# The multiples of a numerical semigroup

Ignacio Ojeda  Departamento de Matemáticas, Universidad de Extremadura, 06071
Badajoz, Spain<EMAIL_ADDRESS>and  José Carlos Rosales  Departamento de
Álgebra. Universidad de Granada, 18010 Granada, Spain<EMAIL_ADDRESS>
###### Abstract.

Given two numerical semigroups $S$ and $T$ we say that $T$ is a multiple of
$S$ if there exists an integer $d\in\mathbb{N}\setminus\\{0\\}$ such that
$S=\\{x\in\mathbb{N}\mid dx\in T\\}$. In this paper we study the family of
multiples of a (fixed) numerical semigroup. We also address the open problem
of finding numerical semigroups of embedding dimension $e$ without any
quotient of embedding dimension less than $e$, and provide new families with
this property.

###### Key words and phrases:

Numerical semigroup; quotient of numerical semigroup; rooted tree; monoids,
Frobenius number; genus; pseudo-Frobenius numbers, type.

###### 2020 Mathematics Subject Classification:

Primary: 20M14 Secondary: 05C05, 11D07

##  1\. Introduction

Let $\mathbb{Z}$ be the set of integers and $\mathbb{N}=\\{x\in\mathbb{Z}\mid
x\geq 0\\}$. A _submonoid_ of $(\mathbb{N},+)$ is a subset of $\mathbb{N}$
containing $0$ and closed under addition.

If $A$ is a non-empty subset of $\mathbb{N}$, then we write $\langle A\rangle$
for the submonoid of $(\mathbb{N},+)$ generated by $A$, that is,

$\langle A\rangle=\left\\{u_{1}a_{1}+\ldots+u_{n}a_{n}\mid
n\in\mathbb{N}\setminus\\{0\\},\\{a_{1},\ldots,a_{n}\\}\subseteq A\
\text{and}\ \\{u_{1},\ldots,u_{n}\\}\subset\mathbb{N}\right\\}.$

If $M$ is a submonoid of $(\mathbb{N},+)$ and $M=\langle A\rangle$ for some
$A\subseteq\mathbb{N}$, we say that $A$ is a _system of generators_ of $M$.
Moreover, if $M\neq\langle B\rangle$ for every $B\subsetneq A$, then we say
that $A$ is a _minimal system of generators_ of $M$. In [19, Corollary 2.8] it
is shown that every submonoid of $(\mathbb{N},+)$ has a unique minimal system
of generators which is also finite. We write $\operatorname{msg}(M)$ for the
minimal system of generators of $M$. The cardinality of
$\operatorname{msg}(M)$ is called the _embedding dimension_ of $M$ and is
denoted $\operatorname{e}(M)$.

A _numerical semigroup_ , $S$, is a submonoid of $(\mathbb{N},+)$ such that
$\mathbb{N}\setminus S$ is finite. The set of numerical semigroups is denoted
by $\mathscr{S}$. In [19, Lemma 2.1] it is shown that a submonoid $M$ of
$(\mathbb{N},+)$ is a numerical semigroup if and only if
$\gcd(\operatorname{msg}(M))=1$.

If $S$ is a numerical semigroup, then both
$\operatorname{F}(S)=\max(\mathbb{Z}\setminus S)$ and
$\operatorname{g}(S)=\\#(\mathbb{N}\setminus S)$, where $\\#$ stands for
cardinality, are two important invariants of $S$ called _Frobenius number_ and
_genus_ , respectively.

The Frobenius problem consists in finding formulas for the Frobenius number
and the genus of a numerical semigroup in terms of its minimal system of
generators (see [13]). This problem is solved in [21] for numerical semigroups
of embedding dimension equal to two, remaining open, in general, for numerical
semigroups of embedding dimension greater than or equal to three.

Let $T$ be a numerical semigroup. Given $d\in\mathbb{N}\setminus\\{0\\}$, we
write

$\frac{T}{d}=\\{x\in\mathbb{N}\mid d\,x\in T\\}.$

In [16] it is proved that $\frac{T}{d}$ is a numerical semigroup. This
semigroup is called the _quotient of $T$ by $d$_. In this case, we also say
that $T$ is a _$d-$ multiple of $\frac{T}{d}$_.

Correspondingly, given two numerical semigroups $S$ and $T$, we say that $T$
is a multiple of $S$ if there exists $d\in\mathbb{N}\setminus\\{0\\}$ such
that $\frac{T}{d}=S$; in particular, $S$ is an arithmetic extension of $T$
(see [12]).

This paper is devoted to the study of multiples of a numerical semigroup. The
interest of this study is partially motivated by the problem that we present
below. In [17] it is shown that a numerical semigroup is proportionally
modular if and only if it has a multiple of embedding dimension two; this
result together with those in [18] allow us to give an algorithmic procedure
for deciding whether a numerical semigroup has a multiple of embedding
dimension two. In [4] the problem of finding a numerical semigroup that has no
multiples of embedding dimension three arises. In [11] the existence of such
semigroups is proved although no concrete example is given. Recently, in [2,
Theorem 3.1], examples of numerical semigroups that have no multiples of
embedding dimension $k$ are given for (fixed) $k\geq 3$.

Today it is still an open problem to give an algorithmic procedure that
determines whether a given numerical semigroup has a multiple with embedding
dimension greater than or equal to three. This problem is part of the more
ambitious goal of determining

$\min\\{\operatorname{e}(T)\mid T\ \text{is a multiple of}\ S\\}$

for a given numerical semigroup $S$. This number is called _quotient rank of
$S$_ (see [2]). It is clearly bounded above by $\operatorname{e}(S)$, since
$S/1=S$. Consequently, $S$ is said to have _full quotient rank_ when the
quotient rank of $S$ is equal to $\operatorname{e}(S)$.

Let $S$ be a numerical semigroup and $d\in\mathbb{N}\setminus\\{0\\}$. We
define

$\operatorname{M}_{d}(S):=\left\\{T\in\mathscr{S}\mid\frac{T}{d}=S\right\\}$

and write $\max\operatorname{M}_{d}(S)$ for the set of maximal elements of
$\operatorname{M}_{d}(S)$ with respect to the inclusion.

In Section 2, we prove that $\max\operatorname{M}_{d}(S)$ is finite. Moreover,
we prove that there is a surjective map
$\Theta_{S}^{d}:\operatorname{M}_{d}(S)\to\max\operatorname{M}_{d}(S)$. Thus,
to compute $\operatorname{M}_{d}(S)$ it is enough to determine the fibers of
$\Theta_{S}^{d}$ as long as $\max\operatorname{M}_{d}(S)$ is known. The task
of computing $\max\operatorname{M}_{d}(S)$ can be performed following [14],
the details are briefly discussed in Section 2 and a draft code for computing
$\max\operatorname{M}_{d}(S)$ is also provided. We finish this section by
showing that the necessary and sufficient condition for $S$ to be irreducible
is that all elements of $\max\operatorname{M}_{d}(S)$ are irreducible
(independently of $d$, in fact).

In Section 3, we deal with the fibers of the map $\Theta_{S}^{d}$. Although
the number of fibers is finite, their cardinality is not necessarily finite
(see [20]). So, we arrange the elements in each fiber of $\Theta_{S}^{d}$ is
form of a rooted tree with root the corresponding element in
$\max\operatorname{M}_{d}(S)$. This tree is locally finite, that is, the
number of children of a given node is finite. Then, after characterizing the
children of a given node, we can formulate a recursive process that will allow
us to theoretically compute the elements of a fiber of $\Theta_{S}^{d}$ from
the corresponding element in $\max\operatorname{M}_{d}(S)$. We also show that
leaves (childless nodes) and fibers of cardinality one can appear. We
emphasize that our recursive process can be easily truncated so as not to
exceed some given Frobenius number or genus, thus producing a true algorithm.

