Apéry Sets and the Ideal Class Monoid of a Numerical Semigroup

The aim of this article is to study the ideal class monoid Cℓ(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathscr {C}}\ell (S)$$\end{document} of a numerical semigroup S introduced by V. Barucci and F. Khouja. We prove new bounds on the cardinality of Cℓ(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathscr {C}}\ell (S)$$\end{document}. We observe that Cℓ(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathscr {C}}\ell (S)$$\end{document} is isomorphic to the monoid of ideals of S whose smallest element is 0, which helps to relate Cℓ(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathscr {C}}\ell (S)$$\end{document} to the Apéry sets and the Kunz coordinates of S. We study some combinatorial and algebraic properties of Cℓ(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathscr {C}}\ell (S)$$\end{document}, including the reduction number of ideals, and the Hasse diagrams of Cℓ(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathscr {C}}\ell (S)$$\end{document} with respect to inclusion and addition. From these diagrams, we can recover some notable invariants of the semigroup. Finally, we prove some results about irreducible elements, atoms, quarks, and primes of (Cℓ(S),+)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$({\mathscr {C}}\ell (S),+)$$\end{document}. Idempotent ideals coincide with over-semigroups and idempotent quarks correspond to unitary extensions of the semigroup. We show that a numerical semigroup is irreducible if and only if Cℓ(S)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathscr {C}}\ell (S)$$\end{document} has at most two quarks.


INTRODUCTION
The ideal class group of a Dedekind domain is a classical mathematical object that has been extensively studied, and has proven to be a useful tool to retrieve information about the underlying domain.This concept can be generalized to the ideal class monoid of any integral domain, which is defined as the set of fractional ideals modulo principal ideals, endowed with multiplication of classes.
The ideal class monoid C (S) of a numerical semigroup S is defined analogously by considering the set of ideals modulo principal ideals of S, endowed with addition.This object is strictly related to the ideal class monoid of the associated semigroup ring K S via the valuation map v : C (K S ) → C (S) [5].The latter object is not very well-understood.However, by working on C (S) it might be possible to interpret C (K S ) from a combinatorial (and hence easier) point of view.
The aim of this article is to extend the study of C (S) which was initiated by V. Barucci and F. Khouja in [2].We prove new estimations of the cardinality of C (S).The most relevant results in this direction are Propositions 3.5 and 3.8, which ensure that if g is the genus of S, t its type and m its multiplicity, then Moreover, the upper bound is reached if and only if either S = {0, m, →} or S = 〈m, m +1, . . ., 2m −2〉.
We then observe that C (S) is isomorphic to the monoid I 0 (S) of ideals of S whose smallest element is 0. This will allow us to relate C (S) to the Apéry sets and the Kunz coordinates of S (see Theorem 4.4).Thanks to this new setting, we will be able to prove further results on the cardinality of C (S) (Corollary 4.5) and to retrieve well-known results (Proposition 4.11).In particular, if m is the multiplicity of S and (k 1 , . . ., k m−1 ) are the Kunz coordinates of S, then The third author is supported by the grant number ProyExcel_00868 (Proyecto de Excelencia de la Junta de Andalucía) and by the Junta de Andalucía Grant Number FQM-343.
Equality holds if and only if S = 〈m〉 ∪ (c + N), with c a positive integer greater than m, or, equivalently, (1) Next, we study some combinatorial and algebraic properties of C (S). Proposition 4.10 provides a description of the canonical ideal of S in terms of its Apéry set.Then we study the reduction number of an ideal I , which is defined, for I ∈ I 0 , as the smallest r such that (r + 1)I = r I .For an ideal I in I 0 generated by 0 and a single gap of S, we compute the reduction number of I in Proposition 4.12.In Proposition 4. 16 we give an upper bound of the reduction number of ideals generated by gaps that are smaller than the multiplicity of the semigroup.
We show that I 0 (S) has a unique maximal element with respect to inclusion, N, and a minimal element, S. The number of maximal elements strictly contained in N equals the type of the semigroup, while the number of minimal elements strictly containing S is precisely the mulitiplicity minus one.The maximal strictly ascending chain of ideals in I 0 (S) has length equal to the genus of the semigroup plus one.Thus, from the Hasse diagram (with respect to inclusion) of I 0 (S) we recover the multiplicity, type and genus of S. In Remark 4.19, we also give a lower bound for the width of this Hasse diagram.
In Section 5, we focus on the study of irreducible elements, atoms, primes and quarks of the ideal class monoid of a numerical semigroup.Minimal non-zero ideals are always quarks, and for every gap g , the (class of the) ideal {0, g } + S is an irreducible element in C (S).Also, the set of unitary extensions of S corresponds with the set of idempotent quarks of its ideal class monoid.We show that the semigroup S is irreducible (it cannot be expressed as the intersection of two numerical semigroups properly containing it) if and only if its ideal class monoid has at most two quarks.
The last section is devoted to several open problems that may serve as a motivation to continue studying the ideal class monoid of a numerical semigroup.
Throughout this paper we present a series of examples meant to illustrate the results proven.
For the development of most of these examples we used the GAP [17] package numericalsgps [6] (in fact, some of our results were stated after analysing a series of computer experiments).The functions used in this manuscript, together with a small tutorial, can be found at https://github.com/numerical-semigroups/ideal-class-monoid.

RECAP ON NUMERICAL SEMIGROUPS AND IDEALS
Let N denote the set of non-negative integers.A numerical semigroup S is a submonoid of (N, +) with finite complement in N. The set N \ S is known as the set of gaps of S, denoted G(S), and its cardinality is the genus of S, g(S).Given a subset A of N, the submonoid generated by then we say that A is a generating set of S. Every numerical semigroup admits a unique minimal generating set (whose elements we refer to as minimal generators), which is S * \ (S * + S * ), where S * = S \ {0}, and its cardinality is known as the embedding dimension of S, denoted e(S) (see for instance [12,Chaper 1]).The smallest positive integer in S is called the multiplicity of S, m(S).Clearly, two minimal generators of S cannot be congruent modulo the multiplicity of S and, consequently, e(S) ≤ m(S).
