The isomorphism problem for ideal class monoids of numerical semigroups

From any poset isomorphic to the poset of gaps of a numerical semigroup $S$ with the order induced by $S$, one can recover $S$. As an application, we prove that two different numerical semigroups cannot have isomorphic posets (with respect to set inclusion) of ideals whose minimum is zero. We also show that given two numerical semigroups $S$ and $T$, if their ideal class monoids are isomorphic, then $S$ must be equal to $T$.


INTRODUCTION
A numerical semigroup S is a submonoid of (N, +) such that N \ S has finitely many elements, where N denotes the set of non-negative integers.A set of integers I is said to be an ideal of S if I +S ⊆ I and I has a minimum (see for instance [2,Chapter 3] for some basic background on ideals of numerical semigroups).On the set of ideals of S, we define the following relation: I ∼ J if there exists an integer z such that I = z + J .The set of ideals modulo this equivalence relation is known as the ideal class monoid of S, denoted C ℓ(S).Addition of two classes [I ] and [J ] is defined in the natural way: The ideal class monoid of a numerical semigroup was introduced in [3] inspired by the definition of ideal class group of a Dedekind domain.In [5], we proved that from some combinatorial properties of the ideal class monoid of a numerical semigroup we can recover relevant information of the numerical semigroup like, for instance, its genus, multiplicity, type, and number of unitary extensions.
We say that an ideal I of S is normalized if min(I ) = 0; we denote by I 0 (S) the set of normalized ideals I of S. The map C ℓ(S) → I 0 (S), [I ] → − min(I ) + I is bijective.Moreover, for I , J ∈ I 0 (S), the ideal I + J is also in I 0 (S).Thus, the mapping [I ] → − min(I ) + I is a monoid isomorphism.
In [5,Section 4.3], we studied some of the properties of the poset (I 0 (S), ⊆).It is natural to wonder if (I 0 (S), ⊆) completely determines S in the following sense: if T is a numerical semigroup and (I 0 (S), ⊆) is isomorphic to (I 0 (T ), ⊆), then S = T ?Recall that two posets (P, ≤ P ) and (Q, ≤ Q ) are isomorphic if there exists an order isomorphism f from P to Q, that is, f is bijective and for every a, b ∈ P , a ≤ P b if and only if f (a) ≤ Q f (b).We translate this problem of poset isomorphism of normalized ideals of a numerical semigroup to an isomorphism problem of posets of gaps with respect to the order induced by the semigroup.
For a numerical semigroup S, the order induced by S on the set of integers, denoted ≤ S , is defined as a ≤ S b if b − a ∈ S. The poset (Z, ≤ S ) (with Z the set of integers) has been studied for several families of numerical semigroups, and more particularly the Möbius function associated to ≤ S (see [7] or [6] for a generalization to affine semigroups).
The set G(S) = N \ S is the gap set of S; its elements are called gaps of S. It was already shown in [3,Proposition 2.6] that the set C ℓ(S) is in one-to-one correspondence with the set of antichains of gaps with respect to ≤ S (these antichains are called S-Leans in [11]).For every gap g of S, the set {0, g } + S is an ideal of S, and so G(S) is embedded naturally in I 0 (S).Moreover, if g ′ is another gap of S, then g ≤ S g ′ if and only if {0, g ′ } + S ⊆ {0, g } + S (Lemma 8).If we are able to characterize the ideals of the form {0, g } + S from their properties in the poset (I 0 (S), ⊆), then we can extract a poset isomorphic to (G(S), ≤ S ) and thus recover S.This is actually the strategy we use to prove that if the posets (I 0 (S), ⊆) and (I 0 (T ), ⊆) are isomorphic, then S and T must be equal.
The ideal class monoid of a numerical semigroup is a monoid.Thus, it is natural to ask if two different numerical semigroups will have isomorphic ideal class monoids [5,Question 6.1].The answer is no.Theorem 18 states that if S and T are numerical semigroups, and their ideal class monoids are isomorphic, then S and T must be equal.
It order to solve the isomorphism problem for ideal class monoids of numerical semigroups, we study what are the consequences of having an isomorphism between (I 0 (S), +) and (I 0 (T ), +), with S and T numerical semigroups.In particular, we show that over-semigroups of S are in correspondence with over-semigroups of T , and their corresponding ideal class monoids must be isomorphic.
Most of the computations in the examples presented in this manuscript where performed using the GAP [9] package numericalsgps [8].The code used for these calculations can be found at https://github.com/numerical-semigroups/ideal-class-monoid.The package numericalsgps was also used to draw Hasse diagrams of the posets mentioned above for several numerical semigroups, providing in this way clues on what where the results needed to proof our main theorems.

