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Sets of values of fractional ideals of rings of algebroid curves

Edison Marcavillaca Niño de GuzmánTese de Doutorado do Programa de Pós-Graduação emMatemática (PPG-Mat)

SERVIÇO DE PÓS-GRADUAÇÃO DO ICMC-USP

Data de Depósito:

Assinatura: ______________________

Edison Marcavillaca Niño de Guzmán

Sets of values of fractional ideals of rings of algebroid curves

Doctoral dissertation submitted to the Institute ofMathematics and Computer Sciences – ICMC-USP, inpartial fulfillment of the requirements for the degree ofthe Doctorate Program in Mathematics. FINALVERSION

Concentration Area: Mathematics

Advisor: Prof. Dr. Abramo HefezCo-advisor: Prof. Dr. Marcelo José Saia

USP – São CarlosJune 2018

Ficha catalográfica elaborada pela Biblioteca Prof. Achille Bassie Seção Técnica de Informática, ICMC/USP,

com os dados fornecidos pelo(a) autor(a)

Guzmán, Edison Marcavillaca Niño deG634s Sets of values of fractional ideals of rings

of algebroid curves / Edison MarcavillacaNiño de Guzmán; orientador Abramo Hefez;co-orientador Marcelo José Saia. – São Carlos –SP, 2018.

64 p.

Tese (Doutorado - Programa de Pós-Graduação emMatemática) – Instituto de Ciências Matemáticas e deComputação, Universidade de São Paulo, 2018.

1. Fractional ideal. 2. sets of values.3. symmetry. 4. Gorenstein rings. I. Hefez, Abramo,orient. II. Saia, Marcelo José, coorient. III.Título.

Edison Marcavillaca Niño de Guzmán

Conjunto de valores de ideais fracionários de anéis decurvas algebroides

Tese apresentada ao Instituto de CiênciasMatemáticas e de Computação – ICMC-USP,como parte dos requisitos para obtenção do títulode Doutor em Ciências – Matemática. VERSÃOREVISADA

Área de Concentração: Matemática

Orientador: Prof. Dr. Abramo HefezCoorientador: Prof. Dr. Marcelo José Saia

USP – São CarlosJunho de 2018

Dedicated to the memory of my brother Walter.

ACKNOWLEDGEMENTS

Começo agradecendo à minha mãe Maria del Carmen a quem devo grande partedo que sou.

Ao meu orientador Prof. Dr. Abramo Hefez pela orientação, paciência, amizade epela confiança que sempre me deu ao longo da realização deste trabalho.

Ao professor Marcelo José Saia, por ter aceitado ser meu co-orientador.

Aos professores Marcelo Escudeiro, Fernando Torres e Renato Vidal Martins queestiveram na banca por se disporem a ler este trabalho e por terem contribuído com críticase sugestões valiosas e pertinentes.

À minha família que sempre me apoiou. Em especial à minha irmã Mariluz e meussobrinhos-filhos Jose Vicente e Luis Enrique.

Ao professor Percy Fernandez, pelo seu apoio e a confiança depositada para começaro doutorado.

Aos meus colegas e amigos do ICMC e da UFF.

À Elizabeth que foi essencial para muito do que se passou nesses anos de doutorado.

Ao ICMC e a pós-graduação em Matemática da UFF.

À CAPES, pelo apoio financeiro, sem o qual não seria possível a realização destetrabalho.

Finalmente agradeço a todas as pessoas que direta ou indiretamente participaramda realização deste trabalho.

ABSTRACT

NIÑO DE GUZMÁN, M. E. Sets of values of fractional ideals of rings of algebroidcurves. 2018. 64 p. Tese (Doutorado em Ciências – Matemática) – Instituto de CiênciasMatemáticas e de Computação, Universidade de São Paulo, São Carlos – SP, 2018.

The aim of this work is to study rings of algebroid Gorenstein rings. We explore moredeeply the symmetry that exists among the sets of values of a fractional ideal and that ofits dual and also to express the codimension of a fractional ideal in terms of the maximalpoints of the value set of the ideal. We apply the formulas we obtained to express theTjurina number of a complete intersection curve in terms of invariants of its componentsand the maximal points of the set of values of the Kähler differentials on the curve.

Keywords: Fractional ideal, sets of values, symmetry, Gorenstein rings.

RESUMO

NIÑO DE GUZMÁN, M. E. Conjunto de valores de ideais fracionários de anéisde curvas algebroides. 2018. 64 p. Tese (Doutorado em Ciências – Matemática) –Instituto de Ciências Matemáticas e de Computação, Universidade de São Paulo, SãoCarlos – SP, 2018.

O objetivo desse trabalho é o estudo dos anéis de curvas algebróides de Gorenstein.Expolramos mais aprofundadamente a simetria que existe entre os conjuntos de valoresde um ideal fracionário e de seu dual e também expressar a codimensão de um idealfracionário em função dos pontos maximais de seu conjunto de valores. Aplicamos asfórmulas obtidas para relacionar o número de Tjurina de uma curva de interseção completacom certos invariantes de suas componentes e dos pontos maximais do conjunto de valoresdas diferenciais de Kähler sobre a curva.

Palavras-chave: Ideais fracionários, conjunto de valores, simetria, anéis Gorenstein.

CONTENTS

1 SYMMETRY FOR GORENSTEIN RINGS . . . . . . . . . . . . . . 171.1 Algebroid curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.2 Fractional ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.3 Value sets of fractional ideals . . . . . . . . . . . . . . . . . . . . . . . 231.3.1 Maximal points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.3.2 Apéry Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.4 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 COLENGTHS OF FRACTIONAL IDEALS . . . . . . . . . . . . . . 312.1 Case r=2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2 Case 𝑟 ≥ 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.3 Case r=3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3 SYMMETRY OF MAXIMALS . . . . . . . . . . . . . . . . . . . . . 43

4 KÄHLER DIFFERENTIALS ON COMPLETE INTERSECTIONS . . 49

5 EXAMPLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

BIBLIOGRAPHY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

15

PREFACE

Semigroup of values of rings of algebroid plane branches were studied by Zariskiin [18] and they constitute a complete set of discrete invariants for their topologicalclassification. These semigroups live in N and they are determined by a finite set ofelements, since, as shown by Zariski, they have a conductor, that is, a number 𝑐 such that𝑐 − 1 is not in the semigroup, but from 𝑐 on all natural numbers belong to the semigroup.From the work of Apéry [1], it follows also that this semigroup is symmetric, that is in theinterval [0, 𝑐 − 1] there as many elements in the semigroup as in its complement. Manyyears later, E. Kunz, in [14], showed that the symmetry of the semigroup of values forbranches in a higher dimensional space is equivalent to the property of the branch to beGorenstein.

For a plane curve singularity with several branches, O. Zariski in [18] characterizedits topological type through the semigroups of its branches and their mutual intersectionmultiplicities. This characterization was shown by R. Waldi in [17] to be equivalent tothe knowledge of the semigroup of values of the curve, this time in N𝑟, where 𝑟 is thenumber of branches of the curve. Although not finitely generated, this semigroup, for𝑟 = 2, was shown by A. Garcia in [9] to be determined in a combinatorial way by afinite set of points that he called maximal points. Garcia also showed that these maximalpoints of the semigroup of a plane curve have a certain symmetry. These results weregeneralized later for any value of 𝑟 by F. Delgado in [7]. Delgado discovered several typesof maximals among them the relative and absolute maximals and showed that the relativemaximals determine the semigroup of values in a combinatorial way and extended forthese maximals Garcia’s symmetry. Some time later, Delgado, in [8], generalizing the workof Kunz, introduced a concept of symmetry for the semigroup of values in N𝑟 and showedthat this symmetry is equivalent to the property of the ring of the curve to be Gorenstein.

The computation of the codimension of an ideal, using elementary paths in its setof values was also considered by Garcia and Delgado, being generalized for fractional idealsby D’Anna [6] and by Barucci, D’Anna and Fröberg and in [2]. These authors observedthat for sets of values of fractional ideals there exist maximal points that determine inthe same combinatorial way such sets, as for semigroups of values. What were missing inthe literature and we provided was a formula relating the codimension of an ideal withthe maximals of its set of values. We also refined this formula for curves in three-space,proving a formula conjectured by M. Hernandes.

16 CONTENTS

On the other hand, D. Pol, in the work [16], showed that the property of thesingularity to be Gorenstein is equivalent to the fact that the set of values of the dualof any fractional ideal is determined by the set of values of the ideal. This motivated usto describe the property of a curve singularity to be Gorenstein by means of Apéry setsfollowing the original steps by Apéry. For this, we defined the Apéry set for a set of valuesof a fractional ideal and characterized Gorenstein rings of algebroid curves by means ofsome symmetry property between the Apéry set of a fractional ideal and that of its dual.We also showed that for Gorenstein curve singularities one has a symmetry between theabsolute maximals of the set of values of a fractional ideal and the relative maximals ofthe set of values of its dual ideal, generalizing a result by A. Campillo, F. Delgado andK. Kiyek in [5] for semigroups of values. Finally, we apply our formulas for codimensionsof ideals to express the Tjunina number of a complete intersection curve in term of theTjurina numbers of its components and the set of values of the Kähler differentials on thecurve, generalizing a work by Bayer, Hefez and Hernandes in [4], where the case of planecurves was treated.

17

CHAPTER

1SYMMETRY FOR GORENSTEIN RINGS

In this chapter we will find some new relations between the set of values of afractional ideal and that of its dual ideal for Gorenstein rings of algebroid curves over analgebraically closed ground field of arbitrary characteristic, generalizing the works of F.Delgado in [7] and [8] and by D. Pohl in [16].

1.1 Algebroid curvesThis introductory section is based on the work of Gorenstein [11] , Hironaka [13]

and Garcia-Lax [10].

Throughout this work we will denote by Z the set of integers and by N the set ofnon-negative integers.

A reduced algebroid curve is given by 𝐶 = Spec(𝒪) where 𝒪 is a finitely generatedlocal, reduced and complete of pure dimension one 𝑘-algebra and 𝑘 is an algebraicallyclosed field. The ring 𝒪 will be called the ring of 𝐶.

Let ℘1, . . . , ℘𝑟 be the minimal primes of 𝒪. We will use the notation 𝐼 = {1, . . . , 𝑟}.The branches of 𝐶 are the integral schemes 𝐶𝑖 = Spec(𝒪𝑖), 𝑖 ∈ 𝐼, where 𝒪𝑖 = 𝒪/℘𝑖. Wewill denote by 𝜋𝑖 : 𝒪 → 𝒪/℘𝑖 the canonical surjection, which corresponds to an inclusion𝐶𝑖 →˓ 𝐶.

Since 𝒪 is reduced, we have that ⋂𝑟𝑖=1 ℘𝑖 =

√(0) = (0), so the homomorphism

𝜋 : 𝒪 →˓ 𝒪1 × · · · × 𝒪𝑟

ℎ ↦→ (𝜋1(ℎ), . . . , 𝜋𝑟(ℎ))

is an injection.

More generally, if 𝐽 = {𝑖1, . . . , 𝑖𝑠} is any subset of 𝐼, we may consider 𝐶𝐽 =Spec(𝒪𝐽), where 𝒪𝐽 = 𝒪/ ∩𝑠

𝑗=1 ℘𝑖𝑗. The scheme 𝐶𝐽 represents the curve 𝐶𝑖1 ∪ · · · ∪ 𝐶𝑖𝑠 .

18 Chapter 1. Symmetry for Gorenstein rings

We will consider the natural surjection

𝜋𝐽 : 𝒪 −→ 𝒪𝐽 .

Notice that we have 𝒪𝐼 = 𝒪 and 𝐶𝐼 = 𝐶.

We will denote by 𝒦 the total ring of fractions of 𝒪 and by 𝒦𝐽 the total ring offractions of the ring 𝒪𝐽 , when 𝐽 ⊂ 𝐼.

If 𝐽 = {𝑖}, then 𝒪{𝑖} is equal to the above defined domain 𝒪𝑖 whose field offractions will be denoted by 𝒦𝑖. Let 𝒪 be the integral closure of 𝒪 in 𝒦 and 𝒪𝐽 be thatof 𝒪𝐽 in 𝒦𝐽 . By the splitting principle, one has that 𝒪𝐽 ≃ 𝒪𝑖1 × · · · × 𝒪𝑖𝑛 , which in turnis the integral closure of 𝒪𝑖1 × · · · × 𝒪𝑖𝑛 in its total ring of fractions.

We have the following diagram:

𝒦𝐽 ≃ 𝒦𝑖1 × · · · × 𝒦𝑖𝑛

↑ ↑𝒪𝐽 ≃ 𝒪𝑖1 × · · · × 𝒪𝑖𝑛

↑ ↑𝒪𝐽 →˓ 𝒪𝑖1 × · · · × 𝒪𝑖𝑛

where the up arrows are inclusions.

Since each 𝒪𝑖 is a complete DVR over the field 𝑘 with valuation 𝑣𝑖, one has that𝒪𝑖 ≃ 𝑘[[𝑡𝑖]], where 𝑣𝑖(𝑡𝑖) = 1. It follows that 𝒦𝑖 ≃ 𝑘((𝑡𝑖)) is a valuated field with theextension of the valuation 𝑣𝑖. We also have that 𝒦 ≃ 𝑘((𝑡1)) × · · · × 𝑘((𝑡𝑟)).

We will denote by 𝜙𝑖 the map 𝒪 → 𝒪𝑖 and by 𝜙 the injective map 𝒪 → 𝒪1×· · ·× 𝒪𝑟,𝜙(ℎ) = (𝜙1(ℎ), . . . , 𝜙𝑟(ℎ)).

It is known that 𝒪𝐽 is a finitely generated 𝒪𝐽 -module, hence the set

𝒞𝐽 = {ℎ ∈ 𝒦𝐽 ; ℎ 𝒪𝐽 ⊂ 𝒪𝐽}

has a regular element (ie. a nonzero divisor). It is easily verified that this set 𝒞𝐽 is an idealof both 𝒪𝐽 and 𝒪𝐽 and, in fact, it contains any simultaneous ideal of these two rings.This ideal is called the conductor ideal of 𝒪𝐽 in 𝒪𝐽 . When 𝐽 = 𝐼, the conductor 𝒞𝐼 willbe denoted by 𝒞.

An important class of rings 𝒪 are the Gorenstein rings, for which one has, bydefinition that, as 𝑘-vector spaces,

dim𝑘 𝒪/𝒞 = dim𝑘𝒪/𝒪.

The next result will characterize the ideals of 𝒪.

1.1. Algebroid curves 19

Proposition 1.1. If ℐ is an ideal of 𝒪 ≃ 𝑘[[𝑡1]] × · · · × 𝑘[[𝑡𝑟]], then ℐ = (𝑡𝛾11 , . . . , 𝑡𝛾𝑟

𝑟 ) 𝒪,for some 𝛾1, . . . , 𝛾𝑟 ∈ N.

Proof Let

𝛾𝑖 = min{𝑣𝑖(ℎ); ℎ ∈ ℐ}.

We will prove that (𝑡𝛾11 , . . . , 𝑡𝛾𝑟

𝑟 ) generates ℐ. First of all let us observe that(𝑡𝛾1

1 , . . . , 𝑡𝛾𝑟𝑟 ) ∈ ℐ. Indeed, let ℎ𝑖 ∈ ℐ be such that 𝑣𝑖(𝜋𝑖(ℎ𝑖)) = 𝛾𝑖, then if we take

𝜆1, . . . , 𝜆𝑟 ∈ 𝑘 general enough, then ℎ = 𝜆1ℎ1 + · · · + 𝜆𝑟ℎ

𝑟 ∈ ℐ and ℎ = (𝑡𝛾11 𝑢1, . . . , 𝑡𝛾𝑟

𝑟 𝑢𝑟),where each 𝑢𝑖 is a unit in 𝑘[[𝑡𝑖]] , hence ℎ = 𝑢(𝑡𝛾1

1 , . . . , 𝑡𝛾𝑟𝑟 ), where 𝑢 is a unit in 𝒪. This

shows that (𝑡𝛾11 , . . . , 𝑡𝛾𝑟

𝑟 ) ∈ ℐ. Now, any element 𝑔 ∈ ℐ may be written as

𝑔 = (𝑡𝛾1+𝛼11 𝑢1, . . . , 𝑡𝛾𝑟+𝛼𝑟

𝑟 𝑢𝑟) = (𝑡𝛼11 𝑢1, . . . , 𝑡𝛼𝑟

𝑟 𝑢𝑟)(𝑡𝛾11 , . . . , 𝑡𝛾𝑟

𝑟 ).

It then follows that there is some 𝛾 = (𝛾1, . . . , 𝛾𝑟) ∈ N𝑟 such that 𝒞 = 𝑡𝛾 𝒪, where𝑡𝛾 = (𝑡𝛾1

1 , . . . , 𝑡𝛾𝑟𝑟 ).

Example 1.2. (Apéry [1]) If Spec(𝒪) is a plane branch, then one has that 𝒞 = 𝑡2𝛿𝑘[[𝑡]],where 𝛿 = dim𝑘

𝒪𝒪 = dim𝑘

𝒪𝒞 . The number 𝑐 = 2𝛿 is the degree of the conductor 𝒞.

Example 1.3. (Gorenstein [11]) If Spec(𝒪) is a reduced plane curve given by an equa-tion 𝑓 = 𝑓1 · · · 𝑓𝑟, then the conductor of 𝒪 in 𝒪 is a principal ideal generated by(𝑡𝐼1+𝑐1

1 , . . . , 𝑡𝐼𝑟+𝑐𝑟𝑟 ), where the 𝑐𝑖 denote the degrees of the conductors 𝒞𝑖 of the branches

determined by the 𝑓𝑖 and 𝐼𝑖 = ∑𝑗 =𝑖 𝐼(𝑓𝑗, 𝑓𝑖).

These two examples will be stated in greater generality later in Theorems 6 and 7.

Let us denote by 𝑍(𝐴) the set of zero divisor of a ring 𝐴. One has that 𝑍(𝒪) = ∪𝑖℘𝑖

and 𝑍(𝐴) = {(𝑎1, . . . , 𝑎𝑟) ∈ 𝐴; 𝑎𝑖 = 0, for some 𝑖}, if 𝐴 = 𝒪 or 𝐴 = 𝒦. So, one has that𝑔 ∈ 𝑍(𝒪) if and only if 𝜋𝑖(𝑔) = 0, for some 𝑖 = 1, . . . , 𝑟 (if and only if 𝜙𝑖(𝑔) = 0).

If 𝐽1 and 𝐽2 are two disjoint subset of 𝐼, then we define the intersection multiplicityof 𝐶𝐽1 and 𝐶𝐽2 as being

𝐼(𝐶𝐽1 , 𝐶𝐽2) = dim𝑘𝒪⋂

𝑗∈𝐽1 ℘𝑗 + ⋂𝑗∈𝐽2 ℘𝑗

.