The largest integer not belonging to a numerical semigroup S (this integer exists as we are assuming G(S) to have finitely many elements) is known as the Frobenius number of S and will be denoted by F(S).In particular, F(S) + 1 + N ⊆ S. The integer F(S) + 1 is the conductor of S, c(S).
A numerical semigroup S induces the following ordering on Z: a ≤ S b if b − a ∈ S. The set of maximal elements in G(S) with respect to ≤ S is denoted by PF(S), and its elements are the pseudo-Frobenius numbers of S. Observe that by the maximality of these elements, if f ∈ PF(S), then for every non-zero s ∈ S, we have that f + s ∈ S. The cardinality of PF(S) is known as the type of S, and will be denoted by t(S).
Recall that a gap f of a numerical semigroup S is a special gap if S ∪{ f } is a numerical semigroup.We denote the set of special gaps of S by SG(S).It is well known (see for instance [12,Section 3.3 In particular, the cardinality of SG(S) corresponds with the number of unitary extensions of a numerical semigroup (numerical semigroups T containing S such that |T \ S| = 1).Given a numerical semigroup S and a non-zero element n ∈ S, the Apéry set of n in S is the set This set contains n elements, one per congruence class modulo n.In fact, if w i is the least element in S congruent to i modulo n, i ∈ {0, . . ., n−1}, then Ap(S, n) = {w 0 , w 1 , . . ., w n−1 }, and clearly, w 0 = 0. Recall that (see for instance [12,Proposition 2.20]) for a numerical semigroup S with multiplicity m, we have that A numerical semigroup S is symmetric if for every integer x ∈ S, F(S) − x ∈ S.This is equivalent to g(S) = (F(S) + 1)/2, or to the fact that F(S) is odd and S is maximal (with respect to set inclusion) in the set of numerical semigroups not containing F(S).If F(S) is even, then S is said to be pseudosymmetric if for every integer x ∈ S, x = F(S)/2, we have that F(S) − x ∈ S.This is equivalent to g(S) = (F(S) + 2)/2, or to the fact that S is maximal in the set of numerical semigroups not containing F(S).A numerical semigroup S is irreducible if it cannot be expressed as the intersection of two numerical semigroups properly containing it, or equivalently, it is maximal in the set of numerical semigroups not containing F(S).Thus a numerical semigroup S is irreducible if and only if it is either symmetric (and its Frobenius number is odd) or pseudo-symmetric (and its Frobenius number is even).These equivalences and other characterisations of irreducible numerical semigroups can be found in [12,Chapter 3].
Let S be a numerical semigroup.We say that E ⊆ Z is an ideal of S if S + E ⊆ E and there exists s ∈ S such that s+E ⊆ S.This last condition implies that there exists min(E ), which is usually known as the multiplicity of E .Notice that if E is an ideal, so is − min(E ) + E .
An ideal E of S is said to be integral if E ⊆ S. The set S \ {0} is known as the maximal ideal of S. Given a set of integers, {x 1 , . . ., x r }, the set {x 1 , . . ., x r } + S = r i =1 (x i + S) is an ideal of S, known as the ideal generated by {x 1 , . . ., x r }.When r = 1, we write x 1 + S instead of {x 1 } + S, and we say that x 1 + S is a principal ideal.
Let E be an ideal of a numerical semigroup S. Let m be the multiplicity of S and set Clearly, {x 1 , x 2 } + S ⊆ E .Moreover, x 1 and x 2 are not congruent modulo m, since x 2 ∈ x 1 + S.This process must stop after a finite number of steps, since the x i s obtained are not congruent modulo m.Thus, E = {x 1 , . . ., x r } + S for some {x 1 , . . ., x r } ⊆ Z, r ≤ m.This set is not only a generating set of E , but a minimal generating set of E in the sense that none of its proper subsets X verifies that X +S = E .The cardinality of the minimal generating set of E is known as the embedding dimension of E and it is denoted as ν(E ).Clearly, ν(E ) ≤ m.
Given two ideals I and J of S, the following sets are also ideals of S: For an ideal E of S, we define F(E ) = max(Z \ E ).

THE IDEAL CLASS MONOID OF A NUMERICAL SEMIGROUP
Let S be a numerical semigroup.The set of all ideals of S, is a monoid with respect to addition, since for every ideals I and J of S, I + J is also an ideal of S, and I + S = I for every I (and so S is the identity element of this monoid).
Remark 3.1.The monoid (I (S), +) is not cancellative (I + J = I + K for I , J , K ∈ I (S) implies J = K ).As a matter of fact, it is not even unit-cancellative (I + J = J implies I is a unit).To see this, let F be the Frobenius number of S, and set I = {0, F } + S. Then it is easy to see that I + I = I .If there exists J ∈ I (S) such that I + J = S, then J = 0 + J ⊆ S. Also F + J ⊆ S, which forces 0 ∈ J .As 0 ∈ S = J + I , 0 = j + i for some j ∈ J and i ∈ I .But this is impossible, since j > 0 (J ⊆ S \ {0}) and i ≥ 0.
Observe also that if we consider only integral ideals of S (ideals contained in S), then the resulting monoid is unit-cancellative.Let I and J be two ideals of S with I ⊆ S and J ⊆ S. Set i = min(I ) and j = min(J ).If I + J = I , then i = x + y for some x ∈ I and j ∈ J .Thus i ≥ i + j ≥ i , which forces j = 0.But then 0 ∈ J , and J + S = S yields S ⊆ J , whence S = J .
