) S is pseudo-symmetric if and only if g(S) =

F(S)+2 . 2

Hence irreducible numerical semigroups are those with the least possible genus once the Frobenius number is fixed. Recall that the Frobenius number and genus for every embedding dimension two numerical semigroup are known; as a consequence, we get the following. Corollary 7 Let S be a numerical semigroup. If e(S) = 2, then S is symmetric. The rest of the section is devoted to characterizations in terms of the Apéry sets (confirming in this way their ubiquity). First we show that Apéry sets are closed under summands.

2.1 Characterizations of Irreducible Numerical Semigroups

21

Lemma 8 Let S be a numerical semigroup and let n ∈ S ∗ . If x, y ∈ S and x + y ∈ Ap(S, n), then x, y ∈ Ap(S, n). Proof Assume to the contrary, and without loss of generality, that y − n ∈ S. Then x + y − n ∈ S, and consequently x + y ∈ / Ap(S, n). This in particular means that Ap(S, n) is fully determined by the set of maximal elements in Ap(S, n) with respect to ≤ S . Proposition 16 (Apéry) Let S be a numerical semigroup and let n ∈ S ∗ . Let Ap(S, n) = {0 = a0 < a1 < · · · < an−1 }. Then S is symmetric if and only if ai + an−1−i = an−1 for all i ∈ {0, . . . , n − 1}. Proof Suppose that S is symmetric. From Proposition 5, we know that F(S) = an−1 − n. Let 0 ≤ i ≤ n − 1. Since ai − n ∈ / S, we get F(S) − ai + n = an−1 − ai ∈ S. Let s ∈ S be such that an−1 − ai = s. Since an−1 = ai + s ∈ Ap(S, n), by Lemma 8, s ∈ Ap(S, n). Hence s = a j for some 0 ≤ j ≤ n − 1. As this is true for any i, we deduce that j = n − 1 − i. Conversely, the hypothesis implies that Maximals≤S Ap(S, n) = {an−1 }. Hence PF(S) = {F(S)} (Proposition 8). Also, by Proposition 7, {F(S)} = Maximals≤S (Z \ S). If x ∈ / S, then x ≤ S F(S), whence F(S) − x ∈ S. As a consequence of the many invariants that can be computed using Apéry sets, we get the following characterizations of the symmetric property. Corollary 8 Let S be a numerical semigroup. The following conditions are equivalent. (i) (ii) (iii) (iv)

S is symmetric. PF(S) = {F(S)}. If n ∈ S, then Maximals≤S (Ap(S, n)) = {F(S) + n}. t(S) = 1.

Notice that every symmetric numerical semigroup is almost symmetric. Indeed, let S be a numerical semigroup. If S is symmetric, its type is one, and by Corollary 6, we obtain g(S) = (F(S) + 1)/2 = (F(S) + t(S))/2. Example 16 Let S = 5, 7, 9, 11. Figure 2.2 represents the Hasse diagram of Ap(S, 5) (with respect to ≤ S ). One can see the symmetry in the shape forced by Proposition 16, and how the properties in Corollary 8 hold for this semigroup.

Fig. 2.2 Hasse diagram of Ap(5, 7, 9, 11, 5)

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2 Irreducible Numerical Semigroups

Example 17 Recall that for a and b integers greater than one with gcd(a, b) = 1, Ap(a, b, a) = {0, b, . . . , (a − 1)b}. By Corollary 8, this implies that a, b is symmetric, recovering again Corollary 7. Now, we are going to obtain the analogue for pseudo-symmetric numerical semigroups. The first step is to deal with one half of the Frobenius number. Lemma 9 Let S be a numerical semigroup and let n ∈ S ∗ . If S is pseudo-symmetric, + n ∈ Ap(S, n). then F(S) 2 Proof Clearly F(S) ∈ / S. If F(S) +n ∈ / S, then F(S) − F(S) − n ∈ S. This implies that 2 2 2 F(S) F(S) − n ∈ S and thus ∈ S, which is a contradiction. 2 2 Proposition 17 Let S be a numerical semigroup and let n ∈ S ∗ . Let Ap(S, n) =   + n . Then S is pseudo-symmetric if and only {0 = a0 < a1 < · · · < an−2 } ∪ F(S) 2 if ai + an−2−i = an−2 for all i ∈ {0, . . . , n − 2}. + Proof Suppose that S is pseudo-symmetric and let w ∈ Ap(S, n). If w = F(S) 2 F(S) n, then w − n ∈ / S and w − n = 2 . Hence F(S) − (w − n) = F(S) + n − w = max Ap(S, n) − w ∈ S. Since F(S) − w ∈ / S, F(S) + n − w = max Ap(S, n) − + n (otherwise w = F(S) , a contraw ∈ Ap(S, n). But max A p(S, n) − w = F(S) 2 2 diction). Now we use the same argument as in the symmetric case (Proposition 16). ,x∈ / S. Take w ∈ Ap(S, n) such that w ≡ x (mod n). Conversely, let x = F(S) 2 There exists k ∈ N∗ such that x = w − kn (compare with Proposition 4). + n, then F(S) − x = F(S) + (k − 1)n. But x = F(S) . Hence k ≥ 2, 1. If w = F(S) 2 2 2 and consequently F(S) − x = w + (k − 2)n ∈ S. + n, then F(S) − x = F(S) + n − w + (k − 1)n = an−2 − w + 2. If w = F(S) 2 (k − 1)n ∈ S, because an−2 − w ∈ S. Again, by using the properties of the Apéry sets, we get several characterizations for pseudo-symmetric numerical semigroups. Corollary 9 Let S be a numerical semigroup. The following conditions are equivalent. (i) S is pseudo-symmetric.   . (ii) PF(S) = F(S), F(S) 2 (iii) If n ∈ S, then Maximals≤S (Ap(S, n)) =



F(S) 2

 + n, F(S) + n .

Example 18 Let S be a numerical semigroup. If S is pseudo-symmetric, then t(S) = 2. The converse is not true in general. Take for instance S = 5, 6, 8 from Example 8. We have Ap(S, 5) = {0, 6, 12, 8, 14}, PF(S) = {7, 9}, and t(S) = 2. However S is not pseudo-symmetric. The Hasse diagram depicting Ap(S, 5) is in Fig. 1.1. GAP Example 8 Let us see how many numerical semigroups with Frobenius number 16 and type 2 are not pseudo-symmetric.

2.1 Characterizations of Irreducible Numerical Semigroups

23

gap> l:=NumericalSemigroupsWithFrobeniusNumber(16);; gap> Length(l); 205 gap> Filtered(l, s->Type(s)=2); [ , , , , , , , , , , , , , ] gap> Filtered(last,IsPseudoSymmetricNumericalSemigroup); [ , , , , , , ] gap> Difference(last2,last); [ , , , , , , ] gap> List(last, MinimalGenerators); [ [ 3, 14, 19 ], [ 3, 17, 19 ], [ 5, 7, 18 ], [ 5, 9, 12 ], [ 6, 7, 11 ], [ 6, 9, 11, 13 ], [ 7, 10, 11, 12, 13 ] ]

Every irreducible numerical semigroup is almost symmetric. We already know that if S is symmetric, then it is irreducible. Let S be a pseudo-symmetric numerical semigroup. By Corollary 9, we know that t(S) = 2, and in light of Corollary 6, we have g(S) = (F(S) + 2)/2 = (F(S) + t(S))/2. Corollary 10 Every irreducible numerical semigroup is almost symmetric. We have seen in this section that if S is irreducible, then its type is either one (symmetric case) or two (pseudo-symmetric case). Hence t(S) + 1 ≤ e(S) (we do not have pseudo-symmetric numerical semigroups with embedding dimension two, since they are all symmetric). This, together with Corollary 5 proves that Wilf’s conjecture holds for irreducible numerical semigroups (as mentioned in the last chapter).

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2 Irreducible Numerical Semigroups

Fig. 2.3 The Hasse diagram of the irreducible oversemigroups of 5, 7, 9

2.2 Decomposition of a Numerical Semigroup into Irreducible Semigroups Recall that a numerical semigroup S is irreducible if it cannot be expressed as the intersection of two numerical semigroups properly containing it. We show in this section that every numerical semigroup can be expressed as a finite intersection of irreducible numerical semigroups. Theorem 2 Let S be a numerical semigroup. There exists a finite set of irreducible numerical semigroups {S1 , . . . , Sr } such that S = S1 ∩ · · · ∩ Sr . Proof If S is not irreducible, then there exist two numerical semigroups S 1 and S 2 such S = S 1 ∩ S 2 and S ⊂ S 1 and S ⊂ S 2 . If S 1 is not irreducible, then we restart with S 1 , and so on. We construct this way a sequence of oversemigroups of S. This process will stop, because O(S) has finitely many elements.

2.2 Decomposition of a Numerical Semigroup into Irreducible Semigroups

25

Example 19 Figure 2.3 represents the Hasse diagram of irreducible oversemigroups of 5, 7, 9 with respect to set inclusion (we have included 5, 7, 9 in the diagram). Since the minimal irreducible oversemigroups of 5, 7, 9 are 5, 7, 9, 11 and 5, 7, 8, 9, we have that 5, 7, 9 = 5, 7, 9, 11 ∩ 5, 7, 8, 9 is a decomposition of 5, 7, 9 into irreducibles. The next step is to find a way to compute an “irredundant” decomposition into irreducible numerical semigroups. The key result to accomplish this task is the following proposition. Proposition 18 Let S be a numerical semigroup and let S1 , . . . , Sr ∈ O(S). The following conditions are equivalent. (i) S = S1 ∩ · · · ∩ Sr . (ii) For all h ∈ SG(S), there is i ∈ {1, . . . , r } such that h ∈ / Si . Proof (i) implies (ii) Let h ∈ SG(S). Then h ∈ / S, which implies that h ∈ / Si for some i ∈ {1, . . . , r }. (ii) implies (i) Suppose that S ⊂ S1 ∩ · · · ∩ Sr , and let h = max((S1 ∩ · · · ∩ Sr ) \ S). In light of Lemma 6, h ∈ SG(S), and for all i ∈ {1, . . . , r }, h ∈ Si , contradicting the hypothesis. Remark 5 Let I(S) be the set of irreducible numerical semigroups of O(S), and let Min⊆ (I (S)) be the set of minimal elements of I(S) with respect to set inclusion. / Si }. We Assume that Min⊆ (I (S)) = {S1 , . . . , Sr }. Define C(Si ) = {h ∈ SG(S) : h ∈ have S = S1 ∩ · · · ∩ Sr if and only if SG(S) = C(S1 ) ∪ · · · ∪ C(Sr ). This gives a procedure to compute a (nonredundant) decomposition of S into irreducibles. This decomposition might not be unique, and not all might have the same number of irreducibles involved. GAP Example 9 Let us decompose the semigroup S = 7, 9, 11, 17 into irreducibles. gap> s:=NumericalSemigroup(7,9,11,17);; gap> DecomposeIntoIrreducibles(s); [ , , ] gap> List(last, MinimalGenerators); [ [ 7, 8, 9, 10, 11, 12 ], [ 7, 9, 10, 11, 12, 13 ], [ 7, 9, 11, 13, 15, 17 ] ]

There are exactly 17 irreducible oversemigroups of S. gap> Length(Filtered(OverSemigroups(s), IsIrreducible)); 17

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2 Irreducible Numerical Semigroups

There are some (inefficient) bounds for the number of irreducible numerical semigroups appearing in a minimal decomposition of a numerical semigroup into irreducibles. Actually, a numerical semigroup might have different minimal decompositions (in the sense that they cannot be refined to other decompositions) with different cardinalities. So it is an open problem to know the minimal cardinality among all possible minimal decompositions. Also it is not clear how this decomposition translates to K[[S]], or if the study of the curve associated to S can benefit from this decomposition.

2.3 Free Numerical Semigroups We present in this section a way to construct easily symmetric numerical semigroups. This idea was originally exploited by Bertin, Carbonne, and Watanabe among others (see [17, 36, 76]) and goes back to the 70s. As mentioned above this is a way to produce one dimensional local Gorenstein rings. The semigroups appearing in Chap. 4 are of this form, and this is why we pay special attention to them. Let S be a numerical semigroup and let {r0 , . . . , rh } be its minimal set of generators. Let d1 = r0 and for all k ∈ {2, . . . , h + 1}, set dk = gcd(dk−1 , rk−1 ). Define k , for k ∈ {1, . . . , h}. ek = ddk+1 We say that S is free for the arrangement (r0 , . . . , rh ) if for all k ∈ {1, . . . , h}: (i) ek > 1, (ii) ek rk belongs to the semigroup generated by {r0 , . . . , rk−1 }. We say that S is telescopic if r0 < r1 < · · · < rh and S is free for the arrangement (r0 , . . . , rh ). Example 20 Let S = 4, 6, 9. Then h = 2, and d1 = 4, d2 = 2 and d3 = 1. Hence, e1 = 2 = e2 . As 2 × 6 ∈ 4 and 2 × 9 ∈ 4, 6, we have that S is telescopic. Free numerical semigroups were named in this way by Bertin and Carbonne [17]; the name telescopic was used by Kirfel and Pellikaan in [54]. The motivation of Bertin and Carbonne was finding families of complete intersection numerical semigroups (that are always symmetric); while Kirfel and Pellikaan interest was determining numerical semigroups with associated algebraic–geometric codes with nice properties. There is an alternative way of introducing free semigroups with the use of gluings (more modern notation), see for instance [69, Chap. 8]. Example 21 The easiest example of free numerical semigroup (apart from N) is a, b (and as this is a numerical semigroup other than N, a and b coprime integers greater than one). Example 22 We know that 4, 6, 9 is telescopic. Let us see how to construct another telescopic numerical semigroup from this one. Take any positive integer, for this

2.3 Free Numerical Semigroups

27

example, we choose 2, and multiply the sequence (4, 6, 9) by 2, obtaining (8, 12, 18). In this way the old di ’s are multiplied by 2 as well. Now we need another positive integer coprime with 2 so that the whole sequence has greatest common divisor one. Also 2 times this integer must be in 8, 12, 18, or equivalently, our integer must be in 4, 6, 9. Since we are looking for a telescopic numerical semigroup, we need an integer greater than 18. We can for instance choose 19, which is in 4, 6, 9 and it is coprime with 2. Thus 8, 12, 18, 19 is a telescopic numerical semigroup. We can repeat this process as many times as desired. Any telescopic numerical semigroup can be obtained in this way. One of the advantages of dealing with free numerical semigroups is that every integer admits a unique representation in terms of its minimal generators if we impose some bounds on the coefficients. Lemma 10 Assume that S is free for the arrangement (r0 , . . . , rh ), and let x ∈ Z. There exist unique λ0 , . . . , λh ∈ Z such that the following holds h (i) x = k=0 λk r k , (ii) for all k ∈ {1, . . . , h}, 0 ≤ λk < ek . Proof Existence. hThe group generated by S is Z, and so there exist α0 , . . . , αh ∈ Z αk rk . Write αh = qh eh + λh , with 0 ≤ λh < eh . Now we use such that x = k=0 h−1 that eh rh = i=0 βi ri , with βi ∈ N for all i ∈ {1, . . . , h − 1}. Hence x=

h−1  (αk + qh βk )rk + λh rh , k=0

follows an easy induction on h. and 0 ≤ λh < eh . Now the by result h h αk rk = k=0 βk rk be two distinct such representaUniqueness. Let x = k=0 tions, and let j ≥ 1 be the greatest integer such that α j = β j . We have (α j − β j )r j =

j−1  (βk − αk )rk . k=0

In particular, d j divides (α j − β j )r j . But gcd(d j , r j ) = d j+1 , whence rj dj . As gcd(d j /d j+1 , r j /d j+1 ) = 1, this implies that d j+1 (α j − β j ) d j+1 dj , yielding a contradiction. However |α j − β j | < e j = d j+1

dj d j+1

divides

divides α j − β j .

An expression of x like in the preceding lemma is called a standard representation. Example 23 Let S = 4, 6, 9, and let us consider the integer 30. There are several ways to represent 30 as a linear combination of {4, 6, 9} with nonnegative integer coefficients:

28

2 Irreducible Numerical Semigroups gap> FactorizationsIntegerWRTList(30,[4,6,9]); [ [ 6, 1, 0 ], [ 3, 3, 0 ], [ 0, 5, 0 ], [ 3, 0, 2 ], [ 0, 2, 2 ] ]

The first one in the list corresponds with the standard representation of 30 with respect to S (recall that in this example e1 = 2 = e2 ). As a consequence of this representation we obtain the following characterization for membership to a free numerical semigroup. Lemma  11 Suppose that S is free for the arrangement (r0 , . . . , rh ) and let x ∈ N. h λk rk be the standard representation of x. We have x ∈ S if and only Let x = k=0 if λ0 ≥ 0. h Proof If λ0 ≥ 0 then clearly x ∈ S. Suppose that x ∈ S and write x = k=0 αk rk with α0 , . . . , αh ∈ N. As in Lemma 10, whenever αi ≥ ei , with i > 0, we can replace ei ri with its expression in terms of r0 , . . . , ri−1 . At the end we will have the standard representation of x, and by construction the coefficient of r0 will be nonnegative (will be greater than or equal to α0 ). We will come back to the rewriting procedure used in the above lemmas in Sect. 5.2. With all this information, it is easy to describe the Apéry set of the first generator in the arrangement that makes the semigroup free. Corollary 11 Suppose that S is free for the arrangement (r0 , . . . , rh ). Then  Ap(S, r0 ) =

h 

   λk rk  0 ≤ λk < ek for all k ∈ {1, . . . , h} .

k=1

h Proof Let x ∈ S and let x = k=0 λk rk be the standard representation of x. Clearly h x − r0 = (λ0 − 1)r0 + k=1 λk rk is the standard representation of x − r0 . Hence / S if and only if λ0 = 0. This proves our assertion. x − r0 ∈ As we have seen in this last result, the shape of the Apéry set of a free numerical semigroup is rectangular. D’Anna, Micale and Sammartano have studied recently a generalization of these semigroups by considering Apéry sets with these shapes [29]. As usual, once we know an Apéry set, we can derive many properties of the semigroup. Proposition 19 Let S be free for the arrangement (r0 , . . . , rh ). h (i) F(S) = k=1 (ek − 1)rk − r0 . (ii) S is symmetric. . (iii) g(S) = F(S)+1 h 2 (iv) r0 = i=1 ei .

2.3 Free Numerical Semigroups

29

Proof h We have F(S) = max Ap(S, r0 ) − r0 , by Proposition 5. As max Ap(S, r0 ) = k=1 (ek − 1)r k , (i) follows easily. Assertion (ii) is a consequence of Corollary 11 and Proposition 16. Assertion (iii) is a consequence of (ii) and Corollary 6. We know from Lemma 1 that #Ap(S, r0 ) = r0 . In light Corollary 11, #Ap(S, r0 ) = e1 × · · · × eh . This proves (iv). Example 24 Let us revisit the semigroup S = 8, 12, 18, 19, which we know that is a telescopic numerical semigroup. In this setting, (d1 , d2 , d3 , d4 ) = (8, 4, 2, 1), whence (e1 , e2 , e3 ) = (2, 2, 2). By Corollary 11, this means that 

Ap(S, 8) = a × 12 + b × 18 + c × 19 | (a, b, c) ∈ {0, 1}3 . Hence Ap(S, 8) = {0, 12, 18, 19, 30, 31, 37, 49}. Also, we have that F(S) = 49 − 8 = 3k=1 (ei − 1)ri − r0 = 12 + 18 + 19 − 8 = 41, and g(S) = (41 + 1)/2 = 21. GAP Example 10 The proportion of free numerical semigroup compared with symmetric numerical semigroups with fixed Frobenius number (or genus) is small. gap> List([1,3..51], i -> > [Length(FreeNumericalSemigroupsWithFrobeniusNumber(i)), > Length(IrreducibleNumericalSemigroupsWithFrobeniusNumber(i))]); [ [ 1, 1 ], [ 1, 1 ], [ 2, 2 ], [ 3, 3 ], [ 2, 3 ], [ 4, 6 ], [ 5, 8 ], [ 3, 7 ], [ 7, 15 ], [ 8, 20 ], [ 5, 18 ], [ 11, 36 ], [ 11, 44 ], [ 9, 45 ], [ 14, 83 ], [ 17, 109 ], [ 12, 101 ], [ 18, 174 ], [ 24, 246 ], [ 16, 227 ], [ 27, 420 ], [ 31, 546 ], [ 21, 498 ], [ 35, 926 ], [ 38, 1182 ], [ 27, 1121 ] ]

Chapter 3

Ideals

As it happens for rings and, in particular, for one-dimensional domains, it is possible to study ideals for numerical semigroups, to get relevant information on the semigroup itself. The case of symmetric and pseudo-symmetric numerical semigroups (that is, irreducible numerical semigroups) has been extensively deepened in [15], where the authors deal, in particular, with duality properties of their ideals. It turns out that these are particular cases of the duality with respect to a canonical ideal. This concept was introduced for numerical semigroups by Jäger in [51], where he proves the existence of a class of ideals that plays a relevant role for numerical semigroups, in analogy to what happens for rings. In fact, as we will see, many concepts and invariants connected to ideals are borrowed from ring theory, but their study at semigroup level allows to make them somehow more clear, seeing them from a different point of view; and it often happens that a property proved for semigroup ideals is true for ideals of one dimensional rings, even if the proof in the second case is not a straightforward translation of the proof given at semigroup level. It is worth noticing that any value set of a fractional ideal of K [[S]] or, more generally, of an analytically irreducible and residually rational local one-dimensional domain is a semigroup ideal of S, and this is the reason why many multiplicative properties of ring ideals have their analogue additive properties at numerical level.

3.1 Basic Definitions and Relevant Invariants Let S be a numerical semigroup; a subset E of Z is a relative ideal of S if S + E = {s + e | s ∈ S, e ∈ E} ⊆ E and there exists s ∈ S such that s + E = {s + e | e ∈ E} ⊆ S. The last condition means that E has a minimum, called multiplicity, that we will always denote by m(E). If E ⊆ S, we will say that E is a proper ideal or, simply, an ideal of S. It is straightforward that S is an ideal of itself; also M = S ∗ is an ideal of S that is called the maximal ideal of S. There are two other relevant ideals: the conductor and the standard canonical ideal. The conductor (ideal) of S is S − N = {z ∈ Z | © Springer Nature Switzerland AG 2020 A. Assi et al., Numerical Semigroups and Applications, RSME Springer Series 3, https://doi.org/10.1007/978-3-030-54943-5_3

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z + N ⊆ S} = {C(S), →} = C(S) + N and it is not difficult to check that it is the biggest ideal shared by S and N. The standard canonical ideal of S is defined as K(S) = {x ∈ Z | F(S) − x ∈ / S}. It is also easy to verify that K(S) is a relative ideal such that S ⊆ K(S) ⊆ N, and that S = K(S) if and only if S is symmetric, while K(S) = N if and only if S = N. If no confusion arises, in the sequel we will write simply K instead of K(S). Two relative ideals E and H are said to be equivalent if there exists x ∈ Z such that E = x + H = {x + h | h ∈ H }. Hence every relative ideal is equivalent to some proper ideal. From the fact that S + E ⊆ E, it is straightforward, as it happens for S, that the set Z \ E has a maximum, which we will call Frobenius number of E and will denote by F(E). Hence, if we consider  = E + F(S) − F(E), E we obtain an equivalent relative ideal with the same Frobenius number of S. Using this notation we have the following result. Proposition 20 Let S be a numerical semigroup and let K be its standard canonical  such that ideal. Every relative ideal E of S is equivalent to a relative ideal E,  S − N ⊆ E ⊆ K. ˜ For the other inclusion,  it is obvious that S − N ⊆ E. Proof By the definition of E,  assume that x ∈ E and x ∈ / K . This latter condition is equivalent to F(S) − x ∈  contradicting that the Frobenius S. Thus, we obtain F(S) = (F(S) − x) + x ∈ E,  number of E is F(S). ˜ ≥ 0 and equality holds if and only if E˜ ⊇ S. We notice that m( E) The previous proposition shows one aspect of the importance of the standard canonical ideal. However, it was defined to solve the problem of duality and to give a representation of the canonical ideal for numerical semigroup rings (see [51]), as we will see later from the numerical point of view. GAP Example 11 Let S = 3, 5, 7 . gap> s:=NumericalSemigroup(3,5,7);; gap> k:=CanonicalIdeal(s);

gap> SmallElements(k); [ 0, 2, 3, 5 ] gap> SmallElements(s); [ 0, 3, 5 ] gap> SmallElements(MaximalIdeal(s)); [ 3, 5 ]

3.1 Basic Definitions and Relevant Invariants

33

SmallElements of an ideal shows the elements of the ideal up to the Frobenius plus one. All integers greater than the last element in this set are in the ideal. If we pick a symmetric numerical semigroup, for instance S = 4, 6, 9 , then its canonical ideal must be the semigroup itself. gap> gap> [ 0, gap> [ 0,

s:=NumericalSemigroup(4,6,9);; SmallElements(CanonicalIdeal(s)); 4, 6, 8, 9, 10, 12 ] SmallElements(s); 4, 6, 8, 9, 10, 12 ]

We introduce now some more notations and definitions. Given two relative ideals E, H we can set E + H = {e + h | e ∈ E, h ∈ H }, and E − H = {z ∈ Z | z + H ⊆ E}; both E + H and E − H are relative ideals. We recall some properties of the second operation that are not difficult to check for any E, G, H relative ideals and for any x ∈ Z: • • • •

(x + E) − H = x + (E − H ) and E − (x + H ) = −x + (E − H ); if G ⊆ H , then E − G ⊇ E − H and G − E ⊆ H − E; E − (G ∩ H ) ⊇ (E − G) ∪ (E − H ) and E − (G ∪ H ) = (E − G) ∩ (E − H ); (E − G) − H = E − (G + H ).