DETERMINING A NUMERICAL SEMIGROUP FROM THE ORDER INDUCED IN ITS GAP SET
Suppose that we are given a numerical semigroup S as a sequence {s 0 , s 1 , . . ., s n , . . .} of which the only data we know is whether s i ≤ S s j for i , j ∈ N. The ν sequence ν i = |{ j ∈ N : s j ≤ S s i }| completely determines S (see [4]; here |X | denotes the cardinality of the set X ).Thus, if S and T are numerical semigroups whose respective posets (S, ≤ S ) and (T, ≤ T ) are isomorphic, then S = T .Now, suppose that what we have is an enumeration H = {h 1 , . . ., h g } of the gap set of S, G(S), and how these elements are arranged with respect to ≤ S .We want to recover S from this information.
Recall that the multiplicity of S is the least positive integer in S. From [3, Lemma 2.5(1)], we know that multiplicity of S is the cardinality of Minimals ≤ S (H ) plus one.The argument used in the proof of that lemma also shows that the maximal number of elements in an antichain (with respect to ≤ S ) is precisely the multiplicity of the semigroup minus one.
For h ∈ H , and inspired by the ν sequence described above, define and set nd As a consequence of the following result, the map nd H is non-decreasing.
In particular, nd The second assertion follows from the first.

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Let S be a numerical semigroup.A (finite) run of elements in S is an interval {s, s + 1, . . ., s + k} of elements of S such that s − 1 ̸ ∈ S and s + k + 1 ̸ ∈ S. Analogously, a run of gaps of S, or desert, is an interval {h, h + 1, . . ., h + l } of gaps of S such that h − 1 ∈ S and k + l + 1 ∈ S. Let C(S) be the conductor of S, that is, the least integer c such that c + N ⊆ S. The numerical semigroup S can be expressed as ) for all i , that is, all the elements in S i are smaller than those in S i +1 and there is at least a gap of S between these two sets.If S ̸ = N, then S 0 = {0}.Theorem 2. Let S be a numerical semigroup, S ̸ = N.

Corollary 5. Let S and T be numerical semigroups. If the posets (G(S), ≤ S ) and (G(T ), ≤ T ) are isomorphic, then S = T .
Let PF(S) = Maximals ≤ S (Z \ S), which is known as the set of pseudo-Frobenius numbers of S. The cardinality of PF(S) is the type of S, denoted t(S).The Frobenius number of S, defined as F(S) = max(Z \ S), is always a pseudo-Frobenius number, and so the type of a numerical semigroup is a positive integer.Clearly, C(S) = F(S) + 1.
Notice that if we consider the Hasse diagram of (G(S), ≤ S ) as an undirected graph, then this graph has at most t(S) connected components.The type of S is three and the undirected graph has two connected components.
Recall that an affine semigroup is a finitely generated submonoid of (N n , +) for some positive integer n.The poset of the set of gaps does not uniquely determine an affine semigroup as the following example shows.

THE POSET OF NORMALIZED IDEALS OF A NUMERICAL SEMIGROUP UNDER INCLUSION
Let S be a numerical semigroup.Recall that the set of normalized ideals of S is For I ∈ I 0 (S), set I * = I \ {0}; in particular, S * = S \ {0}.Lemma 8. Let S be a numerical semigroup and let g and g ′ be gaps of S.Then, g ≤ S g ′ if and only if Proof.Notice that {0, g ′ } + S ⊆ {0, g } + S if and only if g ′ ∈ {0, g } + S, or equivalently, g ′ = g + s for some s ∈ S, and this means that g ≤ S g ′ .
In light of Lemma 8, the poset (P 0 (S), ⊇) is isomorphic to (H , ≤ S ).Thus, if we find a way to recover the set P 0 (S) from (I 0 (S), ⊆), we will be able to recover S from (I 0 (S), ⊆) by using Remark 3.
Observe that for I ∈ I 0 (S), we have that I = Minimals ≤ S (I ) + S, and that every X ⊆ N , for which

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Given two ideals I and J in I 0 (S) we say that I covers J if J ⊊ I and there is no other K ∈ I 0 (S) such that J ⊊ K ⊊ I .Lemma 10.Let S be a numerical semigroup and let I , J ∈ I 0 (S), with J ⊊ I .Then, I covers J if and Proof.Suppose that there is K ∈ I 0 (S) such that J ⊊ K ⊊ I .Take x ∈ I \ K and y ∈ K \ J .Then, x ̸ = y and x, y ∈ I \ J , which forces |I \ J | ≥ 2.