On I (S) we define the following equivalence relation: I ∼ J if there exists z ∈ Z such that I = z+J .Clearly, if I ∼ J and I ∼ J , then I + I ∼ J + J , and consequently ∼ is a congruence.This makes a monoid, which is known as the ideal class monoid of S.
Notice that (I 0 (S), +) is also a monoid (with identity element S), and it is isomorphic to C (S); the isomorphism is E → [E ], since every class [I ] of C (S) contains − min(I ) + I , which is in I 0 (S).Observe also that, for all I and J in I 0 (S), ).Let S be a numerical semigroup.The only invertible element of In particular, C (S) is a group if and only if S = N.
Proof.Let I be an ideal of I 0 (S) such that there exists J ∈ I 0 (S) with I + J = S.We have I ⊆ I + J = S = 0 + S ⊆ I , which forces I = S.
Clearly, for every I ∈ I 0 (S), since 0 ∈ I , there is no other minimal generator of I contained in S. Hence, there exists X ⊆ N \ S such that I = ({0} ∪ X ) + S. In particular, (2) |C (S)| ≤ 2 g , which was already shown in [2] (| • | denotes cardinality).It is also clear that for every g ∈ N \ S, the ideal {0, g } + S ∈ I 0 (S), and so (the plus one corresponds to the ideal S).
We now recall the notion of Hasse diagram of a poset.Let (P, ≤) be a finite poset, and let a, b be two elements of P .We say that b covers a if a < b and there is no c ∈ P such that a < c < b.The Hasse diagram of P is the graph whose set of vertices is P and whose edges are the covering relations.An antichain of P is a set of pairwise incomparable elements of P .The width of P is the biggest cardinality of an antichain of P .
Let S be a numerical semigroup and consider the Hasse diagram of the poset (G(S), ≤ S ).It has the following properties: • The width of (G(S), ≤ S ) is m(S)−1 because Minimals ≤ S (G(S)) = {1, . . ., m(S)−1} is a maximal antichain (a set with at least m(S) gaps will contain at least two elements congruent modulo m(S) and thus they will be comparable via ≤ S ).According to Dilworth's Theorem [7], there exists a partition of G(S) into m(S) − 1 chains.This partition is easily achieved with the chains The Hasse diagram of (G(S), ≤ S ) is related to the study of the ideal class monoid of S: if I ∈ I 0 (S), then its minimal generating set is of the form {0}∪X , with X a set of gaps incomparable with respect to ≤ S .Thus, we obtain the following.

Proposition 3.4. Let S be a numerical semigroup. The cardinality of C (S) equals the number of antichains of gaps of S with respect to ≤ S .
Having in mind that the pseudo-Frobenius numbers are incomparable gaps with respect to the ordering induced by the numerical semigroup, we can find a lower bound for the cardinality of the ideal class monoid.Proposition 3.5.Let S be a numerical semigroup with genus g and type t , S = N.Then Proof.Let f 1 , . . ., f r be pseudo-Frobenius numbers of S.Then, by (1), { f 1 , . . ., f r } is an antichain of gaps of S with respect to ≤ S , and thus {0, f 1 , . . ., f r } is a minimal generating set of the ideal {0, f 1 , . . ., f r } + S ∈ I 0 (S).This in particular means that 2 t ≤ |C (S)|.
For the other inequality, let P (N \ S) be the set of subsets of gaps of S. Consider the following map: and so G is injective.Let us prove that if E = S, then E \ S contains at least a pseudo-Frobenius number of S. Let x be in E \ S. Since PF(S) = Maximals ≤ S (Z \ S), there exists f ∈ FP(S) such that x ≤ s f .Hence there exists s ∈ S such that x + s = f , and consequently f ∈ E \ S.This means that the image of G is included in the set of subsets of gaps of S with at least one pseudo-Frobenius number, and the cardinality of this set is 2 Remark 3.6.Let us see that the upper bound in Proposition 3.5 for m ≥ 3 is attained if and only if either S = {0, m, →} (here → means that every integer greater than m is in the semigroup) or S = 〈m, m + 1, . . ., 2m − 2〉.
Example 3.7.Let c be a positive integer and let S = {0} ∪ (c + N).In this case, the type of S equals the genus of S, and every set X of pseudo-Frobenius numbers gives rise to an ideal I = ({0}∪ X )+S.Then the size of the class monoid is precisely two to the power of the type of S and the minimal bound is attained for S.
We can improve a little bit the lower bound given in Proposition 3.5.Proposition 3.8.Let S be a numerical semigroup with multiplicity m and genus g .Then Proof.The proof follows by considering antichains in the Hasse diagrams of G(S) of the form {x} with x a gap larger than m, and the antichains formed by gaps in {1, . . ., m − 1}.This lower bound is better since, by (1), the type of a numerical semigroup is at most its multiplicity minus one.Example 3.9.There are three numerical semigroups of genus 10 attaining this bound: 〈11, . . ., 21〉, 〈10, 11, 12, 13, 14, 15, 16, 17, 18〉, and 〈2, 21〉.and that E = Ap(E , m) + S. Notice that the definition of Apéry set for an ideal is slightly different to that given in [11].Notice that S is itself an ideal of S, and that Ap(S, m) coincides with the Apéry set of m in S in the usual sense.It is well known that every element s in S can be expressed uniquely as s = km + w with k ∈ N and w ∈ Ap(S, m).The following result characterizes those sets that are Apéry sets of ideals of a numerical semigroup (see [13,Lemma 8] for a similar result for numerical semigroups).

Lemma 4.3. Let S be a numerical semigroup with multiplicity m and let A
Proof.For sake of simplicity, denote i + j = (i + j ) mod m.