In Section 4, we see that intersection of finitely many elements in
$\operatorname{M}_{d}(S)$ is an element of $\operatorname{M}_{d}(S)$; this is
not true for infinitely many elements, giving rise the notion of
$\operatorname{M}_{d}(S)-$monoid. We prove that this particular family of
submonoids of $(\mathbb{N},+)$ are precisely those generated by $\\{\langle
X\rangle+d\,S\mid X\subset S,X\ \text{is finite and}\ \langle
X\rangle\cap(d(\mathbb{N}\setminus S))=\varnothing\\};$ moreover, if
$\gcd(X\cup\\{d\\})=1$, then a numerical semigroup is obtained, and vice
versa. Then, given a $\operatorname{M}_{d}(S)-$monoid, $M$, we prove there
exists a unique $X\subset S$ such that $M=\langle X\rangle+d\,S$ and
$M\neq\langle Y\rangle+dS$ for every $Y\subsetneq X$, and we introduce the
notion of the $\operatorname{M}_{d}(S)-$embedding dimension of $M$ as the
cardinality of $X$.

In Section 5, we focus our attention on the elements of
$\operatorname{M}_{d}(S)$ of $\operatorname{M}_{d}(S)-$embedding dimension
one, we prove that these are a kind of generalization of a gluing of $S$ and
$\mathbb{N}$. (see [19, Chapter 8] for more details on gluings). We solve the
Frobenius problem for this family of numerical semigroups, as well as
providing formulas for their pseudo-Frobenius numbers and type.

Finally, in section 6 we give a sufficient condition for a numerical semigroup
to have full quotient rank. We also give new examples of families with this
property.

##  2\. The set $\max\operatorname{M}_{d}(S)$

######  Definition 1.

Let $S$ and $T$ be numerical semigroups and $d\in\mathbb{N}\setminus\\{0\\}$.
We say that $T$ is a _$d-$ multiple of $S$_ if $\frac{T}{d}=S$.

Alternatively, we have the following equivalent and easy-to-prove definition
that one semigroup is a multiple of another.

######  Lemma 2.

Let $S$ and $T$ be two numerical semigroups. Given
$d\in\mathbb{N}\setminus\\{0\\}$, one has that $T$ is $d-$multiple of $S$ if
and only if

$d\,(\mathbb{N}\setminus S)\subseteq\mathbb{N}\setminus
T\subseteq\mathbb{N}\setminus d\,S.$

Specifically, in this case, $\operatorname{F}(T)\geq d\operatorname{F}(S)$.

Let $S$ and $T$ be two numerical semigroups. Given
$d\in\mathbb{N}\setminus\\{0\\}$, we write $\operatorname{M}_{d}(S)$ for the
set of all numerical semigroups $d-$multiples of $S$, which, by Lemma 2, is
equal to

$\left\\{T\in\mathscr{S}\ \mid\ d\,(\mathbb{N}\setminus
S)\subseteq\mathbb{N}\setminus T\subseteq\mathbb{N}\setminus d\,S\right\\}.$

It is known that there are infinitely many elements in the set
$\operatorname{M}_{d}(S)$ (see, e.g., [20]).

######  Notation 3.

Given $S\in\mathscr{S}$ and $d\in\mathbb{N}\setminus\\{0\\}$, we write
$\max\operatorname{M}_{d}(S)$ for the set of maximal elements of
$\operatorname{M}_{d}(S)$ with respect to inclusion.

Let us prove that $\max\operatorname{M}_{d}(S)$ is a non-empty set with
finitely many elements, for every $S\in\mathscr{S}$ and
$d\in\mathbb{N}\setminus\\{0\\}$. But first, we need to see some previous
results on $\operatorname{M}_{d}(S)$.

######  Proposition 4.

Let $S$ be a numerical semigroup and $d\in\mathbb{N}\setminus\\{0\\}$. If
$\\{T,T^{\prime}\\}\subseteq\operatorname{M}_{d}(S),T\subsetneq T^{\prime}$
and $f=\max(T^{\prime}\setminus T)$, then
$T\cup\\{f\\}\in\operatorname{M}_{d}(S)$.

###### Proof.

By [19, Lemma 4.35], we have that $T\cup\\{f\\}$ is a numerical semigroup. So,
it suffices to see that $\frac{T}{d}=S$. Now, since $T\subsetneq
T\cup\\{f\\}\subseteq T^{\prime}$, we have that $\mathbb{N}\setminus
T^{\prime}\subseteq\mathbb{N}\setminus(T\cup\\{f\\})\subseteq\mathbb{N}\setminus
T$. Thus, by Lemma 2, we conclude that

$S=\frac{T}{d}\subseteq\frac{T\cup\\{f\\}}{d}\subseteq\frac{T^{\prime}}{d}=S$

and we are done. ∎

######  Lemma 5.

Let $S\neq\mathbb{N}$ be a numerical semigroup and
$d\in\mathbb{N}\setminus\\{0\\}$. If $T\in\operatorname{M}_{d}(S)$ and
$T\not\in\max\operatorname{M}_{d}(S)$, then there exists

(1) $\theta_{S}^{d}(T):=\max\left\\{x\in\mathbb{N}\setminus T\mid
T\cup\\{x\\}\in\operatorname{M}_{d}(S)\right\\}.$

###### Proof.

Since $T\not\in\max\operatorname{M}_{d}(S),$ there exists
$T^{\prime}\in\operatorname{M}_{d}(S)$ such that $T\subsetneq T^{\prime}$. Let
$f=\max(T^{\prime}\setminus T)$, by Proposition 4, we have that
$f\in\left\\{x\in\mathbb{N}\setminus T\mid
T\cup\\{x\\}\in\operatorname{M}_{d}(S)\right\\}$. So,
$\left\\{x\in\mathbb{N}\setminus T\mid
T\cup\\{x\\}\in\operatorname{M}_{d}(S)\right\\}$ is non-empty and it has a
maximum because it is a subset of $\mathbb{N}\setminus T$ which is a finite
set. ∎

Observe that, by Lemma 2, we have that $\theta_{S}^{d}(T)\not\in d\,S,$ for
every $T\in\operatorname{M}_{d}(S),S\in\mathscr{S}$ and
$d\in\mathbb{N}\setminus\\{0\\}$. In the next section we will discuss the
properties of $\theta_{S}^{d}(-)$ in more detail.

######  Proposition 6.

The set $\max\operatorname{M}_{d}(S)$ is not empty, for every
$S\in\mathscr{S}$ and $d\in\mathbb{N}\setminus\\{0\\}$.

###### Proof.

Let $S\in\mathscr{S}$. If $T\in\operatorname{M}_{d}(S)$, then we define the
following sequence: $T_{0}=T$ and

(2)
$T_{i+1}=\left\\{\begin{array}[]{cc}T_{i}\cup\\{\theta_{S}^{d}(T_{i})\\}&\text{if}\
T_{i}\not\in\max\operatorname{M}_{d}(S);\vskip 3.0pt plus 1.0pt minus 1.0pt\\\
T_{i}&\text{otherwise},\end{array}\right.$

for each $i\in\mathbb{N}$. Then, since $T=T_{0}\subseteq\ldots\subseteq
T_{i}\subseteq\ldots$ and $\mathbb{N}\setminus T$ is finite, we conclude that
there exits $k\in\mathbb{N}$ such that $T_{k}=T_{k+i},$ for every
$i\in\mathbb{N}$. Necessarily, by construction,
$T_{k}\in\max\operatorname{M}_{d}(S)$. ∎

Let us prove now that $\max\operatorname{M}_{d}(S)$ has finite cardinality for
every $S\in\mathscr{S}$ and $d\in\mathbb{N}\setminus\\{0\\}$.