ISOMORPHIC IDEAL CLASS MONOIDS
In this section, we prove that if S and T are numerical semigroups, then the existence of an isomorphism between (I 0 (S), +) and (I 0 (T ), +) forces S and T to be equal.
We start by proving that some notable elements of the ideal class monoid of a numerical semigroup are preserved under isomorphisms.To this end, we recall some definitions given in [5, Section 5]; for the definitions on a general monoid, please refer to [13].
Given I , J ∈ I 0 (S), we write I ⪯ J if there exists K ∈ I 0 (S) such that I + K = J .We use the notation I ≺ J when I ⪯ J and I ̸ = J (in general this is not the usual definition, though in [5, Section 5] it is shown that in our setting the usual definition is equivalent to this one).
We say that I ∈ I 0 (S) is irreducible if I ̸ = J +K for all J , K ∈ I 0 (S)\{S} such that J ≺ I and K ≺ I .By [5,Lemma 5.4], I is irreducible if and only if I ̸ = J + K for all J , K ∈ I 0 (S) \ {I }.Irreducible elements are important since they generate (I 0 (S), +) as a monoid [5,Proposition 5.5].Clearly, if f : I 0 (S) → I 0 (T ) is a monoid isomorphism, then it sends irreducible elements to irreducible elements.
An ideal I ∈ I 0 (S) is a quark if there is no ideal J ∈ I 0 (S) \ {S} such that J ≺ I , that is, there is no J ∈ I 0 (S) \ {I , S} and K ∈ I 0 (S) such that I = J + K .Every quark is irreducible but the converse does not hold in general (see for instance [5,Example 5.3]).Again, quarks go to quarks under monoid isomorphisms of ideal class monoids.
The concepts of over-semigroup and irreducible numerical semigroup are crucial in the proof of the main result of this section.So, next we spend some time recalling the basic facts associated to these notions.
Let S and T be numerical semigroups.We say that T is an over-semigroup of S if S ⊆ T .By [5, Proposition 5.14], T is an idempotent of I 0 (S) if and only if T is an over-semigroup of S. Denote by O (S) the set of over-semigroups of S.Then, (O (S), +) is a submonoid of (I 0 (S), +).
If T is an over-semigroup of S with |T \ S| = 1, then we say that T is a unitary extension of S. In this setting, T = S ∪ {x}, and x must be a special gap of S, that is x ∈ FP(S) and 2x, 3x ∈ S. The set of special gaps of S is denoted by SG(S) and its cardinality coincides with the set of unitary extensions of S (see for instance [12,Section 3.3

]).
A numerical semigroup S is irreducible if S cannot be expressed as the intersection of two numerical semigroups properly containing S. Every irreducible numerical semigroup is either symmetric or pseudo-symmetric.A numerical semigroup S is symmetric if for every z ∈ Z \ S, the integer F(S) − z is in S.And it is pseudo-symmetric if F(S) is even and for every z ∈ Z \ (S ∪ {F(S)/2}), we have that F(S) − z ∈ S. If S is not irreducible, then it can be expressed as the intersection of finitely many irreducible over-semigroups of S (for basic characterizations of irreducibility, symmetry and pseudo-symmetry, see [2,Chapter 2] or [12,Chapter 3]).
It is well known that S is irreducible if and only if the cardinality of SG(S) is at most one [12,Corollary 4.38].If S ̸ = N, then F(S) ∈ SG(S).Thus, for S ̸ = N, if S is irreducible, then S ∪ {F(S)} is the only unitary-extension of S, and thus every proper over-semigroup of S contains S ∪ {F(S)}.
Quarks are relevant since they can be used to decide if the semigroup is symmetric or pseudosymmetric, and thus to determine if the semigroup is irreducible; see Propositions 5.17 and 5.19, and Theorem 5.20 in [5].Unitary extensions of a numerical semigroup S are precisely the idempotent quarks of I 0 (S) [5,Propostion 5.13].
For every idempotent E ∈ I 0 (S), define