Suppose that E is an ideal of S with min(E ) = 0. Then w i (E ) + w j (S) ∈ E , and consequently w i (E ) + w j (S) ≥ w i + j (E ).Now suppose that A is a set fulfilling the inequalities of the statement.Let E = A +S. Then w i (E ) ≤ w i for all i ∈ {0, . . ., m −1}.As w i (E ) ∈ E , we have that w i (E ) = w j + s for some j ∈ {0, . . ., m − 1} and s ∈ S.There exists t ∈ N and k ∈ {0, . . ., m − 1} such that s = t m + w k (S), whence w i (E ) = w j + t m + w k (S).By the standing hypothesis w j + w k (S) ≥ w j +k .Notice that j + k = i , and so w i (E ) ≥ t m + w i ≥ w i , forcing w i = w i (E ).Let S be a numerical semigroup with multiplicity m, and let E be an ideal of S with min(E ) = 0. Recall that the Kunz coordinates of S are the (m −1)-uple (k 1 (S), . . ., k m−1 (S)) such that that w i (S) = k i (S)m +i for all i ∈ {1, . . ., m −1}.We can proceed similarly with E .For every i ∈ {0, . . ., m −1}, there exists The inequalities in Lemma 4.3 imply that for all i , j ∈ {0, . . ., m − 1}, w i (E ) + w j (S) ≥ w i + j (E ), and so The first inequality follows from w 0 (E ) + w j (S) ≥ w j (E ).Notice that we have the correspondence an ideal, its Kunz coordinates will fulfill the above inequalities, and if we have a tuple (k 1 , . . ., k m−1 ) fulfilling these inequalities, then the set {0, k 1 m+1, . . ., k m−1 m+m−1} is the Apéry set of an ideal by Lemma 4.3 (actually the ideal {0, k 1 m + 1, . . ., k m−1 m + m − 1} + S).In light of the above discussion we can state the following result.Theorem 4.4.Let S be a numerical semigroup with multiplicity m and Kunz coordinates (k 1 , . . ., k m−1 ).The set of ideals E of S with min(E ) = 0 are in one-to-one correspondence with the set K (S) of solutions of the following system of inequalities over the set of non-negative integers: In particular, from the first set of inequalities we get the following bound.
Equality holds if and only if S = 〈m〉 ∪ (c + N), with c a positive integer greater than m, that is, Proof.The inequality follows directly from 0 ≤ k i (E ) ≤ k i (S).So, let us focus on when the equality holds.
Necessity.Notice that |I 0 (S)| = m−1 i =1 (k i + 1) forces every tuple (k 1 , . . ., k m−1 ) with 0 ≤ k i ≤ k i for all i to be the Kunz coordinates of an ideal of S. In particular, the tuple (0, . . ., 0, k i +1 , 0, . . ., 0) (k i +1 in the (i +1)th coordinate and zero in the rest) corresponds with the Kunz coordinates of some ideal E of S. Then k 1 (E ) As (k 1 , 0, . . ., 0) are the Kunz coordinates of some ideal of S, say E , we deduce that proving in this way the second assertion.
Example 4.9.Let S = 〈3, 5, 7〉.The following table is the addition table of I 0 (S).We represent each ideal by its Apéry set with respect to the multiplicity of S. The identity element (and the minimum) is 0 + S = {0, 7, 5} + S, while {0, 1, 2} + S = N is the maximal ideal and acts as sink or infinity, that is, every time we add an ideal to N we obtain again N.
The following result recovers the description of the Apéry set of the canonical ideal of a numerical semigroup given in [15, Proposition 1.3.9].Proposition 4.10.Let S be a numerical semigroup with multiplicity m, and let E be in I 0 (S).Let f = F(S) mod m.Then E = K(S) if and only if w i (E ) + w j (S) = w f (S) = w f (E ) for all i , j ∈ {0, . . ., m − 1} with i + j ≡ f (mod m).
As w i (K )−m ∈ K , we have that F(S)−(w i (K )−m) = w f (S)−w i (K ) ∈ S, which means that w f (S)− w i (K ) ≥ w j (S), i + j ≡ f mod m, or equivalently, w i (K ) + w j (S) ≤ w f (S), and by Lemma 4.3, we know that w i (K ) + w j (S) ≥ w f (S).Hence w i (K ) + w j (S) = w f (S).Now suppose that E is an ideal in I 0 (S) with w f (E ) = w f (S) and w i (E ) + w j (S) = w f (E ) for all i , j ∈ {0, . . ., m−1} with i + j ≡ f (mod m).Let x ∈ K .Let i = x mod m, and j = ( f −i ) mod m.Then F(S)−x ∈ S, which means that w f (S)−m −x < w j (S), and so w j (S)+x +m > w f (S) = w i (E )+w j (S).It follows that x + m > w i (E ), and so x ≥ w i (E ), yielding x ∈ E .For the other inclusion, let us prove that w i (E ) is in K for all i .Let j be as above.We have to show that F(S) − w i (E ) ∈ S, or equivalently, w f (S) − m − w i (E ) < w j (S), which translates to w i (E ) + w j (S) + m > w f (E ).But this trivially holds, since by hypothesis w i (E ) + w j (S) = w f (E ).
With this, we retrieve the following result, that is most probably known, but for which we do not find an appropriate reference in the literature.Corollary 4.11.Let S be a numerical semigroup.Then the canonical ideal of S is generated by {F(S)− g : g ∈ PF(S)}.