In the above situation we also have that 𝐶𝐽1 ∪ 𝐶𝐽2 = 𝐶𝐽1∪𝐽2 .


1.2 Fractional idealsAn 𝒪-submodule ℐ of 𝒦 will be called a fractional ideal of 𝒪 if it contains a regular

element of 𝒪 and there is a regular element 𝑑 in 𝒪 such that 𝑑 ℐ ⊂ 𝒪. Such an element 𝑑

will be called a common denominator for ℐ.

Since 𝑑 ℐ is an ideal of 𝒪, which is a noetherian ring, one has that ℐ ⊂ 𝒦 is anontrivial fractional ideal if and only if it contains a regular element of 𝒪 and it is afinitely generated 𝒪-module.

Examples of fractional ideals of 𝒪 are 𝒪 itself, 𝒪, the conductor 𝒞 of 𝒪 in 𝒪, orany ideal of 𝒪 or of 𝒪 that contains a regular element. Also, if ℐ is a fractional ideal of 𝒪,then for all ∅ = 𝒥 ⊂ 𝐼 one has that 𝜋𝐽(ℐ) is a fractional ideal of 𝒪𝐽 .

Remark 1.4. Any element ℎ ∈ 𝒦 will be written as ℎ = ℎ1ℎ2

, where ℎ1, ℎ2 ∈ 𝒪, ℎ2 /∈ 𝑍(𝒪).If ℎ = ℎ1

ℎ2∈ ℐ, then ℎ1 ∈ ℐ. And, ℎ ∈ 𝑍(𝒦) if and only if ℎ1 ∈ 𝑍(𝒪).

So, we may assume that the regular element contained in a fractional ideal is indeedin 𝒪. In the set of fractional ideals of 𝒪 there are defined the operations +, · and : . Thefirst two operations are clear and the last one is defined, for ℐ = {0}, as follows:

(ℐ : 𝒥 ) = {𝑥 ∈ 𝒦; 𝑥𝒥 ⊂ ℐ}.

This defines a fractional ideal. Indeed, it is an 𝒪-module which contains a regularelement: 𝑑𝑑′, where 𝑑 is a common denominator for ℐ and 𝑑′ is a common denominatorfor 𝒥 . On the other hand, the element 𝑏𝑑, where 𝑏 is a regular element in 𝒥 ∩ 𝒪, is acommon denominator for (ℐ : 𝒥 ).

As an example of this construction we have that 𝒞 = (𝒪 : 𝒪).

Proposition 1.5. Given a fractional ideal ℐ of 𝒪 there exist 𝛽 ∈ N𝑟 and 𝛼 ∈ Z𝑟 suchthat

𝑡𝛽 𝒪 ⊂ ℐ ⊂ 𝑡𝛼 𝒪.

Proof From the definition of a fractional ideal we know that there exists a regular element𝑑 in 𝒪 such that 𝑑 ℐ ⊂ 𝒪 ⊂ 𝒪. Therefore, as subsets of 𝒦, one has that ℐ ⊂ 1

𝑑𝒪. Since

we may write 1𝑑

= 𝑡𝛼𝑢, where 𝛼 ∈ Z𝑟 and 𝑢 is a unit of 𝒪, we get that

ℐ ⊂ 𝑡𝛼𝑢 𝒪 = 𝑡𝛼 𝒪.

On the other hand, we now that there is a regular element of 𝒪 in ℐ. Write 𝑎 = 𝑡𝜖𝑣,where 𝜖 ∈ N𝑟 and 𝑣 is a unit of 𝒪. Let 𝑡𝛾 be a generator of the conductor ideal of 𝒪 in 𝒪,then

𝑡𝜖+𝛾 𝒪 = 𝑡𝜖𝑣𝑡𝛾 𝒪 ⊂ 𝑎𝒪 ⊂ ℐ.

1.2. Fractional ideals 21

The remaining inclusion we are looking for follows taking 𝛽 = 𝜖 + 𝛾.

We define the dual module of ℐ as being the 𝒪-module ℐ∨ = 𝐻𝑜𝑚𝒪(ℐ, 𝒪).

If 𝜙 ∈ ℐ∨ and ℎ = ℎ1ℎ2

∈ ℐ, then it is easy to verify that

𝜙

(ℎ1

ℎ2

)= 𝜙(ℎ1)

ℎ2.

Our aim now is to realize ℐ∨ as a fractional ideal of 𝒪. To do this, we begin withthe following result.

Lemma 1.6. Let 𝜙 ∈ ℐ∨. If 𝑔 ∈ ℐ ∖ 𝑍(𝒦), then 𝜙(𝑔)𝑔

is independent of 𝑔.

Proof Let 𝑔 = 𝑔1𝑔2

and 𝑔′ = 𝑔′1

𝑔′2

in ℐ ∖ 𝑍(𝒦), then

𝜙(𝑔)𝑔

= 𝑔2𝜙(𝑔)𝑔1

= 𝜙(𝑔2𝑔)𝑔1

= 𝜙(𝑔1)𝑔1

,

𝜙(𝑔′)𝑔′ = 𝑔′

2𝜙(𝑔′)𝑔′

1= 𝜙(𝑔′

2𝑔′)𝑔′

1= 𝜙(𝑔′

1)𝑔′

1,

which make sense since 𝑔1 and 𝑔′1 are not in 𝑍(𝒪).

But, since 𝜙(𝑔1𝑔′1) = 𝜙(𝑔′

1𝑔1) and 𝜙 is an 𝒪-modules homomorphism, it followsthat 𝑔1𝜙(𝑔′

1) = 𝑔′1𝜙(𝑔1), which implies our result.

For 𝑔 ∈ ℐ ∖ 𝑍(𝒦), the map

Φ: 𝐻𝑜𝑚𝒪(ℐ, 𝒪) → 𝒦𝜙 ↦→ 𝜙(𝑔)

𝑔,

which is independent of 𝑔, is a homomorphism of 𝒪-modules.

Given a fractional ideal ℐ of 𝒪, we will use the notation

ℐ* = (𝒪 : ℐ) = {𝑥 ∈ 𝒦; 𝑥ℐ ⊂ 𝒪}.

Proposition 1.7. The map Φ induces an 𝒪-module isomorphism between ℐ∨ and ℐ*.

Proof We first show that Φ is injective. Let 𝜙 ∈ 𝐻𝑜𝑚𝒪(ℐ, 𝒪) be such that 𝜙(𝑔)𝑔

= 0, forsome 𝑔 ∈ (ℐ ∩ 𝒪) ∖ 𝑍(𝒪). Let ℎ1 ∈ ℐ ∩ 𝒪, then

𝜙(ℎ1) = 𝑔𝜙(ℎ1)𝑔

= ℎ1𝜙(𝑔)

𝑔= 0,

which implies that 𝜙(ℎ1ℎ2

) = 𝜙(ℎ1)ℎ2

= 0, for ℎ1ℎ2

∈ ℐ, so Φ is injective.


Now we show that 𝐼𝑚(Φ) ⊂ ℐ*. Indeed, let 𝑎 ∈ 𝐼𝑚(Φ), then 𝑎 = 𝜙(𝑔)𝑔

for some𝜙 ∈ ℐ* and some 𝑔 ∈ (ℐ ∩ 𝒪) ∖ 𝑍(𝒪). Now, let ℎ = ℎ1

ℎ2∈ ℐ, then

𝑎ℎ1

ℎ2= 𝜙(𝑔)

𝑔

ℎ1

ℎ2= 𝜙(𝑔ℎ1)

𝑔ℎ2= 𝑔

𝑔ℎ2𝜙(ℎ1) = 𝜙(ℎ1)

ℎ2= 𝜙

(ℎ1

ℎ2

)∈ 𝒪.

Finally, let us show that ℐ* ⊂ 𝐼𝑚(Φ). Take 𝑎 in ℐ* and define

𝜙𝑎 : ℐ → 𝒪ℎ ↦→ 𝑎ℎ.

It is easy to verify that 𝜙𝑎 is a homomorphism of 𝒪-modules, and

𝑎 = 𝑎𝑔

𝑔= 𝜙𝑎(𝑔)

𝑔∈ 𝐼𝑚(Φ).

We will always view ℐ∨ = 𝐻𝑜𝑚𝒪(ℐ, 𝒪), as the fractional ideal ℐ*, which will becalled the dual ideal of the fractional ideal ℐ. It is easy to verify that taking duals is orderreversing with respect to inclusions; that is,

ℐ ⊂ 𝒥 =⇒ ℐ* ⊃ 𝒥 *.

As examples of duals of fractional ideals one has that 𝒪* = 𝒪 and 𝒪* = 𝒞.

One always has an inclusion ℐ ⊂ (ℐ*)*, which is almost tautological, on the otherhand, it is not always true that (ℐ*)* ⊂ ℐ. But we have the following result.

Theorem 1.8. ([3, Theorem 6.3]) The following statemens are equivalent:

i) 𝒪 is a Gorenstein ring;

ii) (ℐ*)* ⊂ ℐ for all fractional ideal ℐ of 𝒪;

iii) dim𝑘𝒥ℐ = dim𝑘

ℐ*

𝒥 * , for all fractional ideals ℐ and 𝒥 such that ℐ ⊂ 𝒥 .

If 𝐽 ⊂ 𝐼 we define𝛿𝐽 = dim𝑘

𝒪𝐽/𝒪𝐽 ,

and if we put 𝐽 ′ = 𝐼 ∖ 𝐽 , then one defines 𝐼𝐽 = 𝐼(𝐶𝐽 , 𝐶𝐽 ′), we have the following results:

Theorem 1.9. ([10, Corollary 3.10]) Let 𝒪 be a Gorenstein ring, 𝐽 ⊂ 𝐼 and 𝑇1, . . . , 𝑇𝑡 apartition of 𝐼. Then one has

𝛿𝐽 + 12𝐼𝐽 =

𝑡∑𝑗=1

𝛿𝑇𝑗+ 1

2

𝑡∑𝑗=1

𝐼𝑇𝑗.

Theorem 1.10. ([10, Theorem 3.13]) Let 𝒪 be a Gorenstein ring, then the conductor 𝒞of 𝒪 in 𝒪 is given by

𝒞 = (𝑡𝐼1+2𝛿11 , . . . , 𝑡𝐼𝑟+2𝛿𝑟

𝑟 ) 𝒪.

1.3. Value sets of fractional ideals 23

1.3 Value sets of fractional idealsFor any 𝑛, we will consider on Z𝑛 the partial order given by

(𝑎1, . . . , 𝑎𝑛) ≤ (𝑏1, . . . , 𝑏𝑛) ⇐⇒ 𝑎𝑖 ≤ 𝑏𝑖, 𝑖 = 1, . . . , 𝑛.

We will write (𝑎1, . . . , 𝑎𝑛) < (𝑏1, . . . , 𝑏𝑛) when 𝑎𝑖 < 𝑏𝑖, for all 𝑖 = 1, . . . , 𝑛.

Consider the value map

𝑣 : 𝒦 ∖ 𝑍(𝒦) → Z𝑟

ℎ ↦→ (𝑣1(𝜋1(ℎ)), . . . , 𝑣𝑟(𝜋𝑟(ℎ))),

where 𝜋𝑖 here denotes the projection 𝒦 → 𝒦𝑖, which is the extension of the previouslydefined projection map 𝜋𝑖 : 𝒪 → 𝒪𝑖. This map 𝑣 can be extended to a map 𝑣 : 𝒦 → 𝑍

𝑟 =(Z ∪ {∞})𝑟.

If ℐ is a fractional ideal of 𝒪, we define the value set of ℐ as being

𝐸 = 𝑣(ℐ ∖ 𝑍(𝒦)) ⊂ Z𝑟,

and the extended value set of ℐ as being

𝐸 = 𝑣(ℐ) ⊂ Z𝑟.

We will denote by 𝐸* the value set of ℐ*. The value set of 𝒪 will be denoted by𝑆(𝒪), or simply by 𝑆, and it is a subsemigroup of N𝑟, called the semigroup of values of 𝒪.The value sets 𝐸 of fractional ideals are not necessarily closed under addition, but they aresuch that 𝑆 + 𝐸 ⊂ 𝐸. For this reason they are called monomodules over the semigroup 𝑆.

From the definition of the ideal ℐ*, if 𝐸 is the value set of ℐ and 𝐸* is the valueset of ℐ*, it follows immediately that

𝐸 + 𝐸* ⊂ 𝑆(𝒪). (1.1)

Let 𝐼 = {1, . . . , 𝑟} and 𝐽 ⊂ 𝐼. If 𝐽 = {𝑖1, . . . , 𝑖𝑛}, then we denote by 𝜋𝐽 theprojection 𝒦 → 𝒦𝑖1 × · · · × 𝒦𝑖𝑛 and by 𝑝𝑟𝐽 the corresponding projection Z𝑟 → Z𝑛,(𝛼1, . . . , 𝛼𝑟) ↦→ (𝛼𝑖1 , . . . , 𝛼𝑖𝑛).

Let us define

𝐸𝐽 = 𝑣(𝜋𝐽(ℐ) ∖ 𝑍(𝒦𝐽)) and 𝐸𝐽 = 𝑣(𝜋𝐽(ℐ)).

Caution If 𝑗 ∈ 𝐽 = {𝑖1, . . . , 𝑖𝑡, . . . 𝑖𝑠} ⊂ 𝐼, with 𝑖𝑡 = 𝑗, for 𝛽 = (𝛽𝑖1 , . . . , 𝛽𝑖𝑠) ∈ 𝐸𝐽 ,we will define

𝑝𝑟𝑗(𝛽) = 𝛽𝑖𝑡 = 𝛽𝑗,


instead of 𝑝𝑟𝑗(𝛽) = 𝛽𝑖𝑗, as it would be natural.

From Proposition 2 one has that, given a set of value 𝐸, there exist 𝛽 ∈ 𝐸 and𝛼 ∈ Z𝑟 such that

𝛽 + N𝑟 ⊂ 𝐸 ⊂ 𝛼 + N𝑟.

Value sets of fractional ideals have the following fundamental properties:

Property 1.11 (A). If 𝛼 = (𝛼1, . . . , 𝛼𝑟) and 𝛽 = (𝛽1, . . . , 𝛽𝑟) belong to 𝐸, then

𝑚𝑖𝑛(𝛼, 𝛽) = (𝑚𝑖𝑛(𝛼1, 𝛽1), . . . , 𝑚𝑖𝑛(𝛼𝑟, 𝛽𝑟)) ∈ 𝐸.

Property 1.12 (B). If 𝛼 = (𝛼1, . . . , 𝛼𝑟), 𝛽 = (𝛽1, . . . , 𝛽𝑟) belong to 𝐸, 𝛼 = 𝛽 and𝛼𝑖 = 𝛽𝑖 for some 𝑖 ∈ {1, . . . , 𝑟}, then there exists 𝛾 ∈ 𝐸 such that 𝛾𝑖 > 𝛼𝑖 = 𝛽𝑖 and𝛾𝑗 ≥ 𝑚𝑖𝑛{𝛼𝑗, 𝛽𝑗} for each 𝑗 = 𝑖, with equality holding if 𝛼𝑗 = 𝛽𝑗.

Since there is an 𝛼 ∈ Z𝑟 such that 𝐸 ⊂ 𝛼+N𝑟, then from Property (A) there existsa unique 𝑚𝐸 = min(𝐸), that is, for all (𝛽1, . . . , 𝛽𝑟) ∈ 𝐸, one has 𝛽𝑖 ≥ 𝑚𝐸𝑖, 𝑖 = 1, . . . , 𝑟.

On the other side, one has the following

Lemma 1.13. If 𝛽, 𝛽′ ∈ 𝐸 are such that 𝛽 + N𝑟 ⊂ 𝐸 and 𝛽′ + N𝑟 ⊂ 𝐸, then min(𝛽, 𝛽′) +N𝑟 ⊂ 𝐸.

Proof Let 𝜃 = min(𝛽, 𝛽′), we want to prove that 𝜃 + N𝑟 is a subset of 𝐸.

Let 𝑎 ∈ N𝑟. Since 𝛽 + 𝑎 ∈ 𝐸 and 𝛽′ + 𝑎 ∈ 𝐸, from Property (A), we have that

min(𝛽, 𝛽′) + 𝑎 = min(𝛽 + 𝑎, 𝛽′ + 𝑎) ∈ 𝐸.

Therefore, 𝑚𝑖𝑛(𝛽, 𝛽′) + N𝑟 ⊂ 𝐸.

The above lemma guarantees that there is an unique least element 𝛾 ∈ 𝐸 withthe property that 𝛾 + N𝑟 ⊂ 𝐸. This element is what we call the conductor of 𝐸 and willdenote it by 𝑐(𝐸).

Observe that one always have

𝑐(𝐸𝐽) ≤ 𝑐(𝑝𝑟𝐽(𝐸)), ∀ 𝐽 ⊂ 𝐼.

One has the following result:

Lemma 1.14. If ℐ be a fractional ideal of 𝒪 and 𝐽 ⊂ 𝐼, then 𝑝𝑟𝐽(𝐸) = 𝐸𝐽 .

Proof One has obviously that 𝑝𝑟𝐽(𝐸) ⊂ 𝐸𝐽 . On the other hand, let 𝛼𝐽 ∈ 𝐸𝐽 . Take ℎ ∈ ℐsuch that 𝑣𝐽(𝜋𝐽(ℎ)) = 𝛼𝐽 . If ℎ ∈ 𝑍(𝒦) we are done. Otherwise, choose any ℎ′ ∈ ℐ ∖ 𝑍(𝒦)


such that 𝑝𝑟𝐽(𝑣(ℎ′)) > 𝛼𝐽 , which exists since 𝐸 has a conductor. Hence, 𝑣𝐽(ℎ + ℎ′) = 𝛼𝐽 ,proving the other inclusion.

Let 𝑆 = 𝑆(𝒪). We will use the following notation:

f(𝑆) = 𝑐(𝑆) − (1, . . . , 1),

and call f(𝑆) the Frobenius vector of 𝑆 or of 𝒪.

When 𝒪 is Gorenstein, from Theorem 1.10, one has that

𝑐(𝑆) = (𝐼1 + 2𝛿1, . . . , 𝐼𝑟 + 2𝛿𝑟),

where 𝐼𝑗 = ∑𝑖 =𝑗 𝐼(𝐶𝑗, 𝐶𝑖) and 𝛿𝑗 is the singularity index of 𝐶𝑗.

1.3.1 Maximal points

We now introduce the important notion of a fiber of an element 𝛼 ∈ 𝐸 with respectto a subset 𝐽 ⊂ 𝐼 = {1, . . . , 𝑟} that will play a central role in what follows.