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Let S be a numerical semigroup.Observe that if f is an isomorphism between (I 0 (S), +) and (I 0 (T ), +), then from Proposition 13 (and its proof ), we obtain the following consequences.Unfortunately, from the poset (O (S), ⊆) it is not possible to recover S as the next example shows.As usual, for a set A of non-negative integers, we denote by which is a submonoid of (N, +), and it is a numerical semigroup if and only if gcd(A) = 1 (see for instance [12,Lemma 2.1]).Example 14.The numerical semigroups 〈3, 5, 7〉 and 〈2, 7〉 have isomorphic posets of over-semigroups with respect to set inclusion.
Notice that if E is an idempotent of I 0 (S), then (C E , +) is a monoid, but it is not a submonoid of (I 0 (S), +) unless E = S.There is a dual construction that allows us to construct submonoids of (I 0 (S), +) associated to its idempotents.Let T be an over-semigroup of S.Then, T ∈ I 0 (S) and T is idempotent.Define T ↓ = {I ∈ I 0 (S) : I ⊆ T }.
Proposition 15.Let S be a numerical semigroup, and let T be an over-semigroup of S. For every I ∈ I 0 (S), I ∈ T ↓ if and only if I + T = T .In particular, (T ↓ , +) is a submonoid of (I 0 (S), +).We are now ready to prove that if (I 0 (S), +) is isomorphic to (I 0 (T ), +), with S and T numerical semigroups, then S = T .To this end, we proceed by induction on the genus of S (which must be the same as the genus of T by [5,Corollary 5.2]).Once we know (I 0 (S), +) all the unitary extensions of S will be uniquely determined by the induction hypothesis.
Lemma 17.Let S be a numerical semigroup, S ̸ = N.If S is irreducible, then the intersection of all its unitary extensions is S ∪ {F(S)}.Otherwise, this intersection is S.
Proof.Recall that as S ̸ = N, we have that F(S) ∈ SG(S).Notice that S is irreducible if and only if | SG(S)| = 1 [12,Corollary 4.38].Also, every unitary extension of S is of the form S ∪ {h} with h ∈ SG(S).Hence, if S is irreducible, then the intersection of all its unitary extensions (it has only one), is S ∪ {F(S)}.If S is not irreducible, then take h ∈ SG(S) \ {F(S)}.Clearly, S = (S ∪ {F(S)}) ∩ (S ∪ {h}), and thus S = g ∈SG(S) (S ∪ {g }).
Proof.Denote by ϕ the isomorphism between I 0 (S) and I 0 (T ).
Notice that S = N if and only if I 0 (S) is trivial.Thus, we may assume that S and T are different from N. Notice also that the number of quarks of I 0 (S) and I 0 (T ) must be the same.Thus, S is irreducible if and only if T is irreducible.Also, by [5, Corollary 5.2], g(S) = g(T ).
We proceed by induction on the genus of S (which is the same as the genus of T ).For g(S) = 0 there is nothing to prove, since in this case S = T = N.So, suppose that the assertion is true for all semigroups having genus g and let us prove it for genus g + 1.
Unitary extensions of S correspond to idempotent quarks in I 0 (S) [5,Proposition 5.13].Thus, for every unitary extension O of S, ϕ(O) is also unitary extension of T , and by Proposition 13, I 0 (O) = C O is isomorphic to C ϕ(O) = I 0 (ϕ(O)).Unitary extensions of S have genus g (the same holds for T ).Thus, by induction hypothesis S and T have the same unitary extensions.possible from the poset with respect to ⪯ to distinguish between {0, 1} + S and {0, 2} + S; the latter being idempotent, while the first is not.Notice that in this case the genus is two, and there are only two numerical semigroups with this genus.The posets of the corresponding set of normalized ideals are different.

Lemma 9 .
holds, contains Minimals ≤ S (I ).Thus, Minimals ≤ S (I ) is a minimal generating system of I and it is included in G(S)∪{0}.Also, Minimals ≤ S (I ) = I \(I +S * ).The elements of Minimals ≤ S (I ) are called the minimal generators of I .Let S be a numerical semigroup and let I ∈ I 0 (S).For x ∈ I * , the set I \ {x} ∈ I 0 (S) if and only if x is a minimal generator of I .Proof.If x ̸ ∈ Minimals ≤ S (I ), then x = g + s, with g ∈ Minimals ≤ S (I ) and s ∈ S * .Hence, x = g + s ̸ ∈ I \ {x}, and consequently I \ {x} is not an ideal of S.If I \ {x}, with x ∈ Minimals ≤ S (I ), is not an ideal (notice that 0 = min(I \ {x})), then there exists y ∈ I \ {x} and s ∈ S, such that y + s ̸ ∈ I \ {x}.But y ∈ I , and thus y + s ∈ I , which forces y ̸ = y + s = x, contradicting that x is minimal in I with respect to ≤ S .