GAP Example 12 Let S = 10, 13, 21, 22 . gap> s:=NumericalSemigroup(10,13,21,22);; gap> e:=[10,11]+s;; gap> h:=[-1,2,3]+s;; gap> e+h;

gap> SmallElements(e+h); [ 9, 10, 12, 13, 14, 19, 20, 22, 23, 24, 25, 26, 27, 29 ] gap> SmallElements(e-h); [ 21, 31, 34, 38, 41, 42, 43, 44, 47, 48, 50 ] gap> g:=5+s;; gap> (e-g)-h=e-(g+h); true gap> SmallElements(Intersection(e,h)); [ 20, 21, 23, 24, 30, 31, 32, 33, 34, 36, 37, 40, 41, 42, 43, 44, 45, 46, 47, 49 ]

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We have defined already the multiplicity of E as its smallest element, m(E); the genus is g(E) = |(m(E) + N) \ E| and the type is t(E) = |(E − M) \ E|. We notice that the last two definitions generalize the corresponding definition for numerical semigroups, in the sense that genus and type of a numerical semigroup S correspond to genus and type of S as ideal of itself. Notice, however, that the multiplicity of S as ideal of itself is 0. Moreover, it is not difficult to check that t(M) = t(S) + 1. As for the standard canonical ideal we have that K − M = K ∪ {F(S)}: in fact, the inclusion K ∪ {F(S)} ⊆ K − M is trivial, so let x ∈ (K − M) \ K ∪ {F(S)}); then 0 = F(S) − x ∈ S and, thus, F(S) = (F(S) − x) + x ∈ M + (K −M) ⊆ K , which yields a contradiction. Therefore t(K ) = 1. We call a set of generators of E any set of elements {e1 , . . . , eh } ⊂ E such that every e ∈ E can be written as s + ei , for some i ∈ {1, . . . , h} and for some s ∈ S, that is, E = {e1 , . . . , eh } + S. Is it easy to check that, for every relative ideal E, there exists a unique minimal set of generators, namely E \ (M + E). It is also clear that m(S) + E ⊆ (M + E); hence |E \ (M + E)| ≤ |E \ (m(S) + E)| = m(S), as it happens for numerical semigroups. We also notice that we can think of E \ (m(S) + E) as the Apéry set of E with respect to m(S), generalizing in an obvious way the definition of Apéry set given for numerical semigroups. GAP Example 13 Let S = 4, 6, 9 , and let I = {10, 18, 21} + S. gap> S:=NumericalSemigroup(4,6,9);; gap> I:=[10,18,21]+s;; gap> M:=MaximalIdeal(S);

gap> Difference(I,M+I); [ 10, 21 ] gap> MinimalGenerators(I); [ 10, 21 ] gap> AperyList(I,Minimum(I)); [ 10, 21, 22, 23, 14, 25, 16, 27, 18, 19 ] gap> Difference(I,Minimum(I)+I); [ 10, 14, 16, 18, 19, 21, 22, 23, 25, 27 ] gap> Set(last)=Set(last2); true

Observe that the output of AperyList differs from the difference of I and m(I ) + I (as lists, though they are the same sets). This is because, AperyList is a list in which the ith element is congruent with i modulo m(I ). We notice that if E is a proper ideal of S, then its complement X = S \ E is a set with the following property: if x ∈ X , for any y ∈ S, y ≤ S x, it holds y ∈ X (where, as defined in the first chapter, y ≤ S x if and only if x − y ∈ S); in fact if y ∈ E, then, by x = y + s, for some s ∈ S, it follows x ∈ E. A set X with this property will be called closed under divisors. It is not difficult to check that if X ⊂ S is closed under divisors, then E = S \ X is an ideal: in fact for

3.1 Basic Definitions and Relevant Invariants

35

any s ∈ S and for any e ∈ E, s + e ∈ E, otherwise also e should belong to X being a divisor of an element s + e ∈ X . Hence we have proved the following statement. Lemma 12 A subset E of a numerical semigroup S is an ideal if and only if it is the complement of a subset X of S, closed under divisors. Remark 6 If X is a subset of a numerical semigroup S closed under divisors, then we can compute a system of generators of the ideal S \ X in the following way. Let m be the multiplicity of S (any nonzero element would also work), and let A = Ap(S, m). We know that A = {w0 , . . . , wm−1 } with wi the minimum element in S congruent with i modulo m. If for some i ∈ {0, . . . , m − 1} we have that wi ∈ X , then we replace wi with wi + km with k the minimum positive integer such that / X (since X is a finite set, this integer always exists). In this way, the wi + km ∈ resulting set A is a generating system of the ideal S \ X (not necessarily minimal). GAP Example 14 Let S = 10, 11, 15, 19 and E = {20, 21, 25} + S. gap> gap> gap> [ 0, gap> true gap> true

S:=NumericalSemigroup(10,11,15,19);; I:=[20,21,25]+s;; D:=Difference(0+s,i); 10, 11, 15, 19, 22, 26, 29, 33, 34, 37, 38, 48 ] IsComplementOfIntegralIdeal(D,S); IdealByDivisorClosedSet(D,S)=I;

Let us turn our attention back to the invariants of ideals; we start noticing that genus, type and minimal number of generators coincide for all the relative ideals in the same equivalence class, that is, these definitions are invariant under translations. There are some inequalities that relate these invariants and the invariants of S. Lemma 13 Let E be a relative ideal of a numerical semigroup S. Then F(S) + 1 −  + m( E)  and equality holds if and only if E  = K. g(S) ≤ g( E)  + m( E)  is the number of elements in N \ E  (m( E)  is a Proof The integer g( E) nonnegative integer). The integer F(S) + 1 − g(S) is the number of elements s in S,  the thesis follows smaller than F(S) + 1. Since s ∈ S implies F(S) − s ∈ / K ⊇ E, by the definition of K . The inequality of the previous lemma is equivalent to another inequality introduced in the contest of linear codes connected to Weierstrass semigroups. More precisely, in [19] it is proved that if E is a proper ideal of S, F(E) ≤ 2 g(S) − 1 + |S \ E| and the ideals attaining the equality are called maximum sparse ideals. We can rewrite this inequality observing that, for proper ideals, |S \ E| = |N \ E| − |N \  + [m( E)  + F(E) − F(S)] − g(S). Hence F(E) S| = g(E) + m(E) − g(S) = g( E) ≤ 2 g(S) − 1 + |S \ E| becomes

36

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 + m( E)  − F(S) + F(E) − g(S), F(E) ≤ 2 g(S) − 1 + g( E)  + m( E).  Hence the maximum that is in turn equivalent to F(S) + 1 − g(S) ≤ g( E)  = K. sparse ideals are exactly those proper ideals E such that E The following proposition introduces a new class of ideals.  + m( E)  ≥ F(S) − g(S) + Proposition 21 Let E be a relative ideal of S. Then g( E)  − M = K − M. t(E) and equality holds if and only if E  and g( E)  + m( E)  − t( E)  = |N \ E|  − |( E  − M) \ Proof Clearly, t(E) = t( E)   E| = |N \ ( E − M)|. Moreover, since F(S) + 1 − g(S) is the number of the elements of S smaller than F(S) + 1, it holds that F(S) − g(S) = |N \ K | − 1 = |N \ (K ∪ F(S))|.  − M ⊆ K ∪ {F(S)}. First of all, since F(S) = Hence we need to show that E    − M) \ {F(S)} ⊆ K : in fact, if F( E), this integer is always in E − M. Moreover, ( E   = F(S) = (F(S) − x ∈ ( E − M) \ {F(S)} and x ∈ / K , then F(S) − x ∈ M and F( E)  x) + x ∈ E, a contradiction. The last assertion follows by the equality K − M = K ∪ {F(S)}, that we have seen above. Notice that for a numerical semigroup S = N, the inequality in the previous proposition becomes 2 g(S) ≥ F(S) + t(S) (see Proposition 10), whence the equality is satisfied if and only if S is almost symmetric. Notice also that the equivalent condi − M = K − M, for E = S, becomes S − M = K ∪ {F(S)}, which is in turn tion E equivalent to K ⊆ S − M; this last one is the original definition of almost symmetric semigroup given in [12]. Going back to ideals, in analogy to almost symmetric semigroups, the ideals satisfying the equality in the previous statement have been called almost canonical [30].  = K. Corollary 12 If E is almost canonical and t(E) = 1, then E Proof By the previous proposition and the hypothesis we immediately obtain that  + m( E),  which means E  = K , by Lemma 13. F(S) + 1 − g(S) = g( E) We next prove a better result, that is the ideals in the equivalence class of K are characterized by the property of having type equal to 1, without assuming that E is almost canonical. This fact reflects, at numerical level, a property that holds for the canonical module of a Cohen Macaulay ring. Moreover, in a one dimensional Cohen-Macaulay local ring R possessing a canonical ideal, any two canonical ideals are obtained one from the other by multiplication by a non zerodivisor in the total quotient ring of R. In the same way we will say that a relative ideal E of a numerical  = K . In the next section we will see other semigroup S is a canonical ideal if E interesting properties of canonical ideals in connection with duality.  = K. Proposition 22 A relative ideal E has t(E) = 1 if and only if E

3.1 Basic Definitions and Relevant Invariants

37

= K Proof We have observed that t(K ) = 1, hence any relative ideal such that E   K and let y be the largest integer in has type equal to 1. Conversely, assume that E  If x ∈ M then, since K is a relative ideal, y + x ∈ K ; moreover y < x + y, K \ E.  It follows that ( E  − M) \ E ⊇ hence, by the choice of y, we obtain y + x ∈ E.  = t(E) ≥ 2, a contradiction. {y, F(S)}, that means t( E) GAP Example 15 Let S = 3, 5, 7 . gap> S:=NumericalSemigroup(3,5,7);

gap> MinimalGenerators(CanonicalIdeal(S)); [ 0, 2 ] gap> IsCanonicalIdeal(2+CanonicalIdeal(S)); true

3.2 Duality In [15], the authors study duality properties of ideals for symmetric and pseudosymmetric numerical semigroups. In particular, they show that, as it happens for one dimensional Gorenstein local rings, every relative ideal of a symmetric numerical semigroup is bi-dual, in the sense that S − (S − E) = E. The same happens for ideals in a pseudo-symmetric semigroup, provided that E − E contains F(E)/2. Moreover symmetric and pseudo-symmetric can be characterized by means of these properties (see [15]). Following Jäger [51] we can recover these results form a different point of view, using canonical ideals. Lemma 14 Let S be a numerical semigroup and let K be its canonical ideal. For any relative ideal E, K − E = {x ∈ Z | F(S) − x ∈ / E}. Proof If x + E ⊆ K , then F(S) − x ∈ / E, since F(K ) = F(S) ∈ / K . Conversely, if F(S) − x ∈ / E, then F(S) − (x + e) ∈ / S for any e ∈ E, otherwise F(S) − x = F(S) − (x + e) + e ∈ E; this means that x + E ⊆ K , whence x ∈ K − E. Using the previous lemma other duality properties can be easily proved, for any two relative ideals E, H , using the map x → F(S) − x: E ⊆ H if and only if K − H ⊆ K − E;

(3.1)

E = H is equivalent to K − H = K − E; if E ⊆ H, then |H \ E| = |(K − E) \ (K − H )|.

(3.2) (3.3)

Moreover, just applying the previous lemma to the ideal E = K , we immediately get K − K = S. It is also worth noticing that K − (E ∩ H ) = (K − E) ∪ (K − H ), for any E, H relative ideals (recall that for an ideal other than K we only had an

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inclusion). In fact, dualizing both sides of the equality with respect to K we get (E ∩ H ) = K − [(K − E) ∪ (K − H )] = [K − (K − E)] ∩ [K − (K − H )], that is clearly true. The proof we give of the next theorem is inspired by the characterization of symmetric numerical semigroups presented in [15]. First we need a lemma. Lemma 15 Let E ⊆ F be two relative ideals of a numerical semigroup S. Then |F \ E| = n if and only if there exists a chain of relative ideals E = E 0 ⊂ E 1 ⊂ · · · ⊂ E n = F, such that |E i \ E i−1 | = 1 for any i = 1, . . . , n. In particular, |F \ E| = 1 if and only if for any relative ideal H , E ⊆ H ⊆ F, either E = H or H = F. Proof One implication is trivial, so we have only to prove that |F \ E| = n implies the existence of such a chain. If we denote the elements of F \ E in increasing order by {x1 , x2 , . . . , xn }, it is clear that E ∪ {xn } is again a relative ideal; hence we set E 1 = E ∪ {xn } and we proceed recursively. Theorem 3 (Jäger) Let E be a relative ideal of S. Then E is a canonical ideal if and only if, for any relative ideal H , the equality E − (E − H ) = H holds.  = E, since Proof Notice that we can always assume that F(E) = F(S), that is E (z + E) − H = z + (E − H ) for any z ∈ Z and H relative ideal; hence (z + E) − ((z + E) − H ) = E − (E − H ). Applying two times Lemma 14, we immediately get K − (K − H ) = H , for any relative ideal H . Conversely, by Proposition 22, we need to prove that t(E) = |(E − M) \ E| = 1 or, equivalently (because of the previous lemma), that we cannot find any relative ideal H such that E  H  E − M. So let H be such that E  H ⊆ E − M. Hence we get M = E − (E − M) ⊆ E − H ⊆ E − E = S, where the last equality follows by E = (E − S), that, together with the hypothesis on E, implies S = E − (E − S) = E − E. Now, E  H , hence 0 ∈ / E − H , which means M = E − H . By the hypothesis on E we finally get E − M = E − (E − H ) = H , and this concludes the proof. GAP Example 16 Let us consider again S and I as in the GAP example 13. gap> gap> gap> gap> true gap> true

S:=NumericalSemigroup(4,6,9);; I:=[10,18,21]+S;; K:=CanonicalIdeal(S);; K-(K-I)=I; (2+K)-((2+K)-I)=I;

Corollary 13 A numerical semigroup S is symmetric if and only, for every relative ideal H , S − (S − H ) = H .

3.2 Duality

39

Proof It is enough to observe that S is symmetric if and only if S = K and then apply the previous theorem. To study the pseudo-symmetric case we need the following lemma. Observe that, by definition, for any relative ideal E, the inclusion E ⊆ S − (S − E) holds (and can be strict). By applying this fact to the ideal S − E, we obtain S − E ⊆ S − (S − (S − E)); and the inclusion E ⊆ S − (S − E) yields S − E ⊇ S − (S − (S − E)). Thus, when we dualize three times, we always get S − (S − (S − E)) = S − E.

(3.4)

Lemma 16 Let S be a numerical semigroup and H be a relative ideal such that H − H ⊇ K . Then S − (S − H ) = H . Proof The inclusion S − (S − H ) ⊇ H always holds, hence assume by contradiction that S − (S − H )  H and let x ∈ S − (S − H ) \ H . It follows that F(S) − x ∈ / S − (S − (S − H )) (otherwise F(S) = (F(S) − x) + x would belong to S). In light of (3.4), F(S) − x ∈ / S − H. Now, for any e ∈ H we have x − e ∈ / H − H , since x ∈ / H ; moreover, by hypothesis H − H ⊇ K , therefore x − e ∈ / K , which means that F(S) − (x − e) = (F(S) − x) + e ∈ S, and this yields F(S) − x ∈ S − H , a contradiction. Observe that S is pseudo-symmetric if and only if S ∪ {F(S)/2} = K . With this and the preceding lemma, we have the following consequence. Corollary 14 If a numerical semigroup S is pseudo-symmetric then, for every relative ideal H such that F(S)/2 ∈ H − H , the equality S − (S − H ) = H holds. Proof Notice that the hypothesis F(S)/2 ∈ H − H implies that S ∪ {F(S)/2} = K ⊆ H − H . The proof now follows from Lemma 16. To get the converse of the previous corollary we need to assume some extra hypotheses. Proposition 23 A numerical semigroup S is pseudo-symmetric if and only F(S) is even, F(S)/2 ∈ S − M and, for every relative ideal H such that F(S)/2 ∈ H − H , the equality S − (S − H ) = H holds. Proof The necessity was shown in the preceding corollary. For the sufficiency, assume that F(S)/2 ∈ S − M and that for every relative ideal H , such that F(S)/2 ∈ H − H , the equality S − (S − H ) = H holds. We need to prove that S \ (S − M) = {F(S)/2, F(S)}. By assumption H = S ∪ {F(S)/2, F(S)} is a numerical semigroup containing S and, therefore, it is a relative ideal of S and F(S)/2 ∈ H = H − H . Moreover S  H ⊆ S − M. Dualizing with S and using the fact that M − M = S − M, the hypothesis implies that M = S − (S − M) ⊆ S − H  S − S = S (the last containment is strict, otherwise H = S − (S − H ) = S − S = S, a contradiction). It immediately follows that M = S − H , which yields S − M = S − (S − H ) = H , and we are done.

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3.3 Irreducibility In this section we want to study irreducibility of ideals, in analogy to the case of numerical semigroups. It turns out that we can consider two different notions of irreducibility: the irreducibility of a relative ideal (as relative ideal), that will be clearly dependent only on its equivalence class, and the irreducibility of a proper ideal as an ideal of a numerical semigroup, that can change if we take different ideals in the same class. In the first case, the type of the ideal counts the number of its irreducible components and thus it measures how much the ideal is close to be irreducible; on the other hand, in the case of proper ideals, the type of the semigroup measures the number of irreducible components (as proper ideals) of any principal ideal, in perfect analogy to one–dimensional ring theory. The results in this section appeared mainly in [16]. Let us start with irreducibility for relative ideals. Fix a numerical semigroup S; we will say that a relative ideal E of S is Z-irreducible if it cannot be expressed as a finite intersection of other relative ideals properly containing it. It is straightforward that any canonical ideal is irreducible, since any other relative ideal with the same Frobenius number is contained in it. Moreover, being the only maximal relative ideal with fixed conductor, we expect that canonical ideals are the only Z-irreducible ideals. We can prove a more deep result. Proposition 24 Let S be a numerical semigroup and let E be a relative ideal of S. Let x1 , . . . , x h be the minimal generators of the ideal K − E. Then E = (−x1 + K ) ∩ · · · ∩ (−x h + K ), and this decomposition is irredundant. In particular, the relative ideal E is Zirreducible if and only if it is a canonical ideal. Proof By assumption K − E = (x1 + S) ∪ · · · ∪ (x h + S) and xi − x j ∈ / S for any i, j ∈ {1, . . . , h}, i = j. It follows that E = K − (K − E) = (K − (x1 + S)) ∩ · · · ∩ (K − (x h + S)). Since K − (xi + S) = −xi + (K − S) and K − S = K , the first part of the statement follows immediately.  The decomposition is irredundant because, i (−x j + K ),  if (−xi + K ) ⊇ j = then xi ∈ xi + S = K − (−xi + K ) ⊆ j =i K − (−x j + K ) = j =i x j + S, against the minimality of the generators of K − E. In particular, if E is irreducible, then h = 1, that is, E = −x1 + K , which means that E is a canonical ideal. The converse is obvious, since, as we mentioned above, canonical ideals are Z-irreducible. GAP Example 17 Let S = 3, 4, 5 , and let us consider its ideal {4, 5} + S. gap> S:=NumericalSemigroup(3,5,7);; gap> I:=[4,5]+S;; gap> K:=CanonicalIdeal(S);;

3.3 Irreducibility

41

gap> MinimalGenerators(K-I); [ -2, 2 ] gap> MinimalGenerators(Intersection(-2+K,2+K)); [ 4, 5 ]

And so {4, 5} + S = (−2 + K ) ∩ (2 + K ). Using the argument of the proof of the previous proposition, and the fact that the concept of Z-irreducible and canonical ideal coincide, it is not difficult to check that the irredundant decomposition of E presented above is unique. We will refer to the ideals −xi + K as the irreducible Z-components of E. Corollary 15 The number of irreducible Z-components of E coincides with t(E). Proof By the previous proposition we know that the number of irreducible components of E is the cardinality of the minimal set of generators of K − E, which equals |(K − E) \ (K − E) + M|. Dualizing with respect to K , we get that this number coincides with |K − ((K − E) + M) \ E| (see (3.3)). Moreover K − ((K − E) + M) = (K − (K − E)) − M = E − M, hence we obtain |(K − E) \ (K − E) + M| = |(E − M) \ E| = t(E). We observe that with this result we recover immediately the irreducibility of symmetric numerical semigroups. In fact, given a numerical semigroup S, any numerical semigroup containing S is a relative ideal of S (while the converse is obviously false). If S is symmetric, then it coincides with its standard canonical ideal, whence it is irreducible as relative ideal and, a fortiori, as numerical semigroup. We turn now our attention to irreducibility of proper ideals. We will say that a proper ideal E of a numerical semigroup S is irreducible if it cannot be expressed as finite intersection of proper ideals of S strictly containing it. Obviously S is irreducible as ideal of itself, so in the following we will always assume that 0 ∈ / E. Recall that in the first chapter we defined the partial order relation on Z as follows: a ≤ S b if b − a ∈ S. With this notation we can set, for any x ∈ S, D(x) = {s ∈ S | s ≤ S x}. This set is known as the set of divisors of x in S. We will come back to this set in Sect. 6.4. It is obvious by definition that D(x) is closed under divisors; therefore we know, by Lemma 12, that S \ D(x) is a proper ideal of S. GAP Example 18 The function in numericalsgps to compute the set of divisors of an element in a numerical semigroup is DivisorsOfElementInNumerical Semigroup. gap> S:=NumericalSemigroup(3,5,7);; gap> DivisorsOfElementInNumericalSemigroup(S,10); [ 0, 3, 5, 7, 10 ]

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Lemma 17 Let S be a numerical semigroup, and let x ∈ S. Then, for any proper ideal E of S, the following conditions are equivalent: 1. x ∈ / E; 2. E ⊆ S \ D(x). Proof Assume that E is a proper ideal of S and that x ∈ / E. For every s ∈ E, we have s ≤ S x, because otherwise x would be in E, and so s ∈ S \ D(x). The other implication is obvious, since x ∈ D(x). GAP Example 19 Let S = 3, 5, 7 and x = 10; 10 ∈ / 10 + S, and thus 10 + S ⊆ S \ D(10). gap> gap> [ 0, gap> gap> true

S:=NumericalSemigroup(3,5,7);; D:=DivisorsOfElementInNumericalSemigroup(S,10); 10 ] H:=IdealByDivisorClosedSet(D,S);; IsSubset(H,11+S);

It turns out that irreducible ideals are exactly of the form S \ D(x) (compare with the more general setting given in [70]). Proposition 25 Let E be a proper ideal of a numerical semigroup S. The following conditions are equivalent: (i) E is irreducible; (ii) E is not the intersection of any set of proper ideals strictly containing it; (iii) E = S \ D(x), for some x ∈ S. Proof Conditions (i) and (ii) are equivalent, since E has finite complement in S. For the implication (ii) implies (iii), let H be the intersection of all proper ideals containing E. By hypothesis E  H , so there exists x ∈ H \ E. As x ∈ / E, Lemma 17 asserts that E ⊆ S \ D(x). Since S \ D(x) is an ideal of S, if the previous inclusion is proper, then E  H ⊂ S \ D(x), contradicting that x ∈ H . Finally, assume that (iii) holds and let us prove that (ii) also holds true. Every proper ideal strictly containing E = S \ D(x), has x as an element (Lemma 17). It follows immediately that E cannot be the intersection of a family of proper ideals strictly containing it. Lemma 18 Let E be a proper ideal of a numerical semigroup S. Then: (1) every irreducible ideal containing E is of the form S \ D(x), with x ∈ S \ E; (2) every irreducible ideal containing E and minimal with respect to inclusion is of the form S \ D(x), with x ∈ Maximals≤S (S \ E). Proof 1. Follows Lemma 17 and Proposition 18.

3.3 Irreducibility

43

2. If x, y ∈ S, then S \ D(x) ⊇ S \ D(y) if and only if D(x) ⊆ D(y), that is in turn equivalent to say x ∈ D(y), or, in other words, x ≤ S y. The assertion now follows immediately. GAP Example 20 Let S = 3, 5, 7 and I = {3, 5} + S. gap> gap> gap> [ 0,

S:=NumericalSemigroup(3,5,7);; I:=[3,5]+S;; Difference(0+S,I); 7 ]

According to Lemma 18, there are two irreducible ideals containing I , namely S \ D(0) (the maximal ideal of S) and S \ D(7) = I . Thus, I is irreducible. gap> d:=DivisorsOfElementInNumericalSemigroup(7,S);; gap> IdealByDivisorClosedSet(d,S)=I; true

Since the set of proper ideals containing a proper ideal is finite, every proper ideal can be expressed as a finite intersection of irreducible proper ideals. This expression can have redundant ideals, and thus can be eventually refined to another intersection in which all ideals are needed. We can say that these irredundant ideals are the irreducible components of our original ideal. We are going to see next that, in the case of proper ideals, any irredundant decomposition is unique, and we will be able to count the number of irreducible components in it (in contrast with what happens for numerical semigroups; see Sect. 2.2). Let E and F be proper ideals of a numerical semigroup S. It is easy to check that the set E − S F = {s ∈ S | s + F ⊆ E} is again a proper ideal of S. GAP Example 21 Let S = 3, 5, 7 , I = {3, 5} + S and J = {5, 7} + S. We can compute I − S J as the intersection of I − J with S. gap> S:=NumericalSemigroup(3,5,7);; gap> I:=[3,5]+S;; gap> J:=[5,7]+S;; gap> SmallElements(I-J); [ -2, 1, 3 ] gap> SmallElements(Intersection(0+S,I-J)); [ 3, 5 ]

Theorem 4 Let S be a numerical semigroup, M be its maximal ideal and let E be a proper ideal of S. Assume that (E − S M) \ E = {x1 , . . . , x h }. Then

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E = (S \ D(x1 )) ∩ · · · ∩ (S \ D(x h )), and this is the unique irredundant decomposition of E into proper irreducible ideals. Proof We first prove that the elements in (E − S M) \ E = Maximals≤S (S \ E). In fact, x ∈ (E − S M) \ E if and only if x ∈ S \ E and x + y ∈ E for any y ∈ M, which is equivalent to saythat x is maximal. \ D(x), for Now we see that E = x∈S\E (S \ D(x)). In fact, E is contained in S any x ∈ / E (Lemma 17) and, conversely, if y ∈ / E, then y ∈ D(y), so y ∈ / x∈S\E (S \ D(x)). The proof now follows from Lemma 18 (2). We could reformulate the previous theorem taking complements in S; in this way we obtain S \ E = D(x1 ) ∪ · · · ∪ D(x h ) (with xi varying in the set (E − S M) \ E). GAP Example 22 Let S = 3, 5, 7 and I = 10 + S. Let us use Theorem 4 to see how I decomposes into irreducibles. gap> S:=NumericalSemigroup(3,5,7);; gap> I:=10+S;; gap> Difference(Intersection(0+S,I-M),I); [ 12, 14 ]

Hence I = (S \ D(12)) ∩ (S \ D(13)). gap> d:=x->DivisorsOfElementInNumericalSemigroup(x,S);; gap> MinimalGenerators(IdealByDivisorClosedSet(d(12),S)); [ 8, 10 ] gap> MinimalGenerators(IdealByDivisorClosedSet(d(14),S)); [ 10, 12 ]

That is, I = ({8, 10} + S) ∩ ({10, 12} + S). gap> Intersection(IdealByDivisorClosedSet(d(12),S), IdealByDivisorClosedSet(d(14),S)) = I; true

Let us also note that (E − S M) ⊆ (E − M), whence |(E − S M) \ E| ≤ t(E) and that, in general, the inequality can be strict. For example, M is always irreducible as proper ideal of S, but it is never Z-irreducible unless S = N, because, if S = N, M − M  S, so t(M) ≥ 2. Corollary 16 Let S be a numerical semigroup, and let s be a nonzero element of S. Assume that {x1 , . . . , xt } = Maximals≤S Ap(S, s). Then

3.3 Irreducibility

45

s + S = (S \ D(x1 )) ∩ · · · ∩ (S \ D(xt )) is the unique decomposition of s + S into irreducible proper ideals. In particular, the number of its irreducible components equals the type of S. Proof Set E = s + S. The first part of the statement is an immediate consequence of Theorem 4, since, by the definition of Apéry set, Ap(S, s) = S \ E, and in the proof of the above mentioned theorem we noticed that its maximal elements coincide with the elements of (E − S M) \ E. By Proposition 8, the cardinality of the maximal elements of Ap(S, s) with respect to ≤ S is precisely t(S). GAP Example 23 If we go back to the gap Example 22, the number of irreducible components is precisely the type of S. gap> Type(NumericalSemigroup(3,5,7)); 2

3.4 Reduction Number, Blowup and Multiplicity Sequence; Arf Semigroups In this section we will present, for semigroups, some classical results on the sequence of blowups; these results are strictly connected with the problem of understanding the equisingularity class of an algebroid branch. The original connection between this problem and semigroups goes back to Zariski and Arf. Zariski’s studies in equisingularity are famous papers; on the other hand, the work of Arf was used by Lipman who proposed, in his honour, the notion of Arf ring. To Arf is due the proof of the fact that any algebroid branch has the same multiplicity sequence of its Arf closure (as it is called now). The same concept can be translated at numerical semigroup level, and it is proved in [15] that an Arf analytically irreducible domain can be characterized by the Arf property of its value semigroup, provided that they share the same multiplicity sequence. The blowup of the maximal ideal (both for rings and semigroups) is also a key ingredient to investigate the Cohen–Macaulay, Gorenstein and complete intersection properties for the associated graded ring of a numerical semigroup ring or, more generally, of an analytically irreducible one-dimensional local domain (see [13, 14, 22, 28]). In this context, the notion of reduction number of an ideal plays an important role. So we start the section illustrating this concept. Let H ⊆ E be two ideals of a numerical semigroup S. We say that H is a reduction of E if, for some positive integer h, the equality (h + 1)E = H + h E holds. Proposition 26 Let E be an ideal of a numerical semigroup S. The principal ideal m(E) + S is a reduction of E. Proof Let us consider the following chain of relative ideals:

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S ⊆ E − m(E) ⊆ 2E − 2 m(E) ⊆ · · · ⊆ j E − j m(E) ⊆ · · · ⊆ N.