Definition 1.15. Given 𝛼 ∈ Z𝑟 and ∅ = 𝐽 ⊂ 𝐼, we define:

𝐹𝐽(𝛼) = {𝛽 ∈ Z𝑟; 𝑝𝑟𝐽(𝛽) = 𝑝𝑟𝐽(𝛼) and 𝑝𝑟𝐼∖𝐽(𝛽) > 𝑝𝑟𝐼∖𝐽(𝛼)},

𝐹 𝐽(𝛼) = {𝛽 ∈ Z𝑟; 𝑝𝑟𝐽(𝛽) = 𝑝𝑟𝐽(𝛼), and 𝑝𝑟𝐼∖𝐽(𝛽) ≥ 𝑝𝑟𝐼∖𝐽(𝛼)},

𝐹𝐽(𝐸, 𝛼) = 𝐹𝐽(𝛼) ∩ 𝐸 and 𝐹 𝐽(𝐸, 𝛼) = 𝐹 𝐽(𝛼) ∩ 𝐸,

𝐹 (𝛼) =𝑟⋃

𝑖=1𝐹{𝑖}(𝛼), 𝐹 (𝐸, 𝛼) = 𝐹 (𝛼) ∩ 𝐸.

The last set above will be called the fiber of 𝛼.

The sets 𝐹{𝑖}(𝛼) and 𝐹 {𝑖}(𝛼) will be denoted simply by 𝐹𝑖(𝛼) and 𝐹 𝑖(𝛼). Noticethat 𝐹𝐼(𝛼) = {𝛼}.

Definition 1.16. Let 𝛼 ∈ 𝐸. We will say that 𝛼 is a maximal point of 𝐸 if 𝐹 (𝐸, 𝛼) = ∅.

This means that there is no element in 𝐸 with one coordinate equal to thecorresponding coordinate of 𝛼 and the other ones bigger.

By convention, when 𝑟 = 1, the maximal points of 𝐸 are exactly the gaps of 𝐸.

From the fact that 𝐸 has a minimum 𝑚𝐸 and a conductor 𝛾 = 𝑐(𝐸), one hasimmediately that all maximal elements of 𝐸 are in the limited region

{(𝑥1, . . . , 𝑥𝑟) ∈ Z𝑟; 𝑚𝐸𝑖 ≤ 𝑥𝑖 < 𝛾𝑖, 𝑖 = 1, . . . , 𝑟}.

This implies that 𝐸 has finitely many maximal points.


Definition 1.17. We will say that a maximal point 𝛼 of 𝐸 is an absolute maximal if𝐹𝐽(𝐸, 𝛼) = ∅ for every 𝐽 ⊂ 𝐼, 𝐽 = 𝐼. If a maximal point 𝛼 of 𝐸 is such that 𝐹𝐽(𝐸, 𝛼) = ∅,for every 𝐽 ⊂ 𝐼 with #𝐽 ≥ 2, then 𝛼 will be said to be a relative maximal of 𝐸.

Notice that when 𝑟 = 1, as we said before, the maximals coincide with the gaps,in the case where 𝑟 = 2 the notions of maximal, relative maximal and absolute maximalcoincide. For 𝑟 = 3 we only may have relative maximals or absolute maximals.

We will denote by M(𝐸), RM(𝐸) and AM(𝐸) the sets of maximals, of relativemaximals and absolute maximals of the set 𝐸, respectively.

The theorem below says that the set of relative maximal of 𝐸 determines 𝐸 in acombinatorial sense as follows:

Theorem 1.18 (generation). Let 𝛼 ∈ Z𝑟 be such that 𝑝𝐽(𝛼) ∈ 𝐸𝐽 for all 𝐽 ⊂ 𝐼 with#𝐽 = 𝑟 − 1. Then

𝛼 ∈ 𝐸 ⇐⇒ 𝛼 /∈ 𝐹 (Z𝑟, 𝛽), ∀𝛽 ∈ RM(𝐸).

We will omit the proof since this result is a slight modification of [7, Theorem 1.5 ]with essentially the same proof.

The following result is the content of [8, Corollary 1.9] and of [8, Corollary 2.7].

Theorem 1.19. Let 𝒪 be the ring of an algebroid curve and let 𝑆 = 𝑆(𝒪). One has

i) 𝐹 (𝑆, f(𝑆)) = ∅;

ii) If 𝒪 is Gorenstein, then f(𝑆) is a relative maximal of 𝑆.

Part (i) of the above theorem implies the following result:


Corollary 1.20. Let 𝒪 be the ring of an algebroid curve, and let 𝑆 = 𝑆(𝒪). Then

𝛼 ∈ 𝑆 =⇒ 𝐹 (𝑆, f(𝑆) − 𝛼) = ∅.

Proof Indeed, if 𝛼 ∈ 𝑆 and 𝐹 (𝑆, f(𝑆) − 𝛼) = ∅, then if 𝜃 ∈ 𝐹 (𝑆, f(𝑆) − 𝛼), then𝜃 + 𝛼 ∈ 𝐹 (𝑆, f(𝑆)), which is a contradiction.

The following lemma generalizes [7, Lemma 2.9] with the same proof.

Lemma 1.21. Let 𝛼 ∈ 𝐸 be such that 𝐹𝑖(𝐸, 𝛼) = ∅. Then, there exists 𝐽 ⊂ 𝐼 with#𝐽 ≥ 2, containing 𝑖 and a relative maximal element 𝛽 of 𝐸𝐽 , such that 𝛽𝑖 = 𝛼𝑖 and𝛽𝑗 ≤ 𝛼𝑗, for all 𝑗 ∈ 𝐽 .

The following two lemmas give us other characterizations of the relative andabsolute maximal points.

Lemma 1.22. Given a value set 𝐸 ⊂ Z𝑑 and 𝛼 ∈ Z𝑑 with the following properties:

i) there is 𝑖 ∈ 𝐼 such that 𝐹𝑖(𝐸, 𝛼) = ∅,

ii) 𝐹𝑖,𝑗(𝐸, 𝛼) = ∅ for all 𝑗 ∈ 𝐼 ∖ {𝑖}.

Then 𝛼 is a relative maximal of 𝐸.

Proof Follows the same steps of [7, Lemma 1.3].

Lemma 1.23. Given a value set 𝐸 ⊂ Z𝑑 and 𝛼 ∈ 𝐸, assume that there exists an index𝑖 ∈ 𝐼 such that 𝐹𝐽(𝐸, 𝛼) = ∅ for every 𝐽 ⊂ 𝐼 with 𝑖 ∈ 𝐽 . Then 𝛼 is an absolute maximalof 𝐸.

Proof We have to prove that 𝐹𝐾(𝐸, 𝛼) = ∅ for all 𝐾 ⊂ 𝐼 with 𝑖 /∈ 𝐾.

Assume, by reductio ad absurdum, that there exists some 𝐾 ⊂ 𝐼 with 𝑖 /∈ 𝐾

such that 𝐹𝐾(𝐸, 𝛼) = ∅. Let 𝛽 be an element in 𝐹𝐾(𝐸, 𝛼), then 𝛽𝑗 = 𝛼𝑗, ∀𝑗 ∈ 𝐾 and𝛽𝑘 > 𝛼𝑘, for all 𝑘 /∈ 𝐾. Applying the Property (B) for the index 𝑘 ∈ 𝐾, there exists𝜃 ∈ 𝐸 such that 𝜃𝑘 > 𝛽𝑘 = 𝛼𝑘, 𝜃𝑙 ≥ min{𝛼𝑙, 𝛽𝑙}, ∀𝑙 ∈ 𝐾 ∖ {𝑘} and 𝜃𝑗 = 𝛼𝑗 for all𝑗 /∈ 𝐾 . If we take the indices 𝑙 for which we have equality, then 𝐹𝐵(𝐸, 𝛼) = ∅, where𝐵 = (𝐼 ∖ 𝐾) ∪ {𝑙 ∈ 𝐾, 𝜃𝑙 = 𝛼𝑙} is such that 𝑖 ∈ 𝐵. This contradicts the hypothesis of ourlemma.


1.3.2 Apéry Sets

Let ℐ be a fractional ideal of 𝒪 and let 𝐸 be its set of values. For any 𝛼 ∈ 𝐸 ∖ {0},we define the Apéry set of 𝐸 with respect to 𝛼 to be the set

𝐴𝛼(𝐸) = {𝛽 ∈ 𝐸; 𝛽 − 𝛼 /∈ 𝐸}.

We will denote 𝐴𝛼(𝑆) by 𝐴𝛼.

Lemma 1.24. Let 𝑆 be the set of values of an algebroid curve, let 𝛼 ∈ 𝑆 ∖ {0} and let𝐴𝛼 be its corresponding Apéry set. For 𝛽 ∈ Z𝑟 one has that 𝛽 /∈ 𝑆 if only if 𝛽 = 𝑎 − 𝜌𝛼,for some 𝑎 ∈ 𝐴𝛼 and some 𝜌 ∈ N ∖ {0}.

Proof

(⇒) Since 𝑆 ⊂ N𝑟, we have that 𝛼 > 0. Suppose that 𝛽 /∈ 𝑆 and let 𝜌 be thepositive integer such that 𝑎 = 𝛽 + 𝜌𝛼 ∈ 𝑆, but 𝛽 + (𝜌 − 1)𝛼 /∈ 𝑆. This implies 𝑎 ∈ 𝐴𝛼,giving the result.

(⇐) We have that 𝛽 = 𝑎 − 𝜌𝛼 for 𝑎 ∈ 𝐴𝛼 and 𝜌 > 0. Suppose that 𝛽 ∈ 𝐸. Since𝛼 ∈ 𝑆 and 𝜌 > 0, we have that

𝑎 − 𝛼 = 𝛽 + (𝜌 − 1)𝛼 ∈ 𝑆,

which contradicts the fact that 𝑎 ∈ 𝐴𝛼.

1.4 SymmetryIn what follows we will denote by 𝒪 the ring of an algebroid curve and by 𝑆 = 𝑆(𝒪)

its semigroup of values. Recall that we defined f(𝑆) as being 𝑐(𝒪) − (1, . . . , 1). A fractionalideal of 𝒪 will be denoted by ℐ, its dual by ℐ*, while 𝐸 and 𝐸* will denote the value setsof ℐ and 𝐼*, respectively.

The property of the ring 𝒪 of an algebroid curve to be Gorenstein is equivalent tothe symmetry of the semigroup of values 𝑆(𝒪) of 𝒪 as pointed out by E. Kunz in [14] inthe irreducible case and generalized by F. Delgado de la Mata in [8] as follows:

Theorem 1.25. ([8, Theorem 2.8]) The ring 𝒪 is Gorenstein if and only if

∀𝛼 ∈ N𝑟, 𝛼 ∈ 𝑆 ⇐⇒ 𝐹 (𝑆, f(𝑆) − 𝛼) = ∅. (1.2)

A semigroup 𝑆 satisfying the property (1.2) above, will be called a symmetricsemigroup.

The above result was generalized by D. Pol in [16] as follows:

1.4. Symmetry 29

Theorem 1.26. ([16, Theorem 2.4]) The ring 𝒪 is Gorenstein if and only if, for allfractional ideal ℐ of 𝒪, the following property is verified

∀𝛽 ∈ Z𝑟, 𝛽 ∈ 𝐸* ⇐⇒ 𝐹 (𝐸, f(𝑆) − 𝛽) = ∅. (1.3)

Property (1.3) says that for a Gorenstein ring 𝒪, the values set 𝐸 of any fractionalideal, determines the values set 𝐸* of its dual ℐ*, and conversely.

We will show in this section that the Gorenstein property of 𝒪 may be detected inseveral other ways using the Apéry sets, generalizing the above two results.

Motivated by the classical theory of semigroups of values of irreducible curves, wehave the following result:

Theorem 1.27. Let 𝑆 = 𝑆(𝒪) be a semigroup of values in N𝑟 and let 𝛼 ∈ 𝑆 ∖ {0}. Thefollowing conditions are equivalent:

i) 𝒪 is Gorenstein;

ii) 𝑆 is symmetric;

iii) ∀𝑎 ∈ N𝑟, 𝑎 ∈ 𝐴𝛼 =⇒ ∅ = 𝐹 (𝑆, f(𝑆) + 𝛼 − 𝑎) ⊆ 𝐴𝛼.

Proof 𝑖) ⇔ 𝑖𝑖) This is the content of Theorem 1.25.

𝑖𝑖) ⇒ 𝑖𝑖𝑖) Suppose that 𝑆 is symmetric. Let 𝑎 ∈ 𝐴𝛼, then 𝑎 − 𝛼 /∈ 𝑆 and since 𝑆 issymmetric, we have 𝐹 (𝑆, f(𝑆)− (𝑎−𝛼)) = ∅. Now we will show that 𝐹 (𝑆, f(𝑆)− (𝑎−𝛼)) ⊆𝐴𝛼.

Let 𝛾 be in 𝐹 (𝑆, f(𝑆) − (𝑎 − 𝛼)). If 𝛾 − 𝛼 belonged to 𝑆, then we would have𝛾 −𝛼 ∈ 𝐹 (𝑆, f(𝑆)−𝑎), then 𝐹 (𝑆, f(𝑆)−𝑎) = ∅. But, since 𝑎 ∈ 𝐴𝛼 ⊂ 𝑆 and 𝑆 is symmetric,it follows that 𝐹 (𝑆, f(𝑆) −𝑎) = ∅, which is a contradiction. So, 𝛾 −𝛼 /∈ 𝑆 and consequently𝛾 ∈ 𝐴𝛼.

𝑖𝑖𝑖) ⇒ 𝑖𝑖) Assuming (iii), let us prove that 𝑆 is symmetric, i.e., 𝛾 ∈ 𝑆 ⇔ 𝐹 (𝑆, f(𝑆)−𝛾) = ∅.

We know from Corollary 1.20 that the implication 𝛾 ∈ 𝑆 ⇒ 𝐹 (𝑆, f(𝑆) − 𝛾) = ∅ isalways satisfied.

To prove the other direction, we show that if 𝛾 /∈ 𝑆 then 𝐹 (𝑆, f(𝑆) − 𝛾) = ∅. So, let𝛾 /∈ 𝑆, then by Lemma 1.24 we have that 𝛾 = 𝑎 + 𝜌𝛼, for some 𝑎 ∈ 𝐴𝛼 and some negativeinteger 𝜌. From our hypothesis we know that 𝐹 (𝑆, f(𝑆) + 𝛼 − 𝑎) = ∅, hence take 𝜃 in thisset.

Since 𝜃 ∈ 𝑆 and −(𝜌 + 1)𝛼 ∈ 𝑆, we have that

𝜃 − (𝜌 + 1)𝛼 ∈ 𝐹 (N𝜌, f(𝑆) − (𝑎 + 𝜌𝛼)) ∩ 𝑆 = 𝐹 (𝑆, f(𝑆) − 𝛾),


showing that 𝐹 (𝑆, f(𝑆) − 𝛾) = ∅ and therefore 𝑆 is symmetric.

Observe that the condition 𝐹 (𝑆, f(𝑆) + 𝛼 − 𝑎) = ∅ implies that 𝑎 − 𝛼 /∈ 𝑆, hence if𝑎 ∈ 𝑆, then 𝑎 ∈ 𝐴𝛼. It follows that condition (iii) in the above proposition may be readedas

∀𝑎 ∈ N𝑟, 𝑎 ∈ 𝐴𝛼 ⇐⇒ 𝑎 ∈ 𝑆 and ∅ = 𝐹 (𝑆, f(𝑆) + 𝛼 − 𝑎) ⊆ 𝐴𝛼.

Theorem 1.28. Let 𝒪 be the ring of an algebroid curve. We denote by ℐ a fractionalideal of 𝒪 and by 𝐸 its value set. For 𝛼 ∈ 𝐸 ∩ 𝐸*, the following conditions are equivalent:

i) 𝒪 is Gorenstein;

ii) ∀ ℐ, ∀ 𝛽 ∈ Z𝑟, 𝛽 ∈ 𝐸* ⇐⇒ 𝐹 (𝐸, f(𝑆) − 𝛽) = ∅.

iii) ∀ ℐ, ∀ 𝑎 ∈ Z𝑟, 𝑎 ∈ 𝐴𝛼(𝐸) =⇒ ∅ = 𝐹 (𝐸*, f(𝑆) + 𝛼 − 𝑎) ⊆ 𝐴𝛼(𝐸*).

Proof 𝑖) ⇐⇒ 𝑖𝑖) This is the content of Theorem 1.26.

𝑖𝑖) ⇒ 𝑖𝑖𝑖) Let us assume that (ii) is fulfilled. Let 𝑎 ∈ 𝐴𝛼(𝐸), then 𝑎 − 𝛼 /∈ 𝐸, sofrom (ii), we have 𝐹 (𝐸*, f(𝑆) + 𝛼 − 𝑎) = ∅ (recall that (ii) is equivalent to 𝒪 is Gorensteinand, in this situation, (𝐸*)* = 𝐸).

Now, suppose that 𝛾 ∈ 𝐹 (𝐸*, f(𝑆) + 𝛼 − 𝑎), then 𝛾 ∈ 𝐸*. It remains to showthat 𝛾 − 𝛼 /∈ 𝐸*. Suppose that the opposite holds, that is, 𝛾 − 𝛼 ∈ 𝐸*, then 𝛾 − 𝛼 ∈𝐹 (𝐸*, f(𝑆) − 𝑎), hence 𝐹 (𝐸*, f(𝑆) − 𝑎) = ∅. But, since 𝑎 ∈ 𝐸, from (ii) one gets that𝐹 (𝐸*, f(𝑆) − 𝑎) = ∅, a contradiction.

𝑖𝑖𝑖) ⇒ 𝑖) Let us assume that (iii) holds for all fractional ideal ℐ. In particular itholds for ℐ = 𝒪. Now, Theorem 1.27 implies that 𝒪 is Gorenstein.

31

CHAPTER

2COLENGTHS OF FRACTIONAL IDEALS

In this chapter we will give a formula relating the codimension of a fractional idealwith the maximal elements in its set of values.

Let 𝒪 be the ring of an algebroid curve and 𝒥 ⊂ ℐ are two fractional ideals withsets of values 𝐷 and 𝐸, respectively. Since 𝒥 ⊂ ℐ, one has that 𝐷 ⊂ 𝐸, hence 𝑐(𝐸) ≤ 𝑐(𝐷).Our aim in this section is to find a formula for the length ℓ𝒪(ℐ/𝒥 ) of ℐ/𝒥 as 𝒪-modules,called the colength of 𝒥 in ℐ, in terms of the value sets 𝐷 and 𝐸.

The motivation comes from the case in which 𝒪 is a domain, when one has that

ℓ𝒪(ℐ/𝒥 ) = #(𝐸 ∖ 𝐷).

For 𝛼 ∈ Z𝑟 and ℐ a fractional ideal of 𝒪, with value set 𝐸, we define

ℐ(𝛼) = {ℎ ∈ ℐ ∖ 𝑍(𝒦); 𝑣(ℎ) ≥ 𝛼}.