######  Lemma 7.

Let $S$ be a numerical semigroup and $d\in\mathbb{N}\setminus\\{0\\}$. If
$T\in\operatorname{M}_{d}(S)$, then $T\cup\\{d\operatorname{F}(S)+i\mid
i\in\mathbb{N}\setminus\\{0\\}\\}\in\operatorname{M}_{d}(S)$.

###### Proof.

Clearly, $T^{\prime}=T\cup\\{d\operatorname{F}(S)+i\mid
i\in\mathbb{N}\setminus\\{0\\}\\}$ is a numerical semigroup. Now, since
$\operatorname{F}(S)=\max(\mathbb{N}\setminus S)$, we have that
$d\,(\mathbb{N}\setminus
S)\subseteq\mathbb{N}\setminus\\{d\operatorname{F}(S)+i\mid
i\in\mathbb{N}\setminus\\{0\\}\\}$. Therefore, by Lemma 2, we obtain that
$\mathbb{N}\setminus T^{\prime}\subseteq\mathbb{N}\setminus
T\subseteq\mathbb{N}\setminus d\,S$ and that

$\displaystyle\mathbb{N}\setminus T^{\prime}$
$\displaystyle=\mathbb{N}\setminus(T\cup\\{d\operatorname{F}(S)+i\mid
i\in\mathbb{N}\setminus\\{0\\}\\})$ $\displaystyle=(\mathbb{N}\setminus
T)\cap(\mathbb{N}\setminus\\{d\operatorname{F}(S)+i\mid
i\in\mathbb{N}\setminus\\{0\\}\\})\supseteq d\,(\mathbb{N}\setminus S).$

Now, by Lemma 2 again, we conclude that
$T^{\prime}\in\operatorname{M}_{d}(S)$. ∎

######  Proposition 8.

Let $S\in\mathscr{S}\setminus\mathbb{N}$ and $d\in\mathbb{N}\setminus\\{0\\}$.
If $T\in\max\operatorname{M}_{d}(S)$ then
$\operatorname{F}(T)=d\operatorname{F}(S)$. In particular,
$\max\operatorname{M}_{d}(S)$ has finite cardinality.

###### Proof.

If $T\in\max\operatorname{M}_{d}(S)$, by Lemma 7, we have that
$\mathbb{N}\setminus T\subseteq\\{1,\ldots,d\operatorname{F}(S)\\}$. Now
since, by Lemma 2, $\operatorname{F}(T)\geq d\operatorname{F}(S)$, we conclude
both that $\operatorname{F}(T)=d\operatorname{F}(S)$ and that
$\max\operatorname{M}_{d}(S)$ is finite. ∎

We note that the converse of the first statement in the above proposition is
not true.

######  Example 9.

Let $S=\langle 3,4,5\rangle$ and $d=3$. If $T=\langle 4,7,9,10\rangle$ and
$T^{\prime}=\langle 4,5,7\rangle$, then one can see that $T\subsetneq
T^{\prime}$, that $\frac{T}{d}=\frac{T^{\prime}}{d}=S$ and that
$\operatorname{F}(T)=\operatorname{F}(T^{\prime})=d\operatorname{F}(S)=6.$
Therefore, $T\not\in\max\operatorname{M}_{d}(S)$ although has the minimum
possible Frobenius number among the elements of $\operatorname{M}_{d}(S)$.

Moreover, one can easily check that $\max\operatorname{M}_{d}(S)=\\{\langle
4,5,7\rangle\\}$ because there are only four numerical semigroups with
Frobenius number equal to $6$:

$\langle 4,5,7\rangle,\langle 4,7,9,10\rangle,\langle 5,7,8,9,11\rangle\
\text{and}\ \langle 7,8,9,10,11,12,13\rangle,$

all of them contained in $\langle 4,5,7\rangle$.

Once we know that $\max\operatorname{M}_{d}(S)$ is a finite set, we can take
advantage of the algorithms in [14] to compute it, simply by noting, by Lemma
2, that

$\max(\operatorname{M}_{d}(S))=\max\\{T\in\mathscr{S}\mid d\,S\subseteq T\
\text{and}\ T\cap(d(\mathbb{N}\setminus S))=\varnothing\\}$

and that the right hand side is nothing but
$\mathscr{M}(d\,S,d(\mathbb{N}\setminus S))$ in the notation of [14].

######  Remark 10.

Since $\max\operatorname{M}_{d}(S)$ is contained in the set of numerical
semigroups with Frobenius number $d\operatorname{F}(S)$, we can use the GAP
([6]) package NumericalSgps ([5]) to compute $\max\operatorname{M}_{d}(S)$ for
given $S$ and $d$. This method is not very efficient since it requires the
computation of all numerical semigroups with Frobenius number
$d\operatorname{F}(S)$.

For example, if we want to compute $\max\operatorname{M}_{d}(S)$ for
$S=\langle 3,5,7\rangle$ and $d=3$ we can use the following GAP code:

       LoadPackage("NumericalSgps");
       S:=NumericalSemigroup(3,5,7);
       d:=Int(3);
       FS:=FrobeniusNumber(S);
       L:=NumericalSemigroupsWithFrobeniusNumber(d*FS);;
       L:=Filtered(L,T->QuotientOfNumericalSemigroup(T,d)=S);;
       M:=[];
       for T in L do
          aux:=Difference(L,[T]);
          control:=Set(aux,R->IsSubsemigroupOfNumericalSemigroup(R,T));
          if control=[false] then Append(M,[T]); fi;
       od;
       M;
       List(M,i->MinimalGenerators(i));

In this way, we obtain that $\max\operatorname{M}_{d}(S)$ is equal to

$\\{\langle 5,8,9,11\rangle,\langle 7,8,9,10,11,13\rangle\\}$

for $S=\langle 3,5,7\rangle$ and $d=5$.

Now, given $S\in\mathscr{S},d\in\mathbb{N}\setminus\\{0\\}$ and
$T\in\operatorname{M}_{d}(S)$, we write $\Theta_{S}^{d}(T)$ for the element of
$\max\operatorname{M}_{d}(S)$ produced by the sequence (2) for $T_{0}=T$, and
define the map

(3)
$\Theta_{S}^{d}:\operatorname{M}_{d}(S)\longrightarrow\max\operatorname{M}_{d}(S);\
T\mapsto\Theta_{S}^{d}(T)$

which is clearly surjective, giving rise the equivalence relation

$T\sim
T^{\prime}\Longleftrightarrow\Theta_{S}^{d}(T)=\Theta_{S}^{d}(T^{\prime})$

on $\operatorname{M}_{d}(S)$, so that
$\max\operatorname{M}_{d}(S)=\operatorname{M}_{d}(S)/\sim$. Therefore, to
compute $\operatorname{M}_{d}(S)$, it is enough to know what the fibers of
$\Theta_{S}^{d}$ are like.

Before dealing with the fibers of $\Theta_{S}^{d}$, let us look at an
interesting result that relates the irreducibility of $S$ to that of the
elements in $\max\operatorname{M}_{d}(S)$.

######  Remark 11.

Recall that a numerical semigroup is _irreducible_ if it cannot be written as
the intersection of two numerical semigroups properly containing it. Recall
also that a numerical semigroup is symmetric (pseudo-symmetric, resp.) if it
is irreducible with odd (even, resp.) Frobenius number (see [19, Chapter 4]
for more details).