(3.5)

It eventually stabilizes, since |N \ S| is finite; hence there exists an integer h such that h E − h m(E) = (h + 1)E − (h + 1) m(E), or equivalently, (h + 1)E = h E + m(E) = h E + m(E) + S, proving in this way that m(E) + S is a reduction for E. GAP Example 24 Take S = 3, 5, 7 , and its ideal I = {2, 4} + S. Let us see with gap how the chain (3.5) stabilizes. gap> gap> gap> [ 0, gap> [ 0, gap> [ 0,

S:=NumericalSemigroup(3,5,7);; I:=[2,4]+S;; SmallElements(-2+I); 2, 3, 5 ] SmallElements(-2*2+2*I); 2 ] SmallElements(-2*3+3*I); 2 ]

It is easy to check that if H is a reduction of E, then m(H ) = m(E). Hence H is a reduction of E if and only if m(E) + S ⊆ H ⊆ E. We will denote by r(E) = min{h ≥ 1 | (h + 1)E = m(E) + h E}, and call it reduction number of E. An ideal is said to be stable if r(E) = 1, that is, 2E = m(E) + E or, equivalently, E − E = E − m(E). We also note that the stability for the maximal ideal M = S ∗ is equivalent to the maximal embedding dimension for S. It is well known, by ring theory, that r(M) ≤ m(S) − 1. We give here a nice numerical argument due to A. Moscariello, that we publish here with his permission. Proposition 27 Let S = N be a numerical semigroup. Then r(M) ≤ m(S) − 1. Proof Set m = m(S). We need to prove that m M = (m − 1)M + m. Pick s = s1 + · · · + sm ∈ m M, and consider the partial sums ti = s1 + · · · + si . If ti = hm for some i, then i ≤ h, since ti ≥ im. Hence s = hm + si+1 + · · · + sm ∈ (m − 1)M + m. If for every i, ti ≡ 0 (mod m), then we have m elements in m − 1 residue classes, whence there exist two indices i, j ∈ {1, . . . , m}, with i < j and t j − ti ≡ 0(modm), that is, si+1 + · · · + s j = hm and, as noted above, j − i ≤ h. Again we obtain that s = s1 + · · · + si + hm + s j+1 + · · · + sm ∈ (m − 1)M + m, that is the thesis. GAP Example 25 Let us go back to S = 3, 5, 7 , and its ideal I = {2, 4} + S. gap> S:=NumericalSemigroup(3,5,7);; gap> I:=[2,4]+S;; gap> First([1..3], j->(j+1)*I=2+j*I); 2

3.4 Reduction Number, Blowup and Multiplicity Sequence; Arf Semigroups

47

This means that the reduction number for I is in this case two. For a relative ideal E we can consider also another chain: S ⊆ E − E ⊆ 2E − 2E ⊆ · · · ⊆ j E − j E ⊆ · · · ⊆ N. In this case the elements of the chain are not only relative ideals of S, but also numerical semigroups themselves. Again, since |N \ S| is finite, the chain eventually stabilizes. It is easy to check that j E − j E ⊆ j E − j m(E), for every j. Moreover the following equivalences hold. Proposition 28 Let E be any relative ideal of a numerical semigroup S and let j be a positive integer. The following statements are equivalent. 1. j E − j E = j E − j m(E). 2. j E − j m(E) is a numerical semigroup. 3. j ≥ r(E). Proof We start proving the equivalence between the first two statements; for simplicity of notations we denote m(E) by m. Assume that j E − j E = j E − jm, and take x − jm, y − jm ∈ j E − jm (that is x, y ∈ j E). Then (x − jm) + y ∈ j E, since x − jm ∈ j E − jm = j E − j E and y ∈ j E. Hence (x − jm) + (y − jm) = (x − jm + y) − jm ∈ j E − jm, and it follows that j E − jm is a numerical semigroup. Now assume that j E − jm is a numerical semigroup. We know that j E − j E ⊆ j E − jm, so in order to prove the equality, we only need to show that j E − jm ⊆ j E − j E. Take x − jm ∈ j E − jm, and y ∈ j E; we must prove that x − jm + y ∈ j E, which is equivalent to x − jm + y − jm ∈ j E − jm, and this follows because j E − jm is a numerical semigroup. Observe that j ≥ r(E) if and only if m + j E = ( j + 1)E, and this is equivalent to jm + j E = 2 j E. The equality jm + j E = j E + j E holds if and only if j E − jm = j E − j E, and we are done. From now on, we will refer to r(E)E − r(E) m(E) = r(E)E − r(E)E as the blowup of E, that we will denote by B(E); the blowup of M is also called the blowup of S. This name comes from ring theory: for a regular ideal I of a one-dimensional  ring R the blowup of I (also known as blowing-up) is R[I x −1 ] = j∈N (I j : I j ), where x is a minimal reduction of I . Under proper hypotheses this union coincides with (I j : I j ), for j grater than or equal to the reduction number of I , and, as Rmodule, it is equal to I j x − j . It is clear from the previous results, that B(E) is a numerical semigroup such that S ⊆ B(E) ⊆ N. The first inclusion is always proper, provided that E is not principal, in fact by (3.5), E − m(E) ⊆ B(E), and E − m(E) = S means that E is principal. In particular, if S = N, then B(M) always contains S properly. On the other hand, if S = N, the blowing up of its maximal ideal produces N again. It is worth noticing that, if S = n 1 , . . . , n d , with m(S) = n 1 , then B(M) = n 1 , n 2 − n 1 , . . . , n d − n 1 . In fact, a generic element of B(M) is of the form s − hn 1 ,

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with s ∈ h M and some nonnegative integer h. Writing s as a linear combination of the  generators of S, that is s= i λi n i and setting L = i λi , we immediately get that L ≥ h and s − hn 1 = i λi (n i − n 1 ) + (L − h)n 1 . The other inclusion is trivial. However, in general, {n 1 , n 2 − n 1 , . . . , n d − n 1 } is not a minimal set of generators for B(M). Setting S 1 = B(M), we can consider the blowup of its maximal ideal that we will denote by S 2 . Assuming we have constructed the chain of successive blowups S = S 0 ⊆ S 1 ⊆ S 2 ⊆ · · · ⊆ S i , we will denote by S i+1 the blowup of the maximal ideal of S i . In this way we get a chain that stabilizes at N. If we consider the multiplicity m i = m(S i ) of each S i , we get a non increasing sequence (m 0 , m 1 , m 2 , . . . ), that eventually stabilizes at 1 and is called multiplicity sequence of S. In order to characterize those sequences that can be realized as multiplicity sequences of some numerical semigroup S, we need to introduce the notion of Arf semigroup. First we define an ideal E of S to be integrally closed if it is of the form S(s) = (s + N) ∩ S, for some s ∈ S; notice that, in particular, the maximal ideal is integrally closed. GAP Example 26 For a numerical semigroup S and a positive integer s in S, the set S ∩ [0, s − 1] is a subset of S closed under divisors. Thus, we can compute S(n) as S \ [0, s − 1] = S \ (S ∩ [0, s − 1]). gap> S:=NumericalSemigroup(3,5,7);; gap> S9:=IdealByDivisorClosedSet(Intersection(S,[0..9]),S);

gap> SmallElements(S9); [ 10 ] gap> MinimalGenerators(S9); [ 10, 11, 12 ]

A numerical semigroup S is said to be Arf if every integrally closed ideal of S is stable. The stability of S(s) means that 2S(s) = s + S(s), and so it is straightforward to check that S is Arf if and only if for any s, t, u ∈ S, with s ≤ t ≤ u, the integer t + u − s is in S (this is also equivalent to imposing that for all s, t ∈ S, with s ≤ t, the integer 2t − s is in S, see for instance [24, Proposition 1]). Using the last characterization, we immediately get that the intersection of two Arf numerical semigroups is Arf; therefore, if we fix any numerical semigroup S, since the number of numerical semigroups containing S is finite, there exists the smallest Arf numerical semigroup containing S (the intersection of all Arf numerical semigroups containing it). We will call this semigroup the Arf closure of S and we will denote it by Arf(S). Our next goal will be to prove that a numerical semigroup S and its Arf closure, Arf(S), share the same multiplicity sequence. To this aim, we first observe that the blowup of an Arf semigroup S is again Arf: in fact, since M is stable, B(M) = M − m(S); taking any s − m(S) ≤ t − m(S) ≤ u − m(S) ∈ M − m(S), we have (t − m(S)) + (u − m(S)) − (s − m(S)) = (t + u − s) − m(S) that belongs to M −

3.4 Reduction Number, Blowup and Multiplicity Sequence; Arf Semigroups

49

m(S), since S is Arf. From this fact, it follows that the multiplicity sequence of an Arf semigroup can be computed very easily from its elements. Proposition 29 Let S = {0 = s0 , m(S) = s1 , s2 , s3 , . . . } be an Arf numerical semigroup. Then the multiplicity sequence of S is (s1 , s2 − s1 , s3 − s2 , . . . ). Proof The multiplicity of S = S 0 is, by definition, s1 . The maximal ideal M = S(s1 ) of S is stable, hence S 1 = S(s1 ) − s1 and its multiplicity equals s2 − s1 . We noted above that S 1 is again an Arf semigroup. It follows that S 2 , that is the blowup of S 1 , coincides with S 1 (s2 − s1 ) − (s2 − s1 ) and its multiplicity is (s3 − s1 ) − (s2 − s1 ) = s3 − s2 . Moreover S 2 is again Arf, being the blowing up of an Arf semigroup; now it is clear that we can proceed recursively to get the thesis. By the previous proposition it is obvious that we can reconstruct the elements of an Arf semigroup by its multiplicity sequence as follows: S = {0, m 0 , m 0 + m 1 , m 0 + m 1 + m 2 , . . . }. GAP Example 27 Let us compute the multiplicity sequence and the Arf closure of 4, 6, 9 . gap> S:=NumericalSemigroup(4,6,9);; gap> SmallElements(S); [ 0, 4, 6, 8, 9, 10, 12 ] gap> MultiplicitySequence(S); [ 4, 2, 2, 1 ] gap> MinimalGenerators(BlowUp(S)); [ 2, 5 ] gap> MinimalGenerators(BlowUp(BlowUp(S))); [ 2, 3 ] gap> MinimalGenerators(BlowUp(BlowUp(BlowUp(S)))); [ 1 ] gap> A:=ArfClosure(S);; gap> SmallElements(A); [ 0, 4, 6, 8 ] gap> MultiplicitySequence(A); [ 4, 2, 2, 1 ]

The Arf over–semigroups of S are the following (we list their generating systems). gap> List(Filtered(OverSemigroups(S),IsArf),MinimalGenerators); [ [ 1 ], [ 2, 3 ], [ 2, 5 ], [ 2, 7 ], [ 2, 9 ], [ 3 .. 5 ], [ 4 .. 7 ], [ 4, 6, 7, 9 ], [ 4, 6, 9, 11 ] ]

The next step will be to prove that Arf closure and blowup commute. For this we need two lemmas.

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Lemma 19 Any numerical semigroup S has the same multiplicity of its Arf closure Arf(S). Proof Since S ⊆ Arf(S), m(Arf(S)) ≤ m(S). So it enough to show that there exists an Arf over–semigroup of S with the same multiplicity; this semigroup is S = {0} ∪ m(S) + N. Lemma 20 If T is an Arf semigroup and t ∈ T \ {0}, then {0} ∪ (t + T ) is an Arf semigroup, whose blowup is T . Proof It is straightforward to check that {0} ∪ (t + T ) is a numerical semigroup. It is also Arf, since, for any s + t ≤ u + t ≤ v + t ∈ {0} ∪ (t + T ), we have (u + t) + (v + t) − (s + t) = (u + v − s) + t, that belongs to {0} ∪ (t + T ), because T is Arf. Since the maximal ideal of {0} ∪ (t + T ) is exactly t + T , its blowup is (t + T ) − t = T . Proposition 30 Blowup and Arf closure commute. More precisely, if S is a numerical semigroup, denoting by Arf(S) its Arf closure and by Arf(M) the maximal ideal of Arf(S), then the Arf closure of the blowup of S is Arf(B(M)) = B(Arf(M)). Proof Using the notation of the statement, we have that M ⊆ Arf(M) and they share the same smallest element, which is the common multiplicity of S and Arf(S) (see Lemma 19); therefore B(M) ⊆ B(Arf(M)) = Arf(M) − m(S), by the stability of Arf(M). The semigroup B(Arf(M)) is Arf, as it is the blowup of an Arf semigroup, thus Arf(B(M)) ⊆ B(Arf(M)). Conversely, let T be any Arf semigroup containing B(M). By the previous lemma {0} ∪ (m(S) + T ) is an Arf semigroup containing S: in fact, for any s ∈ M, s − m(S) ∈ B(M) ⊆ T . Thus {0} ∪ (m(S) + T ) ⊇ Arf(S), and they share the same multiplicity. By taking blowups we obtain T ⊇ B(Arf(M)); the thesis now follows immediately. Corollary 17 Let S be a numerical semigroup and let Arf(S) be its Arf closure. Then they have the same multiplicity sequence. Proof By Proposition 30 the blowup of the Arf closure of S is exactly the Arf closure of the blowup of of S; proceeding recursively we get that the Arf closures of the semigroups S i , in the chain of successive blowups of S, are exactly the semigroups appearing in the chain of successive blowups of Arf(S). The proof now follows from Lemma 19. We can use Proposition 29 in order to characterize all the multiplicity sequences of Arf semigroups; this means, in light of the previous corollary, that we have the characterization of the multiplicity sequences of any numerical semigroup. Theorem 5 A non increasing sequence of positive integers (m 0 , m 1 , m 2 , . . . ), that eventually stabilizes to 1, is the multiplicity sequence of a numerical semigroup if and only if, for every i ≥ 0, there exists a positive integer h i such that

3.4 Reduction Number, Blowup and Multiplicity Sequence; Arf Semigroups

mi =

hi 

51

m i+ j .

j=1

Proof As previously observed it is enough to characterize those sequences that are achieved as multiplicity sequences of an Arf semigroup. Let S be an Arf semigroup. By construction, any S i in the chain of successive blowups of S is an Arf semigroup with multiplicity m i ; so, denoting by M i the maximal ideal of S i , S i+1 = M i − m i . From m i ∈ S i+1 , we get that m i − m i+1 ∈ S i+2 , so m i − m i+1 − m i+2 ∈ S i+3 and so on, until we obtain 0, obtaining the desired condition on the m i ’s. Conversely, pick a non increasing sequence (m 0 , m 1 , m 2 , . . . ) as in the statement; we need to prove that the set S = {0, m 0 , m 0 + m 1 , m 0 + m 1 + m 2 , . . . } is an Arf semigroup. To this aim, set n = min{i | m i = 1} and Tn = N. Then we define Th−1 = {0} ∪ (m h−1 + Th ) = {0, m h−1 , m h−1 + m h , m h−1 + m h + m h+1 , . . . }, for every h ∈ {1, → n − 1}. The condition on the m i ’s implies that m h−1 ∈ Th ; therefore we can apply, at each step, Lemma 20 to deduce that Th−1 is an Arf numerical semigroup. Since T0 = S we have the thesis. GAP Example 28 Proposition 29 together with Theorem 5 can be used to calculate all Arf numerical semigroups with given genus or Frobenius number (see [46]). gap> ArfNumericalSemigroupsWithFrobeniusNumber(7); [ , , , , , , ] gap> List(last,MinimalGenerators); [ [ 8 .. 15 ], [ 6, 8, 9, 10, 11, 13 ], [ 5, 8, 9, 11, 12 ], [ 4, 9, 10, 11 ], [ 4, 6, 9, 11 ], [ 3, 8, 10 ], [ 2, 9 ] ] gap> List(last,MinimalGenerators); [ [ 8 .. 15 ], [ 7, 9, 10, 11, 12, 13, 15 ], [ 6, 9, 10, 11, 13, 14 ], [ 6, 8, 10, 11, 13, 15 ], [ 5, 9, 11, 12, 13 ], [ 5, 8, 11, 12, 14 ], [ 4, 10, 11, 13 ], [ 4, 6, 13, 15 ], [ 3, 11, 13 ], [ 2, 15 ] ] gap> ArfNumericalSemigroupsWithGenusAndFrobeniusNumber(5,7); [ , ] gap> List(last,MinimalGenerators); [ [ 4, 6, 9, 11 ], [ 3, 8, 10 ] ]

Chapter 4

Semigroup of an Irreducible Meromorphic Series

Let K be an algebraically closed field of characteristic zero and let f (x, y) = y n + a1 (x)y n−1 + · · · + an (x) be a nonzero polynomial of K((x))[y] where K((x)) denotes the field of meromorphic series in x. The aim of this chapter is to associate with f , when f is irreducible, a subsemigroup of Z. The construction of this subsemigroup is based on the notion of Newton–Puiseux exponents. These exponents appear when we solve f as a polynomial in y, and it turns out that the roots are elements of K((x 1/n )). Two cases are of interest: the local case, that is, the case when f ∈ K[[x]][y], and the case when f ∈ K[x −1 ][y] with the condition that F(x, y) = f (x −1 , y) has one place at infinity. In the first case, the subsemigroup associated with f is a numerical semigroup. In the second case, this subsemigroup is a subset of −N, and some of its numerical properties have some interesting applications in the study of the embedding of special curves with one place at infinity in the affine plane. In the preceding chapters, we already gave results to better understand these semigroups. We will also need some results related to minimal polynomials and algebraic extensions. The reader can find them for instance in [71].

4.1 Some Notation Let K be a field and x a variable. We denote by K[[x]] the setof formal series on x with coefficients in K, that is, the set of elements of the form i∈N ai x i with ai ∈ K for all i ∈ N.  Let f (x) = i∈N ai x i ∈ K[[x]]. Define the order of f as the smallest i ∈ N such that ai = 0, and we will denote it by ord x ( f (x)). We will write ord x (0) = +∞. We will use K((x)) to refer to the quotient field of K[[x]], that is, the set of fractions f (x) with f (x), g(x) ∈ K[[x]] and g(x) nonzero. g(x) A series f (x) in K[[x]] is a unit if f (0) = 0, or in other words, the independent term of f is nonzero. So every element q(x) = f (x)/g(x) ∈ K((x)) can be (x) where g(x) = x m h(x) and h(0) = 0 (that is, m = ord x (g(x))). expressed as x mf h(x) © Springer Nature Switzerland AG 2020 A. Assi et al., Numerical Semigroups and Applications, RSME Springer Series 3, https://doi.org/10.1007/978-3-030-54943-5_4

53

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Let k(x) be theinverse of h(x) in K[[x]]. Then, q(x) = x −m f (x)k(x). In other words, q(x) = i∈Z bi x i with bi ∈ K and there is a minimum n ∈ Z such that b j = 0 for all j < n and bn = 0. We define ord x (q(x)) = n. One can check that ord x (q(x)) = ord x ( f (x)) − ord x (g(x)), and so this extends in a natural way our definition of order for formal series.  We are theni going to identify the elements in K((x)) with expressions of the form i∈−m+N ai x with ai ∈ K for all i and m ∈ N. These are known in the literature as meromorphic series. Let K and L be two fields. Recall that L is an algebraic extension of K if every element α in L is a root of a polynomial with coefficients in K. We can always choose this polynomial to be monic and irreducible, which is known as the minimal polynomial of α. We will use this and other basic facts of Galois theory. The reader may find these results in [71]. Also we will use Bézout’s theorem, that states that in an algebraically closed field with zero characteristic, the number of intersection points (counting multiplicities) of two curves is the product of the degrees of the curves (of course, if they do not have common branches). This and many other results related to algebraic curves can be found in [27].

4.2 Characteristic Sequences We start by giving the basic steps needed to construct the semigroup from the polynomial f . Some are well known, but we are including them here for completeness. The first of these basic results is Hensel’s lemma, which allows us to lift a factorization of the independent term of f ∈ K[[x]][y], f (0, y), to a factorization of f itself. We give the version of Hensel’s lemma that suits our needs. However, there are more scopes where Hensel’s lifting applies. Theorem 6 (Hensel’s lemma) Let f = y n + a1 (x)y n−1 + · · · + an (x) ∈ K[[x]][y]. Assume that there exist two nonconstant polynomials g, ˜ h˜ ∈ K[y] such that (i) g, ˜ h˜ are monic in y, ˜ = 1, (ii) gcd(g, ˜ h) ˜ (iii) f (0, y) = g˜ h. Then, there exist g, h ∈ K[[x]][y] such that (i) (ii) (iii) (iv)

g, h are monic in y, ˜ g(0, y) = g, ˜ h(0, y) = h, ˜ deg y g = deg g, ˜ deg y h = deg h, f = gh.

˜ and write f (x, y) as Proof Let r (respectively s) be the degree of g˜ (respectively h) a formal series in x with coefficients in K[y], say f (x, y) = q≥0 f q (y)x q . Clearly, f (0, y) = f 0 is monic of degree n in y. Also, deg y f q < n for all q ≥ 1. The idea is

4.2 Characteristic Sequences

55

to build the expression of g and h inductively as formal series on x. For all i ≥ 0, we construct gi , h i ∈ K[y] such that ˜ 1. g0 = g, ˜ h 0 = h, 2. for all i ≥ 1, deg y gi < r and deg y h i < s, q 3. for all q ≥ 1, f q = i=0 gi h q−i . ˜ If i = 0, then we set g0 = g, ˜ h 0 = h. q−1 Suppose that we have g0 , . . . , gq−1 , h 0 , . . . , h q−1 . Let eq = f q − i=1 gi h q−i . Note that deg y eq < n. We need two monic polynomials gq , h q such that eq = h 0 gq + g0 h q , deg y gq < r and deg y h q < s. In order to obtain them, we use Euclid’s extended algorithm for polynomials with coefficients in a field. By hypothesis, gcd(g0 , h 0 ) = 1. Let α, β ∈ K[y] be such that αg0 + βh 0 = 1. If we multiply by eq , we have eq = (eq α)g0 + (eq β)h 0 . Let G q = eq β, Hq = eq α and write G q = Qg0 + R with deg y R < r , using division algorithm. Hence eq = (eq α)g0 + (Qg0 + R)h 0 = (eq α + Qh 0 )g0 + Rh 0 . Finally, define gq = R and h q = eq α + Qh 0 . Since deg y gq < r , it follows that deg y h q < s. This proves that gq and h q fulfill the desired conditions. As an application of Hensel’s lemma, we get the following “parametrization” result. Proposition 31 Let f (x, y) = y n + a1 (x)y n−1 + · · · + an (x) ∈ K((x))[y]. There exist m ∈ N and y(t) ∈ K((t)) such that f (t m , y(t)) = 0. Proof We will start by assuming some conditions that at first sight might seem restrictive. However, they are crucial if we want to apply Hensel’s lemma. The first one has to do with a1 (x). If a1 (x) = 0, let z = y + a1n(x) and F(x, z) =   f x, z − a1n(x) . If we can find m ∈ N and z(t) ∈ K((t)) such that F (t m , z(t)) = 0,   then we have f t m , z(t) − a1n(t) = 0. Hence we will assume that a1 (x) = 0. We use induction on the degree in y of f . If n = 1, then f = y − a1 (x) = y. Hence f (t, 0) = 0. Suppose that n ≥ 2. We shall assume the following condition. (1) ak (x) ∈ K[[x]] for all k ∈ {2, . . . , n} and ak (0) = 0 for some k ∈ {2, . . . , n}. This condition is needed for two reasons. The first one is that we have shown Hensel’s lemma in the context of K[[x]][y], and not in K((x))[y]. The second is that from this condition it follows that f (0, y) = y n + a2 (0)y n−2 + · · · + an (0) is not of the form (y − c)n in K[y]. Hence there exist nonconstant monic polynomi˜ ˜ ˜ als g(y), ˜ h(y) ∈ K[y] such that gcd(g(y), ˜ h(y)) = 1 and f (0, y) = g(y) ˜ h(y). By Hensel’s lemma (Theorem 6), there exist monic polynomials g, h ∈ K[[x]][y] such that deg y g, deg y h < n and f = gh. By induction hypothesis there exist m ∈ N and y(t) ∈ K((t)) such that g(t m , y(t)) = 0. In particular, f (t m , y(t)) = 0. So, to complete the proof it suffices to study what happens when condition (1) does not hold.

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4 Semigroup of an Irreducible Meromorphic Series

 Let f = y n + nk=2 ak (x)y n−k with ak (x) = 0 for some k ∈ {2, . . . , n} (if f (x, y) = y n , then f (t, 0) = 0, and so it is enough to take m = 1 and y(t)  = 0). For  all k ∈ {2, . . . , n} such that ak = 0, let u k = ord x (ak ). Set u = min ukk | ak = 0 . There exists an index r such that u = urr . Let x = wr and z = w −ur y, and let g(w, z) = w −nur f (wr , y). We have ⎛ g(w, z) = w −nu r ⎝wnu r z n +

n

⎞ ak (wr )wu r (n−k) z n−k ⎠ = z n +

k=2

n

ak (wr )w−ku r z n−k .

k=2

Let bk (w) = ak (wr )w −kur . As ordw bk = r u k − ku r ≥ 0, we obtain bk ∈ K[[w]]. Furthermore, ordw br (w) = 0, that is, br (0) = 0. Finally, if m ∈ N and w(t) ∈ K((t)) are such that g(t m , w(t)) = 0, then it can be easily checked that f (t mr , t mur z(t)) = 0. We want to improve Proposition 31 when f is irreducible. In particular, we will prove that we can choose m = n in that proposition. To this end we will make use of minimal polynomials and algebraic extensions. In this way we will have a decomposition of f in an extension field that can be handled easier. Lemma 21 Let m ∈ N∗ . The extension K((t m )) → K((t)) is an algebraic extension of degree m. Proof The field K((t)) is a K((t m ))-vector space with basis {1, t, . . . , t m−1 }. The proof now follows from [71, Theorem 46]. Let y(t) ∈ K((t)) and let F(t m , y) ∈ K((t m ))[y] be the minimal polynomial of y(t) over K((t m )) (with the notation in [71], this is the irreducible polynomial of y(t) over K((t m ))). By abuse of notation we write F(x, y) ∈ K((x))[y] for F(t m , y). Then F(x, y) is a monic irreducible polynomial of K((x))[y], F(t m , y(t)) = 0, for all g(x, y) ∈ K((x))[y], if g(t m , y(t)) = 0, then F(x, y) divides g(x, y), deg y F(x, y) = [K((t m ))(y(t)) : K((t m ))], deg y F(x, y) divides m.  Write y(t) = p c p t p . Define the support of y(t) to be (M1) (M2) (M3) (M4) (M5)

  Supp(y(t)) = p | c p = 0 . The following result allows us to describe F(t m , y) in the case m is coprime with the greatest common divisor of the support of y(t). We will apply this later to our polynomial f , with the parametrization (t m , y(t)) described in Proposition 31. Proposition 32 Let the notations be as above. If gcd(m, gcd(Supp(y(t)))) = 1, then the following holds: (i) F(t m , y) = w,wm =1 (y − y(wt)), and if w1 = w2 , w1m = w2m = 1, then y(w1 t) = y(w2 t),

4.2 Characteristic Sequences

57

(ii) deg y F(x, y) = m. Proof If w m = 1, then F(t m , y(wt)) = F((wt)m , y(wt)) = 0. Hence y(wt) is a root of F(t m , y), which implies that (y − y(wt)) | F(t m , y). Now let us see that all these roots are  different. Let w1 = w2 be such that w1m = p p m w2 = 1. We have y(w1 t) − y(w2 t) = p (w1 − w2 )c p t p . If y(w1 t) = y(w2 t), p p then w1 = w2 for all p ∈ Supp(y(t)). But w1m = w2m , and gcd(m, gcd(Supp(y(t)))) = 1, which yields w1 = w2 ; a contradiction. Hence we have m different roots and by (M5), deg y F(x, y) | m, which implies F(t m , y) = w,wm =1 (y − y(wt)) and deg y F(x, y) = m. As announced above, with these results, we can sharpen Proposition 31 for the case f is irreducible (as a polynomial in y). Proposition 33 Let f (x, y) = y n + a1 (x)y n−1 + · · · + an (x) ∈ K((x))[y]. Suppose that f (x, y) is irreducible. There exists y(t) ∈ K((t)) such that f (t n , y(t)) = 0. Furthermore, (i) f (t n , y) = w,wn =1 (y − y(wt)), (ii) if w1 = w2 , w1n = w2n = 1, then y(w1 t) = y(w2 t), (iii) gcd(n, gcd(Supp(y(t)))) = 1. If particular, this implies that f (t n , y) is the minimal polynomial of y(t) over K((t n )). Proof We know by Proposition 31 that there exist m ∈ N and y(t) ∈ K((t)) such that f (t m , y(t)) = 0. Let m be the smallest integer with this property, and let d = gcd(m, If d > 1, then y(t) = z(t d )for some z(t) ∈ K((t)). 

m/d

dgcd(Supp(y(t)))). m/d d Hence f (t ) , z(t ) = 0. It follows that, f t , z(t) = 0, contradicting the minimality of m. This shows that d = 1. The polynomial f is monic and irreducible. Thus f is consequently the minimal polynomial of y(t) over K((t m )). In particular m = n. Now we can use Proposition 32 to complete the proof. n n Notice that if f (t n , y) = i=1 (y − yi (t)), then f (x, y) = i=1 (y − yi (x 1/n )), that as a consequence of Proposition 33, the which now lives in K((x 1/n )). Observe  algebraic closure of K((x)) equals n∈N K((x 1/n )). We have now the ingredients to introduce the Newton–Puiseux exponents of an irreducible polynomial. Related to them we define a series of sequences, which will characterize the semigroup associated to the corresponding polynomial.  Suppose that f is irreducible and let y(t) = p c p t p as above. • Let d1 = n = deg y f and let m 1 = min{ p ∈ Supp(y(t)) | d1  p}, d2 = gcd(d1 , m 1 ). Then for all i ≥ 2, if di = 1, let m i = min{ p ∈ Supp(y(t)) | di  p} and di+1 = gcd(di , m i ). By Proposition 33, m i is well-defined and there exists h ≥ 1 such that dh+1 = 1. We set m = (m 1 , . . . , m h ) and d = (d1 , . . . , dh+1 ). i for all i ∈ {1, . . . , h}. • We also set ei = ddi+1

58

4 Semigroup of an Irreducible Meromorphic Series

• We finally define the sequence r = (r0 , . . . , rh ) as follows: r0 = n, r1 = m 1 and for all i ∈ {2, . . . , h}, ri = ri−1 ei−1 + m i − m i−1 . The sequence m is called the set of Newton–Puiseux exponents of f . The sequences m, d, r are called the characteristic sequences associated with f . An easy verification shows that dk = gcd(r0 , . . . , rk−1 ) for all k ∈ {1, . . . , h + 1}. The sequence r will be the set of generators of our semigroup. First, we will prove this fact, and then we will introduce new polynomials to compute them (approximate roots). The following result together with Lemma 26 ensure that the minimum multiple of rk that it is in the monoid generated by {r0 , . . . , rk−1 } is ek rk . Indeed next lemma shows a bit more than we need, and the underlying concept is that of being free in the sense we saw in Sect. 2.3. Lemma 22 Let k ∈ {1, . . . , h} and i ∈ {1, . . . , ek − 1}. Then irk is not in the group generated by {r0 , . . . , rk−1 }.  Proof Assume to the contrary that we can write irk = k−1 j=0 θ j r j , for some integers θ0 , . . . , θk−1 . Since gcd(r0 ,. . . , rk−1) = dk , we get that dk divides irk . Hence ek = dk k k = 1 and i < ek ; a contradiction. divides i drk+1 . But gcd ek , drk+1 dk+1 We can relate the order of the difference of roots of f to the Newton–Puiseux exponents. We will see later that this result is really helpful. Lemma 23 Let the notations be as above. In particular, we are assuming that f is irreducible and y(t) ∈ K((t)) is a root of f (t n , y) = 0. (i) ordt (y(t) − y(wt)) ∈ {m 1 , . . . , m h }. (ii) The cardinality of {y(wt) | ordt (y(t) − y(wt)) > m k } is dk+1 . (iii) The cardinality of {y(wt) | ordt (y(t) − y(wt)) = m k } is dk − dk+1 .  Proof (i) From of y(t) = p∈Z c p t p , we get the equality y(t) −  the expression y(wt) = p (1 − w p )c p t p . Let M = ordt (y(t) − y(wt)). It follows that for every nonnegative integer p smaller than M, we have w p = 1. Hence, if d = gcd({n} ∪ {a ∈ Supp(y(t)) | a < M}), then wd = 1. From the construction of d, there exists some k ∈ {1, . . . , h} such that d = dk . This implies that M = m k−1 . (ii) If the inequality ordt (y(t) − y(wt)) > m k holds, then by (i), ordt (y(t) − y(wt)) = m l with l ∈ {k + 1, . . . , h}. From the expression of ordt (y(t) − y(wt)) we deduce that this means that w p = 1 holds for all p ∈ {n} ∪ {a ∈ Supp(y(t)) | a ≤ m l }. From the definition of dk+1 , we obtain w dk+1 = 1. Now assume that w dk+1 = 1. Then w p = 1 for all p ∈ {n} ∪ {a ∈ Supp(y(t)) | a ≤ m k }. Hence ordt (y(t) − y(wt)) > m k .