It is clear that if 𝑚𝐸 = min 𝐸, then ℐ(𝑚𝐸) = ℐ.

One has the following result:

Proposition 2.1. ([2, Proposition 2.7]) Let 𝒥 ⊆ ℐ be two fractional ideals of 𝒪, withvalue sets 𝐷 and 𝐸, respectively, then

ℓ𝒪

( ℐ𝒥

)= ℓ𝒪

(ℐ

ℐ(𝛾)

)− ℓ𝒪

(𝒥

𝒥 (𝛾)

),

for sufficiently large 𝛾 ∈ N𝑟 (for instance, if 𝛾 ≥ 𝑐(𝐷)).

Let 𝑒𝑖 ∈ Z𝑟 denote the vector with zero entries except the 𝑖-th entry which is equalto 1, and recall the definition we gave in Subsection 1.3.1

𝐹 𝑖(𝐸, 𝛼) = {𝛽 ∈ 𝐸; 𝛽𝑖 = 𝛼𝑖 and 𝛽𝑗 ≥ 𝛼𝑗, ∀𝑗 = 𝑖},

32 Chapter 2. Colengths of fractional ideals

then the following result will give us an effective way to calculate colengths of ideals.

Proposition 2.2. [6, Proposition 2.2] If 𝛼 ∈ Z𝑟, then we have

ℓ𝒪

(ℐ(𝛼)

ℐ(𝛼 + 𝑒𝑖)

)=

⎧⎪⎪⎪⎨⎪⎪⎪⎩1, 𝑖𝑓 𝐹 𝑖(𝐸, 𝛼) = ∅,

0, otherwise.

So, to compute, for instance, ℓ𝒪

(ℐ

ℐ(𝛾)

), one takes a chain

𝑚𝐸 = 𝛼0 ≤ 𝛼1 ≤ · · · ≤ 𝛼𝑚 = 𝛾,

and observe that

ℓ𝒪

(ℐ

ℐ(𝛾)

)= ℓ𝒪

(ℐ(𝛼0)ℐ(𝛾)

)=

𝑚∑𝑗=1

ℓ𝒪

(ℐ(𝛼𝑗−1)ℐ(𝛼𝑗)

).

This shows that the colength of an ideal ℐ is independent from the chain one chooses.So, if we take a saturated chain, that is, a chain such that 𝛼𝑗 − 𝛼𝑗−1 ∈ {𝑒𝑖; 𝑖 = 1, 2, . . . , 𝑟}for all 𝑗 = 1, . . . , 𝑚, then the computation may be done using Proposition 2.2.

In what follows We will denote ℓ𝒪 simply by ℓ.

For the application we have in mind, we will have to compute

ℓ(ℐ1 × · · · × ℐ𝑟

ℐ

),

where ℐ𝑖, 𝑖 = 1, . . . , 𝑟, is the projection of ℐ ⊂ 𝒦1 × · · · × 𝒦𝑟 into 𝒦𝑖.

Notice that if 𝐸 is the value set of ℐ and 𝐸1, . . . , 𝐸𝑟 those of ℐ1, . . . , ℐ𝑟, then it isclear that 𝐸1 × · · · × 𝐸𝑟 is the value set of ℐ1 × · · · × ℐ𝑟, which obviously has no maximalpoints, since no element in it has an empty fiber.

By Proposition 2.1 we have that

ℓ(ℐ1 × · · · × ℐ𝑟

ℐ

)= ℓ

(ℐ1 × · · · × ℐ𝑟

(ℐ1 × · · · × ℐ𝑟)(𝛾)

)− ℓ

(ℐ

ℐ(𝛾)

), (2.1)

for some 𝛾 ∈ 𝐸 sufficiently large (for instance, if 𝛾 ≥ 𝑐(𝐸)).

2.1 Case r=2This simplest case was studied by D’anna, Barucci and Fröberg in [2] and we

reproduce it here because it gives a clue on how to proceed in general.

Let 𝛼0 = 𝑚𝐸 and consider the saturated chain in Z2

𝛼0 ≤ · · · ≤ 𝛼𝑚 = 𝛾 = (𝛾1, 𝛾2)

2.1. Case r=2 33

such that𝛼0 = (𝛼0

1, 𝛼02), 𝛼1 = (𝛼0

1 + 1, 𝛼02), . . . , 𝛼𝑠 = (𝛾1, 𝛼0

2),

𝛼𝑠+1 = (𝛾1, 𝛼02 + 1), 𝛼𝑠+2 = (𝛾1, 𝛼0

2 + 2), . . . , 𝛼𝑚 = (𝛾1, 𝛾2),and consider the following sets

𝐿1 = {𝛼0, 𝛼1, . . . , 𝛼𝑠} and 𝐿2 = {𝛼𝑠, 𝛼𝑠+1, . . . , 𝛼𝑚}.

By Proposition 2.2, we have

ℓ

(ℐ

ℐ(𝛾)

)= #𝐿1 − #{𝛼 ∈ 𝐿1; 𝐹 1(𝐸, 𝛼) = ∅} +

#𝐿2 − #{𝛼 ∈ 𝐿2; 𝐹 2(𝐸, 𝛼) = ∅}.

Now, because of our choice of 𝐿1, we have that

∀ 𝛼 ∈ 𝐿1, 𝐹 1(𝐸, 𝛼) = ∅ ⇐⇒ 𝑝𝑟1(𝛼) ∈ M(𝐸1),

hence#{𝛼 ∈ 𝐿1; 𝐹 1(𝐸, 𝛼) = ∅} = # M(𝐸1).

Observe that not all 𝛼 ∈ 𝐿2 with 𝐹 2(𝐸, 𝛼) = ∅ are such that 𝑝𝑟2(𝛼) ∈ M(𝐸2),hence

#{𝛼 ∈ 𝐿2; 𝐹 2(𝐸, 𝛼) = ∅} = # M(𝐸2) − 𝜉,

where 𝜉 is the number of 𝛼 in 𝐿2 with 𝑝𝑟2(𝛼) ∈ 𝐸2 and 𝐹 2(𝐸, 𝛼) = ∅. But such 𝛼 are inone-to-one correspondence with the maximals of 𝐸, hence 𝜉 = # M(𝐸).

Putting all this together, we get

Proposition 2.3. We have that

ℓ

(ℐ

ℐ(𝛾)

)= (𝛾1 − 𝛼0

1) − # M(𝐸1) + (𝛾2 − 𝛼02) − # M(𝐸2) − # M(𝐸). (2.2)

On the other hand, by Proposition 2.2, we have that

ℓ

(ℐ1 × ℐ2

(ℐ1 × ℐ2)(𝛾)

)= #𝐿1 − #{𝛼 ∈ 𝐿1; 𝐹 1(𝐸1 × 𝐸2, 𝛼) = ∅}+

#𝐿2 − #{𝛼 ∈ 𝐿2; 𝐹 2(𝐸1 × 𝐸2, 𝛼) = ∅}.

This gives

ℓ

(ℐ1 × ℐ2

(ℐ1 × ℐ2)(𝛾)

)= (𝛾1 − 𝛼0

1) − # M(𝐸1) + (𝛾2 − 𝛼02) − # M(𝐸2), (2.3)

Thus, by (2.1), (2.3) and (2.2) we get

Proposition 2.4. One has

ℓ(ℐ1 × ℐ2

ℐ

)= # M(𝐸). (2.4)


2.2 Case 𝑟 ≥ 3

Let us assume that ℐ is a fractional ideal of 𝒪, where 𝒪 has 𝑟 minimal primes.

Let𝑚𝐸 = 𝛼0 ≤ 𝛼1 ≤ · · · ≤ 𝛼𝑚 = 𝛾,

be the saturated chain in Z𝑟, given by the union of the following paths (see Figure 1, for𝑟 = 3):

𝐿1 : 𝛼0, 𝛼1 = 𝛼0 + 𝑒1, . . . , 𝛼𝑠1 = 𝛼0 + (𝛾1 − 𝛼01)𝑒1 = (𝛾1, . . . , 𝛼0

𝑟),

. . .

𝐿𝑟 : 𝛼𝑠𝑟−1 = (𝛾1, . . . , 𝛾𝑟−1, 𝛼0𝑟), 𝛼𝑠𝑟−1+1 = 𝛼𝑠𝑟−1 + 𝑒𝑟, . . . , 𝛼𝑚 = 𝛾.

Figure 1 – The saturated chain for 𝑟 = 3

Let us define 𝐼 ′ = {1, . . . , 𝑟 − 1}. We will need the following result:

Lemma 2.5. For any 𝛼 ∈ 𝐿1 ∪ . . . ∪ 𝐿𝑟−1, and for 𝑖 ∈ 𝐼 ′ = {1, . . . , 𝑟 − 1}, one has

𝐹 𝑖(𝐸, 𝛼) = ∅ ⇐⇒ 𝐹 𝑖(𝐸𝐼′ , 𝑝𝑟𝐼′(𝛼)) = ∅.

Proof (⇒) This is obvious.

(⇐) Suppose that

(𝜃1, . . . , 𝜃𝑟−1) ∈ 𝐹 𝑖(𝐸𝐼′ , 𝑝𝑟𝐼′(𝛼)) = ∅.

Since by Lemma 1.14 one has that 𝑝𝑟𝐼′(𝐸) = 𝐸𝐼′ , then there exists 𝜃 = (𝜃1, . . . , 𝜃𝑟−1, 𝜃𝑟) ∈𝐸. Since 𝛼 ∈ 𝐿𝑖 for some 𝑖 = 1, . . . , 𝑟 − 1, it follows that 𝛼𝑟 = 𝛼0

𝑟 . Then one cannot have𝜃𝑟 < 𝛼𝑟 = 𝛼0

𝑟 , because otherwise

(𝛼01, . . . , 𝛼0

𝑟−1, 𝜃𝑟) = min(𝛼0, 𝜃) ∈ 𝐸,

2.2. Case 𝑟 ≥ 3 35

which is contradiction, since 𝛼0 is the minimum of 𝐸. Hence 𝜃𝑟 ≥ 𝛼𝑟, so 𝜃 ∈ 𝐹 𝑖(𝐸, 𝛼), andthe result follows.

The above lemma allows us to write

ℓ

(ℐ

ℐ(𝛾)

)= ℓ

(𝜋𝐼′(ℐ)

𝜋𝐼′(ℐ)(𝑝𝑟𝐼′(𝛾))

)+ (𝛾𝑟 − 𝛼0

𝑟) − #{𝛼 ∈ 𝐿𝑟; 𝐹 𝑟(𝐸, 𝛼) = ∅}. (2.5)

Hence to get an inductive formula for ℓ(

ℐℐ(𝛾)

), we only have to compute

#{𝛼 ∈ 𝐿𝑟; 𝐹 𝑟(𝐸, 𝛼) = ∅},

and for this we will need the following lemma.

Lemma 2.6. Let 𝛼 ∈ Z𝑟, then 𝐹 𝑗(𝐸, 𝛼) = ∅ if only if there exist some 𝐽 ⊆ 𝐼 with 𝑗 ∈ 𝐽

and a relative maximal 𝛽 of 𝐸𝐽 such that 𝑝𝑟𝑗(𝛽) = 𝛼𝑗 and 𝑝𝑟𝑖(𝛽) < 𝛼𝑖, for all 𝑖 ∈ 𝐽 , 𝑖 = 𝑗.

Proof (⇐) (We prove more, since it is enough to assume 𝛽 is any maximal) Let usassume that there exist 𝐽 ⊆ 𝐼, with 𝑗 ∈ 𝐽 and 𝛽 ∈ M(𝐸𝐽), such that 𝛽𝑗 = 𝑝𝑟𝑗(𝛽) = 𝛼𝑗

and 𝛽𝑖 = 𝑝𝑟𝑖(𝛽) < 𝛼𝑖, for all 𝑖 ∈ 𝐽 , 𝑖 = 𝑗.

Suppose by reductio ad absurdum that 𝐹 𝑗(𝐸, 𝛼) = ∅. Let 𝜃 ∈ 𝐹 𝑗(𝐸, 𝛼) that is𝜃𝑗 = 𝛼𝑗 and 𝜃𝑖 ≥ 𝛼𝑖, ∀𝑖 ∈ 𝐽 ∖ {𝑗}. Now since

𝑝𝑟𝑗(𝑝𝑟𝐽(𝜃)) = 𝜃𝑗 = 𝛼𝑗 = 𝛽𝑗 and 𝑝𝑟𝑖(𝑝𝑟𝐽(𝜃)) = 𝜃𝑗 ≥ 𝛼𝑖 > 𝛽𝑖, ∀𝑖 ∈ 𝐽, 𝑖 = 𝑗,

then 𝑝𝑟𝐽(𝜃) ∈ 𝐹𝑗(𝐸𝐽 , 𝛽), which contradicts the assumption that 𝛽 ∈ M(𝐸𝐽).

(⇒) Since 𝐹 𝑗(𝐸, 𝛼) = ∅ implies 𝐹𝑗(𝐸, 𝛼) = ∅, the proof follows the same lines asthe proof of [8, Theorem 1.5].

Going back to our main calculation, by Lemma 2.6, if 𝛼 ∈ 𝐿𝑟 is such that 𝐹 𝑟(𝐸, 𝛼) =∅, then there exist a subset 𝐽 of 𝐼 = {1, . . . , 𝑟}, with 𝑟 ∈ 𝐽 , and 𝛽 ∈ RM(𝐸𝐽), with𝑝𝑟𝑟(𝛽) = 𝛼𝑟 and 𝑝𝑟𝑖(𝛽) < 𝛼𝑖 for 𝑖 ∈ 𝐽, 𝑖 = 𝑟.

Notice that for 𝛼 ∈ 𝐿𝑟 one has 𝛼𝑖 = 𝛾𝑖 for 𝑖 = 𝑟, so the condition 𝑝𝑟𝑖(𝛽) < 𝛼𝑖

for 𝑖 ∈ 𝐽, 𝑖 = 𝑟 is satisfied, since 𝛽 ∈ M(𝐸𝐽). So, we have a bijection between theset {𝛼 ∈ 𝐿𝑟; 𝐹 𝑟(𝐸, 𝛼) = ∅} and the set ⋃𝑟∈𝐽⊆𝐼 𝑝𝑟𝑟(RM(𝐸𝐽)), where 𝑝𝑟𝑟(𝛽) is the lastcoordinate of 𝛽, that is, 𝛽𝑟.

This gives us the formula

#{𝛼 ∈ 𝐿𝑟; 𝐹 𝑟(𝐸, 𝛼) = ∅} = #( ⋃

𝑟∈𝐽⊆𝐼

𝑝𝑟𝑟(RM(𝐸𝐽)))

.


If 𝐽 = {𝑟}, we have 𝑝𝑟𝑟(M(𝐸𝑟)) = M(𝐸𝑟), and for all 𝐽 ′ ⊆ 𝐼 with 𝑟 ∈ 𝐽 ′ and𝐽 ′ = {𝑟}, also M(𝐸𝑟) ∩ 𝑝𝑟𝑟(RM(𝐸𝐽 ′)) = ∅, hence

#{𝛼 ∈ 𝐿𝑟; 𝐹 𝑟(𝐸, 𝛼) = ∅} = # M(𝐸𝑟) + #( ⋃

𝑟∈𝐽⊆𝐼

𝐽 ={𝑟}


. (2.6)

Now, putting together Equations (2.5) and (2.6), we get

Theorem 2.7. For a fractional ideal ℐ of a ring 𝒪 with 𝑟 minimal primes, with valuesset 𝐸, one has

ℓ

(ℐ

ℐ(𝛾)

)= ℓ

(𝜋𝐼′(ℐ)

𝜋𝐼′(ℐ)(𝑝𝑟𝐼′(𝛾))

)+ (𝛾𝑟 − 𝛼0

𝑟) − # M(𝐸𝑟)−

#(⋃

𝑟∈𝐽⊆𝐼

𝐽 ={𝑟}𝑝𝑟𝑟(RM(𝐸𝐽))

).

(2.7)

We want now to calculate ℓ(ℐ1 × · · · × ℐ𝑟

ℐ

). By Propositions 2.2 and 2.1 we have:

ℓ(ℐ1 × · · · × ℐ𝑟

ℐ

)= ℓ

(ℐ1 × · · · × ℐ𝑟

(ℐ1 × · · · × ℐ𝑟)(𝛾)

)− ℓ

(ℐ

ℐ(𝛾)

), (2.8)

for some 𝛾 ≥ 𝑐(𝐸). Suppose that we know the formula for any fractional ideal in a ringwith 𝑟 − 1 minimal primes. Then, as we argued in the case 𝑟 = 2, we have

ℓ

(ℐ1 × · · · × ℐ𝑟

(ℐ1 × · · · × ℐ𝑟)(𝛾)

)= ℓ

(ℐ1 × · · · × ℐ𝑟−1

(ℐ1 × · · · × ℐ𝑟−1)(𝑝𝑟𝐼′(𝛾))

)+

(𝛾𝑟 − 𝛼0𝑟) − # M(𝐸𝑟).

(2.9)

Putting together Equations (2.7), (2.8) and (2.9), we have

Theorem 2.8. For a fractional ideal ℐ of a ring 𝒪 with 𝑟 minimal primes, with valuesset 𝐸, one has

ℓ(ℐ1 × · · · × ℐ𝑟

ℐ

)= ℓ

(ℐ1 × · · · × ℐ𝑟−1

𝜋𝐼′(ℐ)

)+ #

( ⋃𝑟∈𝐽⊆𝐼

𝐽 ={𝑟}


. (2.10)

2.3 Case r=3

In this subsection, we will give nicer formulas than Formulas (2.7) and (2.10), when𝑟 = 3. To simplify notations, for any 𝐽 ⊂ 𝐼 = {1, 2, 3}, we will denote by RM𝐽 , AM𝐽 andM𝐽 the sets RM(𝐸𝐽), AM(𝐸𝐽) and M(𝐸𝐽), respectively. Notice also that if #𝐽 = 1, 2,then RM𝐽 = AM𝐽 = M𝐽 .

2.3. Case r=3 37

From Formulas (2.2) and (2.5), one has

ℓ

(ℐ

ℐ(𝛾)

)= (𝛾1 − 𝛼0

1) − # M1 +(𝛾2 − 𝛼02) − # M2 − M{1,2} +

(𝛾3 − 𝛼03) − #{𝛼 ∈ 𝐿3; 𝐹 3(𝐸, 𝛼) = ∅}.

Now, from Lemma 2.6, the points 𝛼 = (𝛼1, 𝛼2, 𝛼3) ∈ 𝐿3 such that 𝐹 3(𝐸, 𝛼) = ∅come from maximal points 𝛽 of either 𝐸3, 𝐸{1,3}, 𝐸{2,3}, or 𝐸 such that 𝑝𝑟3(𝛽) = 𝛼3, sowe have

#{𝛼 ∈ 𝐿3; 𝐹 3(𝐸, 𝛼) = ∅} = # M3 +# RM{1,3} +# M{2,3} +# RM −𝜂, (2.11)

where 𝜂 is some correcting term which will take into account the eventual multiple counting.