######  Proposition 12.

Let $S$ be a numerical semigroup and $d\in\mathbb{N}\setminus\\{0\\}$. Then
$S$ is irreducible if and only if every $T\in\max\operatorname{M}_{d}(S)$ is
irreducible.

###### Proof.

First, suppose that $S$ is irreducible and let
$T\in\max\operatorname{M}_{d}(S)$. By Lemma 2,
$\operatorname{F}(T)=d\operatorname{F}(S)$. If $T$ is not irreducible, then,
by [19, Theorem 4.2], there exists a numerical semigroup $T^{\prime}$ with
Frobenius number $d\operatorname{F}(S)$ strictly containing $T$. Now, since
$\frac{T^{\prime}}{d}$ is a numerical semigroup with Frobenius number
$\operatorname{F}(S)$ containing $S$, by [19, Theorem 4.2] again,
$S=\frac{T^{\prime}}{d}$. Therefore $T\subsetneq
T^{\prime}\in\operatorname{M}_{d}(S)$, a contradiction with the maximality of
$T$.

Conversely, if there exists $T\in\operatorname{M}_{d}(S)$ irreducible with
Frobenius number $d\operatorname{F}(S)$, then $S$ is irreducible. Indeed, if
$x\not\in S$ and $x\neq\operatorname{F}(S)/2$, then $dx\not\in T$ and $dx\neq
d\operatorname{F}(S)/2=\operatorname{F}(T)$; therefore, by [19, Proposition
4.4], $d\operatorname{F}(S)-dx\in T$. Thus, $\operatorname{F}(S)-x\in S$ and,
by [19, Proposition 4.4] again, we conclude that $S$ is irreducible. ∎

This result allows us to update the code given in Observation 10 by replacing
functionNumericalSemigroupsWithFrobeniusNumber by

IrreducibleNumericalSemigroupsWithFrobeniusNumber

when $S$ is irreducible (as it is actually the case in Remark 10), thus
obtaining a faster computing time.

##  3\. The fibers of $\Theta_{S}^{d}$

Throughout this section, $S\neq\mathbb{N}$ and $d$ are a numerical semigroup
and a positive integer, respectively.

Let $\theta_{S}^{d}(-)$ and $\Theta_{S}^{d}(-)$ be defined as in (1) and (3),
respectively. We begin by observing that $\Theta_{S}^{d}$ has a finite number
of fibers of (possibly) infinite cardinality. In fact, since by [20]
$\operatorname{M}_{d}(S)$ has infinite cardinality, there exists at least one
$T\in\max\operatorname{M}_{d}(S)$ such that $(\Theta_{S}^{d})^{-1}(T)$ has
infinite cardinality.

Given $R\in\max\operatorname{M}_{d}(S)$, we arrange the elements in the fiber
of $R,(\Theta_{S}^{d})^{-1}(R)$, in the form of a rooted tree with root $R$
where the vertices are the elements of $(\Theta_{S}^{d})^{-1}(R)$ and there is
an edge joining the vertices $T$ and $T^{\prime}$ if and only if
$T^{\prime}=T\cup\\{\theta_{S}^{d}(T)\\}$. We denote this tree by
$\mathcal{G}^{S}_{d}(R)$, or simply by $\mathcal{G}(R)$ if there is no
possibility of confusion.

The following proposition characterizes the set of children of a vertex in
$\mathcal{G}(R)$ for a given $R\in\max\operatorname{M}_{d}(S)$.

######  Proposition 13.

Let $T\in\operatorname{M}_{d}(S)$. If $R=\Theta_{S}^{d}(T),$ then the set of
children of $T$ in $\mathcal{G}(R)$ is equal to

(4) $\\{T\setminus\\{x\\}\subset\mathbb{N}\mid
x\in\operatorname{msg}(T),x\not\in d\,S\ \text{and}\
\theta_{S}^{d}(T\setminus\\{x\\})=x\\}.$

###### Proof.

First of all, we recall that the necessary and sufficient condition for
$T\setminus\\{x\\}$ to be a numerical semigroup is that
$x\in\operatorname{msg}(T)$ (see, e.g. [19, Exercise 2.1]). So, if
$T^{\prime}$ is a child of $T$ in $\mathcal{G}(R)$, that is,
$T=T^{\prime}\cup\\{\theta_{S}^{d}(T^{\prime})\\}$ for some
$T^{\prime}\in(\Theta_{S}^{d})^{-1}(R)$, then
$T^{\prime}=T\setminus\\{\theta_{S}^{d}(T^{\prime})\\}$; consequently,
$\theta_{S}^{d}(T^{\prime})\in\operatorname{msg}(T)$ and
$\theta_{S}^{d}(T\setminus\\{\theta_{S}^{d}(T^{\prime})\\})=\theta_{S}^{d}(T^{\prime})$.
Moreover, since $T\in\operatorname{M}_{d}(S)$, by Lemma 2, we have that
$\theta_{S}^{d}(T^{\prime})\not\in d\,S$, and we conclude that $T^{\prime}$
belongs to (4). Conversely, if $T^{\prime}$ belongs to (4), by our initial
observation, $T^{\prime}$ is a numerical semigroup; moreover, since
$T^{\prime}\subsetneq T^{\prime}\cup\\{x\\}=T$ and $x\not\in d\,S$, by Lemma
2, we have that $T^{\prime}\in\operatorname{M}_{d}(S)$. Finally, since
$T=T^{\prime}\cup\\{\theta_{S}^{d}(T^{\prime})\\},$ then
$\Theta_{S}^{d}(T)=\Theta_{S}^{d}(T^{\prime})=R$ by construction, and we are
done. ∎

Note that the conditions $x\in\operatorname{msg}(T)$ and $x\not\in d\,S$ in
(4) are equivalent to $T\setminus\\{x\\}\in\operatorname{M}_{d}(S)$, for any
$T\in(\Theta_{S}^{d})^{-1}(R)$ and $R\in\max\operatorname{M}_{d}(S)$. So to
have an effective way to check whether $T\setminus\\{x\\}$ is a child of $T$
in $\mathcal{G}(\Theta_{S}^{d}(T))$, we need a criterion to decide whether
$\theta_{S}^{d}(T\setminus\\{x\\})=x$ or not. To do this, we need a few more
concepts and results.

Following the terminology introduced in [15], an integer number $z$ is a
_pseudo-Frobenius number of $T\in\mathscr{S}$_ if $z\not\in T$ and $z+x\in T$
for every $x\in T\setminus\\{0\\}$. We write $\operatorname{PF}(T)$ for the
set of pseudo-Frobenius numbers of $T\in\mathscr{S}$.