4.2 Characteristic Sequences

59

(iii) Observe that ordt (y(t) − y(wt)) = m k if and only if ordt (y(t) − y(wt)) > m k−1 and ordt (y(t) − y(wt)) ≤ m k . Thus, the result follows from (ii). We now introduce the concept of pseudo-approximate roots that theoretically will enable us to construct recursively the Newton–Puiseux exponents and the characteristic sequences of f . We say theoretically, since from the definition it will be hard to calculate these pseudo-approximate roots. We will see later how to overcome this problem.  Let the notations be as above and let k ∈ {1, . . . , h}. Let y¯ (t) = p 1 and let i ∈ {1, . . . , s}. Since deg y Hi < dnk , by Corollary 19, there exist λi0 ∈ Z, λi1 , . . . , λik−1such that int( f, Hi ) = λi0 r0 + s s i λi0 )r0 + · · · + ( i=1 λik−1 )rk−1 . This contradicts · · · + λk−1rk−1 . Hence rk = ( i=1 Lemma 22. And now we determine int(G i , g) in terms of int( f, g). Proposition 40 Let the notations be as above. Let g be a nonzero polynomial of K((x))[y] and let G 1 , . . . , G h be a set of d1 , . . . , dhth pseudo-approximate roots of n for some k ∈ {0, . . . , h − 1}, then int( f, g) = dk+1 int(G k+1 , g). f . If deg y g < dk+1 Proof From the proof of Corollary 19, we know that the expansion of g with respect to the sequence (G 1 , . . . , G h , f ) is of the form g = θ cθ (x)G θ11 · · · G θkk . By Propoθ0

θ0

sition 38, there is a unique monomial cθ 0 (x)G 11 · · · G kk such that   

θ0  θ0 int( f, g) = int f, cθ 0 (x)G 11 · · · G kk = min int f, cθ (x)G θ11 · · · G θkk | cθ = 0 . Now clearly, the expansion of g above is also that of g with respect to (G 1 , . . . , G k+1 ). Furthermore, if cθ (x) = 0 and if θ0 = ord x cθ (x), then int(G k+1 , cθ (x)G θ11 . . . G θkk ) = k θk θ1 ri 1 i=0 θi dk+1 = dk+1 int( f, cθ (x)G 1 · · · G k ). This implies the result. As we commented above, this will enable us to show the following result, which highlights the recursiveness already present in Proposition 34. Proposition 41 Let (G 1 , . . . , G h ) be a set of pseudo-approximate roots of f . For all k ∈ {1, . . . , h − 1}, (G 1 , . . . , G k ) is a set of pseudo-approximate roots of G k+1 . Proof Fix k ∈ {1, . . . , h − 1} and let i ∈ {1, . . . , k}. By Proposition 40, int(G k+1 , G i ) =

1 dk+1

int( f, G i ) =

ri dk+1

.

Furthermore, G i is irreducible by Proposition 39, and we are done by definition. Let d be a divisor of n, and let G ∈ K((x))[y] be a monic polynomial of degree in y. Then the G-adic expansion of f has the form f = G d + α1 G d−1 + · · · + αd ,

n d

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4 Semigroup of an Irreducible Meromorphic Series

with αk ∈ K((x))[y] and deg y αi (x, y) < dn for all i ∈ {1, . . . , dk }. We define the Tschirnhausen transform of G as T(G) = G + αd1 . We already know from Lemma 22 (we have used it extensively) that irk is not in the group spanned by {r0 , . . . , rk−1 } for i ∈ {1, . . . , ek − 1}; whence not either in the submonoid generated by them. The following result states that precisely ek rk is the least multiple of rk in r0 Z + r1 , . . . , rk−1  (and consequently in the group spanned by this monoid). Lemma 26 Let the notations be as above. For all k ∈ {1, . . . , h}, there exist λk0 ∈ Z, k−1 k λi ri . λk1 , . . . , λkk−1 ∈ N such that ek rk = i=0 Proof Let G h be a dhth pseudo-approximate root of f . Let f = G dhh + α1 G dhh −1 + · · · + αdh be the G h -adic expansion of f .   For every k ∈ {0, . . . , dh }, we have int f, αk G dhh −k = int( f, αk ) + (dh − k)rh (where α0 = 1). As for all k ∈ {1, . . . , dh }, deg y αk < dnh , by Corollary 19, there h−1 k k ∈ N, αik < ei , such that int( f, αk ) = i=0 αi ri , whence exist α0k ∈ Z, α1k , . . . , αh−1  h−1 k int( f, αk G dhh −k ) = i=0 αi ri +(dh − k)rk .Since eh = dh , as a consequence of d −j whenever i, j ∈ Lemma 25, we derive that int f, αi G dhh −i = int f, α j G hh {1, . . . , d } and i  = j. For the same reason, we also derive int  h   dh dh −k f, G h = int f, αk G h for k ∈ {1, . . . , dh − 1}.   Let E = int( f, αk G dhh −k ) | k ∈ {0, . . . , dh } and let k0 ∈ {0, . . . , dh } be such that int( f, αk0 G dhh −k0 ) = min(E). If k0 is unique this property, then +∞ =  with h−1 k0 dh −k0 n = i=0 ai ri + (dh − k0 )rh int( f, f ) = ordt f (t , y(t)) = int f, αk0 G h (Lemma 22), which is a contradiction. Hence there is     at least one k1 = k0 such that int dh −k0 dh −k1 f, αk0 (x, y)G h = int f, αk1 (x, y)G h . According to the preceding paragraph, this is possible only for {k0 , k1 } = {0, dh }, in particular int ( f, G dhh ) = int( f, αdh ). Hence eh rh = dh rh ∈ r0 Z + r1 , . . . , rh−1 . This proves the result for k = h. sequences associated with G h are given by     Recall that theset of characteristic m h−1 dh−1 rh−1 m1 d1 r0 , (Proposition 34). Also, by , . . . , , . . . , , 1 and , . . . , dh dh dh dh dh dh Proposition 41, (G 1 , . . . , G h−1 ) is a set of pseudo-approximate roots of G h . From Proposition 40 we know that int(G h , G i ) = ri /dh . Hence, by repeating the above argument, we obtain rh−1 e ∈ dr0h Z + r1 /dh , . . . , rh−2 /dh . Hence eh−1 rh−1 ∈ r0 Z dh h−1 + r1 , . . . , rh−2 . We continue decreasing h until we reach h = 1: e1 r1 = (r0 /d2 )r1 = (r1 /d2 )r0 . The following result will help us to prove that being a pseudo-approximate root is closed under the Tschirnhausen transform. Lemma 27 Let (G 1 , . . . , G h ) be a set of pseudo-approximate roots of f . With the above notations, int( f, T(G k )) = rk for all k ∈ {1, . . . , h}.

4.2 Characteristic Sequences

67

Proof Let k = h and let f = G dhh + α1 G dhh −1 + · · · + αdh be the G h -adic expansion of f . Let us recover the set   E = int( f, αk G dhh −k ) | k ∈ {0, . . . , dh } = {int( f, αk ) + (dh − k)rh ) | k ∈ {0, . . . , dh }} in Lemma 26. As a consequence of the discussion in this lemma, we deduced that  min(E) = int f, G dhh = int( f, αdh ). But due to the fact that these two elements

 (int f, G dhh and int( f, αdh )) are the only ones attaining min(E), we also derive that int( f, αi ) > irh for all i ∈ {1, . . . , dh − 1} such that αi = 0. It follows that int( f, α1 ) > rh , and consequently int( f, T(G h )) = int ( f, G h + α1 = int( f, G h ) = rh . dh Let k < h and let d

d

k+1 k+1 + α1 G k+1 f = G k+1

−1

+ · · · + αdk+1

be the G k+1 -adic expansion of f . Let also G k+1 = G ekk + β1 G kek −1 + · · · + βek be the G k -adic expansion of G k+1 . Easy calculations show that α1 (x, y) = dk+1 β1 (x, y). By repeating the argument above for (G k+1 , G k ) instead of ( f, G h ), one k . Hence, by Proposition 40, can prove that int(G k+1 , β1 ) > int(G k+1 , G k ) = drk+1 int( f, α1 ) = int( f, β1 ) = dk+1 int(G k+1 , β1 ) > rk .  In particular, int( f, T(G k )) = int f, G k +

α1 dk



= int( f, G k ) = rk .

With this, it is easy to prove that the Tschirnhausen transform of pseudoapproximate roots of f yields pseudo-approximate roots of f . This will allow us to define approximate roots, which in contrast to pseudo-approximate roots, are unique. Corollary 20 Let k ∈ {1, . . . , h} and let G k be a dkth pseudo-approximate root of f . Then T(G k ) is a dkth pseudo-approximate root of f . Proof Clearly, T(G k ) is a monic polynomial of degree dnk in y. By Lemma 27, int( f, T (G k )) = rk , and by Proposition 39, T (G k ) is irreducible. This proves the result.

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4 Semigroup of an Irreducible Meromorphic Series

Let d be a divisor of n and let G be a monic polynomial of K((x))[y] of degree in y. Let f = G d + α1 G d−1 + · · · + αd

n d

be the G-adic expansion of f . We say that G is a d th approximate root of f if α1 = 0. This new restriction forces the uniqueness of the approximate roots. Lemma 28 Let the notations be as above. A d th approximate root of f exists and it is unique. n

Proof Let G = y d and let f = G d + α1 G d−1 + · · · + αd be the G-adic expansion of f . If α1 = 0, then G is a d th approximate root of f . Otherwise, we set G 1 = + · · · + αd1 be the G 1 -adic expansion of f . T(G) = G + αd1 . Let f = G d1 + α11 G d−1 1 1 Easy calculations show that deg y α1 < deg y α1 . Again, if α11 = 0, we restart with f and G 2 = T (G 1 ). Since the degrees of the α1k are in strictly decreasing order, + · · · + αdk is the G k -adic expansion there exists k such that if f = G dk + α1k G d−1 k k th of f , then α1 (x, y) = 0. Hence G k is a d approximate root of f . This proves the existence. For the uniqueness, let G and H be two d th approximate roots. Consider f = d G + α2 G d−2 + · · · + αd and f = H d + β2 H d−2 + · · · + βd , the G-adic and H adic expansion of f , respectively. We have

 G d − H d = (G − H ) G d−1 + H G d−2 + · · · + H d−1 

= β2 H d−2 + · · · + βd − α2 G d−2 + · · · + αd .

 If G = H , then deg y (G − H ) ≥ 0, but deg y G d−1 + H G d−2 + · · · + H d−1 =

 (d − 1) dn > deg y β2 H d−2 + · · · + βd − (α2 G d−2 + · · · + αd ) ; and this is impossible. It results from Lemma 28 that, given a divisor d of n, a d th approximate root exists and it is unique. We denote it by App( f ; d). Example 27 Let f = y 6 − 2x 2 y 3 + x 4 − x 5 y. We proceed as in Lemma 28 to compute App( f ; 2). We start with G = y 6/2 = y 3 . The G-adic expansion of f is G 2 + (−2x 2 )G + (x 4 − x 5 y). Hence α1 = −2x 2 = 0. So, we need to compute G 1 = T(G) = y 3 − x 2 . The G 1 -adic expansion of f is f = G 21 + (−x 5 y), and now α11 = 0. This means that g = G 1 = y 3 − x 2 = App( f ; 2). The picture below is in R2 : the dashed line corresponds to g.

4.2 Characteristic Sequences

69 y

x

Thus we have an algorithmic method to calculate approximate roots of f . As a consequence we will be able to compute rk for k ∈ {1, . . . , h}, and thus the semigroup associated to f . Proposition 42 Let the notations be as above. For all k ∈ {1, . . . , h}, int( f, App( f ; dk )) = rk . Proof Let k ∈ {1, . . . , h} and let G k be a dkth pseudo-approximate root of f . Uniqueness in Lemma 28 allows us to repeat the first part of the proof starting with G k instead of y n/dk . By Proposition 40, int( f, G k ) = int( f, T(G k )). But App( f, dk ) can be obtained by applying the operation T finitely many times to G k . Hence the result is a consequence of Proposition 41 and Corollary 20. Corollary 21 For all k ∈ {1, . . . , h}, App( f, dk ) is irreducible. In particular, App ( f, dk ) is a dkth pseudo-approximate root of f . Proof This follows by Propositions 39 and 42. The arithmetic properties of the sequences r , m, and d, combined with the results above give us a recursive method for the calculation of r (hence the m-sequence) of the irreducible polynomial f without calculating a root of f (x, y) = 0. We illustrate this with an example.

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4 Semigroup of an Irreducible Meromorphic Series

Example 28 Let f = y 6 − 2x 2 y 3 + x 4 − x 5 y, as in Example 27. We have r0 = 6 = d1 . Hence App( f ; d1 ) = y. Proposition 42 asserts that r1 = int( f, y) = int(y, x 4 ) = 4, and thus d2 = 2. We know from Example 27 that App( f ; d2 ) = y 3 − x 2 . By Proposition 42 again, r2 = int( f, y 3 − x 2 ) = int(x 5 y, y 3 − x 2 ) = ordt ((t 3 )5 t 2 ) = 17. Since d3 = 1, we get r = (6, 4, 17) and m = (4, 9). Notice also that now for a given irreducible polynomial f , we know how to calculate its approximate roots (G 1 , . . . , G h ) via the Tschirnhausen transform, and then for every g we are able to compute int( f, g) through the expansion of g with respect to (G 1 , . . . , G h , f ).

4.3 Contact In this section, we introduce the notion of contact between two irreducible polynomials of K((x))[y]. The amount measures us how far the parameterizations of these two polynomials are. We will use this concept to give an irreducibility criterion, and in the next section to study the decomposition of the partial derivatives of a polynomial. Recall that f is a monic irreducible polynomial in K((x))[y] of degree n in y. Let g be another monic irreducible polynomial of K((x))[y], of degree p in y, and let z 1 (t), . . . , z p (t) be the set of roots of g(t p , y) = 0. We define the contact of f with g, denoted c( f, g), to be c( f, g) =

 1 max ordt yi (t p ) − z j (t n ) . np i, j

We can relate the contact with intersection multiplicity. This opens the possibility of an effective calculation of the contact via the intersection multiplicity and the characteristic sequences of the polynomial f . Proposition 43 Let g be an irreducible monic polynomial of K((x))[y] and set p = deg y g. Let k ∈ {1, . . . , h}. 1. c( f, g) ≤ mn1 if and only if int( f, g) = np c( f, g). , then int( f, g) = (rk dk + (nc( f, g) − m k )dk+1 ) np . 2. If mnk ≤ c( f, g) < m k+1 n , then 3. If int( f, g) = (rk dk + (nc − m k )dk+1 ) np for some c ∈ Z, with mnk ≤ c < m k+1 n c = c( f, g). 4. Then c( f, g) = mnk if and only if int( f, g) = rk dk np . (With the assumption that m h+1 = +∞.) f, g) = ordt f (t p , z(t)). Proof Let z(t) be a root of g(t p , y) = 0. We have int( p n p n n (y − yi (t)). Hence Note that f (t , z(t)) = f ((t n ) , z(t)). Also, f (t , y) = i=1 n p p n f ((t n )n , y) = i=1 (y − yi (t n )), which implies that f (t p , z(t)) = i=1 (z(t) p n − yi (t )). Then, using Lemma 35,

4.3 Contact

71

1 int( f, g) = ordt f (t p , z(t)) = ordt f (t np , z(t n )) n n  1 n 1 n = ordt (z(t ) − yi (t p )) = ordt (z(t n ) − yi (t p )). i=1 i=1 n n (4.1) 1 Suppose, without loss of generality, that c( f, g) = np ordt (y1 (t p ) − z(t n )). It follows p n that int( f, g) ≤ ordt (y1 (t ) − z(t )) = np c( f, g). If c( f, g) ≤ mn1 , then ordt (y1 (t p ) − z(t n )) ≤ m 1 p. Take i ∈ {2, . . . , n}. We have z(t n ) − yi (t p ) = z(t n ) − y1 (t p ) + y1 (t p ) − yi (t p ), and by Lemma 23, ordt (y1 (t p ) − yi (t p )) = pordt (y1 (t) − yi (t)) ≥ pm 1 . Using ordt (y1 (t p ) − z(t n )) ≤ m 1 p, ordt (y1 (t p ) − yi (t p )) ≥ pm 1 and ordt (z(t n ) − yi (t p )) ≤ ordt (z(t n ) − y1 (t p )), we deduce that ordt (z(t n ) − yi (t p )) = ordt (z(t n ) − y1 (t p )). In light of (4.1), we obtain that int( f, g) = ordt (z(t n ) − y1 (t p )) = npc( f, g). Conversely, if int( f, g) = npc( f, g), then (4.1) and the inequality int( f, g) ≤ ordt (y1 (t p ) − z(t n )) = npc( f, g) forces ordt (yi (t p ) − z(t n )) = ordt (y1 (t p ) − z(t n )) for all i ∈ {2, . . . , n}. Assume that c( f, g) > mn1 and take i ∈ {1, . . . , n} such that ordt (yi (t) − y1 (t)) = m 1 (such an i exists in virtue of Lemma 23 (iii)). As z(t n ) − yi (t p ) = z(t n ) − y1 (t p ) + y1 (t p ) − yi (t p ), we deduce ordt (z(t n ) − yi (t p )) = ord(y1 (t p ) − yi (t p )) = pm 1 . Also ordt (z(t n ) − yi (t p )) = ordt (y1 (t p ) − z(t n )) = npc( f, g) > pm 1 ; a contradiction. This proves (1). Suppose that c( f, g) ≥ mn1 and let k be the greatest element such that mk ≤ c( f, g) < m k+1 . Let i ∈ {2, . . . , n}. We have z(t n ) − yi (t p ) = z(t n ) − y1 (t p ) + n n p p y1 (t ) − yi (t ). Hence

 ordt (z(t ) − yi (t )) = n

p

ordt (z(t n ) − y1 (t p )) ordt (y1 (t p ) − yi (t p ))

if ordt (yi (t) − y1 (t)) > m k , if ordt (yi (t) − y1 (t)) ≤ m k .

By Lemmas 35 and 23, (4.1) and the relations between the characteristic sequences, we have int( f, g) = = = = =

1 n ordt (z(t n ) − yi (t p )) i=1 n k 1 (dk+1 ordt (z(t n ) − y1 (t p )) + p (di − di+1 )m i ) i=1 n 1 (dk+1 npc( f, g) + p(rk dk − m k dk+1 )) n p p (dk+1 nc( f, g)) + (rk dk − m k dk+1 ) n n p (rk dk + (nc( f, g) − m k )dk+1 ). n

With this we have (2). In order to prove (3), suppose that int( f, g) = (rk dk + (nc − m k )dk+1 ) np for some k ≥ 1. If c( f, g) < mn1 , then by (1) we know that int( f, g) = npc( f, g), and this

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is strictly smaller than np mn1 = pm 1 = pr1 = (r1 d1 ) np ≤ (rk dk + (nc − m k )dk+1 ) np , which is a contradiction. Hence c( f, g) ≥ mn1 , and there must be some i such that mni ≤ c( f, g) < mni+1 . Now by (2), int( f, g) = (ri di + (nc( f, g) − m i )di+1 ) np . Assume that i < k. Then i + 1 ≤ k and int( f, g) < (ri di + (m i+1 − m i )di+1 ) np = ri+1 di+1 np ≤ rk dk np ≤ (rk dk + (nc − m k )dk+1 ) np = int( f, g); a contradiction. Thus i ≥ k. If i > k, we have int( f, g) = (ri di + (nc( f, g) − m i )di+1 ) np ≥ ri di np > rk+1 dk+1 np = (rk dk + (m k+1 − m k )dk+1 ) np > (rk dk + (nc − m k )dk+1 ) np = int( f, g); a contradiction again. This forces i = k. The rest of the proof now follows easily. Finally, (4) is a direct consequence of (2) and (3). As a consequence, the contact of f with its pseudo-approximate roots is easy to calculate. Corollary 22 Let k ∈ {1, . . . , h} and let G k be a dkth pseudo-approximate root of f . We have c( f, G k ) = mnk . In particular c( f, App( f ; dk )) = mnk . k Proof By Proposition 36, int( f, G k ) = rk = (rk−1 dk−1 + (n mnk − m k−1 )dk ) n/d . n mk Hence c( f, G k ) = n by Proposition 43.

We are going to present an irreducibility criterion, and for this we need the following technical lemma. Lemma 29 Let g be a monic irreducible polynomial of K((x))[y] of degree p in n divides p. In particular, if y. If c( f, g) > mnk , for some k ∈ {1, . . . , h}, then dk+1 mh c( f, g) > n , then n divides p. Proof Let z(t) be a root of g(t p , y) = 0 and assume, without loss of generality,  1 ordt (z(t n ) − y1 (t p )). Write y1 (t) = i ai t i and z(t) = j b j t j . that c( f, g) = np The hypothesis ordt (z(t n ) − y1 (t p )) > pm k implies that for all i in Supp(y1 (t)) with i ≤ m k , there exist j ∈ Supp(z(t)) such that jn = i p, that is, j = i np ∈ N. But gcd({n} ∪ {i ∈ Supp(y1 (t)) | i ≤ m k }) = dk+1 , whence dk+1 np ∈ N, which implies n divides p. that dk+1 The second assertion now follows from the fact that dh+1 = 1. The following result can be seen as an irreducibility criterion, and also as a sufficient condition for two polynomials of the same degree to share the same characteristic sequences. Proposition 44 Let g be a monic polynomial of K((x))[y] and assume that deg y g = n. If int( f, g) > rh dh , then g is irreducible. Furthermore, g has the same characteristic sequences as f . Proof Let g = g1 · · · gr be the decomposition of g into irreducible components in K((x))[y]. If r > 1 then, deg y gi < n for all i ∈ {1, . . . , r }. Thus by Lemma 29, c( f, gi ) ≤ mnh for all i ∈ {1, . . . , r }. Set n i = deg y gi . We use Proposition 43 distinguishing the following three cases. • If c( f, g1 ) =

mh , n

then int( f, g) = rh dh nni .

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73

• If c( f, gi ) < mnh and there exists k ∈ {1, . . . , h} such that mnk ≤ c( f, gi ) < m k+1 , n then int( f, gi ) = (rk dk + (nc( f, g) − m k )dk+1 ) nni . Hence, we have that int( f, gi ) ≤ (rk dk + (n mnk − m k )dk ) nn1 = rk dk nni ≤ rh dh nni . • If c( f, gi ) < mn1 , then int( f, gi ) = npi c( f, gi ) ≤ nn i mn1 = r1 d1 nni ≤ rh dh nni .  Therefore, int( f, g) = ri=1 int( f, gi ) ≤ rh dh , contradicting the hypothesis. This proves that g is irreducible and also that c( f, g) > mnh . It follows from the definition of contact that g has the same Newton–Puiseux exponents as f . This proves our assertion. Example 29 Let f be as in Example 27. It follows that c( f, y) = 4/6 and c( f, y 3 − x 2 ) = 9/6. Let g = f + x 6 . Then int( f, g) = 36 > r2 d2 = 34. Hence g is irreducible by Proposition 44. But int( f, g) = 36 = r2 d2 + 6c( f, g) − m 2 , whence c( f, g) = 11/6. We finish this section with a result that will be used later for the decomposition of f y . It also shows that depending on the value of c( f, g), the integer int( f, g) will be in a certain “layer” of the monoid associated to f . Proposition 45 Let g be a monic irreducible polynomial of K((x))[y] of degree p for some k ∈ {1, . . . , h} (as above m h+1 = +∞), then in y. If mnk ≤ c( f, g) < m k+1 n int( f, g) ∈ r0 Z + r1 , . . . , rk . Proof Let (G 1 , . . . , G h ) be a set of pseudo-approximate roots of f . If k = h, then this follows from Corollary 18. So assume that k < h and set c = c( f, g). By Proposition 43, int( f, g) = (rk dk + (nc − m k )dk+1 ) np , with p = deg y g. We prove that c(G k+1 , g) = c( f, g). Let yi (t), i ∈ {1, . . . , n}, be the roots of f (t n , y) = 0. We can chose G k+1 to be the pseudo-approximate root of f as defined ¯1 (t 1/dk+1 ), where y¯1 (t) = in a root of G k+1 (t n/dk+1 , y) = 0 is Y 1 (t) = y Page 59. Then p p p 0 and suppose that we have constructed gi and f i for all i ∈ {1, . . . , k − 1} with the properties described above. Weneed gk ∈ K[[y]] and f k ∈ K[y] with deg y f k < n such that Fk = g0 f k + gk f 0 + i+ j=k,i, j≥1 gi f j , or equivalently

g0 f k + gk f 0 = Fk −

gi f j .

i+ j=k,i, j≥1

 Define u k = Fk − i+ j=k,i, j≥1 gi f j . Since g0 (y) is a unit in K[[y]], u k (y)/g0 (y) ∈   K[[y]]. Write u k (y)/g0 (y) = l≥l0 cl y l , and set f k (y) = l m h−1 , Lemma 29 asserts that dnh divides deg y P. Hence we have that dh (deg y n n P/n) is a positive integer. Let Ph be the product of irreducible components of f y with contact mnh with f . It follows from the properties of the intersection multiplicity that int( f, Ph ) = deg P deg P rh dh ny h . By the argument in the above paragraph, dh ny h is a positive integer, that we will denoteby ah . With this notation int( f, Ph ) = ah rh . Recall that (4.2) states that h (ek − 1)rk . Using both equations and the properties of intersection int( f, f y ) = k=1 h−1 multiplicity, we obtain int( f, f y /Ph ) = k=1 (ek − 1)rk + (eh − 1 − ah )rh . Notice deg y P n that ah = n/dh < n/dh = dh = eh , and so eh − 1 − ah ≥ 0. By construction of Ph , for all the irreducible components Q of f y /Ph , we have c( f, Q) < mnh . A similar argument as the employed above, implies that int( f, f y /Ph ) ∈ r0 , . . . , rh−1 . But then (eh − 1 − ah )rh is in the group spanned by {r0 , . . . , rh−1 }, and this is not possible by Lemma 22, unless eh − 1 = ah . Consequently int( f, Ph ) = deg P (eh − 1)rh , and we know already that int( f, Ph ) = rh dh ny h , yielding deg y Ph = (eh − 1) dnh . Now we continue with f y /Ph , with degree in y equal to n − 1 − (eh − 1) dnh = h−1 (ek − 1)rk . If Q is an irren − 1 − (dh − 1) dnh = dnh − 1, and int( f, f y /Ph ) = k=1 ducible component of f y /Ph , then c( f, Q) > m h−1 would lead by Lemma 29 that dnh n n divides deg y Q ≤ dh − 1, which is impossible. So for every irreducible component Q of f y /Ph , we have that c( f, Q) ≤ m h−1 . An induction procedure completes the n proof.