To compute 𝜂 we will analyze in greater detail the geometry of maximal points.

If 𝛼, 𝛽 ∈ M with 𝛼3 = 𝛽3, then 𝛼1 = 𝛽1 and 𝛼2 = 𝛽2. If 𝛼1 < 𝛽1, then necessarily𝛽2 < 𝛼2.

We say that two relative (respectively, absolute) maximals 𝛼 and 𝛽 of 𝐸 with𝛼3 = 𝛽3 and 𝛼1 < 𝛽1 are adjacent, if there is no (𝜃1, 𝜃2, 𝛼3) in RM (respectively, in AM)with 𝛼1 < 𝜃1 < 𝛽1 and 𝛽2 < 𝜃2 < 𝛼2.

We will describe below the geometry of the maximal points of 𝐸

Lemma 2.9. If 𝛼 ∈ AM, then one and only one of the following three conditions isverified:

i) there exist two adjacent relative maximals 𝛽 and 𝜃 of 𝐸 such that 𝑝𝑟{1,3}(𝛽) =𝑝𝑟{1,3}(𝛼) and 𝑝𝑟{2,3}(𝜃) = 𝑝𝑟{2,3}(𝛼);

ii) there exists 𝛽 ∈ RM such that 𝑝𝑟{1,3}(𝛽) = 𝑝𝑟{1,3}(𝛼) and 𝑝𝑟{2,3}(𝛼) ∈ M{2,3}, or𝑝𝑟{2,3}(𝛽) = 𝑝𝑟{2,3}(𝛼) and 𝑝𝑟{1,3}(𝛼) ∈ M{1,3};

iii) 𝑝𝑟{1,3}(𝛼) ∈ M{1,3} and 𝑝𝑟{2,3}(𝛼) ∈ M{2,3}.

Proof Let 𝛼 = (𝛼1, 𝛼2, 𝛼3) ∈ AM, then 𝐹 (𝐸, 𝛼) = ∅. We consider the following sets:

𝑅1 = {𝛽 ∈ Z3; 𝛽3 = 𝛼3, 𝛽1 > 𝛼1, 𝛽2 < 𝛼2}

and𝑅2 = {𝜃 ∈ Z3; 𝜃3 = 𝛼3, 𝜃1 < 𝛼1, 𝜃2 > 𝛼2}.


Then there are four possibilities:

𝑅1 ∩ 𝐸 = ∅ and 𝑅2 ∩ 𝐸 = ∅, 𝑅1 ∩ 𝐸 = ∅ and 𝑅2 ∩ 𝐸 = ∅.

𝑅1 ∩ 𝐸 = ∅ and 𝑅2 ∩ 𝐸 = ∅, 𝑅1 ∩ 𝐸 = ∅ and 𝑅2 ∩ 𝐸 = ∅.

Suppose 𝑅1 ∩ 𝐸 = ∅ and 𝑅2 ∩ 𝐸 = ∅. Choose 𝛽 ∈ 𝑅1 ∩ 𝐸 and 𝜃 ∈ 𝑅2 ∩𝐸, such that 𝛼2 − 𝛽2 and 𝛼1 − 𝜃1 are as small as possible. Then by Property (A),we have min(𝛼, 𝛽), min(𝛼, 𝜃) ∈ 𝐸. Obviously 𝑝𝑟{1,3}(𝛽) = 𝑝𝑟{1,3}(𝛼) and 𝑝𝑟{2,3}(𝜃) =𝑝𝑟{2,3}(𝛼). Moreover, according to Lemma 1.22, these are relative maximals because𝐹3(𝐸, min(𝛼, 𝛽)) = 𝐹3(𝐸, min(𝛼, 𝜃)) = ∅, 𝐹{1,3}(𝐸, min(𝛼, 𝛽)) = ∅, 𝐹{1,3}(𝐸, min(𝛼, 𝜃)) =∅, 𝐹{2,3}(𝐸, min(𝛼, 𝛽)) = ∅ and 𝐹{2,3}(𝐸, min(𝛼, 𝜃)) = ∅. It follows that min(𝛼, 𝛽) andmin(𝛼, 𝜃) are adjacent relative maximals.

Suppose 𝑅1 ∩ 𝐸 = ∅ and 𝑅2 ∩ 𝐸 = ∅. Choose 𝛽 ∈ 𝑅1 ∩ 𝐸 such that 𝛼2 − 𝛽2 is assmall as possible, then, as we argued above, we have that min(𝛼, 𝛽) ∈ RM and 𝑝𝑟{1,3}(𝛽) =𝑝𝑟{1,3}(𝛼). Moreover, as 𝑅2 ∩ 𝐸 = ∅, it follows that the projection 𝑝𝑟{2,3}(𝛼) ∈ M{2,3}.

The case 𝑅1 ∩ 𝐸 = ∅ and 𝑅2 ∩ 𝐸 = ∅ is similar to the above one, giving us thesecond possibility in (ii).

Suppose 𝑅1 ∩ 𝐸 = ∅ and 𝑅2 ∩ 𝐸 = ∅. It is obvious that 𝑝𝑟{1,3}(𝛼) ∈ M{1,3} and𝑝𝑟{2,3}(𝛼) ∈ M{2,3}.

Given two points 𝜃1, 𝜃2 ∈ Z3 such that 𝑝𝑟3(𝜃1) = 𝑝𝑟3(𝜃2), we will denote by ℛ(𝜃1, 𝜃2)the parallelogram determined by the coplanar points 𝜃1, 𝜃2, min(𝜃1, 𝜃2) and max(𝜃1, 𝜃2).We have the following result:

Corollary 2.10. Let 𝜃1, 𝜃2 ∈ AM, such that 𝑝𝑟3(𝜃1) = 𝑝𝑟3(𝜃2). Then one has ℛ(𝜃1, 𝜃2) ∩RM = ∅.

Proof Because 𝜃1, 𝜃2 ∈ AM, it follows immediately that (iii) of Lemma 2.9 cannot happen,therefore, the existence of the relative maximal is ensured by (i) or (ii).

Lemma 2.11. If 𝛽 and 𝛽′ are adjacent relative maximals, with 𝛽3 = 𝛽′3, then max(𝛽, 𝛽′)

is an absolute maximal of 𝐸.

Proof We may suppose that 𝛽1 > 𝛽′1 and 𝛽2 < 𝛽′

2. As 𝛽 and 𝛽′ are adjacent, we havethat 𝐹{1,3}(𝐸, 𝛽) ∩ 𝐹{2,3}(𝐸, 𝛽′) = ∅, because otherwise if we take 𝛼1 ∈ 𝐹{1,3}(𝐸, 𝛽), with𝛼1

2 the greatest possible and 𝛼2 ∈ 𝐹{2,3}(𝐸, 𝛽′), with 𝛼21 the greatest possible. From Lemma

1.23 it follows that 𝛼1 and 𝛼2 are absolute maximals of 𝐸, then by Corollary 2.10 thereexists a relative maximal in the region ℛ(𝛼1, 𝛼2), this contradicts the fact that 𝛽 and 𝛽′

are adjacent relative maximals.

2.3. Case r=3 39

Then, effectively, 𝐹{1,3}(𝐸, 𝛽) ∩ 𝐹{2,3}(𝐸, 𝛽′) = {max(𝛽, 𝛽′)}, which is an absolutemaximal.

We will use the following notation: 𝐿′3 = {𝛼 ∈ 𝐿3; 𝐹 3(𝐸, 𝛼) = ∅}. Recall that the

elements in 𝐿3 are of the form (𝛾1, 𝛾2, 𝛼3), with 𝛼03 ≤ 𝛼3 ≤ 𝛾3.

Lemma 2.12. Let 𝛼 ∈ 𝐿′3 be such that 𝛼3 ∈ (𝑝𝑟3(M{1,3}) ∖ 𝑝𝑟3(M{2,3})) ∩ 𝑝𝑟3(RM), or

𝛼3 ∈ (𝑝𝑟3(M{2,3}) ∖ 𝑝𝑟3(M{1,3})) ∩ 𝑝𝑟3(RM), then there are the same number of relative asabsolute maximals in 𝐸 with third coordinate equal to 𝛼3.

Proof We assume that 𝛼3 ∈ (𝑝𝑟3(M{1,3}) ∖ 𝑝𝑟3(M{2,3})) ∩ 𝑝𝑟3(RM), since the other caseis analogous.

Since 𝛼3 ∈ 𝑝𝑟3(RM), we may assume that there are 𝑠 (≥ 1) relative maximals𝛽1, . . . , 𝛽𝑠 in 𝐸 with third coordinate equal to 𝛼3. We may suppose that 𝛽1

1 < 𝛽21 < · · · < 𝛽𝑠

1,so the 𝛽𝑖’s are successively adjacent relative maximals, hence, by lemma 2.11, we have that

max(𝛽1, 𝛽2), . . . , max(𝛽𝑠−1, 𝛽𝑠) ∈ AM .

This shows that there are at least 𝑠−1 absolute maximals in 𝐸 with third coordinate𝛼3.

Now as 𝑝𝑟3(𝛼) ∈ 𝑝𝑟3(M{1,3}), then there is a (𝜂11, 𝛼3) ∈ M{1,3} with 𝜂1

1 ≤ 𝛼1 (= 𝛾1),because 𝑐(𝐸{1,3}) ≤ 𝑝𝑟{1,3}(𝑐(𝐸)) = (𝛾1, 𝛾3). Because of our hypothesis, the elements 𝛿 inthe fiber 𝐹{1,3}(𝐸, 𝛽𝑠) are such that 𝛽𝑠

1 < 𝛿1 ≤ 𝜂11. But we must have 𝛿1 = 𝜂1

1, because,otherwise, there would be a point 𝜂1 = (𝜂1

1, 𝜂12, 𝛼3) ∈ 𝑝𝑟−1

{1,3}(𝜂11, 𝛼3), with 𝜂1

2 < 𝛽𝑠2, and a

point 𝜂2 ∈ 𝐹{2,3}(𝐸, 𝛽𝑠) with 𝜂21 < 𝜂1

1 and 𝜂22 = 𝛽𝑠

2. These 𝜂1 and 𝜂2 are absolute maximals,due to Lemma 1.23, then from Corollary 2.10, there would exist a relative maximal in theregion ℛ(𝜂1, 𝜂2), which contradicts the fact that we have 𝑠 relative maximals. This impliesthat (𝛽𝑠

1, 𝜂12, 𝛼3) is an absolute maximal of 𝐸.

We have to show that there are no other absolute maximals. If such maximalexisted, then one of the three conditions in Lemma 2.9 would be satisfied. Obviouslyconditions (i) and (iii) cannot be satisfied, but neither condition (ii) can be satisfied,because otherwise 𝛼3 ∈ 𝑝𝑟3(M{2,3}), which is a contradiction.

Lemma 2.13. Let 𝛼 ∈ 𝐿′3 such that 𝛼3 ∈

(𝑝𝑟3(M{1,3}) ∩ 𝑝𝑟3(M{2,3})

)∖ 𝑝𝑟3(RM), then

there exists one and only one absolute maximal of 𝐸 with third coordinate equal to 𝛼3.

Proof As 𝛼3 ∈ 𝑝𝑟3(M{1,3}) ∩ 𝑝𝑟3(M{2,3}), then there exist (𝛽11 , 𝛼3) ∈ M{1,3} and (𝛽2

2 , 𝛼3) ∈M{2,3} such that 𝛽1

1 < 𝛼1(= 𝛾1) and 𝛽22 < 𝛼2(= 𝛾2).

So, there are also elements 𝛿1 = (𝛽11 , 𝛿1

2, 𝛼3) ∈ 𝑝𝑟−1{1,3}(𝛽1

1 , 𝛼3) and 𝛿2 = (𝛿21, 𝛽2

2 , 𝛼3) ∈𝑝𝑟−1

{2,3}(𝛽22 , 𝛼3) in 𝐸 such that 𝛿2

1 ≤ 𝛽11 and 𝛿1

2 ≤ 𝛽22 , because otherwise (𝛽1

1 , 𝛼3) and/or(𝛽2

2 , 𝛼3) would not be maximals of 𝐸{1,3} and/or 𝐸{2,3}.


From our hypothesis, at least one of the equalities 𝛿21 = 𝛽1

1 or 𝛿12 = 𝛽2

2 holds,because otherwise, there would exist absolute maximals 𝜃1 ∈ 𝑝𝑟−1

{1,3}(𝛽11 , 𝛼3) and 𝜃2 ∈

𝑝𝑟−1{2,3}(𝛽2

2 , 𝛼3), hence from Corollary 2.10 there would exist a relative maximal in ℛ(𝜃1, 𝜃2),hence with third projection 𝛼3, a contradiction. Thus we have (𝛿1

1, 𝛽22 , 𝛼3) ∈ AM, because

𝐹 3(𝐸, (𝛿1, 𝛽2, 𝛼3)) = ∅.

Lemma 2.14. Let 𝛼 ∈ 𝐿′3 such that 𝛼3 ∈ 𝑝𝑟3(M{1,3}) ∩ 𝑝𝑟3(M{2,3}) ∩ 𝑝𝑟3(RM). If there

exist 𝑠 relative maximals with third coodinate equal to 𝛼3, then there exist 𝑠 + 1 absolutemaximals with third coordinate equal to 𝛼3.

Proof Following the proof of Lemma 2.12, we have 𝑠−1 absolute maximals obtained takingthe maximum of each pair of adjacent relative maximals. The conditions 𝛼3 ∈ 𝑝𝑟3(M{1,3})and 𝛼3 ∈ 𝑝𝑟3(M{2,3}) give us two extra absolute maximals, and the same argument usedthere, shows that there are no other.

Lemma 2.15. Let 𝛼 ∈ 𝐿′3 such that 𝛼3 ∈ 𝑝𝑟3(RM) ∖

(𝑝𝑟3(M{1,3}) ∪ 𝑝𝑟3(M{2,3})

). If there

exists 𝑠 relative maximals with third coordinate equal to 𝛼3, then we have 𝑠 − 1 absolutemaximals with third coordinate equal to 𝛼3.

Proof The arguments used in the proofs of the last two lemmas give us the result.

Going back to Formula 2.11, we have to calculate 𝜂. From Lemma 2.6 we canensure that 𝛼 ∈ 𝐿′

3 = {𝛼 ∈ 𝐿3; 𝐹 3(𝐸, 𝛼) = ∅}, in each of the following five cases:

i) 𝛼3 ∈ (𝑝𝑟3(M{1,3}) ∖ 𝑝𝑟3(M{2,3})) ∩ 𝑝𝑟3(RM).

If there exist such 𝛼, then they are related to a unique element of M{1,3} and if thereare 𝑠1 relative maximals with third coordinate 𝛼3, then in our formula 𝛼 was counted𝑠1 + 1 times. By Lemma 2.12 we know that there exist 𝑠1 absolute maximals of 𝐸

with third coordinate 𝛼3. So, we subtract 𝑠1 from our counting to partially correctthe formula.

ii) 𝛼3 ∈ (𝑝𝑟3(M{2,3}) ∖ 𝑝𝑟3(M{1,3})) ∩ 𝑝𝑟3(RM).

Analogously to (i), 𝛼 is related to a unique element of M{2,3} and if there are 𝑠2

relative maximals with third coordinate 𝛼3, then 𝛼 was counted 𝑠2 + 1 times in theformula. Again, by Lemma 2.12 we know that there are 𝑠2 absolute maximals of 𝐸

with third coordinate 𝛼3. So, we subtract 𝑠2 from our counting to partially correctthe formula.

iii) 𝛼3 ∈(𝑝𝑟3(M{1,3}) ∩ 𝑝𝑟3(M{2,3})

)∖ 𝑝𝑟3(RM).

2.3. Case r=3 41

In this case, 𝛼 is related to a unique elements in M{1,3} and in M{2,3}, so in theformula we are counting 𝛼 twice. By Lemma 2.13 there is a unique absolute maximalof 𝐸 with third coordinate 𝛼3 such that its projections 𝑝𝑟{1,3} and 𝑝𝑟{2,3} are inM{1,3} and M{1,3}, respectively. So, we correct partially the formula by subtracting 1,which corresponds to this unique absolute maximal.

iv) 𝛼3 ∈ 𝑝𝑟3(M{1,3}) ∩ 𝑝𝑟3(M{2,3}) ∩ 𝑝𝑟3(RM).

In this case, 𝛼 is related to a unique element of M{1,3}, to a unique element of M{2,3}

and, let us say, 𝑠3 elements of RM, so in our counting, 𝛼 was counted 𝑠3 + 2 times.By Lemma 2.14 there exist 𝑠3 + 1 absolute maximals of 𝐸 with third coordinate 𝛼3.In this case, the correcting term is 𝑠3 + 1, equal to the number of these absolutemaximals.

v) 𝛼3 ∈ (𝑝𝑟3(RM) ∖(𝑝𝑟3(M{1,3}) ∪ 𝑝𝑟3(M{2,3})

).

In this case, 𝛼 is related with, let us say, 𝑠4 elements of RM with third coordinateequal to 𝛼3, so we are counting it 𝑠4 times. By Lemma 2.15 there exist 𝑠4 −1 absolutemaximals with third coordinate 𝛼3. This is exactly the correcting term we mustapply to our formula.

Observe that the above cases exhaust all absolute maximals of 𝐸, implying thefollowing results:

Theorem 2.16. Let 𝒪 be the ring of an algebroid curve with three minimal primes andlet ℐ be a fractional ideal of 𝒪, with values set 𝐸. Then one has

ℓ(

ℐℐ(𝛾)

)= (𝛾1 − 𝛼0

1) − # M1 +(𝛾2 − 𝛼02)−

# M2 − M{1,2} +(𝛾3 − 𝛼033) − # M3 −# M{1,3} −# M{2,3} −

# RM +# AM .

Theorem 2.17. Let 𝒪 be the ring of an algebroid curve with three minimal primes andlet ℐ be a fractional ideal of 𝒪, with values set 𝐸. Then one has

ℓ(

ℐ1×ℐ2×ℐ3ℐ

)= # M{1,2} +# RM{1,3} +# RM{2,3} +# RM −# AM .

43

CHAPTER

3SYMMETRY OF MAXIMALS

The central result in this chapter will establish a correspondence between absolutemaximals of the set of values of a fractional ideal of a Gorenstein algebroid curve andthe relative maximals of the set of values of its dual ideal, generalizing the symmetryrelation among relative maximals and absolute maximals in the semigroup of values of aGorenstein ring, discovered by F. Delgado in [8].