######  Proposition 14.

If $T\in\operatorname{M}_{d}(S)$ and $T\not\in\max\operatorname{M}_{d}(S)$,
then

$\theta_{S}^{d}(T)=\max\\{z\in\operatorname{PF}(T)\mid 2z\in T\ \text{and}\
z\not\in d\,(\mathbb{N}\setminus S)\\}.$

###### Proof.

Set $\theta:=\max\\{z\in\operatorname{PF}(T)\mid 2z\in T\ \text{and}\ z\not\in
d\,(\mathbb{N}\setminus S)\\}$. On the one hand, since
$\theta_{S}^{d}(T)=\max\\{x\in\mathbb{N}\setminus T\mid
T\cup\\{x\\}\in\operatorname{M}_{d}(S)\\}$, we have that
$\theta_{S}^{d}(T)+x\in T,$ for every $x\in T\setminus\\{0\\}$ and that
$2\,\theta_{S}^{d}(T)\in T$; moreover, by Lemma 2, $\theta_{S}^{d}(T)\not\in
d\,(\mathbb{N}\setminus S)$. Therefore, $\max\\{x\in\mathbb{N}\setminus T\mid
T\cup\\{x\\}\in\operatorname{M}_{d}(S)\\}$ is non-empty and
$\theta_{S}^{d}(T)\leq\theta.$ On the other hand, since
$\theta\in\operatorname{PF}(T)$ and $2\theta\in T$, we have that
$T\cup\\{\theta\\}$ is a numerical semigroup; moreover, since
$T\in\operatorname{M}_{d}(S)$ and $\theta\not\in d\,(\mathbb{N}\setminus S)$,
we have that $d\,(\mathbb{N}\setminus
S)\subseteq\mathbb{N}\setminus(T\cup\\{\theta\\})\subseteq\mathbb{N}\setminus
T\subseteq\mathbb{N}\setminus d\,S$. Therefore, by Lemma 2,
$T\cup\\{\theta\\}\in M_{d}(S)$ and, consequently,
$\theta\leq\theta_{S}^{d}(T)$. ∎

Using Proposition 14, we can show that there are numerical semigroups $T$
which are leaves of $\mathcal{G}(\Theta_{S}^{d}(T))$ (that is, that have no
children) for every $d\in\mathbb{N}\setminus\\{0\\}$ and $S=\frac{T}{d}$.

######  Example 15.

Let $T=\langle 6,9,11\rangle$. By direct computation, one can check that
$\operatorname{PF}(T\setminus\\{6\\})=\\{6,9,25\\},\operatorname{PF}(T\setminus\\{9\\})=\\{9,16,25\\}$
and $\operatorname{PF}(T\setminus\\{9\\})=\\{11,14,25\\}$. So, for every
$d\in\mathbb{N}\setminus\\{0\\}$ and $S=\frac{T}{d}$, we have that $T$ is a
leaf of $\mathcal{G}(\Theta_{S}^{d}(T))$ by Proposition 14. Moreover, if
$d=5$, then $S=\langle 3,4\rangle$ and $\max\operatorname{M}_{d}(S)=\\{T\\}$;
therefore, in this case, $(\Theta_{S}^{d})^{-1}(T)=\\{T\\}$.

Let us see that in most cases the maximum of the proposition
$\theta_{S}^{d}(T)$ can be determined immediately.

######  Corollary 16.

If $T\in\operatorname{M}_{d}(S)$ and $\operatorname{F}(T)$ is not divisible by
$d$, then $\theta_{S}^{d}(T)=\operatorname{F}(T)$.

###### Proof.

By Proposition 14, it suffices to observe that
$\operatorname{F}(T)=\max\operatorname{PF}(T)$ and that
$2\operatorname{F}(T)\in T$. ∎

######  Corollary 17.

Let $T\in\operatorname{M}_{d}(S)$ and $x\in\operatorname{msg}(T)$ be such that
$x\not\in d\,S$. If $x>\operatorname{F}(T)$, then
$\theta_{S}^{d}(T\setminus\\{x\\})=x$.

###### Proof.

Since $x>\operatorname{F}(T)$, we have that
$\operatorname{F}(T\setminus\\{x\\})=x$. Moreover, by Lemma 2,
$x>\operatorname{F}(T)\geq d\operatorname{F}(S)$; in particular, $x\not\in
d(\mathbb{N}\setminus S)$. So, since
$x=\operatorname{F}(T\setminus\\{x\\})=\max\operatorname{PF}(T\setminus\\{x\\})$,
we have that $x=\max\\{z\in\operatorname{PF}(T\setminus\\{x\\})\mid 2z\in
T\setminus\\{x\\}\ \text{and}\ z\not\in d(\mathbb{N}\setminus S)\\}$ and, by
Proposition 14, we conclude that $\theta_{S}^{d}(T\setminus\\{x\\})=x$. ∎

The following result characterizes the elements of $\operatorname{M}_{d}(S)$
with minimal Frobenius number (see Lemma 2).

######  Lemma 18.

If $T\in\operatorname{M}_{d}(S)$, then $\operatorname{F}(T)$ is divisible by
$d$ if and only if $\operatorname{F}(T)=d\operatorname{F}(S)$.

###### Proof.

If $\operatorname{F}(T)=d\operatorname{F}(S)$, then it is obvious that
$\operatorname{F}(T)$ is divisible by $d$. Conversely, assume the opposite:
$\operatorname{F}(T)\neq d\operatorname{F}(S)$; in particular, by Proposition
8, $T\not\in\max\operatorname{M}_{d}(S)$. So, by Lemma 5,
$T^{\prime}=T\cup\\{\theta_{S}^{d}(T)\\}$ is a well-defined element of
$\operatorname{M}_{d}(S)$. Therefore, by Proposition 13, there exists
$x\in\operatorname{msg}(T^{\prime})$ with $x\not\in
d\,S,\theta_{S}^{d}(T^{\prime}\setminus\\{x\\})=x$ and
$T^{\prime}\setminus\\{x\\}=T$. If $x<\operatorname{F}(T^{\prime})$, then
$\operatorname{F}(T^{\prime})=\operatorname{F}(T)\neq d\operatorname{F}(S)$.
So, we may replace $T$ by $T^{\prime}$ and repeat the above arguments. Thus,
without loss of generality, we may suppose that
$x>\operatorname{F}(T^{\prime})$. In this case,
$x=F(T^{\prime}\setminus\\{x\\})=F(T)$ is divisible by $d$ by hypothesis, that
is, $x=dy$ for some $y\in\mathbb{N}\setminus\\{0\\}$. Now, since $x\not\in
d\,S$, we have that $y\not\in S$. Therefore, $y\in\frac{T^{\prime}}{d}\neq S$,
a contradiction. ∎

Recall that, even though $\operatorname{F}(T)=d\operatorname{F}(S)$ for every
$T\in\max\operatorname{M}_{d}(S)$, we already know that the condition
$\operatorname{F}(T)=d\operatorname{F}(S),\ T\in\operatorname{M}_{d}(S)$, it
does not imply $T\in\max(\operatorname{M}_{d}(S))$ (see Example 9).

The following result give an easier characterization of the set of children of
$T\in\operatorname{M}_{d}(S)$ in $\mathcal{G}(\Theta_{S}^{d}(T))$, when
$\operatorname{F}(T)$ is not minimal.

######  Proposition 19.

Let $T\in\operatorname{M}_{d}(S)$ and $R=\Theta_{S}^{d}(T)$. If
$\operatorname{F}(T)\neq d\operatorname{F}(S)$, then the set of children of
$T$ in $\mathcal{G}(R)$ is equal to

$\\{T\setminus\\{x\\}\subset\mathbb{N}\mid x\in\operatorname{msg}(T),x\not\in
d\,S\ \text{and}\ x>\operatorname{F}(T)\\}.$

In particular, if $\operatorname{F}(T)\neq d\operatorname{F}(S)$ and
$x<\operatorname{F}(T)$ for every $x\in\operatorname{msg}(T)$, then $T$ is a
leaf of $\mathcal{G}(R)$.