4.4.1 Module of Kähler Differentials Let the notations be as above. In particular, f = y n + a1 (x)y n−1 + · · · + an (x) is an irreducible polynomial of K[[x]][y] and S = ( f ) = r0 , r1 , r2 , . . . , rh , with r0 = n and r1 = m. Denote by C( f ) the algebroid curve f = 0. In this section we will introduce the ideal of values of differentials on C( f ), and then we will give, in terms of this ideal, a criterion for f to be equivalent to a quasi-homogeneous polynomial, that is, we will determine when there is an isomorphism σ : K[[x, y]] −→ K[[Z , W ]] such that σ ( f ) = W n − Z m . We first recall the following result. Lemma 31 Let σ : K[[x, y]] −→ K[[Z , W ]] be an isomorphism and let F(Z , W ) = σ ( f ). We have (F) = ( f ). In particular, S = ( f ) does not depend on the equation of C( f ). Proof In fact, given g ∈ K[[x, y]], we have int( f, g) = int(F, σ (g)).  Let x(t) = t n , y(t) = p≥m c p t p be a parametrization of f , and suppose, after possibly a change of variables, that cm = 1. Suppose that y(t) − t m = 0. Let p1 =

4.4 The Local Case

81

h inf(Supp(y(t) − t m )). If p1 ∈ S, then we write p1 = i=0 ai ri with a0 , . . . , ah ∈ N and we consider the change of variables x1 = x, y1 = y − c p1 x a0 y a1 G a22 · · · G ahh , where for all i ∈ {1, . . . , h}, G i is an element of K[[x, y]] such that int( f, G i ) = ri . It follows that f 1 (x1 , y1 ) = f (x1 , y1 + c p1 x1a0 y1a1 G a22 · · · G ahh ) is an  irreducible polynomial of K[[x1 , y1 ]] with parametrization x1 (t) = t n , y1 (t) = t m + p>m c˜ p t p and either y1 (t) − t m = 0 or inf(Supp(y1 (t) − t m )) > p1 . In the second case, we restart with f 1 , x1 (t), y1 (t). The conductor C(S) of S being finite, we get, after repeating possibly the argument above a finite number of times, the following. Proposition 51 Under the standing hypothesis and notations, we can suppose, modulo an automorphism of K[[x, y]], that f has a parametrization of the form x = t n ,  m y = t + q∈G(S),q>m cq t q . Such an expression is called a short parametrization of C( f ). Let A = K[[x(t), y(t)]], and let M = x (t)A + y (t)A. Given g ∈ M, we set o(g) = ordt g(t). Clearly for all g ∈ M and for all h ∈ A, hg ∈ M. It follows that the set I = o(M) = {o(g) | g ∈ M} is a relative ideal of S. Given g ∈ A, we have g (t) ∈ M. In particular, if s ∈ S, then s − 1 ∈ I . We say that s − 1 is an exact degree. We call the other elements of I non exact degrees of M. We denote by NE(M) the set of non exact degrees of M, that is NE(M) = {i ∈ I | i + 1 ∈ / S}. Let ne(M) denote the cardinality of NE(M). Proposition 52 Let the notations be as above. We have ne(M) ≤ g(S) =

C(S) . 2

Proof We are counting non exact degrees, and these are the i ∈ I such that i + 1 ∈ / S. Since there are at most g(S) gaps in S, ne(M) is at most g(S), which is C(S)/2, since our semigroup is symmetric. Let B = K[[x, y]]/( f ), and let x, ¯ y¯ be the images of x, y in B, also let N = B d x¯ + B d y¯ be the B-module generated by {d x, ¯ d y¯ }. Let B¯ be the integral closure of B, and ˜ ¯ ¯ let N = B d x¯ + B d y¯ . Define ν( f ) = dimK

K[[x, y]] B = dimK . ( f, f x , f y ) ( fx , f y )

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If we denote by (·) the length of a module, then we have the following result. Proposition 53 ([20, Corollary 2]) Under the standing hypothesis and notations,   μ( f ) ˜ ν( f ) = N/N + 2 where we recall that μ( f ) = C(S) = rank K (K[[x]][y]/( f x , f y )) is the Milnor number of f . In our setting, B  K[[x(t), y(t)]] = A (this follows by the Isomorphism Theorem ¯ = by taking the map x → x(t), y → y(t), whose kernel is ( f )), whence B¯  A ¯ is the integral closure of A. It follows that N is isomorphic to M = K[[t]], where A ˜ is isomorphic to M ˜ = x (t)K[[t]] + y (t)K[[t]] = x (t)A + y (t)A and also that N {g (t), g(t) ∈ K[[t]]}. Note that {s − 1 | s ∈ S} ⊆ I and o(g (t)) = o( dtd g(x(t), y(t))) is an exact ele˜ ment for all g ∈ K[[x, y]]. In particular, (N/N) is the cardinality of the set {s ∈ G(S) | s − 1 ∈ / S}. This cardinality is nothing but g(S) − ne(M) = μ(2f ) − ne(M), and it follows that ν( f ) = μ( f ) = C(S) if and only if ne(M) = 0, that is, every element of I is exact. Clearly, if x = t n , y = t m , then f = y n − x m , whence ν( f ) = μ( f ) = (n − 1)(m − 1). In this case, I = {s − 1 | s ∈ S}. Next, we focus on proving the converse, that is, if ν( f ) = μ( f ), then C( f ) has an equation of the form y n − x m . The main idea is to prove that, otherwise, I contains a non exact element, which would contradict the hypothesis. We will need the following technical Lemmas (we recall di for that m, d, r are the characteristic sequences associated with f and that ei = di+1 all i ∈ {1, . . . , h}). Lemma 32 With the notations above, if h ≥ 2 then n + m 2 ∈ / S. Proof We have n + m 2 = n + (r2 − r1 e1 + m) = n + r2 − (e1 − 1)m < n + r2 − (e1 − 1)n = r2 − (e1 − 2)n ≤ r2 . Consequently, if n + m 2 ∈ S, then n + m 2 = an + bm for some a, b ∈ N. This implies that d2 divides m 2 , which is a contradiction. Lemma 33 Let the notations be as above and assume, without loss of generality, that x = t n , y = t m + q∈G(S),q>m cq t q . Suppose that y − t m = 0, and let λ = inf(Supp(y − t m )). If n + λ ∈ S, then there exists a new parametrization x1 , y1 such that either x1 = T n , y1 = T m or x1 = T m , y1 = T m + cλ1 T λ1 + . . . , with cλ1 = 0 and λ1 > λ. Proof Suppose that n + λ ∈ S. If h ≥ 2 then, by Lemma 32, λ < m 2 , whence d2 divides λ and n + λ < n + m 2 ≤ r2 . In particular n + λ ∈ n, m. The same conclusion holds if h = 1. Write n + λ = an + bm where a, b ∈ N. It follows that λ = (a − 1)n + bm, but λ ∈ / S, hence a = 0 and λ = −n + bm = −n + ( j + 1)m with j ≥ 0, because λ > m. Let

4.4 The Local Case

83

x1 = x + αy j = T n , where the constant α is to be determined. We have T n = t n + αt m j + . . . , whence T = t (1 + n1 αt −n+m j + . . . ), which implies that T m = t m + mα t −n+m j + . . . . In n ncλ particular, if α = m , then we obtain a new parametrization x1 = T n , y1 = T m + cλ1 T λ1 + . . . , and if y1 − T m = 0, then cλ1 = 0 and λ1 > λ. As a consequence, we get the following corollary. Corollary 23 With the notations above, we can suppose, modulo an automorphism of K[[x, y]], that f has a parametrization of the form x = t n , y = t m +  q m m / q∈G(S),q>m cq t such that if y − t  = 0 and λ = inf(Supp(y − t )), then n + λ ∈ S. Such an expression is called a very short parametrization of f . Proof If n + λ ∈ S then by Lemma 33, there exists a change of variables such that the new parametrization is either x1 = T n , y1 = T m or x1 = T n , y1 = T m + cλ1 T λ1 + . . . , with cλ1 = 0 and λ1 > λ. In the second case, if n + λ1 ∈ S, then we apply the process described in the proof of Lemma 33. Let us prove that this process can not be infinite. / S. The set of integers i, m ≤ i ≤ m 2 Suppose that h ≥ 2. By Lemma 32, n + m 2 ∈ being finite, we get, after a finite number of operations, a parametrization of the expected form. If h = 1, then, after a finite number of operations, we get either a parametrization of the expected form, or a parametrization of the form xk = τ n , yk = τ m + cτ C(S) + . . . . In the second case, as for all s ≥ C(S), s ∈ S, the short parametrization of C( f ) is given by x˜ = τ1n , y˜ = τ1m . This finishes the proof. The following result is a consequence of the existence of a very short parametrization of C( f ). Corollary 24 Let x = t n , y = t m + cλ t λ + . . . be a very short parametrization of C( f ). If cλ = 0, then the differential form w = mx y − nx y is not exact. Proof We have o(w) = n + λ − 1 and n + λ ∈ / S. This proves our assertion. Now we can state the main result of this section. Theorem 7 ([78]) Let f = y n + a1 (x)y n−1 + · · · + an (x) be an irreducible polynomial of K[[x, y]]. If μ( f ) = ν( f ), then there is an automorphism σ of K[[x, y]] such that σ ( f ) is a quasi-homogeneous polynomial. Proof Let the notations be as above. We know that μ( f ) − ν( f ) = ne(M). Let, modulo an automorphism of K[[x, y]], x = t n , y = t m + q∈G(S),q>m cq t q be a very short parametrization of f . If y − t m = 0 and if λ = inf(Supp(y − t m )), then n + λ∈ / S. But this implies that w = mx y − nx y is not exact by Corollary 24, whence ne(M) > 0, contradicting the hypothesis.

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4 Semigroup of an Irreducible Meromorphic Series

4.5 The Case of Curves with One Place at Infinity Let F = y n + a1 (x)y n−1 + · · · + an (x) be a nonzero polynomial of K[x][y] and assume, possibly after a change of variables, that degx ai (x) < i for all i ∈ {1, . . . , n} such that ai (x) = 0 (this restriction will make sense in Lemma 34). Let C = V(F) be the plane algebraic curve in A2K defined by F(x, y) = 0, and let h F (u, x, y) = u n F( ux , uy ). The projective curve V(h F ) is the projective closure of C in P2K . We are using the embedding A2K ⊂ P2K that sends (x, y) to (u : x : y). By hypothesis on the coefficients of F, (0 : 1 : 0) is the unique point of V(h F ) at the line at infinity u = 0. We say that F has one place at infinity if h F is analytically irreducible at (0 : 1 : 0). Set F∞ (u, y) = h F (u, 1, y). This is the local equation of the projective curve at the point (0 : 1 : 0) (and local coordinates (u, y)). Then F has one place at infinity if and only if the formal power series F∞ (u, y) is irreducible in K[[u]][y]. Lemma 34 Let the notations be as above. Let f (x, y) = F(x −1 , y) ∈ K[x −1 , y]. 1. f (x, x −1 y) = x −n F∞ (x, y). 2. F has one place at infinity if and only if f (x, y) is irreducible in K((x))[y].   Proof Write F(x, y) = y n + i+ j rk+1 dk+1 .h (ek − 1)rk . Int(F, Fy ) = Int(Fx , Fy ) + n − 1 = k=1 h The conductor C(∞ (F)) = Int(Fx , Fy ) = ( k=1 (ek − 1)rk ) − n + 1.

Example 33 Sequences fulfilling the Condition (iv) in Proposition 57 are known as δ-sequences. It is not difficult to deduce a recursive method to construct them all. This is done in the numericalsgps package. gap> DeltaSequencesWithFrobeniusNumber(11); [ [ 5, 4 ], [ 6, 4, 9 ], [ 7, 3 ], [ 9, 6, 4 ], [ 10, 4, 5 ], [ 13, 2 ] ] gap> List(last, CurveAssociatedToDeltaSequence); [ yˆ5-xˆ4, yˆ6-2*xˆ2*yˆ3+xˆ4-xˆ3, yˆ7-xˆ3, yˆ9-3*xˆ2*yˆ6+3*xˆ4*yˆ3-xˆ6-yˆ2, yˆ10-2*xˆ2*yˆ5+xˆ4-x, yˆ13-xˆ2 ] gap> List(last, SemigroupOfValuesOfPlaneCurveWithSinglePlaceAtInfinity); [ 1, there exist a polynomial F with N places at infinity and λ ∈ K∗ such that the number of places of F − λ at infinity is not equal to N , [3]. Let the notations be as above and assume that F has one place at infinity. It follows that F∞ (u, y) is a monic irreducible polynomial of K[[u]][y] of degree n in y. We are going to use the machinery introduced in Sect. 4.4 to study the semigroup associated to ∞ (F). to F∞ and relate it n (y − yi (t)). We have, as in the proof of Proposition 54, Let f (t n , y) = i=1 2 n n f (t n , t −n y) = i=1 (t −n y − yi (t)) = t −n i=1 (y − t n yi (t)). Also F∞ (x, y) = x n f (x, x −1 y), and thus n  F∞ (t n , y) = (y − t n yi (t)). i=1

In particular, the roots of F∞ (t n , y) = 0 are given by Yi (t) = t n yi (t). With this we can relate the characteristic sequences and approximate roots of F∞ with those of f ; the first step to see the connection between ∞ (F) and (F∞ ). Proposition 59 Under the standing hypothesis. (i) (ii) (iii) (iv)

The set of Newton–Puiseux exponents of F∞ is given by m¯ k = n + m k . The d-sequence of F∞ is equal to the d-sequence of f . The r -sequence of F∞ is given by r¯k = n dnk − rk For all k ∈ {1, . . . , h}, App(F∞ ; dk ) = h G k (u, 1, y), with G k = App(F; dk ).

Proof (i) The formal power series Y1 (t) = t n y1 (t) is a root of F∞ (t n , y) = 0. Hence Supp(Y1 (t)) = {n + i | i ∈ Supp(y1 (t))} = n + Supp(y1 (t)). Now the proof of (i) follows immediately. (ii) In fact, for all k ∈ {1, . . . , h}, we have gcd(n, m 1 , . . . , m k ) = gcd(n, n + m 1 , . . . , n + m k ).

4.5 The Case of Curves with One Place at Infinity

89

(iii) We prove the result by induction on k ∈ {1, . . . , h}. We have r¯0 = n and r¯1 = n + m 1 = n − r1 = n dn1 − r1 . From the definition of r¯k+1 , we have r¯k+1 dk+1 = r¯k dk + (n + m k+1 − (n + m k ))dk+1 . By induction hypothesis, n − rk )dk + (m k+1 − m k )dk+1 dk = (−rk dk + (m k+1 − m k )dk+1 ) + n 2 = −rk+1 dk+1 + n 2 .

r¯k+1 dk+1 = (n

n − rk+1 . Hence r¯k+1 = n dk+1 (iv) Easy exercise.

With all this we can derive some invariants of (F∞ ), the numerical semigroup associated with F∞ . Proposition 60 Let the notations be as above. (i) The conductor of (F∞ ) is given by C((F∞ )) = ( n + 1. (ii) C((F∞ )) + C(∞ (F)) = (n − 1)(n − 2).

h

k=1 (ek

− 1)(n dnk − rk )) −

Proof From Propositions 19, 48 and 59 we obtain (i). Now, using Proposition 57 and (i), C((F∞ )) + C(∞ (F)) =

 h

 n (ek − 1)(n − rk ) − n + 1 k=1 dk

h + (ek − 1)rk − n + 1  k=1  h n = (ek − 1)(n ) − 2(n − 1) k=1 dk = n(n − 1) − 2(n − 1) = (n − 1)(n − 2). Example 34 Let F(x, y) = y 6 − 2x 2 y 3 + x 4 − x 3 . Then r0 = n = 6 and d1 = 6. Take G 1 (x, y) = y. If we compute Int(F, G 1 ), we obtain r1 = 4 and consequently d2 = 2. Now take G 2 (x, y) = y 3 − x 2 . Then F(x, y) = G 2 (x, y)2 − x 3 , which means that G 2 = App(F; 2). The calculation of Int(F, G 2 ) gives r2 = 9. Hence d3 = 1 and ∞ (F) = 6, 4, 9. In light of Proposition 59, the r -sequence associated with F∞ is r = (6, 2, 9), whence (F∞ ) = 2, 9. Finally, observe that (n − 1)(n − 2) = 20 = C(∞ (F)) + C((F∞ )) = 12 + 8. Remark 10 Let F − λ, λ = 0 be an element of the family of curves (F − λ)λ∈K . By λ Proposition 58, F − λ has one place at infinity and ∞ (F − λ) = ∞ (F). Let F∞ be the local equation of F − λ at the point at infinity. It follows from Propositions 59 λ λ and 60 that (F∞ ) = (F∞ ). In particular the Milnor number μ(F∞ ) = μ(F∞ ) and λ by Proposition 60, μ(F∞ ) + Int(Fx , Fy ) = (n − 1)(n − 2).

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4 Semigroup of an Irreducible Meromorphic Series

Let F − λ0 be a nonsingular element of the family (F − λ)λ∈K (that is, the ideal (F − λ0 , Fx , Fy ) equals K[x, y]) and let g be the geometric genus of the projective curve V(h F−λ0 ) = V(h F − λ0 u n ). The only possible singular point of V (h F−λ0 ) being the point at infinity of F − λ0 . The curve V (F − λ0 ) is smooth in K2 and has one place at infinity. Hence the λ0 ) = (n − 1)(n − 2). genus formula (see [60, Theorem 10.5]) implies that 2g + μ(F∞ Consequently, 2g = Int(Fx , Fy ) = C(∞ (F)). In particular, the geometric genus g coincides with the genus of the semigroup ∞ (F) (by Proposition 19). We will now give some applications of the arithmetic of the semigroup ∞ (F). Let k ∈ {1, . . . , h} and let gk = App( f ; dk ) where f (x, y) = F(x −1 , y). Let G k (x, y) = gk (x −1 , y). Then G k is a polynomial with one place at infinity (Corollary 21 and Lemma 34). Furthermore, we have the following. Proposition 61 Le the notations be as above. For all k ∈ {1, . . . , h},   0 (i) ∞ (G k ) = drk+1 , , dr1k , . . . , rk−1 dk h (ii) Int(Fx , Fy ) = dk Int((G k )x , (G k ) y ) − dk + 1 + i=k (ei − 1)ri . Proof Assertion (i) follows from Proposition 34. For (ii), use Proposition 34 with h (ei − 1)ri + Proposition 19, to derive that C(∞ (F)) = dk C(∞ (G k )) − dk + i=k 1. The proof then follows from Proposition 57. As a consequence, we get the following result that studies the consequences of Int(Fx , Fy ) = 0. Corollary 25 Let the notations be as above. If Int(Fx , Fy ) = 0, then for all k ∈ {1, . . . , h}, rk = dk+1 and Int((G k )x , (G k ) y ) = 0. In particular, ∞ (F) = N and r1 divides n. h Proof By hypothesis, and from Proposition 57(v), we have 0 = i=1 (ei − 1)ri − n + 1. For  all k ∈ {1, . . . , h}, rk ≥ dk+1 . If r > d for some i ∈ {1, . . . , h}, i i+1   h h (ek − 1)rk − n + 1 > k=1 (ek − 1)dk+1 − n + 1 = ( k=1 (dk − we get 0 = k=1 dk+1 )) − n + 1 = 0, a contradiction. h h h (ei − 1)ri = i=k (ei − 1)di+1 = i=k (di − di+1 ) = Now, as rk = dk+1 , i=k Proposition 61 and the hypothesis that 0 = dk Int((G k )x , dk − 1. Hence from h (ei − 1)ri = dk Int((G k )x , (G k ) y ), we deduce Int((G k )x , (G k ) y ) − dk + 1 + i=k (G k ) y ) = 0 for all k. Since rh = dh+1 = 1, ∞ (F) = N. Also, r1 = d2 = gcd(n, r1 ), whence r1 divides r0 = n. Let the notations be as above and assume that F is nonsingular. Let g be the geometric genus of V (h F ). By Remark 10, Int(Fx , Fy ) = 0 if and only if g = 0 (that is, F is rational or equivalently, F is parametrized by rational functions in one variable, [43]). In particular, F is rational and nonsingular if and only if F − λ is rational and nonsingular for all λ = 0. It follows from Corollary 25 that if F is rational and nonsingular then r1 divides n. Furthermore, by Proposition 61, G k − λ is rational and nonsingular for all 1 ≤ k ≤ h and for all λ ∈ K.

4.5 The Case of Curves with One Place at Infinity

91

Theorem 8 (Abhyankar–Moh Theorem) Let x(t) = t n + a1 t n−1 + · · · + an and y(t) = t m + b1 t m−1 + · · · + bm be two polynomials in K[t] and assume that m < n. If K[x(t), y(t)] = K[t], then m divides n. Proof Let F(x, y) = Rest (x − x(t), y − y(t)). Then F(x, y) is a polynomial with one place at infinity, and F(x, y) = y n + α1 (x)y n−1 + · · · + αn (x) with degx αn (x) = m (see [2]). The ring K[x(t), y(t)] is isomorphic to K[x, y]/(F). So for any G ∈ / (F), Int(F, G) = rank K

K[x, y] K[x(t), y(t)] K[t] = rank K = rank K . (F, G) (G(x(t), y(t))) (G(x(t), y(t)))

Hence Int(F, G) = degt G(x(t), y(t)). As t ∈ K[x(t), y(t)], there exists G(x, y) ∈ K[x, y] such that t = G(x(t), y(t)). Hence Int(F, G) = 1. Then ∞ (F) = N, and consequently C(∞ (F)) = 0. By Proposition 57, Int(Fx , Fy ) = 0, whence m divides n by Corollary 25. Remark 11 Let F(x, y) = y n + α1 (x)y n−1 + · · · + αn (x) be a polynomial with one place at infinity and assume that degx αk (x) < k for all k ∈ {1, . . . , n}. If F is rational and nonsingular, it follows from [2] that F is parametrized by polynomials, that is, there exist x(t) = t n + a1 t n−1 + · · · + an and y(t) = t m + b1 t m−1 + · · · + bm in K[t] such that F = Rest (x − x(t), y − y(t)). By Theorem 8, m divides n. Let d = mn . Let x1 (t) = x(t) − y(t)d , y1 (t) = y(t). The isomorphism φ : K[x(t), y(t)] → K[x1 (t), y1 (t)], φ(x(t)) = x1 (t), φ(y(t) = y1 (t) induces an ¯ ¯ = x1 = x − y d , φ(y) = y1 = y, and isomorphism φ¯ : K[x, y] → K[x1 , y1 ], φ(x) ¯ φ(F) = F1 with F1 (x1 (t), y1 (t)) = 0. Hence, again by Theorem 8, F1 (x1 , y1 ) has one place at infinity and, after possibly a change of variables, we may assume that F1 (x1 , y1 ) = y1n 1 + α11 (x1 )y n 1 −1 + · · · + αn11 (x1 ) with degx1 αi1 < i for all i ∈ {1, . . . , n 1 }. Then we restart with F1 . Note that F1 is rational and nonsingular, hence degt y1 (t) divides degt x1 (t) = n 1 . Repeating the process above, we can conclude that there exists an automorphism φ of K[x, y] such that φ(F) = x. Example 35 Let x(t) = t 4 − t and y(t) = t 2 . Then Rest (x − x(t), y − y(t)) = F(x, y) = (y 2 − x)2 − y. The polynomial F has one place at infinity, and Int(Fx , Fy ) = 0. The morphism φ : K[x, y] −→ K[x1 , y1 ], determined by φ(x) = y12 − x1 , φ(y) = y1 is an isomorphism. Also, φ(y 2 − x) = x1 , and thus φ(F) = x12 − y1 . If we swap x1 and y1 we get F1 (x1 , y1 ) = y12 − x1 , which, by the isomorphism φ1 (x1 ) = y22 − x2 , φ1 (y1 ) = y2 , is mapped to a coordinate. We can rewrite the condition C(∞ (F)) = 0 (equivalently Int(Fx , Fy ) = 0) in terms of the invariants of C((F∞ )). Corollary 26 Let the notations be as above. The following are equivalent: (i) C(∞ (F)) = 0, (ii) C((F∞ )) = (n − 1)(n − 2), (iii) for all k ∈ {1, . . . , h}, rk = dk+1 ,

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4 Semigroup of an Irreducible Meromorphic Series

(iv) for all k ∈ {1, . . . , h}, r¯k = n dnk − dk+1 . Proof This follows from Corollary 25 and Proposition 60. Now let us see what happens it r1 does not divide n = r0 . The following proposition shows that in this setting Int(Fx , Fy ) ≥ n − 1 (recall that Int(Fx , Fy ) = C(∞ (F))). Hence either Int(Fx , Fy ) = 0 or Int(Fx , Fy ) ≥ n − 1 (this corresponds with the “global” Milnor number). Proposition 62 Let the notations be as above. If d2 < r1 (that is, r1 does not divide n), then Int(Fx , Fy ) ≥ n − 1 with equality if and only if rk = 2dk+1 for all k ∈ {1, . . . , h}. h Proof We have Int(Fx , Fy ) + n − 1 = ( k=1 (ek − 1)rk ) (Proposition 57). Recall that dk+1 = gcd(r0 , . . . , rk ). As d2 < r1 , and d2 | r1 , we obtain r1 ≥ 2d2 . Now we use induction to prove that dk+1 < rk for all k ∈ {2, . . . , h − 1}. Suppose that rk−1 > dk and let us prove that rk > dk+1 . We have that rk dk = rk−1 dk−1 + (m k − m k−1 )dk > dk dk−1 + (m k − m k−1 )dk . Then rk > dk−1 + (m k − m k−1 ) > dk−1 > dk+1 . Now rk > dk+1 and dk+1 | rk . We conclude that rk ≥ 2dk+1 for all k ∈ {1, . . . h − 1}. Hence h k=1

(ek − 1)rk ≥ 2

h

(ek − 1)dk+1 ≥ 2

k=1

h

(dk − dk+1 ) = 2(d1 − dh+1 ) = 2(n − 1).

k=1

In particular, Int(Fx , Fy ) ≥ n − 1. Clearly, if rk > 2dk+1 for some k ∈ {1, . . . , n}, then Int(Fx , Fy ) > n − 1. This proves our assertion. Example 36 Let F(x, y) = (y 3 − x 2 )3 − y. The r -sequence of F is given by r0 = n = 9, r1 = 6, and r2 = 2. Hence d1 = 9, d2 = 3, d3 = 1. Hence r1 = 2d2 and r2 = 2d3 . The conductor of ∞ (F) is Int(Fx , Fy ) = (e1 − 1)r1 + (e2 − 1)r2 − n + 1 = 8 = n − 1. As a consequence of the last result we obtain the following. Corollary 27 Under the standing hypothesis. (i) Let h = 1, then Int(Fx , Fy ) = n − 1 if and only if ∞ (F) = n, 2. (ii) Let h ≥ 2 and suppose that d2 < r1 . If 2 does not divide n, then Int(Fx , Fy ) > n − 1. Proof (i) follows from Proposition 62. If 2 does not divide n, then r2 > 2d3 . Hence Proposition 62 implies that Int(Fx , Fy ) > n − 1. This proves (ii). In light of Proposition 60, C((F∞ )) + C(∞ (F)) = (n − 1)(n − 2), whence the last results have their counterpart in C((F∞ )).

4.5 The Case of Curves with One Place at Infinity

93

Corollary 28 Let the notations be as above. Then C((F∞ )) ≤ (n − 1)(n − 3) and the following are equivalent. (i) (ii) (iii) (iv)

C((F∞ )) = (n − 1)(n − 3). C(∞ (F)) = n − 1. For all k ∈ {1, . . . , h}, rk = 2dk+1 . For all k ∈ {1, . . . , h}, r¯k = n dnk − 2dk+1 .

Proof This follows from Propositions 60 and 62.

4.5.1 Module of Kähler Differentials on Polynomial Curves Let X (t) = t n + α1 t n−1 + · · · + αn , Y (t) = t m + β1 t m−1 + · · · + βm be two polynomials of K[t], and let A = K[X (t), Y (t)]. Given a non zero polynomial p(t) ∈ K[t], we denote by d( p) the degree of p in the variable t. Let S = d(A) = {d(h) | h ∈ A \ {0}}. We will assume that N \ d(A) is a finite set. In particular S is a numerical semigroup. Let f (X, Y ) be the unique irreducible polynomial of K[X, Y ], monic in Y , such that f (X (T ), Y (T )) = 0. Then f (X, Y ) = Y n + a1 (X )Y n−1 + · · · + an (X ), and f has one place at infinity (see [3]). In particular, if ∞ ( f ) = {Int( f, g) | g ∈ K[X, Y ] \ {0}}, then ∞ ( f ) = S(= d(A)) (we recall that Int( f, g) = rank K (K[x, y]/( f, g))). In the following we will assume, without loss of generality, that n > m and also (by the change of variables t1 = t + βm1 ) that β1 = 0. As in the previous sections, we will associate with f its set of characteristic sequences m, d, e and r , and we recall that S = r0 = n, r1 = m, r2 , . . . , rh . The curve C = V( f ) is called a polynomial curve; it is a rational curve, parametrized by polynomials. Let M = X (t)A + Y (t)A. The set I = d(M) = {d(F) | F ∈ M} is a relative ideal of S = ( f ). Given g ∈ A, we have g (t) = dtd (g(X (t), Y (t))) ∈ M. In particular, if s ∈ S, then s − 1 ∈ I . We say that s − 1 is an exact degree. We call the other elements of I non exact degrees of M. We denote by NE(M) the set of non exact degrees, that is NE(M) = {i ∈ I | i + 1 ∈ / S}. Let ne(M) be the cardinality of NE(M). Proposition 63 Let the notations be as above. We have ne(M) ≤ g(S) =

F(S)+1 . 2

Proof We are counting non exact degrees, and these are the i such that i + 1 ∈ / S. Since there are at most g(S) gaps in S, ne(M) is at most g(S), which is C(S)/2, since S is symmetric. Let B = K[X, Y ]/( f ), and let x, y be the images of X , Y in B. Let N = B d x + B dy be the B-module generated by {d x, dy}, and let B¯ be the integral closure of B. ˜ = B¯ d x + B¯ dy. Let μ( f ) = rank K (K[x, y]/( f x , f y )) and recall that μ( f ) = Let N C(S). Define

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4 Semigroup of an Irreducible Meromorphic Series

ν( f ) = dimK

K[X, Y ] B = dimK . ( f, f X , f Y ) ( f X , fY )

If we denote by (·) the length of a module, then we have the following property. Proposition 64 ([20, Corollary 2]) Under the standing hypothesis and notations, we have   μ( f ) ˜ ν( f ) = N/N + . 2 In our setting, B  K[X (t), Y (t)] = A (this follows by the Isomorphism Theorem by taking the map x −→ x(t), y −→ y(t), whose kernel is ( f )), whence ¯ = K[t], where A ¯ is the integral closure of A. It follows that N is isomorphic to B¯  A

˜ is isomorphic to M ˜ = x (t)K[t] + y (t)K[t] = M = x (t)A + y (t)A and also that N

{g (t), g(t) ∈ K[t]}. Note that if g(X, Y ) ∈ K[X, Y ], then dtd g(X (t), Y (t)) ∈ M, whence d( dtd g(X (t), Y (t))) ∈ I = d(M). It follows that {s − 1 | s ∈ ( f )} ⊆ I and d( dtd g(X (t), Y (t))) ˜ is an exact element. In particular, (N/N) is the cardinality of the set {s ∈ G(S) | s−1∈ / S}. This cardinality is nothing but g(S) − ne(M) = μ(2f ) − ne(M), and it follows that: 1. ν( f ) = μ( f ) = C(S) if and only if ne(M) = 0, that is, every element of I is exact; 2. ν( f ) = μ(2f ) if and only if ne(M) = g(S). In the following, we will prove that, after possibly a change of variables, the curve V( f ) has a parametrization in one of the following forms: 1. X (τ ) = τ n , Y (τ ) = τ m (hence the equation of the curve V( f ) is of the form Y n − X m ), / ( f ) (hence the degree of 2. X (τ ) = τ n + cλ t λ + . . ., Y (τ ) = τ m and m + λ ∈ m X (τ )Y (τ ) − n X (τ )Y (τ ) is a non exact element of I ). We will need to this end the following technical lemma.  Lemma 35 Let q(t) = t + i≥1 ci t −i ∈ K((t)). Define the map l : K((T )) → K((t)), α(T ) → α(q(t)). In particular, l(T ) = q(t). Then l is an isomorphism. Proof We clearly have l(α(T ) + β(T )) = l(α(T )) + l(β(T )) and l(α(T )β(T )) = l(α(T ))l(β(T )) for all α(T ), β(T ) ∈ K((T )). Moreover, l(1) = 1 and ker(l) = {0}. Let us define the inverse of l. We will determine b1 , b2 , . . . ∈ K such that t = l(T + b1 T −1 + b2 T −2 + . . .). Note that for all k ∈ Z, we can write (q(t))k = t k +

i≥1

for some ci(k) ∈ K. In particular,

ci(k) t k−i−1 ,

4.5 The Case of Curves with One Place at Infinity

95

t − l(T ) − b1l(T −1 ) − . . . = t − q(t) − b1 (q(t))−1 − . . . (−1) ci t −i ) − b1 (t −1 + ci t −i−2 = (−c1 − b1 )t −1 + θi t −i , = t − (t + i≥1

i≥1

i≥2

with θi ∈ K for all i ≥ 2. Then we set b1 = −c1 . Suppose that we have b1 , . . . , bk and let us construct bk+1 . By hypothesis, we have t = l(T + b1 T −1 + · · · + bk T −k ) +



ci(k) t −k−i .

i≥1 (k) Then we set bk+1 = c 1 . Let q1 (T ) = T + k≥1 bk T −k and set l1 (γ (t)) = γ (q1 (T )) (in particular l1 (t) = q1 (T )). Since degt (t − l(q1 (T )) ≤ −k for all k ≥ 0, we have t = l(q1 (T )). This proves that l is surjective, whence an ismorphism. Note that l1 = l −1 because l(l1 (t)) = t.