Proposition 3.1. Let ℐ be a fractional ideal of 𝒪 and 𝛼 ∈ Z𝑟. Then

ℓ

(ℐ(𝛼)

ℐ(𝛼 + 𝑒𝑖)

)+ ℓ

(ℐ*(𝑐(𝑆) − 𝛼 − 𝑒𝑖)

ℐ*(𝑐(𝑆) − 𝛼)

)≤ 1 for every 𝑖 ∈ 𝐼.

Proof We know by Proposition 2.2 that ℓ(

ℐ(𝛼)ℐ(𝛼+𝑒𝑖)

)≤ 1 and ℓ

(ℐ*(𝑐(𝑆)−𝛼−𝑒𝑖)

ℐ*(𝑐(𝑆)−𝛼)

)≤ 1. Thus,

it is sufficient to show that if ℓ(


)= 1 then ℓ



)= 0 and vice versa.

The same Proposition 2.2 says that ℓ(


)= 1 if and only if 𝐹 𝑖(𝐸, 𝛼) = ∅ and

that ℓ(

ℐ*(𝑐(𝑆)−𝛼−𝑒𝑖)ℐ*(𝑐(𝑆)−𝛼)

)= 0 if and only if 𝐹 𝑖(𝐸*, 𝑐(𝑆) − 𝛼 − 𝑒𝑖) = ∅.

Let us assume by reductio ad absurdum that 𝐹 𝑖(𝐸, 𝛼) = ∅ and 𝐹 𝑖(𝐸*, 𝑐(𝑆) − 𝛼 −𝑒𝑖) = ∅. Take 𝜃 in the first of these sets and 𝜃′ in the second, then 𝜃 + 𝜃′ ∈ 𝑆; even more,we have that 𝜃 + 𝜃′ ∈ 𝐹𝑖(𝑆, f(𝑆)) because 𝜃𝑖 + 𝜃′

𝑖 = f𝑖(𝑆) and 𝜃𝑗 + 𝜃′𝑗 > f𝑗(𝑆) for all 𝑗 = 𝑖,

hence 𝜃 + 𝜃′ ∈ 𝐹𝑖(𝑆, f(𝑆)), which is a contradiction, since 𝐹𝑖(𝑆, f(𝑆)) = ∅.

The proof of the other case is similar.

The following theorem generalizes [5, Theorem 3.6].

Theorem 3.2. Let 𝒪 be a finitely generated local reduced and complete one-dimensionalalgebra. Then the following statements are equivalent:

i) For all fractional ideal ℐ of 𝒪, one has

ℓ(


)+ ℓ



)= 1, for every 𝛼 ∈ Z𝑟and 𝑖 ∈ 𝐼.

44 Chapter 3. Symmetry of maximals

ii) 𝒪 is a Gorenstein ring.

Proof 𝑖) ⇒ 𝑖𝑖) Assuming (i) and taking ℐ = 𝒪, we will show that 𝑆 is symmetric,proving that 𝒪 is Gorenstein.

Recall that 𝑆 symmetric means that

𝛼 ∈ 𝑆 ⇐⇒ 𝐹 (𝑆, f(𝑆) − 𝛼) = ∅.

One direction of the above equivalence is known to be true from Corollary 1.20,namely

𝛼 ∈ 𝑆 ⇒ 𝐹 (𝑆, f(𝑆) − 𝛼) = ∅.

To prove the other direction, we show firstly that

𝐹 (𝑆, f(𝑆) − 𝛼) = ∅ ⇒ 𝐹 𝑖(𝑆, 𝑐(𝑆) − 𝛼 − 𝑒𝑖) = ∅, for all 𝑖 ∈ 𝐼. (3.1)

Indeed, if 𝐹 (𝑆, f(𝑆) − 𝛼) = ∅ and 𝐹 𝑖(𝑆, 𝑐(𝑆) − 𝛼 − 𝑒𝑖) = ∅, for some 𝑖 ∈ 𝐼, take𝜃 ∈ 𝐹 𝑖(𝑆, 𝑐(𝑆)−𝛼−𝑒𝑖), hence 𝜃𝑖 = 𝑐𝑖(𝑆)−𝛼𝑖−1 = f𝑖(𝑆)−𝛼 and 𝜃𝑗 ≥ 𝑐𝑗(𝑆)−𝛼𝑗 > f𝑖(𝑆)−𝛼𝑗 ,for all 𝑗 = 𝑖. This implies that 𝜃 ∈ 𝐹𝑖(𝑆, f(𝑆) − 𝛼) ⊂ 𝐹 (𝑆, f(𝑆) − 𝛼) = ∅, a contradiction.So, we proved (3.1).

From (3.1) and Proposition 2.2, we get that ℓ(

𝒪(𝑐(𝑆)−𝛼−𝑒𝑖)𝒪(𝑐(𝑆)−𝛼)

)= 0, for all 𝑖 ∈ 𝐼.

This in view of (i) implies that ℓ(

𝒪(𝛼)𝒪(𝛼+𝑒𝑖)

)= 1, for all 𝑖 ∈ 𝐼, which in turn, in view of

Proposition 2.2 implies that 𝐹 𝑖(𝑆, 𝛼) = ∅ for all 𝑖 ∈ 𝐼, from which we conclude that 𝛼 ∈ 𝑆,as we wanted to show.

𝑖𝑖) ⇒ 𝑖) We suppose now that 𝒪 is Gorenstein and ℐ is a fractional ideal of 𝒪.

From Theorem 1.26 we know that

𝛼 /∈ 𝐸 ⇔ 𝐹 (𝐸*, f(𝑆) − 𝛼) = ∅.

To conclude the proof of this part of the theorem, it is enough to show that

∀ 𝑖 ∈ 𝐼, 𝐹 𝑖(𝐸, 𝛼) = ∅ =⇒ 𝐹 𝑖(𝐸*, f(𝑆) − 𝛼) = ∅.

Indeed, because the converse of the above implication is always true (cf. Proposition3.1), item (i) will follow by Proposition 2.2.

Suppose now that 𝐹 𝑖(𝐸, 𝛼) = ∅. This implies that 𝛼 /∈ 𝐸, so, in view of Theorem1.26 we get that 𝐹 (𝐸*, f(𝑆) − 𝛼) = ∅. Hence there exists an element 𝜃 ∈ 𝐹𝑖(𝐸*, f(𝑆) − 𝛼),for some 𝑖 ∈ 𝐼. So, 𝜃𝑖 = f𝑖(𝑆) − 𝛼𝑖 = 𝑐(𝑆) − 1 − 𝛼𝑖 and 𝜃𝑗 > f𝑗(𝑆) − 𝛼𝑗 = 𝑐𝑗(𝑆) − 1 − 𝛼𝑗,for all 𝑗 = 𝑖. This last inequlity implies that 𝜃𝑗 ≥ 𝑐𝑗(𝑆) − 𝛼𝑗, for all 𝑗 = 𝑖. Thus𝜃 ∈ 𝐹 𝑖(𝐸*, 𝑐(𝑆) − 𝛼 − 𝑒𝑖), hence this set is nonempty.

45

In [5, Section 5], the authors define certain integers related to the semigroup ofvalues of an algebroid curve to give another characterization for the property of 𝒪 beingGorenstein. Now, let us define these numbers for fractional ideals.

We denote ℓ(


)by ℓ𝐸(𝛼, 𝑒𝑖) and put 𝑒𝐽 = ∑

𝑖∈𝐽 𝑒𝑖 for ∅ = 𝐽 ⊂ 𝐼, 𝑒∅ = 0 and𝑒 = 𝑒𝐼 = (1, 1, . . . , 1). We will also use the notation 𝐽 𝑐 = 𝐼 ∖ 𝐽 .

Let us define the following sentences:

𝑝(𝛼, 1) : ∀𝐽 ⊂ 𝐼, with #𝐽 = 1, ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 0, ∀𝑖 ∈ 𝐽

...

𝑝(𝛼, 𝑟) : ∀𝐽 ⊂ 𝐼, with #𝐽 = 𝑟, ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 0, ∀𝑖 ∈ 𝐽.

Notice that 𝑝(𝛼, 𝑛) ⇒ 𝑝(𝛼, 𝑛 − 1).

Indeed, let 𝐾 ⊂ 𝐼 with #𝐾 = 𝑛 − 1, now 𝐾 is included in some subset 𝐽 of 𝐼

with #𝐽 = 𝑛. From the assertion 𝑝(𝛼, 𝑛) we have that ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 0 for all 𝑖 ∈ 𝐽 .Suppose by reductio ad absrurdum that 𝑝(𝛼, 𝑛 − 1) is false, hence ℓ𝐸(𝛼 + 𝑒𝐾𝑐 , 𝑒𝑖) = 1 forsome 𝑖 ∈ 𝐾, that is, 𝐹 𝑖(𝐸, 𝛼 + 𝑒𝐾𝑐) = ∅. Take 𝜃 ∈ 𝐹 𝑖(𝐸, 𝛼 + 𝑒𝐾𝑐), then 𝜃 ∈ 𝐹 𝑖(𝐸, 𝛼 + 𝑒𝐽𝑐),because 𝜃𝑖 = 𝛼𝑖 and 𝜃𝑘 ≥ 𝛼𝑘 for 𝑘 ∈ 𝐾, 𝜃𝑗 ≥ 𝛼𝑗 + 1 for 𝑗 ∈ 𝐾𝑐. Hence 𝐹 𝑖(𝐸, 𝛼 + 𝑒𝐽𝑐) = ∅,which contradicts ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 0 for all 𝑖 ∈ 𝐽 .

Thus, we can define

𝑝(𝐸, 𝛼) = max{𝑛; ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 0, ∀𝐽 ⊂ 𝐼, #𝐽 = 𝑛, ∀𝑖 ∈ 𝐽}.

By the definition of 𝑝(𝐸, 𝛼) it follows that

i) 0 ≤ 𝑝(𝐸, 𝛼) ≤ 𝑟, for every 𝛼;

ii) 𝑝(𝐸, 𝛼) = max{𝑛; 𝐹𝐴(𝐸, 𝛼) = ∅, ∀𝐴 ⊂ 𝐼, #𝐴 ≤ 𝑛};

iii) 𝑝(𝐸, 𝛼) ≥ 1 ⇐⇒ 𝐹 (𝐸, 𝛼) = ∅.

We also define the sentences:

𝑞(𝛼, 𝑟 + 1) : 𝐹 (𝐸, 𝛼) = ∅

𝑞(𝛼, 𝑟) : ∀𝐽 ⊂ 𝐼, with #𝐽 = 𝑟, ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 1, ∀𝑖 ∈ 𝐽

...

𝑞(𝛼, 1) : ∀𝐽 ⊂ 𝐼, with #𝐽 = 1, ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 1, ∀𝑖 ∈ 𝐽.


Notice that 𝑞(𝛼, 𝑛 − 1) ⇒ 𝑞(𝛼, 𝑛).

Assume that 𝑞(𝛼, 𝑛 − 1) is true. Let 𝐾 ⊂ 𝐼, #𝐾 = 𝑛 and let 𝐽 be a subset of 𝐾

with #𝐽 = 𝑛 − 1, we have ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 1 for all 𝑖 ∈ 𝐽 , i.e., 𝐹 𝑖(𝐸, 𝛼 + 𝑒𝐽𝑐) = ∅. Take𝜃 ∈ 𝐹 𝑖(𝐸, 𝛼 + 𝑒𝐽𝑐) and notice that 𝜃 ∈ 𝐹 𝑖(𝐸, 𝛼 + 𝑒𝐾𝑐), because 𝜃𝑖 = 𝛼𝑖 and 𝜃𝑗 ≥ 𝛼𝑗 for𝑗 ∈ 𝐾, 𝜃𝑗 ≥ 𝛼𝑗 + 1 for 𝑗 ∈ 𝐾𝑐, hence ℓ𝐸(𝛼 + 𝑒𝐾𝑐 , 𝑒𝑖) = 1, for all 𝑖 ∈ 𝐾.

Thus, we can also define

𝑞(𝐸, 𝛼) = min{𝑛; ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 1, ∀𝐽 ⊂ 𝐼, #𝐽 = 𝑛, ∀𝑖 ∈ 𝐽}.

By definition 𝑞(𝐸, 𝛼) has that following properties;

i) 1 ≤ 𝑞(𝐸, 𝛼) ≤ 𝑟 + 1, for every 𝛼;

ii) 𝑞(𝐸, 𝛼) = min{𝑛; 𝐹𝐴(𝐸, 𝛼) = ∅, ∀𝐴 ⊂ 𝐼, #𝐴 ≥ 𝑛}

iii) 𝑞(𝐸, 𝛼) ≤ 𝑟 ⇐⇒ 𝐹 𝑖(𝐸, 𝛼) = ∅, for 𝑖 = 1, . . . , 𝑟;

iv) 𝑝(𝐸, 𝛼) < 𝑞(𝐸, 𝛼) for every 𝛼.

The following theorem generalizes [5, Theorem 5.3].

Theorem 3.3. With notations as above, for any fractional ideal ℐ of 𝒪 with value set 𝐸

and for any 𝛼 ∈ Z𝑟, we have

𝑝(𝐸, 𝛼) + 𝑞(𝐸*, f(𝑆) − 𝛼) ≥ 𝑟 + 1. (3.2)

Moreover, 𝒪 is a Gorenstein ring if an only if equality holds in (3.2) for every fractionalideal ℐ of 𝒪 and every 𝛼 ∈ Z𝑟.

Proof Take any fractional ideal ℐ and 𝛼 ∈ Z𝑟, and define 𝑛 = 𝑟 + 1 − 𝑞(𝐸*, 𝑓(𝑆) − 𝛼).Take 𝐽 ⊂ 𝐼 with #𝐽 ≤ 𝑛 and any 𝑖 ∈ 𝐽 . Put 𝐾 = 𝐽 𝑐 ∪ {𝑖}. By Proposition 3.1, one has

ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) + ℓ𝐸*(𝑐(𝑆) − 𝑒 − 𝛼 + 𝑒𝐾𝑐 , 𝑒𝑖) ≤ 1, for all 𝑖 ∈ 𝐽. (3.3)

Since #𝐾 ≥ 𝑞(𝐸*, 𝑓(𝑆) − 𝛼) and 𝑖 /∈ 𝐾𝑐, by definition of 𝑞 one has ℓ𝐸*(𝑐(𝑆) −𝑒 − 𝛼 − 𝑒𝐾𝑐 , 𝑒𝑖) = 1 and therefore ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 0. This shows that 𝑛 ≤ 𝑝(𝐸, 𝛼), andtherefore (3.2) is proven.

Now, if 𝒪 is Gorenstein, to prove that (3.2) holds, we must show that 𝑝(𝐸, 𝛼) = 𝑛.Suppose by reductio ad absurdum that 𝑝(𝐸, 𝛼) ≥ 𝑛 + 1. Then, we have that

ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 0, ∀𝑖 ∈ 𝐽, ∀𝐽 ⊂ 𝐼 with #𝐽 = 𝑛 + 1.

47

Consider a set 𝐾 ⊂ 𝐼 with #𝐾 = 𝑟 − 𝑛 and define the set 𝐽 = 𝐾𝑐 ∪ {𝑖}, with 𝑖 ∈ 𝐾.Notice that 𝐽 has 𝑛 + 1 elements, then

ℓ𝐸(𝛼 + 𝑒𝐽𝑐 , 𝑒𝑖) = 0, ∀𝑖 ∈ 𝐽.

Since, that 𝒪 is Gorenstein, we have

ℓ𝐸*(f(𝑆) − 𝛼 + 𝑒𝐾𝑐 , 𝑒𝑖) = 0, ∀𝑖 ∈ 𝐾,

and since, 𝑖 is any element of 𝐾, we have that

ℓ𝐸*(f(𝑆) − 𝛼 + 𝑒𝐾𝑐 , 𝑒𝑖) = 0, ∀𝑖 ∈ 𝐾 and ∀𝐾 ⊂ 𝐼, with #𝐾 = 𝑟 − 𝑛.

As 𝑛 = 𝑟 + 1 − 𝑞(f(𝑆) − 𝛼), we have that #𝐾 < 𝑞(f(𝑆) − 𝛼), this contradicts the definitionof 𝑞(f(𝑆) − 𝛼). Therefore, it follows that 𝑝(𝐸, 𝛼) = 𝑛.

Now, assume that we have equality in (3.2), for every fractional ideal ℐ of 𝒪 andevery 𝛼 ∈ Z𝑟. Let 𝛼 ∈ Z𝑟 and 𝑖 ∈ 𝐼. If 𝐹 𝑖(𝐸, 𝛼) = ∅, then there exists 𝛽 with 𝛽𝑖 = 𝛼𝑖

and 𝛽𝑗 < 𝛼𝑗 for every 𝑗 = 𝑖 such that 𝐹 (𝐸, 𝛽) = ∅. From the condition 𝐹 (𝐸, 𝛽) = ∅we get that 𝑝(𝐸, 𝛽) ≥ 1, hence from Equality (3.2) we get that 𝑞(𝐸*, f(𝑆) − 𝛽) ≤ 𝑟.Therefore, 𝐹 𝑗(𝐸*, f(𝑆) − 𝛽) = ∅ for every 𝑗 ∈ 𝐼. In particular, 𝐹 𝑖(𝐸*, f(𝑆) − 𝛽) = ∅. Onthe other hand, since f𝑖(𝑆) − 𝛽𝑖 = f𝑖(𝑆) − 𝛼𝑖 and f𝑗(𝑆) − 𝛽𝑗 > f𝑗(𝑆) − 𝛼𝑗, it follows thatf(𝑆) − 𝛽 ≥ f(𝑆) − 𝛼, therefore 𝐹 𝑖(𝐸*, f(𝑆) − 𝛼) = ∅. So, we proved that

ℓ𝐸(𝛼, 𝑒𝑖) + ℓ𝐸*(𝑐(𝑆) − 𝛼 − 𝑒𝑖, 𝑒𝑖) = 1, for every 𝛼 ∈ Z𝑟and all 𝑖 ∈ 𝐼.

So, by Theorem 3.2 it follows that 𝒪 is Gorenstein.

Notice that in Theorem 3.3 we have the same result with the inequality

𝑝(𝐸*, 𝛽) + 𝑞(𝐸, f(𝑆) − 𝛽) ≥ 𝑟 + 1.

Remark 3.4. In terms of 𝑝(𝐸, 𝛼) and 𝑞(𝐸, 𝛼), we may characterize the maximals of 𝐸

as follows:

i) 𝛼 is maximal if and only if 𝑝(𝐸, 𝛼) ≥ 1 and 𝑞(𝐸, 𝛼) ≤ 𝑟;

ii) 𝛼 is a relative maximal if and only if 𝑝(𝐸, 𝛼) = 1 and 𝑞(𝐸, 𝛼) = 2; and

iii) 𝛼 is an absolute maximal if and only if 𝑝(𝐸, 𝛼) = 𝑟 − 1 and 𝑞(𝐸, 𝛼) = 𝑟.