###### Proof.

Let $x\in\operatorname{msg}(T)\setminus d\,S$. On the one hand, if
$x<\operatorname{F}(T),$ then
$\operatorname{F}(T\setminus\\{x\\})=\operatorname{F}(T)\neq
d\operatorname{F}(S)$. Thus, by Lemma 18,
$\operatorname{F}(T\setminus\\{x\\})$ is not divisible by $d$ and, by
Corollary 16,
$\theta_{S}^{d}(T\setminus\\{x\\})=\operatorname{F}(T\setminus\\{x\\})=\operatorname{F}(T)>x$;
in particular, $\theta_{S}^{d}(T\setminus\\{x\\})\neq x$. On the other hand,
if $x>\operatorname{F}(T)$, then, by Corollary 17,
$\theta_{S}^{d}(T\setminus\\{x\\})=x$. Therefore,
$\theta_{S}^{d}(T\setminus\\{x\\})=x$ if and only if $x>\operatorname{F}(T)$.
Now, by Proposition 13, we are done. ∎

The condition $\operatorname{F}(T)\neq d\operatorname{F}(S)$ is necessary in
the proposition above.

######  Example 20.

Let $S=\langle 2,3\rangle$ and $d=11$. If $T=\langle 5,7,8,9\rangle$, then
$\frac{T}{11}=S$. In this case,
$\operatorname{F}(T)=11=11\cdot\operatorname{F}(S)$ and both
$T\setminus\\{8\\}=\langle 5,7,9,13\rangle$ and $T\setminus\\{9\\}=\langle
5,7,8\rangle$ are children of $T$ in $\mathcal{G}(\Theta_{S}^{d}(T))$. So, $T$
is not a leaf of $\mathcal{G}(\Theta_{S}^{d}(T))$.

The following example shows that leaves can actually appear in the case
$\operatorname{F}(T)=d\operatorname{F}(S)$.

######  Example 21.

Let $S=\langle 3,5,7\rangle$ and $d=3$. The numerical semigroup $T=\langle
5,8,9\rangle$ belongs to $\operatorname{M}_{d}(S)$ and has Frobenius number
$12=3\operatorname{F}(S)$. By Proposition 14, we have that $T$ is a leaf of
$\mathcal{G}(\Theta_{S}^{d}(T))$, because
$\operatorname{PF}(T\setminus\\{5\\})=\\{5,6,7,11,12\\},\operatorname{PF}(T\setminus\\{8\\})=\\{4,8,11,12\\}$
and $\operatorname{PF}(T\setminus\\{9\\})=\\{9,11,12\\}$.

##  4\. $\operatorname{M}_{d}(S)-$monoides

Hereafter, $S\neq\mathbb{N}$ and $d$ denote a numerical semigroup and a
positive integer, respectively.

We begin by observing that the obvious equality

$\frac{T\cap T^{\prime}}{d}=\frac{T}{d}\cap\frac{T^{\prime}}{d}$

implies that $\operatorname{M}_{d}(S)$ is closed under finite intersections.
This does not occur for intersection of infinitely many elements of
$\operatorname{M}_{d}(S)$, since the intersection of infinite numerical
semigroups is not a numerical semigroup because the condition of having finite
complement with respect to $\mathbb{N}$ fails; for example,
$T(n)=d\,S\cup\\{n+i\mid i\in\mathbb{N}\\}\in\operatorname{M}_{d}(S)$ for all
integer $n>d\operatorname{F}(S)$, however
$\bigcap_{n>d\operatorname{F}(S)}T(n)=d\,S\not\in\operatorname{M}_{d}(S)$.

Despite of above, the intersection of elements of $\mathscr{S}$ is always a
submonoid of $(\mathbb{N},+)$ and, consequently, the same happens to
$\operatorname{M}_{d}(S)$. We say that a submonoid of $(\mathbb{N},+)$ that
can be written as an intersection of elements of $\operatorname{M}_{d}(S)$ is
a _$\operatorname{M}_{d}(S)-$ monoid_.

######  Definition 22.

A _$\operatorname{M}_{d}(S)-$ set_ is a subset of $\mathbb{N}$ which is
contained in some $T\in\operatorname{M}_{d}(S)$.

Observe that every $\operatorname{M}_{d}(S)-$set $X$ is, in particular, a
subset of $S$.

######  Proposition 23.

If $X\subseteq\mathbb{N}$, then the following conditions are equivalent:

  1. (1)

$X$ is a $\operatorname{M}_{d}(S)-$set.

  2. (2)

$\langle X\rangle\cap d(\mathbb{N}\setminus S)=\varnothing$.

  3. (3)

$(\langle X\rangle+d\,S)\cap d(\mathbb{N}\setminus S)=\varnothing$.

  4. (4)

$\langle X\rangle+d\,S$ is a $\operatorname{M}_{d}(S)-$monoid.

###### Proof.

$(1)\Rightarrow(2)$. If $X$ is $\operatorname{M}_{d}(S)-$set, then exists
$T\in\operatorname{M}_{d}(S)$ such that $X\subseteq T$. Now, since $T\cap
d(\mathbb{N}\setminus S)=\varnothing$ by Lemma 2, we are done.

$(2)\Rightarrow(3)$. Suppose on contrary that $(\langle X\rangle+d\,S)\cap
d(\mathbb{N}\setminus S)\neq\varnothing$, that is, there exists
$h\in\mathbb{N}\setminus S,a\in\langle X\rangle$ and $s\in S$ such that
$a+ds=dh$. In this case, $a=d(h-s)$ with $h-s\in\mathbb{N}\setminus S$ and,
consequently, $\langle X\rangle\cap d(\mathbb{N}\setminus S)\neq\varnothing$.

$(3)\Rightarrow(4)$. By Lemma 2, we have that

$T(n):=(\langle X\rangle+d\,S)\cup\\{n+i\mid
i\in\mathbb{N}\\}\in\operatorname{M}_{d}(S),$

for every integer $n>d\operatorname{F}(S)$. Therefore, $\langle
X\rangle+d\,S=\bigcap_{n>d\operatorname{F}(S)}T(n)$ is, by definition, a
$\operatorname{M}_{d}(S)-$monoid.

$(4)\Rightarrow(1)$. Since $\langle X\rangle+d\,S$ is an intersection of
element of $\operatorname{M}_{d}(S)$, there exists
$T\in\operatorname{M}_{d}(S)$ such that $X\subseteq\langle
X\rangle+d\,S\subseteq T$; thus, $X$ is a $\operatorname{M}_{d}(S)-$set. ∎

######  Corollary 24.

If $X$ is a $\operatorname{M}_{d}(S)-$set, then $\langle X\rangle+d\,S$ is the
smallest $\operatorname{M}_{d}(S)-$monoid containing $X$.

###### Proof.

If $T\in\operatorname{M}_{d}(S)$ and $X\subseteq T$, then $\langle
X\rangle\subseteq T$; moreover, by Lemma 2, we already know that
$d\,S\subseteq T$. Therefore, $\langle X\rangle+d\,S\subseteq T$ for every
$T\in\operatorname{M}_{d}(S)$ containing $X$. Now, since, by Proposition 23,
$\langle X\rangle+d\,S$ is a $\operatorname{M}_{d}(S)-$monoid and it clearly
contains $X$, we conclude that $\langle X\rangle+d\,S$ the smallest
$\operatorname{M}_{d}(S)-$monoid containing $X$. ∎

######  Proposition 25.

If $M$ is a $\operatorname{M}_{d}(S)-$monoid, then there exists a finite
$\operatorname{M}_{d}(S)-$set $X$ such that $M=\langle X\rangle+d\,S$.