Proof.Let K be the canonical ideal of S. Let m be the multiplicity of S and f the Frobenius number of S. Let us prove that K = { f − g : g ∈ PF(S)} + S. By Proposition 4.10, we know that w i (K ) = w f (S) − w f −i (S) for all i ; whence w i (K ) = ( f + m) − w f −i (S).We also know that K = Ap(K , m) + S. For every i , there exists m i ∈ Maximals ≤ S (Ap(S, m)) and s i ∈ S such that w f −i (S) (1), m i − m ∈ PF(S), and so w i (K ) ∈ { f − g : g ∈ PF(S)} + S.This proves that K ⊆ { f − g : g ∈ PF(S)} + S. Now let g ∈ PF(S).Then f − ( f − g ) = g ∈ S, and thus f − g ∈ K , and consequentely { f − g : g ∈ PF(S)} + S ⊆ K .4.2.Reduction number.If in Proposition 4.8, we take J = I , we obtain Since I ⊆ I + I = 2I and I 0 (S) has finitely many elements, we deduce that there is some r such that (r + 1)I = r I .This r is known as the reduction number of I , denoted by r(I ).
Notice that for all i , w i ((r + 1)I ) ≤ w i (r I ), and r is the reduction number of I precisely when all these inequalities become equalities.
Some basic properties of the reduction number of an ideal can be found in [2], and we summarize them in the next result.Proposition 4.12.Let S be a numerical semigroup with multiplicity m.
(1) For every ideal E of S and every positive integer j ≤ r(E ), we have that ν( j E ) > j .
Example 4.13.Let S be a numerical semigroup with multiplicity m, and let g ∈ N \ S. Consider the ideal E = {0, g } + S. Observe that w i (E ) = w i (S) for all i ≡ g (mod m), and that w g (E ) = g < w g (S).
Example 4.15.Let n be an odd integer greater than one.Let S = 〈2, n〉.Every ideal E in I 0 (S) \ {S} is of the form {0, k} + S, with k an odd integer smaller than n (a gap of S).Notice that 2k ∈ S, and so according to Proposition 4.14, r(E ) = 1.Note that this also follows from Proposition 4.12 (2), since in this setting m = 2.
Notice that, by Proposition 4.12 (3), if m > 2, then there is an ideal E such that r(E ) > 1.Thus, if every nontrivial ideal has reduction number one, the multiplicity must be at most two.
Observe also that 2E = E for all E ∈ I 0 (S) \ {S} implies that E = 2E + S, and so E is in 2E + I 0 (S) (that is, I 0 (S) is a Clifford semigroup).The converse is also true.If I 0 (S) is a Clifford semigroup, that is, for all E ∈ I 0 (S), E ∈ 2E + I 0 (S), then there exists Next we give an upper bound for the reduction number of ideals generated by sets of the form {0} ∪ X with X a set of gaps smaller than the multiplicity of the semigroup.Proof.We can assume h < m −1, otherwise E = N and there is nothing to prove.Moreover, for h = 1 the thesis follows immediately by Proposition 4.12, so we can assume h > 1.
Set a 0 = 0 and x = m − h + 1.We want to show that xE = (x − 1)E .So take an element t in xE .We have for some i ∈ {0, . . ., h}, then, since a i < m, we deduce u = a i + ym, with y ∈ N, which means means For every k ∈ {2, . . ., x}, set u k = a i 1 + • • • + a i k .As above, if u k ≡ a i (mod m) for some i , then u k = a i + ym, with y ∈ N, since a i < m for every i .Thus, t = u + s = a i + ym Hence we may suppose that u and every u k are not congruent to any a i modulo m.By setting u x = u, since |{u 2 , . . ., u x−1 , u x }| = x −1 = m −h and |{a 0 , . . ., a h }| = h +1, there exist two partial sums u z , u w , with 2 ≤ w < z ≤ x, such that u z ≡ u w (mod m).
we obtain once more that t ∈ (x − 1)E as desired.

4.3.
Hasse diagram of the class monoid.By using Apéry sets, we can obtain some information about the Hasse diagram (with respect to inclusion, see Figure 1) of the ideal class monoid of a numerical semigroup.Proposition 4.17.Let S be a numerical semigroup with multiplicity m.Then Proof.Recall that for I , J ∈ I 0 (S), I ⊆ I if and only if (0, w 1 (J ), . . ., w m−1 (J )) ≤ (0, w 1 (I ), . . ., w m−1 (I )).Notice that whenever w i (J ) ≤ w i (I ), we also have that w i (I ) − w i (J ) is a multiple of m.
The first two assertions follow directly from the definitions.As for the third, notice that for N = {0, 1, . . ., m − 1} + S, we have that w i (N) = i for all i .Thus the potential candidates to be maximal below N are precisely those whose Apéry sets with respect to m differ in one element.Let us prove that for every i , the set {0, 1, . . ., i −1, i +m, i +1, . . ., m−1} is an Apéry list of an ideal in I 0 (S).By using Lemma 4.3, this reduces to showing that for every k ∈ {0, . . ., m − 1}, j + w k (S) ≥ ( j + k) mod m for j = i and that i +m+w k (S) ≥ (i +k) mod m.The first inequality holds because j +w k (S) ≥ j +k and ( j +k) mod m is either j +k or j +k −m.The second inequality also holds, because i +m +w k (S) > i + k.Now let us focus on the fourth assertion of the statement.For the same reason as in the previous paragraph, the candidates to be minimal above S are those with Apéry lists of the form {0, w 1 (S), . . ., w i −1 (S), w i (S) − m, w i +1 (S), . . ., w m−1 (S)}.With the use of Lemma 4.3, let us check which of these sets are Apéry sets of ideals in I 0 (S).For all j = i and for all k, we have that w j (S) + w k (S) ≥ w j +k (S), if ( j + k) mod m is not equal to i , since S is a numerical semigroup.If ( j + k) mod m is i , then we also have w j (S) + w k (S) ≥ w i (S) ≥ w i (S) − m.Also w i (S) − m + 0 ≥ w i (S) − m, so it remains to check whether w i (S) − m + w k (S) ≥ w i +k (S) for all k = 0. Notice that (w i (S) − m + w k (S)) mod m = i + k, and consequently w i (S) − m + w k (S) ≥ w i +k (S) if and only if w i (S)−m + w k (S) ∈ S. Observe that w i (S)−m + w k (S) ∈ S for all k = 0 if and only if w i (S)−m + s ∈ S for all s ∈ S \ {0}.Thus, {0, w 1 (S), . . ., w i −1 (S), w i (S) − m, w i +1 (S), . . ., w m−1 (S)} is an Apéry list of an ideal in I 0 (S) if and only if w i (S) − m ∈ PF(S).Finally, observe that {0, w 1 (S), . . ., w i −1 (S), w i (S) − m, w i +1 (S), . . ., w m−1 (S)} + S = {0, w i (S) − m} + S.