Let us make the following change of variables T = t (1 + β2 t

−2

+ · · · + βm t

−m

  1 −2 ) = t 1 + β2 t + . . . = q(t). m 1 m

This change of variables defines a map l : K((T )) → K((t)), l(T ) = q(t). It follows from Lemma 35 that l is an isomorphism. Let X 1 (T ) = X (l −1 (t)) and Y1 (T ) = Y (l −1 (t)). We have Y1 (T ) = T m and X 1 (T ) = T n +



cp T p,

p −∞. We have X 1 (T ) = T n + cλ T λ + . . . and Y1 (T ) = T m . The hypothesis on λ implies that cλ = 0. Let W (T ) = m X 1 (T )Y1 (T ) − nY1 (T )X 1 (T ). / S, then m + λ − 1 is a We have W (T ) = (mλ − nm)cλ T m+λ−1 + . . . . If m + λ ∈ non exact degree. Suppose that m + λ ∈ S. If h = 1 then m + λ ∈ m, n. Suppose / S. This would imply that λ > −m 2 that h ≥ 2. We shall prove that m + (−m 2 ) ∈ (where −m 2 is the first characteristic exponent). Suppose otherwise that m + (−m 2 ) ∈ S, then m − m 2 = an + bm + cr2 for some a, b, c ∈ N, c = 0. But m − m 2 = m + r2 − (e1 − 1)r1 . Thus, m + r2 − (e1 − 1)r1 = an + bm + cr2 . If c ≥ 1,

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then m − (e1 − 1)r1 = an + bm + (c − 1)r2 , which is a contradiction because m − (e1 − 1)r1 = m − (e1 − 1)n < 0. It follows that c = 0, whence m + r2 − (e1 − 1)r1 = an + bm, and r2 = (a + e1 − 1)n + (b − 1)m, but d2 = gcd(n, m) does not divide r2 . This is again a contradiction. It follows that λ > −m 2 , whence n + λ is in the semigroup generated by n, m. Consequently, m + λ = an + bm for some a, b ∈ N. Since n > m > λ then a ≤ 1. Moreover, if a = 1, then b = 0. Hence one of the following conditions holds. 1. a = 1, b = 0, m + λ = n. Let in this case Y2 = Y1 + α, α ∈ K∗ . We have W¯ (T ) = m X 1 (T )Y2 (T ) − nY2 (T )X 1 (T ) = [(mλ − nm)cλ − αmn]T n−1 + . . . cλ = − mn cλ , then W¯ (T ) has degree strictly less than n − 1. Hence, if α = λ−n n As an example of this case, let X (t) = t 9 + t 5 , Y (t) = t 4 . We have W (t) = 16t 8 and 8 + 1 = 9 ∈ d(A). If Y¯ = t 4 + 49 , then W¯ (t) = m X (t)Y¯ (t) − n Y¯ (t)X (t) = −80 4 t and 4 + 1 ∈ / d(A). 9 2. a = 0, m + λ = bm. In this case, λ = (b − 1)m. The change of variables X 2 = X 1 − Y1b−1 , Y2 = Y1 yields that either (X 2 , Y2 ) = (T n , T m ) or X 2 = T n + cλ1 T λ1 + . . . , Y2 = T m with λ1 < λ. As an example of this case, let X (t) = t 7 , Y (t) = t 4 + t. We have W (t) = −21t 7 and 7 + 1 = 8 = 2 × 4 ∈ d(A). Let Y1 = 1 7 T + . . .. T 4 . Then T 4 = t 4 + t, T = t (t −3 + 1) 4 , and X 1 (T ) = T 7 − 14 T 4 + 16 1 7 7 If X 2 = X 1 + 4 Y1 , Y2 = Y1 , then X 2 = T + 16 T + · · · , Y2 = T 4 and m X 2 (T )Y2 (T ) − nY2 (T )X 2 (T ) = 21 T 4 + . . ., with 4 + 1 = 5 ∈ / d(A). 2 We prove that these two processes will eventually stop. This is clear in the first case since we are constructing a strictly increasing sequence of nonnegative integers. In the second case, if h ≥ 2 then this is clear since the set of integers in the interval [λ, −m 2 ] is finite. Suppose that h = 1, that is, gcd(m, n) = 1. If the process is infinite, then after a finite number of steps we will obtain a new parametrization of the curve of the form X˜ = T n + αT −l + . . . , Y˜ = T m with l > nm, which is a contradiction. It follows that either we get a parametrization (τ n , τ m ) of the curve V( f ) (which means that the equation of this curve is W n − Z m with K[X, Y ]  K[Z , W ] and gcd(n, m) = 1), or we get a new parametrization Z (t) = t n + a1 t α1 + · · · + an , W (t) = t m + b1 t β1 + · · · + bm such that the degree of W (t) = m Z (t)W (t) − nW (t)Z (t) is a non exact element of I . We then get the following result. Theorem 9 (See [1, 10]) Let X (t) = t n + a1 t n−1 + · · · + an , Y (t) = t m + b1 t m−1 + · · · + bm be the equations of a polynomial curve in K2 , and let f (X, Y ) be the minimal polynomial of X (t), Y (t), that is, f (X, Y ) is the (monic in Y ) generator of the Kernel of the map K[X, Y ] → K[X (t), Y (t)] that sends X to X (t) and Y to Y (t). Let M = X (t)A + Y (t)A be the A-module generated by X (t), Y (t). The following conditions are equivalent.

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(i) The equality μ( f ) = ν( f ) holds. (ii) Every element of the ideal I = d(M) is exact. (iii) The integers n and m are coprime and there is an isomorphism K[X, Y ] → K[Z , W ] such that the image of f (X, Y ) is W n − Z m . Proof The equivalence between (i) and (ii) is clear, and (ii) implies (iii) results from the calculations above. Finally (iii) implies (i) because W n − Z m ∈ (W n−1 , Z m−1 ).

Chapter 5

Minimal Presentations

It is usual in Mathematics to represent objects by means of a free object modulo certain relations fulfilled by the generators of the free object. The reader familiar to group theory surely has used many times definitions of groups by means of generators and relations. Relations are usually represented as equalities, or simply words in the free group on the generators (this means that they are equal to the identity element; this is due to the fact that we have inverse in groups). Here we represent relations by pairs. These are pairs of factorizations of certain elements in the semigroup; they will be a crucial tool for studying factorizations in the next chapter. Rédei in [67] proved that every finitely generated monoid is finitely presented, in the sense that it can be defined as the quotient of Nn (n the number of generators) over a congruence that is finitely generated. Actually, his proof relies in showing that every congruence on Nn is finitely generated as a congruence. In this chapter, we see that the proof can be drastically simplified for numerical semigroups (indeed the method we present here works in a much more general setting; see [68]). For a numerical semigroup S, K[S] is the coordinate ring of a curve parametrized by monomials. Finding a (minimal) presentation implies finding the defining equations of the curve, and so solving the implicitation problem for curves parametrized by monomials. This is known as Herzog’s correspondence. Thus, the fact that every numerical semigroup is finitely presented is just a consequence of Hilbert’s Basis Theorem; since through this correspondence congruences translate to ideals.

5.1 Generators and Relations Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p }. Then the monoid morphism

© Springer Nature Switzerland AG 2020 A. Assi et al., Numerical Semigroups and Applications, RSME Springer Series 3, https://doi.org/10.1007/978-3-030-54943-5_5

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ϕ : N p → S, ϕ(a1 , . . . , a p ) =

p 

ai n i ,

i=1

known as the factorization homomorphism of S, is an epimorphism, and consequently S is isomorphic to N p / ker ϕ, where ker ϕ is the kernel congruence of ϕ:   ker ϕ = (a, b) ∈ N p × N p | ϕ(a) = ϕ(b) . Notice that for groups, vector spaces, rings…the kernel is defined by the elements mapping to the identity element. This is because there we have inverses and from f (a) = f (b) we get f (a − b) = 0. This is not the case in numerical semigroups, and this is why the kernel is a congruence, and not a “subobject” of the domain. Given τ ⊂ N p × N p , the congruence generated by τ is the smallest congruence on N p containing τ , that is, it is the intersection of all congruences containing τ . We denote by cong(τ ) the congruence generated by τ . Accordingly, we say that τ is a generating system of a congruence σ on N p if cong(τ ) = σ . The congruence generated by a set is precisely the reflexive, symmetric, transitive closure (this would just make the closure an equivalence relation), to which we adjoin all pairs (a + c, b + c) whenever (a, b) is in the closure; so that the resulting relation becomes a congruence. This can be formally written as follows. Proposition 65 Let ρ ⊆ N p × N p . Define ρ 0 = ρ ∪ {(b, a) | (a, b) ∈ ρ} ∪ {(a, a) | a ∈ N p }, ρ 1 = {(v + u, w + u) | (v, w) ∈ ρ 0 , u ∈ N p } Then cong(ρ) is the set of pairs (v, w) ∈ N p × N p such that there exist k ∈ N and v0 , . . . , vk ∈ N p with v0 = v, vk = w and (vi , vi+1 ) ∈ ρ 1 for all i ∈ {0, . . . , k − 1}. Proof We first show that the set constructed in this way is a congruence. Let us call this set σ . 1. Since (a, a) ∈ ρ 0 ⊆ σ for all a ∈ N p , the binary relation σ is reflexive. 2. If (v, w) ∈ σ , there exist k ∈ N and v0 , . . . , vk ∈ N p such that v0 = v, vk = w and (vi , vi+1 ) ∈ ρ 1 for all i ∈ {0, . . . , k − 1}. Since (vi , vi+1 ) ∈ ρ 1 implies that (vi+1 , vi ) ∈ ρ 1 , by defining wi = vk−i for every i ∈ {0, . . . , k}, we obtain that (w, v) ∈ σ . Hence σ is symmetric. 3. If (u, v) and (v, w) are in σ , then there exist k, l ∈ N and v0 , . . . , vk , w0 , . . . , wl ∈ N p such that v0 = u, vk = w0 = v, wl = w and (vi , vi+1 ), (w j , w j+1 ) ∈ ρ 1 for all suitable i, j. By concatenating these we obtain (u, w) ∈ σ . Thus σ is transitive. 4. Finally, let (v, w) ∈ σ and u ∈ N p . There exists k ∈ N and v0 , . . . , vk ∈ N p such that v0 = v, vk = w and (vi , vi+1 ) ∈ ρ 1 for all i ∈ {0, . . . , k − 1}. By defining wi = vi + u for all i ∈ {0, . . . , k} we have (wi , wi+1 ) ∈ ρ 1 and consequently (v + u, w + u) ∈ σ .

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It is clear that every congruence containing ρ must contain σ and this means that σ is the least congruence on N p that contains ρ, whence, σ = cong(ρ). A presentation for S is a generating system of ker ϕ as a congruence, and a minimal presentation is a presentation such that none of its proper subsets is a presentation. Example 37 For instance, a minimal presentation for S = 2, 3 is {((3, 0), (0, 2))}. This means that S is the commutative monoid generated by two elements, say a and b, under the relation 3a = 2b.

5.2 Free Numerical Semigroups In this section, we recall the concept of free numerical semigroup and show how to compute a presentation for these semigroups. Let S be the semigroup generated by {r0 , . . . , rh } (we are using the same notation as in Chap. 4). Set d1 = r0 and dk = gcd(dk−1 , rk−1 ) for all k ∈ {2, . . . , h + 1} (whence dh+1 = 1), and define ek = dk /dk+1 . Recall that S is free (Sect. 2.3) for the arrangement (r0 , . . . , rh ) of generators if for all k ∈ {1, . . . , h}: (i) ek > 1, (ii) ek rk belongs to the semigroup generated by {r0 , . . . , rk−1 }. Recall also that every s ∈ S has a unique standard representation, thatis, there h exists a0 , . . . , ah ∈ N such that ai < ei for all i ∈ {1, . . . , h} and s = k=0 ai ri (Lemmas 10 and 11). We will use  this idea to find a presentation for S. For k ∈ {1, . . . , h}, let ek rk = k−1 j=0 αkk r j be an expression of ek r k in terms of rk−1 . r0 , . . . , rk−1 . We know that we have such an expression h since ek rk ∈ r0 , . . . , h+1 ai ri . So ker ϕ ⊆ N × In our setting ϕ : Nh+1 → S, ϕ(a0 , . . . , ah ) = k=0 Nh+1 . h+1 with ak = 1 and ai = 0 for Let us denote by e k the element (a0 , . . . , ah ) ∈ N k−1 i = k. Hence (ek ek , j=0 αk j e j ) ∈ ker ϕ. Abusing of notation, we will say that (a0 , . . . , ah ) ∈ Nh+1 is in standard form if ak < ek for all k ∈ {1, . . . , h}. Assume that a = (a0 , . . . , ah ) is not in standard form. Then, there exists k ∈ {1, . . . , h} such that ak ≥ ek . It follows that  (ek ek , k−1 j=0 αk j e j ) + (a − ek ek , a − ek ek )   = (a0 , . . . , ah ), (a0 + αk0 , . . . , ak−1 + αkk−1 , ak − ek , ak+1 , . . . , ah ) ∈ ker ϕ.

Observe that if we repeat this procedure (a finite number of steps), we find a  = (a0 , . . . , ah ) in standard form such that (a, a  ) ∈ ker ϕ. According to Lemma 10, a  is unique, that is, there is no other a  in standard form such that (a, a  ) ∈ ker ϕ. Let us denote a  by sf(a).

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From this construction, we deduce that (a, b) ∈ ker ϕ if and only if sf(a) = sf(b). This also is telling that ker ϕ is generated as a congruence by 

ek ek ,

k−1

j=0 αk j ek



   k ∈ {1, . . . , h} ,

since in order to go from a to sf(a) we are only using these relations; and in order to go from sf(a) = sf(b) to b we use their symmetries. By transitivity we can concatenate both chains of transformations. It can be shown that the minimal cardinality of a presentation for S is h (see [69, Chap. 8]), and consequently the above presentation is indeed a minimal presentation. Example 38 Let S = 6, 4, 17, which already appeared in Example 28. Then d1 = 6, d2 = 2 and d3 = 1. Hence e1 = 3 and e2 = 2. This means that 3 × 4 ∈ 6 and 2 × 17 ∈ 6, 4. Let us compute the expressions of 3 × 4 and 2 × 17 in terms of the corresponding generators. • We have 3 × 4 = 2 × 6, and from here we get our first relation: ((2, 0, 0), (0, 3, 0)). • Now, 2 × 17 = 5 × 6 + 4, obtaining the second relation ((5, 1, 0), (0, 0, 2)). Hence a (minimal) presentation for S is ρ = {((2, 0, 0), (0, 3, 0)), ((5, 1, 0), (0, 0, 2))}. Observe that we can choose other presentations, since we can also write 2 × 17 = 3 × 6 + 4 × 4. If we take two expressions of 65 ∈ S, say 65 = 6 + 2 × 4 + 3 × 17 = 4 × 6 + 6 × 4 + 17, we have that ((1, 2, 3), (4, 6, 1)) ∈ ker ϕ. Let us compute the standard forms of (1, 2, 3) and (4, 6, 1). Since 3 ≥ 2, by using the second relation in σ , we go from (1, 2, 3) to (1 + 5, 2 + 1, 3 − 2) = (6, 3, 1), which is equivalent to (1, 2, 3) modulo cong(σ ). Now the second coordinate of this new tuple is larger than or equal to e1 , and we use the first relation and obtain (6 + 2, 3 − 3, 1) = (8, 0, 1), which is in standard form and ((1, 2, 3), (8, 0, 1)) ∈ cong(σ ). We proceed analogously with (4, 6, 1). As 6 ≥ 3, we use the first relation: (4 + 2, 6 − 3, 1) = (6, 3, 1), and then once more (6 + 2, 3 − 3, 1) = (8, 0, 1). We obtain that ((4, 6, 1), (8, 0, 1)) is in the congruence generated by σ . Hence ((1, 2, 3), (4, 5, 6)) ∈ cong(σ ).

5.3 Graphs of Factorizations and Minimal Presentations Let S be minimally generated by {n 1 , . . . , n p }. For s ∈ S, the set of factorizations of s in S is the set Z(s) = ϕ −1 (s) = {a ∈ N p | ϕ(a) = s}.

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103

Notice that the set of factorizations of s has finitely many elements, and corresponds with the set of expressions of s in terms of the generators n 1 , . . . , n p . The finiteness of this set can be shown in different ways. For instance the ith coordinate of a factorization is smaller than or equal to s/n i . Also, two factorizations are incomparable with respect to the usual partial ordering on N p , and thus Dickson’s lemma ensures that there are finitely many of them. We define, associated to s, the graph ∇s whose vertices are the elements of Z(s) and ab is an edge if a · b = 0 (dot product). We say that two factorizations a and b of s are R-related if they belong to the same connected component of ∇s , that is, there exists a chain of factorizations a1 , . . . , at ∈ Z(s) such that • a1 = a, at = b, • for all i ∈ {1, . . . , t − 1}, ai · ai+1 = 0. Example 39 Let S = 5, 7, 11, 13. We draw ∇26 . (3, 0, 1, 0)

(0, 0, 0, 2)

(1, 3, 0, 0)

This graph has two connected components. As ((3, 0, 1, 0), (1, 3, 0, 0)) ∈ ker ϕ, we also have that removing the common part we obtain a new element in the kernel: ((2, 0, 1, 0), (0, 3, 0, 0)). This new element corresponds to 21 = 2 × 5 + 11 = 3 × 7. If we draw ∇21 , we obtain (0, 3, 0, 0)

(2, 0, 1, 0)

which is another nonconnected graph. As we will see next, these nonconnected graphs will be the key tool to build a minimal presentation. Let τ ⊂ N p × N p . We say that τ is compatible with s ∈ S if either ∇s is connected or if R1 , . . . , Rt are the connected components of ∇s , then for every i ∈ {1, . . . , t} we can choose ai ∈ Ri such that for every j ∈ {1, . . . , t}, i = j, there exists i 1 , . . . , i k ∈ {1, . . . , t} fulfilling • i 1 = i, i k = j, • for every m ∈ {1, . . . , k − 1} either (aim , aim+1 ) ∈ τ or (aim+1 , aim ) ∈ τ . Even though this definition might seem strange, we are going to show next that we only have to look at those ∇n that are nonconnected in order to construct a (minimal) presentation. Recall that ei corresponds to the ith row of the p × p identity matrix (now we are indexing again from 1 to p). Theorem 10 Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p }, and let τ ⊆ N p × N p . Then τ is a presentation of S if and only if τ is compatible with s for all s ∈ S.

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Proof Necessity. If ∇s is connected, then there is nothing to prove. Let R1 , . . . , Rt be the R-classes contained in Z(s). Let i and j be in {1, . . . , t} with i = j. Let a ∈ Ri and b ∈ R j . As a, b ∈ Z(n), (a, b) ∈ ker ϕ. Since cong(τ ) = ker ϕ, by Proposition 65, there exist b0 , b1 , . . . , br ∈ N p , such that a = b0 , b = br and (bi , bi+1 ) ∈ τ 1 for i ∈ {0, . . . , r − 1}. Hence, there exist for all i ∈ {0, . . . , r − 1}, z i ∈ N p and (xi , yi ) ∈ τ such that either (bi , bi+1 ) = (xi + z i , yi + z i ) or (bi , bi+1 ) = (yi + z i , xi + z i ). If z i = 0, then bi Rbi+1 . And if z i = 0, then {bi , bi+1 } / R yield the ai ’s we are looking for. ⊆ Z(s). Hence the pairs (bi , bi+1 ) ∈ Sufficiency. It suffices to prove that for every s ∈ S and a, b ∈ Z(s), (a, b) ∈ cong(τ ). We use induction on s. The result follows trivially for s = 0, since Z(0) = {0}. If aRb, then there exists a1 , . . . , ak ∈ Z(s) such that a1 = a, ak = b and ai · ai+1 = 0 for all i ∈ {1, . . . , k − 1}. Hence for every i, there exists j ∈ {1, . . . , p} such that ai − e j , ai+1 − e j ∈ Z(s − n j ). By induction hypothesis (ai − e j , ai+1 − e j ) ∈ cong(τ ), whence (ai , ai+1 ) ∈ cong(τ ) for all i. By transitivity (a, b) ∈ τ . Assume now that a and b are in different connected components of ∇s . If R1 , . . . , Rt are the connected components of ∇s , we may assume without loss of generality that a ∈ R1 and b ∈ R2 . As τ is compatible with s, there exists a chain a1 , . . . , ak such that either (ai , ai+1 ) ∈ τ or (ai+1 , ai ) ∈ τ , a1 ∈ R1 and ak ∈ R2 . Hence (ai , ai+1 ) ∈ cong(τ ), and by the above paragraph, (a, a1 ), (ak , b) ∈ cong(τ ). By transitivity we deduce that (a, b) ∈ cong(τ ). Observe that as a consequence of this theorem, in order to obtain a presentation for S we only need to choose, for every s ∈ S with nonconnected graph ∇s and for every connected component R of ∇s , a factorization x and pairs of this factorizations having the following property: every two connected components of ∇s are connected by a sequence of these factorizations in a way that the pairs of adjacent elements in the sequence are either the ones selected or their symmetries. The least possible number of edges we need is when we choose the pairs so that we obtain a tree connecting all connected components. Thus the least possible number of pairs for every s ∈ S with associated nonconnected graph is the number of connected components of ∇s minus one. Corollary 29 Let S be a numerical semigroup. The cardinality of any minimal presentation of S equals s∈S (nc(∇s ) − 1), where nc(∇s ) is the number of connected components of ∇s . We now show that this cardinality is finite by proving that only finitely many elements of S have nonconnected associated graphs. Proposition 66 Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p }, and let s ∈ S. If ∇s is not connected, then s = n i + w with i ∈ {2, . . . , p} and w ∈ Ap(S, n 1 ) \ {0}. / {n 1 , . . . , n p }, and thus there exists i ∈ Proof Observe that ∇ni = {ei }. Hence s ∈ {1, . . . , p} such that s − n i ∈ S ∗ . If s ∈ Ap(S, n 1 ), then s − n i ∈ Ap(S, n 1 ), and we are done.

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105

Now assume that s − n 1 ∈ S. There exists an element a ∈ Z(s) with a − e1 ∈ N p . Take b ∈ Z(s) in a different connected component of ∇s than the one containing a. / N p . Since b = 0, there exists i ∈ {2, . . . , p} Clearly a · b = 0, and thus b − e1 ∈ p such that b − ei ∈ N , and consequently s − n i ∈ S. We prove that s − (n i + n 1 ) ∈ / S, and thus s = (s − n i ) + n i with s − n i ∈ Ap(S, n 1 ). Suppose to the contrary that s − (n 1 + n i ) ∈ S. Hence there exists a factorization c of s such that c − (e1 + ei ) ∈ N p . Then a · c = 0 and c · b = 0. This forces a and b to be in the same connected component of ∇s , a contradiction. We say that s ∈ S is a Betti element if ∇s is not connected. Example 40 Let S = 4, 5, 7. It easily follows that Ap(S, 4) = {0, 5, 7, 10}. Hence by Proposition 66, the candidates for Betti elements are in the set {5, 7} + {5, 7, 10} = {10, 12, 14, 15, 17}. • • • • •

Z(10) = {(0, 2, 0)}, whence ∇10 is connected. Z(12) = {(3, 0, 0), (0, 1, 1)}, in this case ∇12 is not connected. Z(14) = {(1, 2, 0), (0, 0, 2)} and so ∇14 is not connected. Z(15) = {(0, 3, 0), (2, 0, 1)}, whence ∇15 is not connected. Z(17) = {(3, 1, 0), (0, 2, 1)} and ∇17 has a unique connected component.

In light of Theorem 10, a minimal presentation (and the only one up to rearrangement of the relations) for S is {((3, 0, 0), (0, 1, 1)), ((1, 2, 0), (0, 0, 2)), ((0, 3, 0), (2, 0, 1))}. Example 41 We continue with the semigroup in Example 39. gap> s:=NumericalSemigroup(5,7,11,13);;

We can use the following to compute a minimal presentation for this semigroup. gap> MinimalPresentation(s); [ [ [ 0, 1, 1, 0 ], [ 1, 0, 0, [ [ 0, 3, 0, 0 ], [ 2, 0, 1, [ [ 1, 3, 0, 0 ], [ 0, 0, 0, [ [ 2, 2, 0, 0 ], [ 0, 0, 1, [ [ 3, 1, 0, 0 ], [ 0, 0, 2, [ [ 4, 0, 0, 0 ], [ 0, 1, 0,

1 0 2 1 0 1

] ] ] ] ] ]

], ], ], ], ], ] ]

Let us have a look at ∇50 . gap> Factorizations(50,s); [ [ 10, 0, 0, 0 ], [ 3, 5, 0, 0 ], [ 5, 2, 1, 0 ], [ 0, 4, 2, 0 ], [ 2, 1, 3, 0 ], [ 6, 1, 0, 1 ], [ 1, 3, 1, 1 ], [ 3, 0, 2, 1 ], [ 2, 2, 0, 2 ], [ 0, 0, 1, 3 ] ]

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gap> Length(RClassesOfSetOfFactorizations(last)); 1

And this means that ∇50 has a single connected component, and thus is not a Betti element. We can compute the set of Betti elements. gap> BettiElements(s); [ 18, 20, 21, 22, 24, 26 ]

So for instance ∇26 has two connected components as we already saw in Example 39. gap> Factorizations(26,s); [ [ 1, 3, 0, 0 ], [ 3, 0, 1, 0 ], [ 0, 0, 0, 2 ] ] gap> RClassesOfSetOfFactorizations(last); [ [ [ 1, 3, 0, 0 ], [ 3, 0, 1, 0 ] ], [ [ 0, 0, 0, 2 ] ] ]

5.4 Presentations and Binomial Ideals We now show an alternative method to compute a presentation based on what is known in the literature as Herzog’s correspondence [50]. }. For K a Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p field, recall that the semigroup ring associated to S is the ring K[S] = s∈S Kt s , where t is a symbol or an indeterminate. We can see the elements in K[S] as polynomials in t whose nonnegative coefficients correspond to exponents in S. Also K[S] = K[t n 1 , . . . , t n p ] ⊆ K[t]. Thus, K[S] can be seen as the coordinate ring of a curve parametrized by monomials. Let x1 , . . . , x p be indeterminates, and K[x1 , . . . , x p ] be the polynomial ring over these indeterminates with coefficients in the field K. For a = (a1 , . . . a p ) ∈ N p write a

X a = x1a1 · · · x p p . Let ψ the ring homomorphism determined by ψ : K[x1 , . . . , x p ] → K[S], xi → t ni . x p ] in the following This can be seen as a graded morphism if we grade K[x1 , . . . , way: a polynomial p is S-homogeneous of degree s ∈ S if p = a∈A ca X a for some A ⊂ N p , with finitely many elements and ϕ(a) = s for all a ∈ A. Observe that K[S] is also S-graded in a natural way, and so ψ is a graded epimorphism. For A ⊆ K[x1 , . . . , x p ], denote by (A) the ideal generated by A.