###### Proof.

On the one hand, since $M$ is a submonoid of $(\mathbb{N},+)$, it is finitely
generated (see, e.g. [19, Lemma 2.3]); thus, there exists a finite subset $Y$
of $\mathbb{N}$ such that $M=\langle Y\rangle$. On the other hand, since
$M=\bigcap_{i\in I}T_{i}$ for some $\\{T_{i}\mid i\in
I\\}\subseteq\operatorname{M}_{d}(S),$ by Lemma 2, we have that
$d(\mathbb{N}\setminus S)\subset\mathbb{N}\setminus T_{i}$ for every $i\in I$.
Therefore, $Y\subseteq\langle Y\rangle=M=\bigcap_{i\in I}T_{i}$, implies
$Y\cap d(\mathbb{N}\setminus S)=\varnothing$. Then, by Proposition 23,
$X=\\{y\in Y\mid dy\not\in S\\}$ is a finite $\operatorname{M}_{d}(S)-$set
such that $M=\langle X\rangle+d\,S$. ∎

Observe that, given a $\operatorname{M}_{d}(S)-$set $X$, we have that $\langle
X\rangle+d\,S$ is $d-$multiple of $S$ if and only if $\gcd(X\cup\\{d\\})=1$
(see [19, Lemma 2.1]). Therefore, taking into account the previous results,
the following characterization of $\operatorname{M}_{d}(S)$ easily follows.

######  Corollary 26.

The necessary and sufficient condition for $T\in\operatorname{M}_{d}(S)$ if
that there exists a finite subset $X$ of $S$ such that $\langle X\rangle\cap
d(\mathbb{N}\setminus S)=\varnothing,\ \gcd(X\cup\\{d\\})=1$ and $T=\langle
X\rangle+d\,S$.

Let $M$ be a $\operatorname{M}_{d}(S)-$monoid. If $X$ is a
$\operatorname{M}_{d}(S)-$set such that $M=\langle X\rangle+d\,S$ and there is
not proper subset of $X$ with that property, we say that $X$ a is _minimal
$\operatorname{M}_{d}(S)-$system of generators of $M$_.

######  Theorem 27.

If $M$ is a $\operatorname{M}_{d}(S)-$monoid, then
$\operatorname{msg}(M)\cap(\mathbb{N}\setminus d\operatorname{msg}(S))$ is the
(unique) minimal $\operatorname{M}_{d}(S)-$system of generators of $M$.

###### Proof.

Set $X=\operatorname{msg}(M)\cap(\mathbb{N}\setminus d\operatorname{msg}(S))$.
Clearly, by Proposition 23, $X$ is a $\operatorname{M}_{d}(S)-$set because
$\langle X\rangle\cap d(\mathbb{N}\setminus S)\subseteq M\cap
d(\mathbb{N}\setminus S)=\varnothing.$ Moreover, since

$\operatorname{msg}(M)=X\cup(\operatorname{msg}(M)\cap
d\operatorname{msg}(S)),$

we have that $M=\langle X\rangle+\langle\operatorname{msg}(M)\cap
d\operatorname{msg}(S)\rangle$. Now, since $d\,S\subseteq M$ and
$\operatorname{msg}(M)\cap d\operatorname{msg}(S)\subseteq d\,S$, we conclude
that $M=\langle X\rangle+d\,S$.

On other hand, if $Y$ is a minimal $\operatorname{M}_{d}(S)-$system of
generators of $M$, then $M=\langle Y\rangle+d\,S$. In particular, $M$ is
generated as submonoid of $\mathbb{N}$ by $Y\cup d\operatorname{msg}(S)$ and,
consequently, $\operatorname{msg}(M)\subseteq Y\cup d\operatorname{msg}(S)$.
Thus,

$X=\operatorname{msg}(M)\cap(\mathbb{N}\setminus
d\operatorname{msg}(S))\subseteq(Y\cup
d\operatorname{msg}(S))\cap(\mathbb{N}\setminus
d\operatorname{msg}(S))\subseteq Y$

and we are done. ∎

######  Definition 28.

If $M$ is a $\operatorname{M}_{d}(S)-$monoide, its _$\operatorname{M}_{d}(S)-$
embedding dimension_ is defined as the cardinality of its minimal
$\operatorname{M}_{d}(S)-$system of generators.

We have that the $\operatorname{M}_{d}(S)-$embedding dimension is less than or
equal to the (usual) embedding dimension for obvious reasons. On other hand, a
$\operatorname{M}_{d}(S)-$monoid $M$ has $\operatorname{M}_{d}(S)-$embedding
dimension zero if and only if $M=d\,S$. So the smallest and non-trivial case
is $\operatorname{M}_{d}(S)-$embedding dimension one.

######  Example 29.

Let $S=\langle 5,7,9\rangle$. By Proposition 23, it is easy to see that the
set $X=\\{9,10\\}$ is a $\operatorname{M}_{2}(S)-$set. In this case, $\\{9\\}$
is the minimal $\operatorname{M}_{2}(S)-$system of generators of the
$\operatorname{M}_{2}(S)-$monoid $T=\langle X\rangle+d\,S$ which is actually
the numerical semigroup $\langle 9,10,14\rangle$. So, $T$ is $2-$multiple of
$S=\langle 5,7,9\rangle$ of $\operatorname{M}_{2}(S)-$embedding dimension one.

##  5\. The $d-$multiples of $S$ with $\operatorname{M}_{d}(S)-$embedding
dimension one

Similar to the previous sections, in the following $S$ denotes a numerical
semigroup and $d$ a positive integer.

The following result is an immediate consequence of Corollary 26 and Theorem
27.

######  Proposition 30.

A subset $T$ of $\mathbb{N}$ is a $d-$multiple of $S$ with
$\operatorname{M}_{d}(S)-$embedding dimension one if and only if there exists
$x\in S$ with $\gcd(x,d)=1$ such that $T=\langle x\rangle+d\,S$. Moreover, in
this case, $x=\min(T\setminus dS)$.

Recall that a numerical semigroup $T$ is a _gluing of $T_{1}$ and $T_{2}$_ if
$T=\lambda T_{1}+\mu T_{2}$ for some $\lambda\in
T_{1}\setminus\operatorname{msg}(T_{1})$ and $\mu\in
T_{2}\setminus\operatorname{msg}(T_{2})$ with $\gcd(\lambda,\mu)=1$ (see [19,
Chapter 8] for more details).

######  Corollary 31.

Given $T\subseteq\mathbb{N}$, there exists $d\in\mathbb{N}\setminus\\{0\\}$
such that $T$ is a $d-$multiple of $S$ with
$\operatorname{M}_{d}(S)-$embedding dimension one and $\min(T\setminus
dS)\not\in\operatorname{msg}(S)$ if and only if $T$ is a gluing of
$\mathbb{N}$ and $S$.

###### Proof.

Let $x=\min(T\setminus dS)$. If there exists $d\in\mathbb{N}\setminus\\{0\\}$
such that $T$ is a $d-$multiple of $S$ with
$\operatorname{M}_{d}(S)-$embedding dimension one and
$x\not\in\operatorname{msg}(S)$, then, by Proposition 30, $T=\langle
x\rangle+d\,S$ and $\gcd(x,d)=1$. Therefore,
$\operatorname{msg}(T)=\operatorname{msg}(x\,\mathbb{N})\cup\operatorname{msg}(d\,S)$
with $x\in S\setminus\operatorname{msg}(S),d\in\mathbb{N}\setminus\\{0,1\\}$
and $\gcd(x,d)=1$, that is, $T$ is a gluing of $\mathbb{N}$ and $S$.

If $T$ is a gluing of $\mathbb{N}$ and $S$ there exist
$d\in\mathbb{N}\setminus\\{0,1\\}$ and $x\in S\setminus\operatorname{msg}(S)$
with $\gcd(x,d)=1$, such that $T=\langle x\rangle+d\,S$. Thus, by Proposition
30, we are done. ∎

The following result follows from [3, Lemma 1] which generalizes [9, Theorem
2] and will be useful to solve the Frobenius problem for $d-$multiples of $S$
with $\operatorname{M}_{d}(S)-$embedding dimension one in terms of
$\operatorname{F}(S)$.