Then the sequence of ideals I i = {0, g 1 , . . ., g i } + S is an increasing sequence of ideals (with respect to inclusion).Remark 4.18.Notice that as consequence of Proposition 4.17 (4), we recover the fact that every ideal in I 0 (S) \ {S} contains a pseudo-Frobenius number (see the proof of Proposition 3.5).It also follows that t(S) = | Minimals ⊆ (I 0 (S) \ {S})|.
Remark 4.19.We now want to find a lower bound for the width of the Hasse diagram of I 0 (S) with respect to inclusion.This can be achieved by looking at the "second level" of the Hasse diagram (minimal non-zero ideals) yielding m(S) − 1, or by using the pigeonhole principle: by removing the maximum (N) and minimum (S) from the Hasse diagram, having g(S) − 1 remaining "levels" we have that the width will be at least (|I 0 (S)| − 2)/(g(S) − 1) .This bound is sharp and is attained for example in the case S = 〈3, 5〉, for which the width is 2.
Example 4.20.Using the description of maximal elements in I 0 (S) \ {N}, we can compute their reduction number.More precisely, let S be a numerical semigroup with multiplicity m, and let E be a maximal element of

IRREDUCIBLE ELEMENTS, ATOMS, QUARKS AND PRIMES
Let H be a monoid (in our setting commutative, and thus we use additive notation).For a, b ∈ H define a b if there exists c ∈ H such that b = a + c.We write a ≺ b whenever a b and b a.We say that a ∈ H is a unit if there exists b ∈ H such that a +b = 0 (the identity element of H ). Following [16] (for our specific , -units are units), a non unit a is said to be • irreducible if a = x + y for all non-units x and y of H such that x ≺ a and y ≺ a; • an atom if a = x + y for all non-units x and y of H ; • a quark if there is no non-unit b with b ≺ a; • a prime if a x + y for some x, y ∈ H implies that a x or a y.
Notice that I 0 (S) is reduced (Proposition 3.3), that is, the only unit is its identity element, which in this case is 0 + S = S. Observe that I 0 (S) is commutative, but it is not cancellative.Also if I J in I 0 (S), then J = I +K for some ideal K ∈ I 0 (S), whence I ⊆ J .Thus I J I implies that I J , since I = J would imply J I .This shows that I ≺ J implies that I J and I = J .Now suppose that I J and that I = J .If J I , then I ⊆ J and J ⊆ I , forcing I = J , a contradiction.Hence, in our setting, I ≺ J if and only if I J and I = J .Remark 5.1.Notice that if we have a chain of ideals , and consequently t ≤ g(S) by Proposition 4.17 (5).Notice also that if are the gaps of S, then ({0, g i +1 } + S) + ({0, g 1 , . . ., g i } + S) = {0, g 1 , . . ., g i +1 } + S.This means that {0, g 1 , . . ., g i } + S ≺ {0, g 1 , . . ., g i +1 } + S for all suitable i , which is precisely the chain used in the proof of Proposition 4.17 (5), meaning that the largest length of a strictly increasing chain for ideals in I 0 (S) with respect to has also length g(S) + 1.
With the help of this remark we can prove the following result.
Proof.Let ϕ : I 0 (S 1 ) → I 0 (S 2 ) be a monoid isomorphism.For every I , J ∈ I 0 (S 1 ), I J if and only if ϕ(I ) ϕ(J ).Thus ϕ maps strictly ascending chains in I 0 (S 1 ) to strictly ascending chains in I 0 (S 2 ).By Remark 5.1, there is a strictly increasing chain in I 0 (S 1 ) of length g(S 1 ) + 1, ) is a maximal strictly ascending chain in I 0 (S 2 ).In particular, this implies that both S 1 and S 2 have the same genus.Observe that in general, there are more quarks than minimal non-zero ideals (with respect to inclusion).There are no prime elements in I 0 (S).For instance, let I = {0, 7} + S and J = {0, 3, 4} + S. Then I + J + J = J + J , and so I J + J , but I J since I is not included in J .Lemma 5.4.Let I be I 0 (S).Then I is irreducible if and only if I = J + K for any non-zero J , K ∈ I 0 (S) \ {I }.
Proof.Suppose that I is irreducible and that I = J + K for some non-zero elements in I 0 (S) with J = I = K .As I = J + K and J = I = K , we have J ≺ I and K ≺ I , contradicting that I is irreducible.
The sufficiency is straightforward.
The following results ensures that we have at least as many irreducible elements in I 0 (S) as gaps in S.Not all minimal factorizations of an ideal in I 0 (S) need to have the same length.For S = 〈5, 6, 8, 9〉, the ideal I = {0, 2, 3, 4} + S has minimal factorizations of length two and three.As I + ({0, 7} + S) = I , one can find factorizations of I as sums of n irreducible elements for any integer n greater than one.