□ Lemma 11 .□Theorem 12 .
Let S be a numerical semigroup and let I ∈ I 0 (S).Then, the number of ideals in I 0 (S) covered by I equals the number of non-zero minimal generators of I .Proof.The proof easily follows from Lemmas 9 and 10.Let S and T be numerical semigroups.If (I 0 (S), ⊆) and (I 0 (T ), ⊆) are isomorphic, then S = T .
This definition is inspired by[10, Section 2].Proposition 13.Let S be a numerical semigroup, and let T be an over-semigroup of S. Then,C T = I 0 (T ).Proof.Let I ∈ C T .Then, min(I ) = 0 and I + T = I , whence I ∈ I 0 (T ).Now, let I ∈ I 0 (T ).Then, min(I ) = 0 and I + T ⊆ I .Hence, I + S ⊆ I + T ⊆ I , and thus I ∈ I 0 (S).As I ⊆ I + T ⊆ I , we get I + T = I , which yields I ∈ C T .

( 1 )
The restriction of f to O (S) is an isomorphism between (O (S), +) and O (T ), +).(2) Also, for O and O ′ over-semigroups of S, O ⊆ O ′ if and only if O + O ′ = O ′ .Thus, we also obtain an isomorphism between the posets (O (S), ⊆) and (O (T ), ⊆).(3)If O is an over-semigroup of S, then f (C O ) = C f (O) .To prove this, take I ∈ f (C O ).Then, there exists J ∈ C O such that I = f (J ).As J + O = J , we deduce that I + f (O) = f (J ) + f (O) = f (J + O) = f (J ) = I , and thus I ∈ C f (O) .For the other inclusion, let J ∈ C f (O) .Then, as f is surjective, there existsI ∈ I 0 (S) such that f (I ) = J .Since J + f (O) = J , we have f (I +O) = f (I )+ f (O) = J + f (O) = J = f (I ), and as f is injective,I + O = I , which means that I ∈ C O and so J = f (I ) ∈ f (C O ).Therefore, the restriction of f to I 0 (O) is an isomorphism between (I 0 (O), +) and (I 0 ( f (O)), +).
Proof.Let I ∈ I 0 (S).If I + T = T , as I ⊆ I + T , we obtain that I ⊆ T .Now, let I ∈ I 0 (S) with I ⊆ T .Then, T ⊆ I + T ⊆ T + T = T (recall that T is idempotent), and so I + T = T .Finally, take I , J ∈ T ↓ .Then (I + J ) + T = I + (J + T ) = I + T = T , and so I + J ∈ T ↓ .The identity element of (T ↓ , +) is S. □ Take T and T ′ two over-semigroups of S (equivalently, two idempotents of I 0 (S)) with T ⊆ T ′ (equivalently, T + T ′ = T ′ ).Then, (T ′ ↓ ∩C T , +) is a monoid with identity element T .Example 16.Let I 1 = {0, 7} + S and I 2 = {0, 2, 3, 5} + S. Both I 1 and I 2 are idempotents, and thus they are over-semigroups of S; moreover, I 1 ⊆ I 2 .The monoid (I 2↓ ∩ C I 1 , +) has eight elements, and its Hasse diagram with respect to ⪯ has height four (the maximal strictly ascending chain with respect to ⪯ has length four).According to [5, Remark 5,1], if (I 2↓ ∩ C I 1 , +) is isomorphic to I 0 (T ) for some semigroup T , then the genus of T should be three.Among the semigroups of genus three, the only one whose ideal class monoid has cardinality eight is T = 〈4, 5, 6, 7〉.However, I 0 (T ) has only three irreducible elements while (I 2↓ ∩ C I 1 , +) has four irreducible elements.Thus, (I 2↓ ∩ C I 1 , +) is not isomorphic to the ideal class monoid of a numerical semigroup.
and this completes the proof by taking into account that R is an interval of elements in S. Remark 3. Notice that Theorem 2 is telling us that if we know nd H : H → N, then we can fully reconstruct S. Example 4. Assume that H = {g 1 , . . ., g 8 } with nd H