5.4 Presentations and Binomial Ideals

107

Proposition 67 ker ψ = (X a − X b | (a, b) ∈ ker ϕ). Proof Clearly ψ(X a ) = t ϕ(a) . Hence for (a, b) ∈ ker ϕ, ψ(X a − X b ) = 0. This implies that (X a − X b | (a, b) ∈ ker ϕ) ⊆ ker ψ. Since ψ is a graded morphism, for the other inclusion, it suffices to proof that if f ∈ ker ψ is S-homogeneous of degree  a s ∈ S, then f ∈ (X a − X b | (a, b) ∈ ker ϕ). Write f = a∈A c a X , with ca ∈ K s and a ∈ Z(s)  for all a ∈ A, and A a finite set. Then ϕ( a∈A ca = 0, and f ) = t consequently a∈A ca = 0. Choose a ∈ A. Then f = a  ∈A\{a} ca  (X a − X a ), and thus f ∈ (X a − X b | (a, b) ∈ ker ϕ). From Proposition 65, it can be easily derived that for any τ ∈ N p × N p (X a − X b | (a, b) ∈ τ ) = (X a − X b | (a, b) ∈ cong(τ )). Hence, we get the following consequence. Corollary 30 Let S be a numerical semigroup and τ a presentation of S. Then ker ψ = (X a − X b | (a, b) ∈ τ ). The converse is also true. We can go from a system of generators of ker ψ consisting of binomials to a presentation of ker φ. Proposition 68 Let S be a numerical semigroup of embedding dimension p. Assume that τ ∈ N p × N p is such that ker ψ = (X a − X b | (a, b) ∈ τ ). Then τ is a presentation of S. Since Proof Assume that (a, b) ∈ ker φ. Let us write any r ∈ N p × N p as (r1 , r2 ).  X a − X b ∈ ker ψ, there exists fr ∈ K[x1 , . . . , x p ] such that X a − X b = r ∈τ fr (X r1 − X r2 ), with all fr zero except for finitely many of them. Notice that X c (X r1 − X r2 ) = X r1 +c − X r2 +c , and that (r1 + c, r2 + c) ∈ τ1 (the ∈ K, with r ∈ τ1 , all definition of τ1 is given in Proposition 65). Hence there are cr  but finitely many of then equal to zero, such that X a − X b = r ∈τ1 cr (X r1 − X r2 ). From this expression it is not difficult to prove that (a, b) ∈ cong(τ ). Observe that the generators of ker ψ can be seen as the implicit equations of the curve whose coordinate ring is K[S]. In this way, we can solve the implicitation problem without the use of elimination theory nor Gröbner bases. Example 42 Let S = 3, 5, 7. Then Ap(S, 3) = {0, 5, 7}. According to Proposition 66, Betti(S) ⊆ {10, 12, 14}. The sets of factorizations of 10, 12 and 14 are {(0, 2, 0), (1, 0, 1)}, {(4, 0, 0), (0, 1, 1)} and {(3, 1, 0), (0, 0, 2)}, respectively. Hence Betti(S) = {10, 12, 14}, and by Theorem 10, {((0, 2, 0), (1, 0, 1)), ((3, 1, 0), (0, 0, 2)), ((4, 0, 0), (0, 1, 1))} is a minimal presentation of S. The implicit equations of the curve parametrized by (t 3 , t 5 , t 7 ) are

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5 Minimal Presentations

⎧ ⎨ x z − y 2 = 0, x 3 y − z 2 = 0, ⎩ 4 x − yz = 0. Let us reproduce this example with the use of polynomials. Take ψ : K [x, y, z] → K [t] be determined by x → t 3 , y → t 5 and z → t 7 . We consider now the ideal (x − t 3 , y − t 5 , z − t 7 ). We now compute a Gröbner basis with respect to any eliminating order on t. We can for instance do this with Singular, [31]. > ring r=0,(t,x,y,z),lp; > ideal i=(x-tˆ3,y-tˆ5,z-tˆ4); > std(i); _[1]=y4-z5 _[2]=xz3-y3 _[3]=xy-z2 _[4]=x2z-y2 _[5]=x3-yz _[6]=tz-y _[7]=ty-x2 _[8]=tx-z _[9]=t3-x

Now we choose those not having t, or we can just type: > eliminate(i,t); _[1]=y7-z5 _[2]=xz-y2 _[3]=xy5-z4 _[4]=x2y3-z3 _[5]=x3y-z2 _[6]=x4-yz

Which by Herzog’s correspondence yields a presentation for S. However this is not a minimal presentation. In order to get a minimal presentation we can use minbase in Singular, but this applies only to homogeneous ideals. To solve this issue, we give weights 3, 5, 7 to x, y, z, respectively. > ring r=0,(t,x,y,z),(dp(1),wp(3,5,7)); > ideal i=(x-tˆ3,y-tˆ5,z-tˆ7); > ideal j=eliminate(i,t); > minbase(j); _[1]=y2-xz _[2]=x4-yz _[3]=x3y-z2

5.5 Shaded Sets and Generating Functions

109

5.5 Shaded Sets and Generating Functions Let S be a numerical semigroup minimally generated by A = {n 1 , . . . , n e }. For n ∈ S, define the following graph Gn associated to n. Let the vertices of Gn be the elements a ∈ A such that n − a ∈ S, and ab is an edge if n − (a + b) ∈ S. Example 43 Let S = 5, 7, 11, 13 as in Example 39. Then G26 looks like: 7

13

5

11

This is because 26 − a ∈ S for all a ∈ {5, 7, 11, 13}. Also 26 − (5 + 7) = 26 − 12 = 14 = 2 × 7 ∈ S; 26 − (5 + 11) = 10 = 2 × 5 ∈ S; while 26 − (7 + 13) = 6∈ / S, 26 − (7 + 11) = 26 − (5 + 13) = 8 ∈ / S, and 26 − (11 + 13) = 2 ∈ / S. Theorem 11 The graphs Gn and ∇n have the same number of connected components. Proof Let V be the set of vertices of Gn and E the set of edges of Gn . Let C1 , . . . , Ck be the connected components of ∇n . Set W j = {n i ∈ A | z i = 0 for some (z 1 , . . . , z e ) ∈ C j }. We are going to prove that W1 , . . . , Wk are the sets of vertices of the connected components of Gn . Take z = (z 1 , . . . , z e ) ∈ C j . If z i = 0, for some i ∈ {1, . . . , e}, then n = z 1 n 1 + · · · + z e n e with z i = 0, and thus n − n i ∈ S. Hence W j is contained in the set of vertices of Gn for all j. Also for every n i ∈ V , we have n − n i ∈ S, and consequently there exists z = (z 1 , . . . , z e ) ∈ Z(n) with z i = 0. If z ∈ C j , then n i ∈ W j . This proves V = W1 ∪ · · · ∪ Wk . Assume that Wi ∩ W j = 0. Let n l ∈ Wi ∩ W j . By definition, there exists z = (z 1 , . . . , z e ) ∈ Ci and z  = (z 1 , . . . , z e ) ∈ Z j such that zl = 0 = zl . But then z · z  = 0, meaning that z and z  are in the same connected component. Hence i = j. This shows that W1 , . . . , Wk is a partition of V . Take z = (z 1 , . . . , z e ) ∈ Cl with z i = 0 = z j . Then n i , n j ∈ Wl and n i n j is in E. Also, for any other z  = (z 1 , . . . , z e ) ∈ C such that z · z  = 0, and any k ∈ {1, . . . , e} such that z k = 0 there is a path connecting n i and n k in Gn . To see this, as z · z  = 0, there exists some m such that z m = 0 = z m . Then n i n m , n m n k ∈ E. Now, let n i , n j ∈ Wl , with i = j. Then there exists z = (z 1 , . . . , z e ) and z  =  (z 1 , . . . , z e ) in C, such that z i = 0 = z j . From the definition of ∇n , there exists a chain u 1 , . . . , u t of factorizations of n such that z = u 1 , z  = u n and u i · u i+1 = 0 for all i ∈ {1, . . . , t − 1}. By using the preceding paragraph we deduce that there is a path joining n i and n j in Gn . Now take n i ∈ Wl and n j ∈ Wm , and assume that n i n j is in E. Then, there is a factorization z = (z 1 , . . . , z e ) of n with z i = 0 = z j . Assume that z ∈ Cr for some r ∈ {1, . . . , k}. Then n i ∈ Wl ∩ Wr , which leads to r = l (W1 , . . . , Wk is a partition of V ). Analogously r = m. Hence there is no edge joining vertices of Wl and Wm ,

110

5 Minimal Presentations

for l = m. This proves that the sets of vertices of the connected components of Gn are precisely W1 , . . . , Wk . Therefore Gn has k connected components. Example 44 The graphs associated to 26 ∈ 5, 7, 9, 11 have two connected components, though they do not have the same number of vertices nor edges (Examples 39 and 43). Observe that the vertices in the connected components of G26 correspond with the support of the connected components of ∇26 . Székely and Wormald defined in [75] a generalization of the graphs Gn . Their definition was introduced for other purposes as we are going to see in this section, and has been used by several authors (probably without knowing their approach) to calculate free resolutions of the semigroup ring associated to a numerical semigroup and to an affine semigroup. Recall that we are assuming that S be a numerical semigroup minimally generated by A = {n 1 , . . . , n e }. For n ∈ S, define     F (n) = V ⊂ A  n −

a∈V

a∈S ,

which is known as the shaded set of n in S. Observe that Gn corresponds with the “layers” of sets of lengths one (vertices) and two (edges). Example 45 Let us go back to Example 43. Recall that for 26 ∈ 5, 7, 11, 13, the graph G26 has two connected components. The set F (26) = {∅, {5}, {7}, {11}, {13}, {5, 7}, {5, 11}}. So in this example F (26) contains no more elements than those inherited from G26 . For 28 ∈ S, we obtain F (28) = {∅, {5}, {5, 7}, {5, 7, 11}, {5, 11}, {5, 13}, {7}, {7, 11}, {11}, {13}}. The graph G28 and the simplicial complex F (28) can be depicted respectively as follows. 13

7

5

13

11

7

5

11

The Euler characteristic of F (n) is χ (F (n)) =



(−1) L .

L∈F (n)

Let x be an indeterminate. Define the Hilbert series (also known as generating function) of S as

5.5 Shaded Sets and Generating Functions

111

H S (x) =



xs.

s∈S

It is not hard to prove that (see for instance [75]) H S (x)(1 − x n 1 ) · · · (1 − x n e ) =



χ (F (s))x s ,

s∈S

and consequently we get the following expression of the Hilbert series of S:  H S (x) =

(1 −

s s∈S χ (F (s))x . x n 1 ) · · · (1 − x n e )

Proposition 69 Let S be a numerical semigroup. Then there are finitely many ele ments in s with χ (F (s)) = 0. In particular, s∈S χ (F (s))x s is a polynomial. Proof Denote as above the minimal generating set of S by A = {n 1 , . . . , n e }. Take s > F(S) + n 1 + · · · + n e (recall that F(S) is  largest the Frobenius number of S; the s − integer no in S). Then for every L ⊆ A, s − a∈L a > F(S), and thus a∈L  a ∈ e (−1)e ei = S. It follows that F (s) consists in all subsets of A. Then χ (F (s)) = i=0 (1 − 1)e = 0. Example 46 Let S = a, b (gcd(a, b) = 1). Then Ha,b (x) =

1 − x ab . (1 − x a )(1 − x b )

To prove this, take s ∈ S. If s − a, s − b, s − (a + b) ∈ S, then we already know from the proof of Proposition 69 that χ (F (s)) = 0. Notice also that if s − (a + b) ∈ S, then s − a and s − b are again in S, and we obtain again zero Euler characteristic. For s ∈ S, with s − a ∈ S and s − b ∈ / S, we have also χ (F (s)) = 1 − 1 = 0 (the same holds swapping a and b). Hence, the only possible elements s in S with χ (F (s)) = 0 are those with Gs not connected, or equivalently, those for which ∇s is not connected (Theorem 11). In view of Proposition 66, s = b + w, with w ∈ Ap(S, a). As gcd(a, b) = 1, if follows easily that Ap(S, a) = {0, b, 2b, . . . , (a − 1)b} (we already mentioned this in Example 5). Hence, the candidates s with χ (F (s)) = 0 are of the form jb with j ∈ {1, . . . , a}. If j ∈ {1, . . . , a − 1}, we have that jb ∈ Ap(S, a), and consequently jb − a ∈ / S, which forces G jb to be a single vertex and thus connected. So j = a and s can only be ab. GAP Example 29 Let S = 3, 5, 7. gap> s:=NumericalSemigroup(3,5,7);; gap> x:=X(Rationals,"x");;

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5 Minimal Presentations

gap> HilbertSeriesOfNumericalSemigroup(s,x); (xˆ5-xˆ4+xˆ3-x+1)/(-x+1) gap> l:=Intersection([1..4+3+5+7],s); [ 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19] gap> List(l,x->ShadedSetOfElementInNumericalSemigroup(x,s)); [ [ [ ], [ 3 ] ], [ [ ], [ 5 ] ], [ [ ], [ 3 ] ], [ [ ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 5 ] ], [ [ ], [ 3 ] ], [ [ ], [ 3 ], [ 3, 7 ], [ 5 ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 5 ] ], [ [ ], [ 3 ], [ 5 ], [ 5, 7 ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 3, 7 ], [ 5 ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 5 ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 3, 5, 7 ], [ 3, 7 ], [ 5 ], [ 5, 7 ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 3, 7 ], [ 5 ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 3, 7 ], [ 5 ], [ 5, 7 ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 3, 5, 7 ], [ 3, 7 ], [ 5 ], [ 5, 7 ], [ 7 ] ], [ [ ], [ 3 ], [ 3, 5 ], [ 3, 7 ], [ 5 ], [ 5, 7 ], [ 7 ] ] ] gap> euler:=function(l) > return Sum(List([1..Length(l)], i->(-1)ˆLength(l[i]))); > end; gap> Filtered(l,x->euler(ShadedSetOfElementInNumericalSemigroup(x,s))0); [ 10, 12, 14, 17, 19 ] gap> HilbertSeriesOfNumericalSemigroup(s,x)*(1-xˆ3)*(1-xˆ5)*(1-xˆ7); xˆ19+xˆ17-xˆ14-xˆ12-xˆ10+1

Hence H S (x) =

1 − x 10 − x 12 − x 14 + x 17 + x 19 . (1 − x 3 )(1 − x 5 )(1 − x 7 )

Notice that here Betti(S) = {10, 12, 14}. There is an alternative way to express the Hilbert series associated to a numerical semigroup S. Let n ∈ S \ {0}. In light of Proposition 4, every s ∈ S can be expressed uniquely as s = kn + w with k ∈ N and w ∈ Ap(S, n). Hence x s = (x n )k x w . Then H S (x) =

 s∈S

xs =





(x n )k x w =

 

k∈N w∈Ap(S,n)

⎛ (x n )k ⎝



⎞ xw⎠ ,

w∈Ap(S,n)

k∈N

and we recover the formula given in [66]:  H S (x) =

w∈Ap(S,n) 1 − xn

xw

.

   1 = n∈N x n = s∈S x s + h∈G(S) x h . As G(S), the set of gaps of S, Also, 1−x  has finitely many elements, P S (x) = (1 − x)H S (x) = 1 − (1 − x) h∈G(S) x h is a polynomial. Its properties have been studied in [61]. In particular, S is symmetric if and only if PS (x) is a palindrome. Also connections with Bernoulli numbers are given in that paper.

5.5 Shaded Sets and Generating Functions

113

Some authors generalized ∇n instead of Gn in the following way, obtaining also a simplicial complex. Let n be an element in a numerical semigroup, and let Z(n) be its set of factorizations. The set N p is a lattice with respect to the partial ordering ≤. Infimum and supremum of a set with two elements is constructed by taking minimum and maximum coordinate by coordinate, respectively. For x = (x1 , . . . , x p ), y = (y1 , . . . , y p ) ∈ N p , inf{x, y} will be denoted by x ∧ y. Thus x ∧ y = (min{x1 , y1 }, . . . , min{x p , y p }). In the same way, we can define ∧a∈A a for A a finite set. We can define the alternative shaded set of n as F  (n) = {A ⊆ Z(n) | ∧a∈A a = 0}. The set F  (n) and F (n) share some nice properties [64], in particular, they have the same Euler characteristic.

Chapter 6

Factorizations and Divisibility

Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p }. For s ∈ S, recall that the set of factorizations of s is Z(s) = ϕ −1 (s), which is the set of nonnegative integer solutions to n 1 x1 + · · · + n p x p = s.  For convenience, given X ⊆ S, we denote by Z(X ) = s∈S Z(x). Observe that if S = N, then there will be always elements with more than one factorization. Hence nontrivial numerical semigroups are never unique factorization monoids. This terminology clearly comes from the concept of unique factorization domain (like Z or K[x]). In a domain (D, +, ·), if we are looking for factorizations we are working in (D, ·), which is a commutative cancellative monoid. This is why many factorization properties are nowadays being studied over monoids. We review some nonunique factorization invariants in the scope of numerical semigroups; the definitions are the same in the more general setting of monoids.

6.1 Length Based Invariants For a factorization x = (x1 , . . . , x p ) of s its length is defined as |x| = x1 + · · · + x p , and the set of lengths of s is L(s) = {|x| | x ∈ Z(s)} . Since Z(s) has finitely many elements, so has L(s). A monoid is half-factorial if the cardinality of L(s) is one for all s ∈ S.

© Springer Nature Switzerland AG 2020 A. Assi et al., Numerical Semigroups and Applications, RSME Springer Series 3, https://doi.org/10.1007/978-3-030-54943-5_6

115

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6 Factorizations and Divisibility

Example 47 Let S = 2, 3. Here 6 factors as 6 = 2 × 3 = 3 × 2, that is, Z(6) = {(3, 0), (0, 2)}. The length of (3, 0) is 3, while that of (0, 2) is 2. So S is not a unique factorization monoid, and it is not either a half-factorial monoid. The only half-factorial numerical semigroup is N. One of the first nonunique factorization invariants that appeared in the literature was the elasticity. It was meant to measure how far is a domain from being halffactorial. The elasticity of a numerical semigroup is a rational number greater than one. Actually, half-factorial monoids are those having elasticity one. Let s ∈ S. The elasticity of s, denoted by ρ(s) is defined as ρ(s) =

max L(s) . min L(s)

The elasticity of S is defined as ρ(S) = sup{ρ(s) | s ∈ S}. The computation of the elasticity in finitely generated cancellative monoids requires the calculation of primitive elements of ker ϕ. However in numerical semigroups, this calculation is quite simple, as the following result shows [48, Example 3.1.6]. Theorem 12 Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p } with n 1 < · · · < n p . Then np ρ(S) = . n1 Proof Let s ∈ S and assume that a = (a1 , . . . , a p ) and b = (b1 , . . . , b p ) are such that |a| = max L(S) and |b| = min L(S). We know that ϕ(a) = ϕ(b), that is, a1 n 1 + · · · + a p n p = b1 n 1 + · · · + b p n p = s. Now by using that n 1 < · · · < n p , we deduce that n 1 |a| ≤ s ≤ n p |b|, and thus ρ(s) = n

np |a| ≤ . |b| n1 n

This implies that ρ(S) ≤ n 1p . Also ρ(n 1 n p ) ≥ n 1p , since n p e1 , n 1 e p ∈ Z(n 1 n p ). Hence np np ≤ ρ(n 1 n p ) ≤ ρ(S) ≤ , n1 n1 and we get an equality. Another way to measure how far we are from half-factoriality, is to determine how distant the different lengths of factorizations are. This is the motivation for the following definition.

6.1 Length Based Invariants

117

Assume that L(s) = {l1 < · · · < lk }. Define the Delta set of s as (s) = {l2 − l1 , . . . , lk − lk−1 }, and if k = 1, (s) = ∅. The Delta set of S is defined as (S) =



(s).

s∈S

So, the bigger (S) is, the farther is S from being half-factorial. A pair of elements (a, b) ∈ N p × N p is in ker ϕ if a and b are factorizations of the same element in S. As a presentation is a system of generators of ker ϕ it seems natural that the information on the factorizations could be recovered from it. We start showing that this is the case with the Delta sets, and will see later that the same holds for other invariants. Let M S = {a − b | (a, b) ∈ ker ϕ} ⊆ Z p . Since ker ϕ is a congruence, it easily follows that M S is a subgroup of Z p . Lemma 36 Let σ be a presentation of S. Then M S is generated as a group by {a − b | (a, b) ∈ σ }. Proof Let z ∈ M S . Then there exists (a, b) ∈ ker ϕ such that z = a − b. From Proposition 65, there exists x1 , . . . , xt such that x1 = a, xt = b, and for all i ∈ {1, . . . , t − 1} there exists (ai , bi ) and ci ∈ N p such that (xi , xi+1 ) = (ai + ci , bi + ci ) with either (ai , bi ) ∈ σ or (bi , ai ) ∈ σ . Then a − b = (x1 − x2 ) + (x2 − x3 ) + · · · + (xt−1 − xt ) =

t−1  (ai − bi ), i=1

and the proof follows easily. Example 48 Let S = 5, 7, 11, 13. The group M S coincides with the kernel of the ¯ y, z, t) = 5x + 7y + 11z + 13t, or equivalently, group morphism ϕ¯ : Z4 → Z, ϕ(x, the group given by the single equation 5x + 7y + 11z + 13t = 0. A basis of this kernel can be computed for instance by using Smith normal form. gap> BasisOfGroupGivenByEquations([[5,7,11,13]],[]); [ [ 1, -1, -1, 1 ], [ 1, -2, 2, -1 ], [ -2, -2, 1, 1 ] ]

Observe that from a basis of M S = ker ϕ¯ one can obtain, in general, a minimal generating system of ker ϕ as a congruence.

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The system of generators of M S in Lemma 36 is: gap> s:=NumericalSemigroup(5,7,11,13);

gap> MinimalPresentation(s); [ [ [ 0, 1, 1, 0 ], [ 1, 0, 0, 1 ] ], [ [ 0, 3, 0, 0 ], [ 2, 0, 1, 0 ] ], [ [ 1, 3, 0, 0 ], [ 0, 0, 0, 2 ] ], [ [ 2, 2, 0, 0 ], [ 0, 0, 1, 1 ] ], [ [ 3, 1, 0, 0 ], [ 0, 0, 2, 0 ] ], [ [ 4, 0, 0, 0 ], [ 0, 1, 0, 1 ] ] ] gap> List(last, x->x[2]-x[1]); [ [ 1, -1, -1, 1 ], [ 2, -3, 1, 0 ], [ -1, -3, 0, 2 ], [ -2, -2, 1, 1 ], [ -3, -1, 2, 0 ], [ -4, 1, 0, 1 ] ]

Observe that it is not a basis. For a given z = (z 1 , . . . , z p ) ∈ Z p , we overload the notation |z| = z 1 + · · · + z p . Lemma 37 Let σ = {(a1 , b1 ), . . . , (at , bt )} be a presentation of S, and set δi = |ai − bi |, i ∈ {1, . . . , s}. Then every element in (S) is of the form λ 1 δ1 + · · · + λ t δt , for some integers λ1 , . . . , λt . Proof The proof follows easily from the proof of Lemma 36. With this we are ready to determine the extreme values of the Delta set of a numerical semigroup. Theorem 13 Let S be a numerical semigroup and let σ be a presentation of S. Then min (S) = gcd{|a − b| | (a, b) ∈ σ }. Proof In order to simplify notation, write d = gcd{|a − b| | (a, b) ∈ σ }. If δ ∈ (S), then by Lemma 37, we know that δ is a linear combination with integer coefficients of elements of the form |a − b| with (a, b) ∈ σ . Hence d | δ, and consequently d ≤ min (S). Now let (a1 , b1 ), . . . , (ak , bk ) ∈ σ and λ1 , . . . , λk ∈ Z be such that λ1 |a1 − b1 | + · · · + λk |ak − bk | = d. If λi < 0, change (ai , bi ) with (bi − ai ), so that we can assume that all λi are nonnegative. The element s = ϕ(λ1 a1 + · · · + λk ak ) = ϕ(λ1 b1 + · · · + λk bk ) has two factorizations z = λ1 a1 + · · · + λk ak and z  = λ1 b1 + · · · + λk bk such that the difference in their lengths is d. Hence min (S) ≤ min (s) ≤ d ≤ min (S), and we get an equality.

6.1 Length Based Invariants

119

This formula holds in a much more general setting (see for instance [48]). Theorem 14 Let S be a numerical semigroup. Then max (S) = max{max (b) | b ∈ Betti(S)}. Proof The inequality maxn∈Betti(S) max (n) ≤ max (S) is clear. Assume to the contrary max (S) > max (b) for all Betti elements b of S. Take x, y factorizations of an element s ∈ S so that |x| − |y| = max (s) with max (s) > max (b) for all b ∈ Betti(S). Consequently no other factorization z of s fulfills |x| < |z| < |y|. As ϕ(x) = ϕ(y), Proposition 65, ensures the existence of x1 , . . . , xt in Z(s) such that x = x1 , xt = y and (xi , xi+1 ) = (ai + ci , bi + ci ), with either (ai , bi ) ∈ σ or (bi , ai ) ∈ σ for all i ∈ {1, . . . , t − 1}. From the above discussion, there exists i ∈ {1, . . . , t − 1}, with |xi | ≤ |x| < |y| ≤ |xi+1 |. Both ai and bi are factorizations of an element n with Z(n) having more than one R-class. So there is a chain of factorizations, say z 1 , . . . , z u , of n such that ai = z 1 , . . . , z u = bi , and |z j+1 | − |z j | ≤ max (n), which we are assuming smaller than (S). But then ϕ(z j + ci ) = ϕ(x) = ϕ(y) for all j, and from the choice of x and y, there is no j such that |x| < |z j + ci | < |y|. Again, we can find j ∈ {1, . . . , u − 1} such that |z j + ci | ≤ |x| < |y| ≤ |z j+1 + ci |. And this leads to a contradiction, since max (S) = |y| − |x| ≤ |z j+1 + ci | − |z j + ci | = |z i+1 − z i | ≤ max (n) < max (S). Again, this description of the maximum of the Delta set is applicable to a wider family of monoids [26]. Example 49 Let us go back to S = 2, 3. We know that the only Betti element of S is 6. The set of factorizations of 6 is Z(6) = {(3, 0), (0, 2)}, and L(S) = {2, 3}. Whence (6) = {1}. The above theorem implies that (S) = {1}. This is actually the closest we can be in a numerical semigroup to be half-factorial. Example 50 Now we do some computations with a numerical semigroup with four generators. gap> s:=NumericalSemigroup(10,11,17,23);; gap> Factorizations(60,s); [ [ 6, 0, 0, 0 ], [ 1, 3, 1, 0 ], [ 2, 0, 1, 1 ] ] gap> LengthsOfFactorizationsElementWRTNumericalSemigroup(60,s); [ 4, 5, 6 ] gap> Elasticity(60,s); 3/2 gap> DeltaSet(60,s); [ 1 ] gap> BettiElements(s); [ 33, 34, 40, 69 ]

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gap> Set(last, x->DeltaSet(x,s)); [ [ ], [ 1 ], [ 2 ], [ 3 ] ] gap> Elasticity(s); 23/10

6.2 Distance Based Invariants We now introduce some invariants that depend on distances between factorizations. These invariants will measure how spread are the factorizations of elements in the monoid. Recall that for x = (x1 , . . . , x p ), y = (y1 , . . . , y p ) ∈ N p , their infimum, inf{x, y}, in (N p , ≤) is x ∧ y = (min{x1 , y1 }, . . . , min{x p , y p }). The distance between x and y is defined as d(x, y) = max{|x − (x ∧ y)|, |y − (x ∧ y)|} (equivalently d(x, y) = max{|x|, |y|} − |x ∧ y|). The distance between two factorizations of the same element is lower bounded in the following way. Lemma 38 Let x, y ∈ N p with x = y and ϕ(x) = ϕ(y). Then   2 + |x| − |y| ≤ d(x, y). Proof We can assume that x ∧ y = 0, since distance is preserved under translations, ||x| − |y|| = ||x − (x ∧ y)| − |y − (x ∧ y)|| and ϕ(x − (x ∧ y)) = ϕ(y − (x ∧ y)). As ϕ(x) = ϕ(y) and x = y, in particular we have that |x| ≥ 2 and the same for |y|. Also, as x ∧ y = 0, d(x, y) = max{|x|, |y|}. If |x| ≥ |y|, then 2 + ||x| − |y|| = |x| + (2 − |y|) ≤ |x| = d(x, y). A similar argument applies for |x| ≤ |y|. Example 51 The factorizations of 66 ∈ 6, 9, 11 are Z(66) = {(0, 0, 6), (1, 3, 3), (2, 6, 0), (4, 1, 3), (5, 4, 0), (8, 2, 0), (11, 0, 0)}. The distance between (11, 0, 0) and (0, 0, 6) is 11. However, we can put other factorizations of 66 between them so that the maximum distance of two consecutive links (sticks in the figure) is at most 4:

6.2 Distance Based Invariants (11, 0, 0)

(8, 2, 0) 3

(3, 0, 0)

121 (5, 4, 0) 3

(0, 2, 0)|(3, 0, 0)

(2, 6, 0) 3

(0, 2, 0)|(3, 0, 0)

(1, 3, 3) 4

(0, 2, 0)|(1, 3, 0)

(0, 0, 6) 4

(0, 0, 3)|(1, 3, 0)

(0, 0, 3)

In the above picture the factorizations are depicted in the top of a post, and they are linked by a “catenary” labeled with the distance between two consecutive sticks. On the bottom we have drawn the factorizations removing the common part with the one on the left and that of the right, respectively. We will say that the catenary degree of 66 in 6, 9, 11 is at most 4. A minimal presentation of S is σ = {((1, 3, 0), (0, 0, 3)), ((3, 0, 0), (0, 2, 0))}. Notice also that this picture is showing us how to go from (11, 0, 0) to (0, 0, 6) by using the relations in σ . For instance, as ((3, 0, 0), (0, 2, 0)) is in σ , we have that ((3 + 8, 0, 0), (0 + 8, 2, 0)) is in the congruence generated by σ , and consequently ((11, 0, 0), (8, 2, 0)) ∈ ker ϕ, that is, (11, 0, 0) and (8, 2, 0) are factorizations of the same element. Since translations preserve distances, in this example, the catenary degree will be at most the maximum of d((1, 3, 0), (0, 0, 3)) = 4 and d((3, 0, 0), (0, 0, 2)) = 3. Given s ∈ S, x, y ∈ Z(s) and a nonnegative integer N , an N -chain joining x and y is a sequence x1 , . . . , xk ∈ Z(s) such that • x1 = x, xk = y, • for all i ∈ {1, . . . , k − 1}, d(xi , xi+1 ) ≤ N . The catenary degree of s, denoted c(s), is the least N such that for any two factorizations x, y ∈ Z(s), there is an N -chain joining them. The catenary degree of S, c(S), is defined as c(S) = sup{c(s) | s ∈ S}. Example 52 Let us compute the catenary degree of 77 ∈ S = 10, 11, 23, 35. We start with a complete graph with vertices the factorizations of 77 and edges labeled with the distances between them. Then we remove one edge with maximum distance, and we repeat the process until we find a bridge. The label of that bridge is then the catenary degree of 77.