There are examples of non-irreducible ideals with a finite number of factorizations.Take S = 〈5, 16, 17, 18, 19〉 and I = {0, 1, 2} + S. Then the only factorization of I as sum of irreducible elements is I = 2({0, 1} + S).This is due to the fact that the only ideal J such that J + I = I is J = 0 + S = S. Corollary 5.8.Let S be a numerical semigroup.Then I 0 (S) is cyclic (there exists I ∈ I 0 (S) such that I 0 (S) = {k I : k ∈ N}) if and only if S = 〈2, 3〉.
Proof.Notice that cyclic implies that there is just one irreducible, and this forces S to have just one gap.The other implication is easy to check.Example 5.9.Let b be an odd integer, and set S = 〈2, b〉.Let us see how I 0 (S) looks like.Notice that G(S) = {1, 3, . . ., b −2}, and that any two gaps are comparable modulo 2. Thus, I 0 (S) = {{0, g }+S : g ∈ G(S)} ∪ {S}, and by Proposition 5.5, every non-zero ideal is irreducible.Inclusion is a total ordering in (I 0 (S), ⊆).Also, notice that if g 1 and g 2 are gaps of S with g 1 ≤ g 2 , then ({0, g 1 }+S)+({0, g 2 }+S) = {0, g 1 } + S. This, in particular implies that every non-zero ideal is also a prime ideal.There are no atoms in I 0 (S).The only quark is {0, b − 2} + S. Lemma 5.10.Let S be a numerical semigroup, and let I be a minimal non-zero ideal in I 0 (S).Then I is a quark.
Proof.If I is not a quark, then there must be another non-zero J ∈ I 0 (S) such that J ≺ I , which would imply that J I , a contradiction.Thus, there are quarks that are not non-zero minimal ideals, and some of them might be generated by elements that are not pseudo-Frobenius numbers.
Next we show that special gaps of the numerical semigroup S are in one-to-one correspondence with idempotent quarks of I 0 (S).Proposition 5.12.Let S be a numerical semigroup.Then Q is an idempotent quark of I 0 (S) if and only if Q = {0, f } + S with f ∈ SG(S).
Proof.Let Q be an idempotent quark.By Remark 4.18, we know that there exists For the converse, let f ∈ SG(S) and set Q = {0, f } + S. As f ∈ PF(S), by Proposition 4.17, Q is a minimal non-zero element of I 0 (S) with respect to inclusion.Thus, in light of Lemma 5.10, Q is a quark.The fact that Q +Q = Q follows easily from the fact that f ∈ PF(S) and 2 f ∈ S.
Observe that another reading of the last result is that unitary extensions of a numerical semigroup S are precisely the idempotent quarks of I 0 (S).
We say that a numerical semigroup T is an over-semigroup of S if S ⊆ T [14].
Proposition 5.13.Let S and T be numerical semigroups.Then T is an over-semigroup of S if and only if T is an idempotent in I 0 (S).
Proof.If T is an over-semigroup of S, then T + S = T and T + T = T , whence T is an idempotent in I 0 (S).Now let I be an idempotent element in I 0 (S).Then I + I = I , which means that I is a semigroup, and S ⊆ I , which means that (1) the finite complement of I in N is finite, and (2) it is an over-semigroup of S.
Observe that idempotent ideals are precisely those with reduction number one.
Remark 5.15.Notice that if T is an over-semigroup of S and I ∈ I 0 (T ), then I + S ⊆ I + T ⊆ I , and thus I ∈ I 0 (S).Thus, I 0 (T ) ⊆ I 0 (S).Not every ideal of I 0 (S) is an ideal of an over-semigroup of S. This is mainly due to the fact that, there might be more pseudo-Frobenius numbers (which correspond to minimal ideals) than special gaps.For instance, for the semigroup S = 〈3, 10, 17〉, we have PF(S) = {7, 14} and SG(S) = {14}.In this case the ideal {0, 7}+S is a minimal non-zero ideal of S (Proposition 4.17) that is not an ideal of any over-semigroup of S (if it were so, it would be an ideal of the unique unitary extension of S, S ∪ {14}, and S ∪ {14} is not included in {0, 7} + S).Lemma 5.16.Let S be a numerical semigroup with Frobenius number f .Let Q be a quark of I 0 (S).
Next, we see that how is the set of quarks of the ideal class monoid of a symmetric numerical semigroup.
Proposition 5.17.Let S be a numerical semigroup, S = N.Then S is symmetric if and only if I 0 (S) has only one quark.
Proof.Let f be the Frobenius number of S. By Proposition 4.17 (4) and Lemma 5.10, we know that {0, f } + S is a quark.
Suppose that S is symmetric.Let Q be a quark.By using Remark 4.18, we have that f ∈ Q, since PF(S) = { f }.But then Lemma 5.16 ensures that Q = {0, f } + S. Now suppose that I 0 (S) has only one quark, which by the first paragraph of this proof must be {0, f }+S.By Lemma 5.10, this means that I 0 (S) has at most one minimal element, and Remark 4.18 ensures that the type of S is at most one, and thus S is symmetric [1, Corollary 8].
Notice that if S is symmetric, being {0, F(S)}+S the only quark of its ideal class monoid, {0, F(S)}+ S is a prime of I 0 (S).
In particular, if S is symmetric, by Remark 4.18, we have One can observe this in the example depicted in Figure 2.
, we deduce that is pseudo-symmetric.