The following result will generalize [7, Theorem 2.10], and is a consequence ofTheorem 3.3 and the previous remark.


Theorem 3.5 (Symmetry of maximals). Suppose that 𝒪 is Gorenstein with value set𝑆 and ℐ is a fractional ideal of 𝒪 with value set 𝐸. Then for 𝛽 ∈ 𝐸*, one has that 𝛽 is amaximal of 𝐸* if and only if f(𝑆) − 𝛽 ∈ 𝐸. Moreover, if 𝛽 ∈ 𝐸* and 𝛼 ∈ 𝐸 are such that𝛼 + 𝛽 = f(𝑆), then 𝛼 is an absolute maximal of 𝐸 if and only if 𝛽 is a relative maximal of𝐸*.

Proof Suppose that 𝛽 is a maximal of 𝐸*, then

𝐹 (𝐸*, f(𝑆) − (f(𝑆) − 𝛽)) = 𝐹 (𝐸*, 𝛽) = ∅,

hence from (1.3) it follows that f(𝑆) − 𝛽 ∈ 𝐸.

Conversely, suppose that f(𝑆) − 𝛽 ∈ 𝐸, then from (1.3) we have that 𝐹 (𝐸*, 𝛽) = ∅and, as by hypothesis, 𝛽 ∈ 𝐸*, it follows that 𝛽 is a maximal of 𝐸*. This proves the firstpart of the theorem.

We now prove the second part of the theorem. Suppose that 𝛽 is a relative maximalof 𝐸* and that 𝛼 ∈ 𝐸 is such that 𝛼 + 𝛽 = f(𝑆). From the first part of the theorem thatwe have just proved, it follows that 𝛼 is a maximal of 𝐸.

By reductio ad absurdum, suppose that 𝛼 is not an absolute maximal of 𝐸, thenthere exists 𝐽 ⊂ 𝐼 with 2 ≤ #𝐽 ≤ 𝑟−1 such that 𝐹𝐽(𝐸, 𝛼) = ∅. On the other hand, since 𝛽

is a relative maximal of 𝐸*, then 𝐹𝐾(𝐸*, 𝛽) = ∅, for all 𝐾 ⊂ 𝐼 with #𝐾 ≥ 2. In particular,take 𝐾 = (𝐼 ∖ 𝐽) ∪ {𝑖}, where 𝑖 ∈ 𝐽 , and choose 𝜃 ∈ 𝐹𝐾(𝐸*, 𝛽) and 𝜃′ ∈ 𝐹𝐽(𝐸, 𝛼). Noticethat because of Equation (1.1) one has 𝜃 + 𝜃′ ∈ 𝑆. On the other hand, our choice of 𝐾

implies that 𝜃 + 𝜃′ ∈ 𝐹 (Z𝑟, f(𝑆)), hence 𝜃 + 𝜃′ ∈ 𝐹 (𝑆, f(𝑆)), which is a contradiction, sincefrom Corollary 1.20 we know that 𝐹 (𝑆, f(𝑆)) = ∅. Thus 𝛼 is an absolute maximal of 𝐸.

It remains to prove that if 𝛼 + 𝛽 = f(𝑆) and 𝛼 is an absolute maximal of 𝐸,then 𝛽 is a relative maximal of 𝐸*. In fact, suppose that 𝛼 is an absolute maximal of 𝐸,which means that 𝑝(𝐸, 𝛼) = 𝑟 − 1 and 𝑞(𝐸, 𝛼) = 𝑟, then by Theorem 3.3 we have that𝑞(𝐸*, 𝛽) = 2 and 𝑝(𝐸*, 𝛽) = 1, i.e., 𝛽 is a relative maximal of 𝐸*.

49

CHAPTER

4KÄHLER DIFFERENTIALS ON COMPLETE

INTERSECTIONS

In this Chapter we generalize for complete intersections curve singularities theresults by Bayer, Hefez and Hernandes in [4], obtained for plane curve singularities.

For the sake of simplicity, we will assume that char(𝑘) = 0, and that 𝐶 = Spec(𝒪)is embedded in Spec(𝑘[[𝑋1, . . . , 𝑋𝑛]]), where 𝑛 is the embedding dimension of 𝐶.

The above embedding of 𝐶 is determined by a surjection 𝑘[[𝑋1, . . . , 𝑋𝑛]] → 𝒪,whose kernel we denote by ℐ. Therefore, one has

𝒪 ≃ 𝑘[[𝑋1, . . . , 𝑋𝑛]]/ℐ.

We define the module of Kähler differentials as being the 𝒪-module

Ω(𝒪) = 𝒪⊕𝑛

⟨𝑓𝑋1𝑒1 + · · · + 𝑓𝑋𝑛𝑒𝑛; 𝑓 ∈ ℐ⟩,

where 𝑒1, . . . , 𝑒𝑛 is the canonical basis of 𝒪⊕𝑛 and 𝑓𝑋𝑖= 𝜕𝑓

𝜕𝑋𝑖, 𝑖 = 1, . . . , 𝑛. This module

does not depend neither on ℐ nor on the embedding of Spec(𝒪).

We will denote the image of 𝑒𝑖 in Ω(𝒪) by 𝑑𝑥𝑖 and for 𝑔 ∈ 𝒪 we consider theuniversal differentiation 𝑑𝑔 = 𝑔𝑋1𝑑𝑥1 + · · · + 𝑔𝑋𝑛𝑑𝑥𝑛 ∈ Ω(𝒪).

The jacobian ideal 𝒥 of 𝒪 is the smallest nonzero Fitting ideal of Ω(𝒪), that is,the nonzero ideal generated by the minors of greatest possible size of the Jacobian matrix

𝐽𝑎𝑐(𝑓1, . . . , 𝑓𝑚) =

⎡⎢⎢⎢⎣𝑓1𝑋1 · · · 𝑓1𝑋𝑛

... ...𝑓𝑚𝑋1 · · · 𝑓𝑚𝑋𝑛

⎤⎥⎥⎥⎦ ,

where 𝑓1, . . . , 𝑓𝑚 are generators of the ideal ℐ.

50 Chapter 4. Kähler differentials on Complete Intersections

The surjection 𝜋𝑖 : 𝒪 → 𝒪𝑖 induces a surjection 𝜋*𝑖 : Ω(𝒪) → Ω(𝒪𝑖), defined by

𝜋*𝑗 (

𝑛∑𝑖=1

𝑔𝑖𝑑𝑥𝑖) =𝑛∑

𝑖=1𝜋𝑗(𝑔𝑖)𝑑𝑥𝑖.

This in turn, induces an 𝒪-modules homomorphism

𝜋* : Ω(𝒪) → Ω(𝒪1) × · · · × Ω(𝒪𝑟)𝜔 ↦→ (𝜋*

1(𝜔), . . . , 𝜋*𝑟(𝜔)).

We also have an 𝒪𝑖-modules homomorphism Ω(𝒪𝑖) → Ω( 𝒪𝑖) ≃ 𝑘[[𝑡𝑖]]𝑑𝑡𝑖, definedby 𝑑𝑥𝑗 ↦→ 𝑥𝑗(𝑡𝑖)′𝑑𝑡𝑖, 𝑗 = 1, . . . , 𝑛, which composed with 𝜋*

𝑖 gives rise to an 𝒪-moduleshomomorphism

𝜙*𝑖 : Ω(𝒪) → Ω( 𝒪𝑖),∑𝑛

𝑗=1 𝑔𝑗𝑑𝑥𝑗 ↦→(∑𝑛

𝑗=1 𝜙𝑖(𝑔𝑗)𝜙𝑖(𝑥𝑗)′)

𝑑𝑡𝑖

where the dash means derivative with respect to 𝑡𝑖.

Now, we also define the 𝒪-modules homomorphism

𝜙* : Ω(𝒪) → Ω( 𝒪1) × · · · × Ω( 𝒪𝑟).𝜔 ↦→ (𝜙*

1(𝜔), . . . , 𝜙*𝑟(𝜔))

The valuation 𝑣𝑖 on 𝒪𝑖 extends to a valuation of Ω(𝒪𝑖) in the following way:

𝑣𝑖(𝑛∑

𝑗=1𝑔𝑗𝑑𝑥𝑗) = 𝑣𝑖

⎛⎝ 𝑛∑𝑗=1

𝜙𝑖(𝑔𝑗)𝜙𝑖(𝑥𝑗)′

⎞⎠+ 1,

where 𝜙𝑖 is the map from 𝒪 to 𝒪𝑖; and we extend the valuation 𝑣 on 𝒪 to Ω(𝒪) as follows:

𝑣(𝜔) = (𝑣1(𝜔), . . . , 𝑣𝑟(𝜔)).

We define the ramification ideal, as the ideal ℛ generated in 𝒪 by the elements

(𝜙1(𝑥𝑖)′, . . . , 𝜙𝑟(𝑥𝑖)′), 𝑖 = 1, . . . , 𝑛,

By choosing properly the coordinates in 𝑘[[𝑋1, . . . , 𝑋𝑛]], we may assume that thehyperplane 𝑋1 is transversal to all the branches 𝐶𝑖 of the curve 𝐶, so the maximal idealℳ of 𝒪 is a fractional ideal and

𝑣(𝑥1) = (mult(𝐶1), . . . , mult(𝐶𝑟)) = min 𝑣(ℳ ∖ 𝑍(𝒪)).

Since char(𝑘) = 0, it follows that ℛ is a fractional ideal of 𝒪 and

𝑣(𝑥1) − (1, . . . , 1) = 𝑣(𝜙1(𝑥1)′, . . . , 𝜙𝑟(𝑥1)′) = min 𝑣(ℛ ∖ 𝑍( 𝒪)).

We will denote by 𝒯 and by 𝒯𝑖 the torsion modules of Ω(𝒪) and of Ω(𝒪𝑖), respec-tively. Concerning these modules we have the following result:

51

Proposition 4.1. Suppose that 𝑋 is irreducible, that is, 𝒪 is a domain, then ker(𝜙*) = 𝒯 .

Proof See [12, Proposition 1, Sect. 7.1].

Proposition 4.2. Suppose that 𝒪 is reduced, then 𝜋*(𝒯 ) ⊂ 𝒯1 × · · · 𝒯𝑟.

Proof Let 𝜔 ∈ 𝒯 , then there exists an element ℎ /∈ 𝑍(𝒪) = ⋃𝑟𝑖=1 ℘𝑖 such that ℎ𝜔 = 0.

Since ℎ /∈ 𝑍(𝒪), it follows that 𝜋𝑗(ℎ) = 0, for all 𝑗 = 1, . . . , 𝑟, hence a nonzero divisor in𝒪𝑗. Since 0 = 𝜋*

𝑗 (ℎ𝜔) = 𝜋𝑗(ℎ)𝜋*𝑗 (𝜔), this implies that 𝜋*

𝑗 (𝜔) ∈ 𝒯𝑖, proving the result.

From now on we will assume that the curve 𝐶 is a complete intersection, this meansthat 𝐶 = Spec 𝑘[[𝑋1, . . . , 𝑋𝑛]]/⟨𝑓1, . . . , 𝑓𝑛−1⟩.

In this situation one has the following theorem:

Theorem 4.3 (Piene [15]). Let 𝒪 be the ring of an algebroid complete intersection curveand let 𝒪 be its normalization. If 𝒥 denotes the Jacobian ideal of Ω(𝒪) and ℛ and 𝒞,respectively, the ramification and conductor ideals of 𝒪 in 𝒪, then one has

𝒥 𝒪 = 𝒞ℛ.

In this situation, one has that 𝒥 = ⟨|𝑀1|, . . . , |𝑀𝑛|⟩, where |𝑀𝑖| is the determinantof the matrix 𝑀𝑖 obtained by deleting the 𝑖-th column of the matrix 𝐽𝑎𝑐(𝑓1, . . . , 𝑓𝑛−1).Since by Theorem 4.3 one has that 𝒥 is a fractional ideal, then by reordering the series𝑓1, . . . , 𝑓𝑛−1, we may assume that

𝑣(|𝑀1|) = inf 𝒥 𝒪 = inf 𝑣(ℛ) + 𝑐 = 𝑣(𝑥1) + f(Γ), (4.1)

where 𝑒 = (1, . . . , 1) and 𝑐 = (𝑐1, . . . , 𝑐𝑟) is the conductor of the semigroup of values Γ.This implies that |𝑀1| is not a zero divisor in 𝒪.

The relation in Ω(𝒪) are𝑛∑

𝑗=1𝑓𝑖𝑋𝑗

𝑑𝑥𝑗 = 0, 𝑖 = 1, . . . , 𝑛 − 1,

which may be written in matricial form as follows

𝑀1

⎡⎢⎢⎢⎣𝑑𝑥2

...𝑑𝑥𝑛

⎤⎥⎥⎥⎦ = −

⎡⎢⎢⎢⎣𝑓1𝑋1

...𝑓𝑛−1𝑋1

⎤⎥⎥⎥⎦ 𝑑𝑥1. (4.2)

If we multiply both sides of equation (4.2) by the adjoint matrix 𝑀*1 of 𝑀1, we get

|𝑀1|𝑑𝑥𝑖 = (−1)𝑖−1|𝑀𝑖|𝑑𝑥1. (4.3)


From Equations (4.1) and (4.3) one gets that

𝑣(|𝑀𝑖|) = 𝑣(𝑥𝑖) + f(Γ). (4.4)

Theorem 4.4. For a complete intersection 𝒪 one has that

𝜋*−1(𝒯1 × · · · × 𝒯𝑟) = 𝒯 .

Proof From Proposition 4.2 one has that 𝒯 ⊂ 𝜋*−1(𝒯1 × · · · × 𝒯𝑟).

For the other inclusion, let 𝜔 = ∑𝑛𝑗=1 𝑔𝑗𝑑𝑥𝑗 ∈ 𝜋*−1(𝒯1 ×· · ·×𝑇𝑟). Hence 𝜋*

𝑖 (𝜔) ∈ 𝒯𝑖,therefore 𝜑*

𝑖 (𝜔) = 0, for all 𝑖 = 1, . . . , 𝑟.

Consider |𝑀1|𝜔, which in view of relations (4.3) may be written as

|𝑀1|𝜔 =𝑛∑

𝑗=1𝑔𝑗|𝑀1|𝑑𝑥𝑗 = (

𝑛∑𝑗=1

(−1)𝑖−1𝑔𝑗|𝑀𝑗|)𝑑𝑥1.

Applying 𝜙*𝑖 to both sides of the above expression, one gets, for all 𝑖 = 1, . . . , 𝑟,

0 = 𝜋𝑖(|𝑀1|)𝜙*𝑖 (𝜔) = 𝜙*

𝑖 (|𝑀1|𝜔) = 𝜋*𝑖 (

𝑛∑𝑗=1

(−1)𝑖−1𝑔𝑗|𝑀𝑗|)𝜙*𝑖 (𝑥1)′.

Since 𝜙*𝑖 (𝑥1)′ = 0, for all 𝑖, then 𝜋*

𝑖 (∑𝑛𝑗=1(−1)𝑖−1𝑔𝑗|𝑀𝑗|) = 0, for all 𝑖, which implies

that ∑𝑛𝑗=1(−1)𝑖−1𝑔𝑗|𝑀𝑗| ∈ ∩𝑖℘𝑖 = (0). So, |𝑀1|𝜔 = 0 and since |𝑀1| /∈ 𝑍(𝒪), we have that

𝜔 ∈ 𝒯 .

Corollary 4.5. Let 𝒪 be the ring of an algebroid complete intersection curve, then onehas 𝜔 = ∑𝑛

𝑗=1 𝑔𝑗𝑑𝑥𝑗 ∈ 𝒯 if and only if

𝑑𝑓1 ∧ · · · ∧ 𝑑𝑓𝑛−1 ∧ 𝜔

𝑑𝑥1 ∧ · · · ∧ 𝑑𝑥𝑛

= det

⎡⎢⎢⎢⎢⎢⎢⎣𝑓1𝑋1 · · · 𝑓1𝑋𝑛

... ...𝑓𝑛−1𝑋1 · · · 𝑓𝑛−1𝑋𝑛

𝑔1 · · · 𝑔𝑛

⎤⎥⎥⎥⎥⎥⎥⎦ = 0.

From Proposition 4.1 and Theorem 4.4 we have shown that if 𝐶 is a completeintersection, then we have inclusions

Ω(𝒪)/𝒯 →˓ Ω(𝒪1)/𝒯1 × · · · × Ω(𝒪𝑟)/𝒯𝑟 →˓ Ω( 𝒪1) × · · · × Ω( 𝒪𝑟) ≃ 𝒪,

which allows us to view Ω(𝒪)/𝒯 and Ω(𝒪1)/𝒯1 × · · · × Ω(𝒪𝑟)/𝒯𝑟 as fractional ideals, onecontained in the other.

Now, it is easy to verify that the map

𝒟 : 𝒪 → 𝒪

𝑔 ↦→ 𝑑𝑓1 ∧ · · · ∧ 𝑑𝑓𝑛−1 ∧ 𝑑𝑔

𝑑𝑥1 ∧ · · · ∧ 𝑑𝑥𝑛

53

is a differential operator on 𝒪, that may be written as

𝒟(𝑔) = det

⎡⎢⎢⎢⎢⎢⎢⎣𝑓1𝑋1 · · · 𝑓1𝑋𝑛

... ...𝑓𝑛−1𝑋1 · · · 𝑓𝑛−1𝑋𝑛

𝑔𝑋1 · · · 𝑔𝑋𝑛

⎤⎥⎥⎥⎥⎥⎥⎦ .

This operator and the universal differentiation on 𝒪 may be extended in a standardway to the total ring of fractions 𝒦 of 𝒪.

Proposition 4.6. One has that

𝑣(𝒟(𝑔)) = 𝑣(𝑔) + f(Γ).

Proof Since 𝜙𝑖(𝑓𝑗)′ = 0, for all 𝑖 = 1, . . . , 𝑟 and 𝑗 = 1, . . . , 𝑛, one has the equalities⎡⎢⎢⎢⎢⎢⎢⎣𝜙𝑖(𝑓1𝑋1) · · · 𝜙𝑖(𝑓1𝑋𝑛)

... ...𝜙𝑖(𝑓𝑛−1𝑋1) · · · 𝜙𝑖(𝑓𝑛−1𝑋𝑛)

𝜙𝑖(𝑔𝑋1) · · · 𝜙𝑖(𝑔𝑋𝑛)

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣𝜙𝑖(𝑥1)′

...𝜙𝑖(𝑥𝑛−1)′

𝜙𝑖(𝑥𝑛)′

⎤⎥⎥⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎢⎣0...0

𝜙𝑖(𝑔)′

⎤⎥⎥⎥⎥⎥⎥⎦ .