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6 Factorizations and Divisibility (1, 4, 1, 0)

(2, 2, 0, 1)

3

3

2

(1, 4, 1, 0)

3 5

2 3

3

5

(2, 2, 0, 1)

3

(0, 7, 0, 0)

6

(2, 1, 2, 0)

(0, 7, 0, 0)

(2, 1, 2, 0)

(1, 4, 1, 0)

3

(2, 2, 0, 1)

(1, 4, 1, 0)

(2, 2, 0, 1)

3

2

3 3

3

(0, 7, 0, 0)

2

(2, 1, 2, 0)

(0, 7, 0, 0)

(2, 1, 2, 0)

Thus the catenary degree of 77 is 3. If one looks at Proposition 65 and Example 51, one sees some interconnection between the transitivity and the way we can move from one factorization to another to minimize distances. This idea is exploited in the following result. Theorem 15 Let S be a numerical semigroup. Then c(S) = max{c(b) | b ∈ Betti(S)}. Proof Set c = maxb∈Betti(S) c(b). Clearly c ≤ c(S). Let us prove the other inequality. Take s ∈ S and x, y ∈ Z(s). Let σ be a minimal presentation of S. Then by Proposition 65, there exists a sequence x1 , . . . , xk such that x1 = x, xk = y, and for every i there exists ci ∈ N p (with p the embedding dimension of S) such that (xi , xi+1 ) = (ai + ci , bi + ci ) for some (ai , bi ) such that either (ai , bi ) ∈ σ or (bi , ai ) ∈ σ . According to Theorem 10, ai , bi are factorizations of a Betti element of S. By using the definition of catenary degree, there is a c-chain joining ai and bi (also bi and ai ). If we add ci to all the elements of this sequence, we have a c-chain joining xi and xi+1 (distance is preserved under translations). By concatenating all these c-chains for i ∈ {1, . . . , k − 1} we obtain a c-chain joining x and y. And this proves that c(S) ≤ c, and the equality follows. Example 53 With the package numericalsgps the catenary degree of an element and of the whole semigroup can be obtained as follows. gap> s:=NumericalSemigroup(10,11,17,23);; gap> Factorizations(60,s);

6.2 Distance Based Invariants

123

[ [ 6, 0, 0, 0 ], [ 1, 3, 1, 0 ], [ 2, 0, 1, 1 ] ] gap> CatenaryDegree(60,s); 4 gap> BettiElements(s); [ 33, 34, 40, 69 ] gap> List(last, x-> CatenaryDegree(x,s)); [ 3, 2, 4, 6 ] gap> CatenaryDegree(s); 6

Let n ∈ S. Assume that n − n i ∈ S for some i ∈ {1, . . . , p}. We define ti (n) = sup{d(z, Z(n) ∩ (ei + N p )) | z ∈ Z(n)} (recall that d(z, A) = mina∈A d(z, a)). Notice that the above supremum is a maximum, since the set Z(n) has finitely many elements. We define the tame degree of n as t(n) = max{ti (n) | n − n i ∈ S, 1 ≤ i ≤ p}, and the tame degree of S as t(S) = sup t(n). n∈S

We will see that this supremum is also a maximum. Notice that we can define ti (s) = sups∈S ti (S), and thus t(S) = sup{ti (S) | i ∈ {1, . . . , p}}. For i ∈ {1, . . . , p}, define Si to be the set of s ∈ S such that Z(s) ∩ Minimals≤ (Z(n i + S)) is not empty. Observe that by Dickson’s lemma, the set Minimals≤ (Z(n i + S)) has finite cardinality, and consequently Si has finitely many elements. The following result is a particularization of [47, Theorem 5.2] for numerical semigroups. Theorem 16 Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p }. Then ti (S) = max{ti (s) | s ∈ Si }. Proof Since Si ⊆ S, max{ti (s) | s ∈ Si } ≤ ti (S). For the other inequality, if ti (S) = 0, we are done. So, assume that ti (S) = 0, and so there exists n ∈ n i + S with ti (n) > 0. Let z, w ∈ Z(n) be such that ti (n) = d(z, w) = d(z, Z(n) ∩ (ei + N p )), with z · ei = 0, and w · ei = 0. As z ∈ Z(n i + S), there exists x ∈ Minimals≤ (Z(n i + S)) such that x ≤ z. Observe that x · ei = 0. Let y = z − x, which is in N p , and set m = ϕ(x), whence m ∈ Si . By definition of ti (m), there exists x  ∈ Z(m) with x  · ei = 0 and d(x, x  ) ≤ ti (m). Denote z  = x  + y. Since z  − ei = (x  − ei ) + y ∈ N p , the choice of w yields d(z, w) ≤ d(z, z  ). Then ti (n) = d(z, w) ≤ d(z, z  ) = d(x, x  ) ≤ ti (m). This implies that ti (n) ≤ max{ti (s) | s ∈ Si }, and consequently ti (S) ≤ max{ti (s) | s ∈ Si }.

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As a consequence of the following result and the above theorem, we recover [25, Theorem 16]. Proposition 70 Let S be a nontrivial numerical semigroup minimally generated by {n 1 , . . . , n p }, and let s ∈ Si . Then there exists j = i, j ∈ {1, . . . , p} such that si = n i + w for some w ∈ Ap(S, n j ). Proof From the definition of Si , s has a factorization z that is minimal in Z(n i + S) and z · ei = 0. As z = 0, we can choose j = i, j ∈ {1, . . . , p} with z · e j = 0. This in particular implies that s − n j ∈ S (and we also have that s − n i ∈ S). Assume that s − (n i + n j ) ∈ S. Then z − e j ∈ Z(s − n j ), s − n j ∈ n i + S, and z − e j < z, contradicting the minimality of z. Hence, s = n i + (s − n i ), and s − n i ∈ Ap(S, n j ). We obtain that the tame degree of a nontrivial numerical semigroup is attained in an element n with associated graph Gn not connected. Theorem 17 Let S be a numerical semigroup other than N, and let n ∈ S be minimal satisfying t(n) = t(S). Then the graph Gn is not complete. Proof Let {n 1 , . . . , n p } be the minimal generating system of S. By definition t(S) = ti (S) for some i ∈ {1, . . . , p}. And by Theorem 16, we have that ti (S) = ti (s) for some s ∈ Si . Now from the proof of Proposition 70 it follows that there exists a / S. It follows minimal generator n j = n i such that s − n j ∈ S, and s − (n i + n j ) ∈ that there is no edge connecting the vertices n i and n j of Gs , and thus this graph is not complete. Remark 12 Let S be a numerical semigroup with Frobenius number f and minimal set of generators {n 1 , . . . , n p }. Then for any n > f + n 1 + · · · + n p , the graph Gn is complete. GAP Example 30 Let S = 5, 6, 7. gap> gap> gap> [ 0, gap> 4

s:=NumericalSemigroup(3,5,7);; c:=Conductor(s);; Set(Intersection(s,[1..c+3+6+7]),x->TameDegree(s,x)); 2, 4 ] TameDegree(s);

6.3 How Far Is an Irreducible from Being Prime As the title suggests, the last invariant we are going to present measures how far is an irreducible from being prime. Recall that a prime element is an element such that if it divides a product, then it divides one of the factors. Numerical semigroups

6.3 How Far Is an Irreducible from Being Prime

125

are monoids under addition, and thus the concept of divisibility must be defined accordingly. Given s, s  ∈ S, recall that we write s ≤ S s  if s  − s ∈ S. We will say that s divides  s . Observe that s divides s  if and only if s  belongs to the ideal s + S = {s + x | x ∈  S} of S. If s ≤ S s  , then t s divides t s in the semigroup rings K[S] and K[[S]], in the “multiplicative” sense. The ω-primality of s in S, denoted ω(S, s), is the least positive integer N such that whenever s divides a1 + · · · + an for some a1 , . . . , an ∈ S, then s divides ai1 + · · · + ai N for some {i 1 , . . . , i N } ⊆ {1, . . . , n}. Observe that an irreducible element in S (minimal generator) is prime if its ωprimality is 1. It is easy to observe that a numerical semigroup has no primes. In the above definition, we can restrict the search to sums of the form a1 + · · · + an , with a1 , . . . , an minimal generators of S as the following lemma shows. Lemma 39 Let S be numerical semigroup and s ∈ S. Then ω(S, s) is the smallest N ∈ N ∪ {∞} with the following property. For all n ∈ N and a1 , . . . , an a sequence of minimal generators of S, if s divides a1 + · · · + an , then there exists a subset Ω ⊂ [1, n] with cardinality less than or equal to N such that  ai . s ≤S i∈Ω

Proof Let ω (S, s) denote the smallest integer N ∈ N0 ∪ {∞} satisfying the property mentioned in the lemma. We show that ω(S, s) = ω (S, s). By definition, we have ω (S, s) ≤ ω(S, s). In order to show that ω(S, s) ≤ ω (S, s), let n ∈ N and a1 , . . . , an ∈ S with s ≤ S a1 + · · · + an . For every i ∈ [1, n] we pick a factorization ai = u i,1 + · · · + u i,ki with ki ∈ N and u i1 , . . . , u i,ki minimal generators of S. Then there is a subset I ∈ [1, n] and, for every i ∈ I , a subset ∅ = Λi ⊂ [1, ki ] such that #I ≤

 i∈I

and hence s ≤ S

 i∈I

#Λi ≤ ω (S, s) and s ≤ S



u i,ν ,

i∈I ν∈Λi

ai .

In order to compute the ω-primality of an element s in a numerical semigroup S, one has to look at the minimal factorizations (with respect to the usual partial ordering) of the elements in the ideal s + S; this is proved in the next result. Proposition 71 Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p }. Let s ∈ S. Then   ω(S, s) = max |m| | m ∈ Minimals≤ (Z(s + S)) .

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6 Factorizations and Divisibility

Proof Notice that by Dickson’s lemma, the set Minimals≤ (Z(s + S))  has finitely  many elements, and thus N = max |m| | m ∈ Minimals≤ (Z(s + S)) is a nonnegative integer. Choose x = (x1 , . . . , x p ) ∈ Minimals≤ (Z(s + S)) such that |x| = N . Since x ∈ Z(s + S), we have that s divides s  = x1 n 1 + · · · + x p n p . Assume that s divides s  = y1 n 1 + · · · + y p n p with (y1 , . . . , y p ) < (x1 , . . . , x p ) (that is, s divides a proper subset of summands of s  ). Then s  ∈ s + S, and (y1 , . . . , y p ) ∈ Z(s + S), contradicting the minimality of x. This proves that ω(S, s) ≥ N . Now assume that s divides x1 n 1 + · · · + x p n p for some x = (x1 , . . . , x p ) ∈ N p . Then x ∈ Z(s + S), and thus there exists m = (m 1 , . . . , m p ) ∈ Minimals≤ (Z(s + S)) with m ≤ x. By definition, m 1 n 1 + · · · + m p n p ∈ s + S, and |m| ≤ N . This proves with the help of Lemma 39 that ω(S, s) ≤ N . Example 54 Let S = 3, 5, 7 and let s = 10 ∈ S. If we want to use Proposition 71, we have to compute the set Minimals≤ (Z(s + S)). To do this, we find the set of nonnegative integer solutions of 3x + 5y + 7z = 10 + 3u + 5v + 7w. In order to do this, we can use Normaliz [21] or NormalizInterface, which is a gap interface to normaliz [49]. gap> cone:=NmzCone(["inhom_equations",[[3,5,7,-3,-5,-7,-10]]]);

gap> Set(NmzModuleGenerators(cone),x->x{[1..3]}); [ [ 0, 0, 3 ], [ 0, 0, 4 ], [ 0, 0, 5 ], [ 0, 1, 2 ], [ 0, 2, 0 ], [ 1, 0, 1 ], [ 4, 1, 0 ], [ 5, 0, 0 ], [ 8, 0, 0 ] ]

The minimal elements here with respect to ≤ are {(0, 0, 3), (0, 2, 0), (5, 0, 0), (1, 0, 1), (4, 1, 0), (0, 1, 2)}. Hence the ω-primality of 10 in S is 5. For numerical semigroups these computations can be performed using Apéry sets (see [18, Remarks 5.9]). gap> s:=NumericalSemigroup(3,5,7);; gap> OmegaPrimality(10,s); 5

For S a numerical semigroup minimally generated by {n 1 , . . . , n p }, the ωprimality of S is defines as ω(S) = max{ω(S, n i ) | i ∈ {1, . . . , p}}.

6.3 How Far Is an Irreducible from Being Prime

127

We are going to relate Delta sets with catenary degree, tame degree and ωprimality. To this end we need the following technical lemmas. For b = (b1 , . . . , b p ) ∈ N p , define Supp(b) = {i ∈ {1, . . . , p} | bi = 0}. Lemma 40 Let S be a numerical semigroups minimally generated by {n 1 , . . . , n p }, and let n ∈ Betti(S). Let a, b ∈ Z(n) in different R-classes. For every i ∈ Supp(b) we have that a ∈ Minimals≤ Z(n i + S). Proof Assume to the contrary that there exists c ∈ Z(n i + S) and x ∈ Nk \{0} such that c + x = a. From c < a, a · b = 0 and i ∈ Supp(b), we deduce that i ∈ / Supp(c). As c ∈ Z(n i + S), there exists d ∈ Z(n i + S) with i ∈ Supp(d) and ϕ(c) = ϕ(d). Hence ϕ(d + x) = ϕ(c + x) = ϕ(a). Moreover (d + x) · (c + x) = (d + x) · a = 0, and (d + x) · b = 0, which leads to aRb, a contradiction. Lemma 41 Let S be a numerical semigroup minimally generated by {n 1 , . . . , n p }. Let a ∈ Minimals≤ (n i + S) \ {ei }. For every b ∈ Z(ϕ(a)) with i ∈ Supp(b), a · b = 0. Proof Assume that there exists b ∈ N p with ϕ(a) = ϕ(b), a · b = 0 and i ∈ Supp(b). Then there exists j ∈ Supp(a) ∩ Supp(b), and since a = ei , j = i. Notice that ϕ(a − e j ) = ϕ(b − e j ), and as ϕ(b) = n i + n j + s for some s ∈ S, we get that ϕ(a − e j ) = n i + s ∈ n i + S, contradicting the minimality of a. Theorem 18 Let S be a numerical semigroup. Then max (S) + 2 ≤ c(S) ≤ ω(S) ≤ t(S). Proof Assume that d ∈ (S). Then there exists s ∈ S and x, y ∈ Z(s) such that |x| < |y|, d = |y| − |x| and there is no z ∈ Z(s) with |x| < |z| < |y|. From the definition of c(S), there is a c(S)-chain z 1 , . . . , z k joining x and y. As in the proof of Theorem 14, we deduce that there exists i such that |z i | < |x| < |y| < |z i+1 |. Then 2 + d = 2 + |y| − |x| ≤ 2 + |z i+2 | − |z i |, and by Lemma 38, 2 + |z i+2 | − |z i | ≤ d(z i , z i+1 ). The definition of c(S)-chain implies that d(z i , z i+1 ) ≤ c(S). Hence 2 + d ≤ c(S), and consequently max (S) + 2 ≤ c(S). Let σ be a minimal presentation of S. For every (a, b) ∈ σ , there exists n i and n j minimal generators such that a ∈ Minimals≤ Z(n i + S) and b ∈ Minimals≤ Z(n j + S) (Lemma 40). From the definition of ω(S), both |a| and |b| are smaller than or equal to ω(S). Set c = max{max{|a|, |b|} | (a, b) ∈ σ }. Then c ≤ ω(S). Now we prove that c(S) ≤ c. Let s ∈ S and x, y ∈ Z(s). Then ϕ(x) = ϕ(y) and as σ is a presentation, by Proposition 65, there exists a sequence x1 , . . . , xk ∈ N p ( p = e(S)) such that x1 = x, xk = y and for every i there exists ai , bi , ci ∈ N p such that (xi , xi+1 ) = (ai + ci , bi + ci ), with either (ai , bi ) ∈ σ or (bi , ai ) ∈ σ . Notice that d(xi , xi+1 ) = d(ai , bi ) = max{|ai |, |bi |} (ai and bi are in different R-classes and thus ai · bi = 0, or equivalently, ai ∧ bi = 0). Hence d(xi , xi+1 ) ≤ c, and consequently x1 , . . . , xk is a c-chain joining x and y. This implies that c(S) ≤ c, and consequently c(S) ≤ ω(S).

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Assume that ω(S) = ω(S, n i ) and that z ∈ Minimals≤ Z(si + S) is such that |z| = ω(S). If |z| = 1, then z = ei , and S is N. Otherwise, from the definition of t(S), there exists z  ∈ N p with ϕ(z) = ϕ(z  ) such that i ∈ Supp(z  ) and d(z, z  ) ≤ t(S). From the preceding lemma, we get that z · z  = 0, and thus d(z, z  ) = max{|z|, |z  |} ≥ |z| = ω(S). Example 55 Let us go back to S = 10, 11, 17, 23. From Example 50 and Theorem 14, we know that max (S) = 3. gap> OmegaPrimality(s); 6

From Theorem 18, we deduce that c(S) ∈ {5, 6}. Recall that by Example 53, we know that c(S) = 6. There are many other nonunique factorization invariants that can be defined on any numerical semigroup. It was our intention just to show some of them and the last theorem that relates these invariants coming from lengths, distances and primality (respectively), and at the same time show how minimal presentations can be used to study them. The reader interested in this topic is referred to [48], and for a review of the computational aspects to [45].

6.4 Divisors and Feng–Rao Distances We have used several times the partial order ≤ S with S a numerical semigroup. In this section we introduce a concept that has been widely used in one point algebraic geometry codes associated to curves. Let s ∈ S. Recall that the set of divisors of s in S can be defined as D(s) = {n ∈ S | s − n ∈ S}. Lemma 42 D(s) = S ∩ (s − S). Proof An integer n ∈ D(S) if and only if n ∈ S and s − n = t for some t ∈ S, and this is equivalent to n ∈ S ∩ (s − S). Let q be a power of a prime number, and let Fq be the finite field with q elements. Let R be the affine coordinate ring of an absolutely irreducible nonsingular curve over Fq with a single rational point at infinity, say Q. Let P = {P1 , . . . , Pn } be a set of n distinct affine Fq -rational points of the curve. Define the evaluation map on P as follows (see [23]) evP : R → Fqn , evP ( f ) = ( f (P1 ), . . . , f (Pn )).

6.4 Divisors and Feng–Rao Distances

129

Set L(m Q) = { f ∈ R | v Q ( f ) ≥ −m}, where v Q is the discrete valuation at Q. The image of −v Q of R ∗ is a numerical semigroup, say S. The function −v Q applied to f ∈ R is measuring the order of the pole Q in f . Since f is in R, if Q is a pole, then it is the only pole of f . Notice also that if f is a meromorphic series whose only pole is Q (the infinity), then f is forced to be a polynomial. Hence S is the Weierstrass semigroup of the curve at Q. Observe that L(0Q) ⊆ L(1Q) ⊆ L(2Q) ⊆ · · · and that equalities in this chain correspond to the gaps in the Weierstrass semigroup. For s ∈ S, denote by Cs the orthogonal (with respect to the usual dot product) of evP (L(s Q)), which is called the one point algebraic code defined by s Q and P. It is well known that the minimum distance of Cs it as least the Feng–Rao distance (or order bound; called after Feng and Rao since it was first introduced in [40]) of s, which is defined as   δ F R (s) = min D(s  ) | s ≤ s  and s  ∈ S . The asymptotic behavior of the Feng–Rao distance is easy to compute thanks to the following result (see [33, Lemma 5]). Lemma 43 Let c and g be the conductor and genus of S, respectively. Let s be an integer with s ≥ 2c − 1. Then D(s) = s + 1 − 2g. Proof Lemma 42 asserts that D(s) = (S ∩ (s − S)). Observe that S ∩ (s − S) ⊆ [0, s], and so at most we have s + 1 divisors of s. Also s ≥ 2c − 1, which means that is larger than the Frobenius number of S. So the gaps of S are not in D(s), and there are exactly g of them in [0, c]. If h is a gap of S, then s − h ∈ / s − S, and thus we are counting g more elements in [s − c, s] that are not in s − S. Any other integer in [0, s] is both in S and in s − S. And this completes the proof. The number s + 2 − 2g is sometimes known as the Goppa bound. Proposition 72 Let S be a numerical semigroup and let s ≥ 2c − 1, with c the conductor of S. Then δ F R (s) = s + 1 − 2g. Proof By definition and Lemma 43, δ F R (s) = min{s  + 1 − 2g | s ≤ s  and s  ∈ S}, which clearly equals s + 1 − 2g. GAP Example 31 Let us see this behavior with S = 3, 5, 7. gap> ndiv:={x,s}-> Length(DivisorsOfElementInNumericalSemigroup(x,s)); function( x, s ) ... end gap> s:=NumericalSemigroup(3,5,7);;

130

6 Factorizations and Divisibility

gap> c:=ConductorOfNumericalSemigroup(s); 5 gap> List(Intersection([0..15],s), x->ndiv(x,s)); [ 1, 2, 2, 3, 2, 4, 4, 5, 6, 7, 8, 9, 10 ]

The advantage of the Feng–Rao distance is that is computed directly on the semigroup. There is a generalization of the Feng–Rao distance, which is known as generalized Feng–Rao distance: δrF R : S −→ N,

  s ≤ s1 ≤ · · · ≤ sr , s → min D(s1 , . . . , sr )  s1 , . . . , sr ∈ S

 where D(s1 , . . . , sr ) = ri=1 D(si ). As expected its computation is harder than the classical Feng–Rao distance (see for instance [34] for the calculation of this function on numerical semigroups with embedding dimension two or [33] for semigroups generated by intervals). For r = 2, we will see next how the asymptotic behavior of this new distance is related with the Apéry sets that we have used many times in this manuscript. Recall that we defined the Apéry set of s ∈ S\{0} as the set of elements n in S such that n − s ∈ / S. We can extend this definition to any positive integer x, though we already saw that the resulting set will no longer have x elements (GAP example 1). So let x ∈ N\{0}. Define the Apéry set of x in S as Ap(S, x) = {s ∈ S | s − x ∈ / S}. Selmer’s formula for the Frobenius number still holds (Proposition 5). Lemma 44 Let S be a numerical semigroup and let n be a positive integer. Then F(S) + n = max(Ap(S, n)). Proof Let x > F(S) + n. Then by definition of Frobenius number x ∈ S, and x − n > F(S), which implies that x ∈ / Ap(S, n). Clearly, F(S) + n ∈ Ap(S, n). A desert of S is a maximal interval of gaps of S (counting also the set of all nonnegative integers). The following curiosity is a straight consequence of the definition. Proposition 73 The number of deserts of S is Ap(S, 1). In order to compute the generalized Feng–Rao distances of order two, we make use of the following characterization [34, Proposition 11]. Lemma 45 Let m ≥ 2c − 1, with c the conductor of S, and let n be a positive integer. Then D(m + n)\D(m) = m + n − Ap(S, n).

6.4 Divisors and Feng–Rao Distances

131

Proof Take s ∈ D(m + n)\D(m). Then m + n − s ∈ S and m − s ∈ / S. This implies that m + n − s ∈ Ap(S, n), whence s ∈ m + n − Ap(S, n). Now take s ∈ m + n − Ap(S, n), say s = m + n − w with w ∈ Ap(S, n). Notice / S. So it that m + n − (m + n − w) = w ∈ S, and m − (m + n − w) = w − n ∈ remains to prove that m + n − w ∈ S. In light of Lemma 44, w ≤ F(S) + n, and consequently n − w ≥ −F(S). Thus m + n − w ≥ 2c − 1 − F(S) = 2c − 1 − (c − 1) = c, which yields m + n − w ∈ S. Hence (D(m + n)\D(m)) = Ap(S, n). This implies that for every pair of integers m and n such that m ≥ 2c − 1 and n > 0, we have that D(m, m + n) = m + 1 − 2g + Ap(S, n). With this we get the following result describing the asymptotic behavior of δ 2F R . Proposition 74 Let S be a numerical semigroup with conductor c. Then for all m ≥ 2c − 1,   δ 2F R (m) = m + 1 − 2g + min Ap(S, n) | n ∈ N∗ . Observe that the amount min { Ap(S, n) | n ∈ N∗ } is fixed, that is, δ 2F R (m) is a translation of m for m big enough. Example 56 Let S = 3, 5. We know that for n ∈ S, Ap(S, n) = n. So in order to calculate the minimum of all Ap(S, n) with n a positive integer, it suffices to calculate this amount for n ∈ {1, . . . , C(S)}. gap> s:=NumericalSemigroup(3,5);; gap> GenusOfNumericalSemigroup(s); 4 gap> ConductorOfNumericalSemigroup(s); 8 gap> Minimum(List([1..8],x-> Length(AperyListOfNumericalSemigroupWRTInteger(s,x)))); 3

Hence for m  0, δ 2F R (m) = m + 1 − 8 + 3 = m − 4.

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Index

A Apéry set, 3 Apéry set of an integer, 130 Approximate root, 68 Arf closure, 48 semigroup, 48

B Betti element, 105 Blowup, 47

C Catenary degree, 121 Chain of factorizations, 121 Characteristic sequences, 58 Conductor of a numerical semigroup, 6 Congruence compatible, 103 generated by a set, 100 Contact, 70 Curve with one place at infinity, 84

D Degree of singularity, 2 Delta set, 117 Desert, 130 Distance factorizations, 120 Divide, 125

E Elasticity, 116

Embedding dimension of a numerical semigroup, 5 Exact degree for algebroid curves, 81 for polynomial curves, 93 Expansion, 62

F Factorization homomorphism, 100 Feng–Rao distance, 129 generalized, 130 Frobenius number, 6 of an ideal, 32

G G-adic expansion, 62 Gap of a numerical semigroup, 2 Generating function, 110 Genus of an ideal, 34 Genus of a numerical semigroup, 2 Graph associated to an element, 103

H Half-factorial monoid, 115 Hilbert series, 110

I Ideal, 31 almost canonical, 36 canonical, 36 conductor, 31 irreducible, 41

© Springer Nature Switzerland AG 2020 A. Assi et al., Numerical Semigroups and Applications, RSME Springer Series 3, https://doi.org/10.1007/978-3-030-54943-5

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138 maximal, 31 maximum sparse, 35 proper, 31 relative, 31 stable, 46 standard canonical, 32 Z-irreducible, 40 Integrally closed ideal, 48 Intersection multiplicity, 59 Irreducible components, 41

L Length of a factorization, 115

M Milnor number, 79 Minimal set of generators, 5 Multiplicity of an ideal, 31 sequence, 48 Multiplicity of a numerical semigroup, 3

N Newton–Puisieux exponents, 58 Numerical semigroup, 1 almost symmetric, 11 associated to a curve, 86 associated to a polynomial, 77 free, 26 irreducible, 17 maximal embedding dimension, 12 pseudo-symmetric, 18 symmetric, 18 telescopic, 26

O ω-primality, 125 One point algebraic code, 129

Index Order, 53 Oversemigroup of a numerical semigroup, 15

P Presentation of a numerical semigroup, 101 Pseudo-approximate root, 59 Pseudo-Frobenius numbers, 8

R Reduction, 45 number, 46

S Semigroup ring, 1 Set closed under divisors, 34 Set of divisors, 41, 128 Set of factorizations, 102 Set of lengths, 115 Shaded set, 110 Short parametrization, 81 Special gaps of a numerical semigroup, 14 Standard representation in a free numerical semigroup, 27 Support, 56

T Tame degree, 123 Tschirnhausen transform, 66 Type of an ideal, 34 Type of a numerical semigroup, 8

V Very short parametrization, 83

W Weirstrass semigroup, 2