Let S be a numerical semigroup, and let I , J ∈ I 0 (S).Recall that J is said to cover I if I ≺ J and there is no K ∈ I 0 (S) such that I ≺ K ≺ J .We denote by I the set of ideals in I 0 (S) covering I .In this way, the set of quarks is precisely S .Analogously, define I to be the set of ideals covered by I .A natural question (dual to determining S ) would be to characterize those ideals belonging to N .We describe this set in the next result.• If I ∈ I 0 (S) is a quark and it is not an idempotent (I = I + I ; the reduction number of I is greater than one), then I is an atom.Notice that I = J + K with J , K ∈ I 0 (S) \ {0} forces J I and K I .As I is a quark, J = I = K , contradicting that I = I + I .• Let I ∈ I 0 (S) be an atom, and suppose that there exists J ∈ I 0 (S) \ {S} such that J ≺ I .By definition, there exists K ∈ I 0 (S) \ {S} such that J + K = I , contradicting the fact that I is an atom.Thus, every atom is a quark.• If I ∈ I 0 (S) is a quark, then I is an irreducible, since I = J + K with S = J = I = K = S forces J ≺ I , a contradiction.Not every irreducible is a quark (see Example 5.3).• Notice also that every atom is irreducible.
• In I 0 (S), being a prime implies being irreducible.Assume that I is prime and that I = J + K with S = J ≺ I , S = K ≺ I .Then J I and K I .As I is prime and I J + K , either I J or I K , but this implies that either I ⊆ J or I ⊆ K ; thus either I ⊆ J I or I ⊆ K I , which in both cases is impossible.This proves that I is irreducible.
To summarise, being an atom implies being a quark, which in turn implies being irreducible, and every prime element is irreducible.Any other implication between these concepts does not hold in light of Examples 5.3 and 5.9.These examples also show that there are numerical semigroups for which its ideal class monoid has no primes, and numerical semigroups having ideal class monoid with no atoms.

OPEN QUESTIONS
We know that if two numerical semigroups have isomorphic ideal class monoids, then they have the same genus (Corollary 5.2).Proposition 5.21 forces their multiplicities to be the same, and Proposition 5.12 tells us that they have the same number of unitary extensions.This leads to the following conjecture.Question 6.1.Let S 1 , S 2 be two numerical semigroups such that C (S 1 ) is isomorphic to C (S 2 ).Does S 1 = S 2 hold?Notice that if C (S 1 ) and C (S 2 ) are isomorphic, then their respective posets with respect to are isomorphic.Thus, we could also conjecture something stronger.Question 6.2.Let S 1 , S 2 be two numerical semigroups such that (C (S 1 ), ) and (C (S 2 ), ) are isomorphic as posets.Does S 1 = S 2 hold?
We can also reformulate this question taking inclusion instead of .If the above question has a positive answer, it would also be interesting to determine if there exists an algorithm to recover a semigroup from the Hasse diagram of its ideal class monoid.
In Remark 4.19, we proved that a lower bound for the width of the Hasse diagram of I 0 (S) is (|I 0 (S)|−2)/(g(S)−1).We also showed that the bound is sharp.However, for numerical semigroups with big ideal class monoids, this bound is not very good: for S = 〈5 In general, an element in a monoid might admit different expressions as a sum of irreducible elements.These expressions are known as factorizations.There are a many invariants that measure how far these factorization are from being unique (as happens in unique factorization monoids) or have the same length (half-factorial monoids); see [9] for a detailed description of this theory in the cancellative setting or [16] for a more general scope.Question 6.4.Given an ideal I of I 0 (S), can we say something about the number of its factorizations in terms of irreducible elements in I 0 (S)?Or even about the lengths of these factorizations?Is this set of lengths an interval?Section 5 was mainly motivated by previous works, started in [8], on the power monoid P fin,0 (N), the set of subsets of N containing 0 and with finitely many elements (and P fin,0 (S) with S a numerical semigroup), endowed with addition A + B = {a + b : a ∈ A, b ∈ B }.Moreover, [4] shows the abundance of atoms in P fin,0 (S); however, this is far from being the case for I 0 (S), since we have examples with no atoms at all.In P fin,0 (S), irreducible elements, atoms and quarks are the same [16,Proposition 4.11(iii) and Theorem 4.12] (notice that these three concepts, irreducible element, atom and quark, differ in our setting), and so it would make more sense to propose the following question instead.Question 6.5.Let S be a numerical semigroup.What is the ratio between the cardinality of irreducible elements in I 0 (S) and the cardinality of I 0 (S)= We know that we have at least as many irreducibles as the genus of S, which in turn is the height of the Hasse diagram of I 0 (S) (minus one).

Remark 3 . 2 .
The ideal class monoid of S does not have to be confused with the class semigroup of S, [9, Section 2.8].As S lives in the free monoid N, we can define the class semigroup, C (S, N), as the quotient of N modulo the relation x ∼ y if (−x +S)∩N = (−y +S)∩N.Notice that if c is the conductor of S and x ≥ c, then (−x + S) ∩ N = N.Thus the class of c under the relation ∼ is equal to c + N. If x is a non-negative integer less than c, then max(N \ (−x + S)) = F − x, with F the Frobenius number of S. So if (−x + S) ∩ N = (−y + S) ∩ N, then their complements in N are equal, and so F − x = F − y, yielding x = y.Thus C (S, N) = {{0}, {1}, . . ., {c − 1}, c + N}.
4. APÉRY SETS Denote by m the multiplicity of S and define Ap(E , m) = {w 0 (E ), . . ., w m−1 (E )}, where w i (E ) is the minimum element in E congruent with m modulo i .It follows easily that Ap(E , m) = {e ∈ E : e − m ∈ S}

Remark 5 .
18. Let S be a numerical semigroup with Frobenius number f .Then I ∈ I 0 (S ∪ {F }) if and only if f ∈ I and I ∈ I 0 (S).The necessity is easy, since I + (S ∪ { f }) ⊆ I implies that I + S ⊆ I and f ∈ I .For the sufficiency, I +
The canonical ideal.Let S be a numerical semigroup.The standard canonical ideal of S is