Calling 𝑀 the above square matrix on the left, and multiplying both sides of theabove equality by the adjoint matrix 𝑀* of 𝑀 , one gets that

𝜙𝑖(𝒟(𝑔))𝜙𝑖(𝑥𝑗)′ = (−1)𝑗+𝑛|𝑀𝑗|𝜙𝑖(𝑔)′, 𝑖 = 1, . . . , 𝑟; 𝑗 = 1, . . . , 𝑛.

From this we get that

𝑣𝑖(𝜙𝑖(𝒟(𝑔))) = 𝑐𝑖 + 𝑣𝑖(𝜙𝑖(𝑔)′) = 𝑐𝑖 + 𝑣𝑖(𝑔) − 1.

which implies the result.

Proposition 4.7. One has that 𝒟( 𝒪) ⊂ 𝒪.

Proof If 𝑔ℎ

∈ 𝒪, with 𝑔, ℎ ∈ 𝒪, then 𝑣(𝑔) ≥ 𝑣(ℎ). Since

𝒟(

𝑔

ℎ

)= ℎ 𝒟(𝑔) − 𝑔 𝒟(ℎ)

ℎ2 ,

and𝑣

(𝒟(𝑔)

ℎ

)= 𝑣

(𝑔 𝒟(ℎ)

ℎ2

)= 𝑣(𝑔) − 𝑣(ℎ) + f(Γ),

one has that𝑣(

𝒟(

𝑔

ℎ

))≥ 𝑣(𝑔) − 𝑣(ℎ) + f(Γ) ≥ f(Γ) ≥ 0.


Using the determinantal definition of 𝒟(𝑢) and the relations (4.3) one gets that

|𝑀1|𝑑𝑢 = (−1)𝑛 𝒟(𝑢)𝑑𝑥1. (4.5)

So, one has that 𝒟(𝑢) is a zero divisor if and only if |𝑀1|𝑑𝑢 is torsion, which meansthat 𝑑𝑢 is torsion, since in caracteristic zero, no exact non-zero differential is torsion, itfollows that the differential

𝜔 = 𝜙*(𝑑𝑢)𝜙(𝒟(𝑢)) = (−1)𝑛

𝜙(|𝑀1|)𝜙*(𝑑𝑥1) ∈ Ω(𝒦),

is independent from 𝑢 = 0 and 𝑣(𝜔) = −𝑐 + 1. We consider the 𝒪-modules homomorphism“multiplication by 𝜔”:

𝑚𝜔 : 𝒪 → Ω(𝒦)ℎ ↦→ 𝜙(ℎ)𝜔

Recalling that 𝒥 and 𝒞 are respectively the jacobian and conductor ideals of 𝒪,we have the theorem:

Theorem 4.8. The following assertions are true

i) 𝑚𝜔 is injective;

ii) 𝑚𝜔(𝒞) = Ω( 𝒪);

iii) 𝑚𝜔(𝒥 ) = 𝜙*(Ω(𝒪)).

Proof (i) This is so, because the homomorphism 𝜙 : 𝒪 → 𝒪 is injective, 𝜔 = 0 andΩ(𝒦) ≃ 𝑘((𝑡1))𝑑𝑡1 × · · · × 𝑘((𝑡𝑟))𝑑𝑡𝑟 is free.

(ii) The elements of Ω( 𝒪) are those elements 𝜌 ∈ Ω(𝒦) for which 𝑣(𝜌) ≥ 𝑒.

Since 𝑣(𝜔) = f(Γ), it follows that for ℎ ∈ 𝒞, 𝑚𝜔(ℎ) = 𝜔𝜙(ℎ) is such that 𝑣(𝑚𝜔(ℎ)) =𝑣(𝜔) + 𝑣(ℎ) ≥ 𝑒, hence 𝑚𝜔(ℎ) ∈ Ω( 𝒪).

Conversely, let 𝜌 = (𝜉1𝑑𝑡1, . . . , 𝜉𝑟𝑑𝑡𝑟) ∈ Ω( 𝒪), hence 𝑣𝑖(𝜉𝑖) ≥ 0. We want to findℎ = (ℎ1, . . . , ℎ𝑟) ∈ 𝒞 such that 𝑚𝜔(ℎ) = 𝜌. The elements

ℎ𝑖 = 𝜉𝑖𝜙𝑖(|𝑀1|)𝜙𝑖(𝑥1)′ ,

will do the job.

iii) Observe that any element in 𝒥 may be written in the form

ℎ =𝑛∑

𝑖=1(−1)𝑖−1𝑔𝑖|𝑀𝑖|, for ℎ ∈ 𝒪

and any element of 𝜙*(Ω(𝒪)) is of the form𝑛∑

𝑖=1𝜙*(𝑔𝑖𝑑𝑥𝑖), for 𝑔𝑖 ∈ 𝒪.

55

The result follows by observing that

𝑚𝜔(ℎ) = (−1)𝑛∑𝑛𝑖=1

(−1)𝑖−1𝜛𝑖(𝑔𝑖|𝑀𝑖|)|𝑀1| 𝜙*

𝑖 (𝑑𝑥1)

= (−1)𝑛∑𝑛𝑖=1 𝜙(𝑔𝑖)𝜙*(𝑑𝑥𝑖)

= (−1)𝑛𝜙*(∑𝑛𝑖=1 𝑔𝑖𝑑𝑥𝑖).

Corollary 4.9 (cf. D. Pol [16]). One has the following 𝒪-module isomorphisms

𝒥 ≃ 𝜙*(Ω(𝒪)) ≃ Ω(𝒪)/𝒯 ,

and𝑣(𝒥 ∖ 𝑍(𝒪)) = 𝑣(Ω(𝒪) ∖ 𝒯 ) + f(Γ).

Proof This is just Theorem 4.8 (iii) and a computation of values.

Let us define the Tjurina number of 𝒪 as being

𝜏(𝒪) = dim𝑘𝒪𝒥

.

Corollary 4.10. Let 𝒪 be the ring of a complete intersection curve, then one has

𝜏(𝒪) = ℓ𝒪

(𝒪𝒞

)+ ℓ𝒪

(Ω( 𝒪)

𝜙*(Ω(𝒪))

).

Proof By computing values, it is easy to check that 𝒥 ⊂ 𝒞, so


(𝒪𝒥

)= ℓ𝒪

(𝒪𝒞

)+ ℓ𝒪

( 𝒞𝒥

).

Now, from Theorem 4.8 one gets

ℓ𝒪

( 𝒞𝒥

)= ℓ𝒪

(𝑚𝜔(𝒞)𝑚𝜔(𝒥 )

)= ℓ𝒪

(Ω( 𝒪)

𝜙*(Ω(𝒪))

),

which proves the result.

Since 𝒪 is Gorenstein, it follows that

ℓ𝒪

(𝒪𝒞

)= ℓ𝒪

( 𝒪𝒪

)= 𝛿.

On the other hand, from the chain of monomorphisms

𝜙*(Ω(𝒪)) = Ω(𝒪)𝒯

−→ ⊕𝑟𝑖=1

Ω𝑖

𝒯𝑖

−→ Ω( 𝒪),


we get that

ℓ𝒪

(Ω( 𝒪)

𝜙*(Ω(𝒪))

)= ℓ𝒪

⎛⎝ Ω( 𝒪)⊕𝑟

𝑖=1Ω𝑖

𝒯𝑖

⎞⎠+ ℓ𝒪

⎛⎝⊕𝑟𝑖=1

Ω𝑖

𝒯𝑖

Ω(𝒪)𝒯

⎞⎠ .

Since

ℓ𝒪

⎛⎝ Ω( 𝒪)⊕𝑟

𝑖=1Ω𝑖

𝒯𝑖

⎞⎠ =𝑟∑

𝑖=1ℓ𝒪𝑖

⎛⎝Ω( 𝒪𝑖)Ω𝑖

𝑇𝑖

⎞⎠ ,

andℓ𝒪𝑖

(Ω(𝒪𝑖)

Ω𝑖𝒯𝑖

)= #

(𝑣(Ω( 𝒪𝑖)) ∖ 𝑣(Ω𝑖

𝒯𝑖))

= #N* ∖ Λ𝑖

= #N ∖ Γ𝑖 − #Λ𝑖 ∖ Γ𝑖

= 𝛿𝑖 − 𝑟𝑖,

where Γ𝑖 and Λ𝑖 are the semigroup of values of the branch 𝒪𝑖 and the set of values of theKähler differentials on the branch 𝒪𝑖, respectively. The integer 𝑟𝑖 = #Λ𝑖 ∖ Γ𝑖 is equal tothe number of nonexact linearly independent differentials modulo the exact differentials ofthe branch 𝒪𝑖. This number 𝑟𝑖 may be made more explicit when 𝒪𝑖 is also a completeintersection, because, according Corollary 51, in this case, one has

𝑟𝑖 = 𝑐𝑖 − 𝜏𝑖,

where 𝑐𝑖 is the conductor of the semigroup of values of 𝒪𝑖 and 𝜏𝑖 is its Tjurina number.

So, to compute 𝜏(𝒪), it remains only to compute ℓ𝒪

(⊕𝑟

𝑖=1Ω𝑖𝒯𝑖

Ω(𝒪)𝒯

), which may be done

with Theorem 2.7, in general, and with Propositition 2.4 for 𝑟 = 2 and Theorem 2.16 for𝑟 = 3.

57

CHAPTER

5EXAMPLE

Consider the plane curve 𝐶 given by

𝑓(𝑥, 𝑦) = (𝑦3 − 𝑥7)(𝑦3 − 3𝑥5𝑦 − 𝑥7 − 𝑥8)(𝑦4 − 2𝑥5𝑦2 − 𝑥7𝑦 − 𝑥9 + 𝑥10)

and let 𝒪 = 𝑘[[𝑥, 𝑦]]⟨𝑓⟩

the ring of the curve.

Let’s calculate the Tjurina number of 𝒪, given by the following formula

𝜏(𝒪) = dim𝑘𝒪𝒥

(5.1)

where 𝒥 = ⟨𝑓𝑥, 𝑓𝑦⟩ is the Jacobian ideal of 𝒪 , since, 𝒪 is Gorenstein we have


( 𝒪𝒪

)+ ℓ𝒪

( 𝒪Ω𝒯

)

andℓ𝒪

( 𝒪𝒪

)=

3∑𝑖=1

𝛿𝑖 +∑

1≤𝑖,𝑗≤3𝐼𝑖,𝑗

where ℓ𝒪𝑖

( 𝒪𝑖

𝒪𝑖

)= 𝛿𝑖 and 𝐼𝑖,𝑗 is the intersection multiplicity of 𝐶𝑖 and 𝐶𝑗 with 1 ≤ 𝑖, 𝑗 ≤ 3.

Let 𝐶1, 𝐶2 and 𝐶3 be the branches of 𝐶 given by 𝑓1(𝑥, 𝑦) = 𝑦3 − 𝑥7, 𝑓2(𝑥, 𝑦) =𝑦3 − 3𝑥5𝑦 − 𝑥7 − 𝑥8 and 𝑓3(𝑥, 𝑦) = 𝑦4 − 2𝑥5𝑦2 − 4𝑥7𝑦 − 𝑥9 + 𝑥10 respectively.

On the other hand,

ℓ𝒪

( 𝒪Ω𝒯

)= ℓ𝒪

⎛⎝ 𝒪⊕3

𝑖=1Ω𝑖

𝒯𝑖

⎞⎠+ ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠= ℓ𝒪

⎛⎝⊕3𝑖=1

𝒪𝑖

⊕3𝑖=1

Ω𝑖

𝒯𝑖

⎞⎠+ ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠=

3∑𝑖=1

(𝜏𝑖 − 𝛿𝑖) + ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠

58 Chapter 5. Example

Substituting in 5.1 we have

𝜏(𝒪) =3∑

𝑖=1𝛿𝑖 +

∑1≤𝑖<𝑗≤3

𝐼𝑖,𝑗 +3∑

𝑖=1(𝜏𝑖 − 𝛿𝑖) + ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠=

3∑𝑖=1

𝜏𝑖 +∑

1≤𝑖<𝑗≤3𝐼𝑖,𝑗 + ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠ .

So to calculate 𝜏(𝒪), just calculate

ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠ = ℓ𝒪

⎛⎝ ⊕3𝑖=1

Ω𝑖

𝒯𝑖(𝛼0)

⊕3𝑖=1

Ω𝑖

𝒯𝑖(𝑐(Λ))

⎞⎠− ℓ𝒪

( Ω𝒯 (𝛼0)

Ω𝒯 (𝑐(Λ))

)

By Theorem 2.17 we have

ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠ = # M{1,2} +# RM{1,3} +# RM{2,3} +# RM −# AM, (5.2)

where AM and RM are the set of absolute maximals and relative maximals of the set Λ,RM{1,2}, RM{1,3} and RM{2,3} are the set of maximals of the set Λ{1,2}, Λ{1,3} and Λ{2,3},Λ1, Λ2 and Λ3 respectively.

Now, we will compute the terms in 5.2 by using a script in Maple developped byM. E. Hernandes.

The figure below shows the elements of Λ{1,2} with conductor 𝑐(Λ{1,2}) = (20, 20),where the set maximals points M{1,2} = {(3, 3), (6, 6), (7, 7), (9, 9), (10, 10), (12, 12), (13, 13),(15, 15), (16, 16), (19, 19)} of Λ{1,2} are marked with circles. Notice that # M{1,2} = 10.

Λ2

Λ13 10 16 19

3

10

16

19

59

The figure below shows the elements of Λ{1,3} with conductor 𝑐(Λ{1,3}) = (22, 29),where the set maximals points M{1,3} = {(3, 4), (6, 8), (7, 9), (9, 12), (10, 14), (12, 16), (13, 19),(14, 18), (15, 20), (17, 23), (18, 24), (21, 28)} of Λ{1,3} are marked with circles. Notice that# M{1,3} = 12.

Λ3

Λ13 10 18 21

4

24

14

28

The figure below shows the elements of Λ{2,3} with conductor 𝑐(Λ{2,3}) = (22, 29),where the set maximals points M{2,3} = {(3, 4), (6, 8), (7, 9), (9, 12), (10, 14), (11, 13), (12, 16),(13, 19), (14, 18), (15, 20), (17, 23), (18, 24), (21, 28)} of Λ{2,3} are marked with circles. Noticethat # M{2,3} = 13.

60 Chapter 5. Example

Λ3

Λ23 10 18 21

4

24

14

28

We have that, # RM = 14, where RM = {(14, 14, 17), (17, 17, 21), (18, 18, 22), (20, 20, 25),(21, 21, 26), (22, 22, 27), (23, 23, 29), (24, 24, 30), (25, 25, 32), (26, 26, 33), (27, 27, 34), (29, 29, 37),(30, 30, 38), (33, 33, 42)} is the set of relative maximals of Λ.

We also have, # AM = 12, where AM = {(3, 3, 4), (6, 6, 8), (7, 7, 9), (9, 9, 12), (10, 10, 14),(12, 12, 16), (13, 13, 19), (14, 14, 18), (15, 15, 20), (17, 17, 23), (18, 18, 24), (21, 21, 28)} is theset of absolute maximals of Λ.

Substituting in 5.2 we have

ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠ = 37

Moreover, we have tha 𝜏1 = 12, 𝜏2 = 11, 𝜏3 = 21, 𝐼1,2 = 22, 𝐼1,3 = 27 and 𝐼2,3 = 27.Therefore,

𝜏(𝒪) =3∑

𝑖=1𝜏𝑖 +

∑1≤𝑖<𝑗≤3

𝐼𝑖,𝑗 + ℓ𝒪

⎛⎝⊕3𝑖=1

Ω𝑖

𝒯𝑖

Ω𝒯

⎞⎠𝜏(𝒪) = 12 + 11 + 21 + 22 + 27 + 27 + 37𝜏(𝒪) = 157.

61

This same result is obtained using the routine implemented in Singular to calculatethe Tjurina number.

LIB "sing.lib";ring r=0,(x,y),ds; //local ringpoly f=(y3-x7)*(y3-3*x5y-x7-x8)*(y4-2*x5y2-x7y-x9+x10);tjurina(f);//-> 157 //Tjurina number at 0.

63

BIBLIOGRAPHY

[1] Apéry, R.; Sur les branches superlinéaires des courbes algébriques, C.R.A.S. Paris,vol. 222, pp. 1198-1200 (1946).

[2] Barucci, V.; D’anna, M.; Fröberg, R., The Semigroup of Values of a One-dimensional Local Ring with two Minimal Primes, Communications in Algebra,28(8), pp. 3607-3633 (2000).

[3] Bass, H.; On the ubiquity of Gorenstein rings, Math. Zeitsch. 82, pp. 8-28 (1963).

[4] Bayer, V.; Hefez, A.; Hernandes, M.E., O Módulo dos Diferenciais de Kählere o Ideal Jacobiano de uma Curva Analítica Plana, manuscript (2010).

[5] Campillo, A.; Delgado de la Mata, F.; Kiyek, K., Gorenstein property andsymmetry for one-dimensional local Cohen-Macaulay rings., Manuscripta Math.83, pp. 405-423 (1994).

[6] D’anna, M., The Canonical Module of a One-dimensional Reduced Local Ring,Communications in Algebra, 25, pp. 2939-2965 (1997).

[7] Delgado de la Mata, F., The semigroup of values of a curve singularity withseveral branches, Manuscripta Math. 59, pp. 347-374 (1987).

[8] Delgado de la Mata, F., Gorenstein curves and symmetry of the semigroupof values, Manuscripta Math. 61, pp. 285-296 (1988).

[9] Garcia, A., Semigroups associated to singular points of plane curves, J. Reine.Angew. Math. 336, 165-184 (1982).

[10] Garcia, A.; Lax, F., On canonical ideals, Intersection numbers, and Weierstrasspoints on Gorenstein curves, Journal of Algebra. 178, pp. 807-832 (1995).

[11] Gorenstein, D., An arithmetic theory of adjoint plane curves, Trans. Amer.Math. Soc. 72, pp. 414-436 (1952).

[12] Hefez, A.; Hernandes, M. E., Computational Methods in the Local Theory ofCurves. Publicações Matemáticas, IMPA (2001).

[13] Hironaka, H.,On the arithmetic genera and the effective genera of algebraiccurves, Mem. College Sci. Univ. Kyoto Ser. A 30, pp. 178-195, (1957).

64 Bibliography

[14] Kunz, E., The value semigroup of a one-dimensional Gorenstein ring, Proc. Amer.Math. Soc., 25, pp. 748-751 (1970).

[15] Piene, R.; Polar classes of singular varieties, Ann. Scient. Ec. Norm. Sup., 11(1978).

[16] Pol, D., On the values of logarithmic residues along curves, arXiv:1410.2126v3(2015).

[17] Waldi, R., Wertehalbgruppe und Singularität einer ebenen algebraischen Kurve,Dissertation. Regensburg (1972).

[18] Zariski, O., Studies in Equisingularity I, Amer. Jour. Math. 87, pp. 507-536(1965).

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