######  Lemma 32.

Let $\\{a_{1},\ldots,a_{k}\\}$ be a set of positive integers such that
$\gcd(a_{1},\ldots,a_{k})=1$. If $\gcd(a_{2},\ldots,a_{k})=b$, then

$\operatorname{F}(\langle
a_{1},a_{2},\ldots,a_{k}\rangle)=(b-1)a_{1}+b\,\operatorname{F}\left(\left\langle
a_{1},\frac{a_{2}}{b},\ldots,\frac{a_{k}}{b}\right\rangle\right)$

and

$\operatorname{g}(\langle
a_{1},a_{2},\ldots,a_{k}\rangle)=\frac{(b-1)(a_{1}-1)}{2}+b\,\operatorname{g}\left(\left\langle
a_{1},\frac{a_{2}}{b},\ldots,\frac{a_{k}}{b}\right\rangle\right).$

######  Proposition 33.

If $T$ is a $d-$multiple of $S$ with $\operatorname{M}_{d}(S)-$embedding
dimension one, then

$\operatorname{F}(T)=(d-1)\min(T\setminus dS)+d\operatorname{F}(S)$

and

$\operatorname{g}(T)=\frac{(d-1)\min(T\setminus
dS)-1)}{2}+d\operatorname{g}(S).$

###### Proof.

Let $x=\min(T\setminus dS)$. Since $x\in S$, if $\\{a_{2},\ldots,a_{k}\\}$ is
a system of generators of $S$, then $\\{x,a_{2},\ldots,a_{k}\\}$ is a system
of generators of $S$, too. So, by Proposition 30, we have that
$\\{x,da_{2},\ldots,da_{k}\\}$ is a system of generators of $T$, and the
desired result follows directly from the Lemma 26. ∎

Notice that if $T$ is a $d-$multiple of $S$ with
$\operatorname{M}_{d}(S)-$embedding dimension one, then, by Proposition 8,
$T\not\in\max\operatorname{M}_{d}(S)$ whenever $d\neq 1$.

######  Corollary 34.

Let $T$ be a $d-$multiple of $S$ with $\operatorname{M}_{d}(S)-$embedding
dimension one, if $x=\min(T\setminus dS),$ then

$T\cup\theta_{S}^{d}(T)=\langle x,(d-1)x+d\operatorname{F}(S)\rangle+d\,S.$

###### Proof.

By Proposition 33, $\operatorname{F}(T)=(d-1)x+d\operatorname{F}(S)$. Since,
$\gcd(x,d)=1$, then $d$ does not divide $\operatorname{F}$. So, by Corollary
16, $\theta_{S}^{d}(T)=\operatorname{F}(T)$ and we are done. ∎

Let us compute the pseudo-Frobenius of the $d-$multiples of $S$ with
$\operatorname{M}_{d}(S)-$embedding dimension one in terms of $S$. To do this,
it is convenient to recall that given a numerical semigroup $\Delta$, by [19,
Proposition 2.19], we have that

$\operatorname{PF}(\Delta)=\operatorname{maximals}_{\preceq\Delta}(\mathbb{Z}\setminus\Delta),$

where $\preceq_{S}$ is the partial order on $\mathbb{Z}$ such that
$x\preceq_{\Delta}y$ if and only if $y-x\in\Delta$. Recall also that the type
of a numerical semigroup $\Delta$, denoted $\operatorname{t}(\Delta)$, is the
cardinality of $\operatorname{PF}(\Delta)$.

######  Theorem 35.

If $T$ is a $d-$multiple of $S$ with $\operatorname{M}_{d}(S)-$embedding
dimension one, then

$\operatorname{PF}(T)=\\{df+(d-1)\min(T\setminus dS)\mid
f\in\operatorname{PF}(S)\\}.$

In particular, $\operatorname{t}(T)=\operatorname{t}(S)$.

###### Proof.

Let $x=\min(T\setminus dS)$. By Theorem 27 and Proposition 30, if
$\\{a_{1},\ldots,a_{k}\\}$ is a system of generators of $S$, then
$\\{x,da_{1},\ldots,da_{k}\\}$ is a system of generators of $T$ and
$\gcd(d,x)=1$. Moreover, since $dx\in T,$ we have that $x\in S$, so
$\\{x,a_{1},\ldots,a_{k}\\}$ is a system of generators of $S$, too; thus, by
[19, Lemma 2.16],

(5) $\\{y+x\in T\mid y\not\in T\\}=d\\{z+x\in S\mid z\not\in S\\}.$

Now, if $g\in\operatorname{PF}(T)=\operatorname{maximals}_{\preceq
T}(\mathbb{Z}\setminus T)$ then, in particular, $g+x\in T$ and $g\not\in T$.
Therefore, by (5), $g=dz+(d-1)x$, for some $z\not\in S$ with $z+x\in S$. If
$z\not\in\operatorname{PF}(S)=\operatorname{maximals}_{\preceq
S}(\mathbb{Z}\setminus S)$, then there exists $f\in\operatorname{PF}(S)$ such
that $z\prec_{S}f$; in this case, we have that $df+(d-1)x\not\in T$ because
$df\not\in T$ and $\gcd(x,d)=1$. Therefore, $g=dz+(d-1)x\prec_{T}df+(d-1)x$,
in contradiction to the maximality of $g$.

Conversely, if $f\in\operatorname{PF}(S)=\operatorname{maximals}_{\preceq
S}(\mathbb{Z}\setminus S)$, in particular, $f+x\in S$ and $f\not\in S$;
therefore, by (5), $d(f+x)\in T$ and $y:=df+(d-1)x=d(f+x)-x\not\in T$. If
$y\not\in\operatorname{PF}(T)=\operatorname{maximals}_{\preceq
T}(\mathbb{Z}\setminus T)$, there exists $g\in\operatorname{PF}(T)$ such that
$y\prec_{T}g$; in this case, there exists $z\not\in S$ with $z+x\in S$ such
that $g=dz+(d-1)x$. Now, since $d(z-f)=g-y\in T$, we conclude that $z-f\in S$,
that is, $f\prec_{S}z$, in contradiction to the maximality of $f$. ∎

######  Example 36.

Let $S=\langle 5,7,9\rangle$ and $T=\langle 9\rangle+2S$. Since
$\operatorname{PF}(S)=\\{11,13\\}$, then $\operatorname{PF}(T)=\\{31,35\\}$ by
Theorem 35.

Numerical semigroups of type one are called symmetric, these are exactly the
irreducible number semigroups with odd Frobenius number (see Remark 11). The
study of symmetric numerical semigroups has its own interest since they define
arithmetically Gorenstein monomial curves (see [1]).

The following result is an immediate consequence of Theorem 35

######  Corollary 37.

Let $T$ be a $d-$multiple of $S$ with $\operatorname{M}_{d}(S)-$embedding
dimension one. Then $S$ is symmetric if and only if $T$ is symmetric.

##  6\. On numerical semigroups with full quotient rank

Let $S$ be a numerical semigroup. In this section we give a sufficient
condition for $S$ to have full quotient rank, that is, for $S$ to verify the
following equality:

$\min\\{\operatorname{e}(T)\mid T\ \text{is a multiple of}\
S\\}=\operatorname{e}(S).$

To do this, let us first recall the notion of Apéry set of $S$ with respect to
$x\in S\setminus\\{0\\}$.

######  Definition 38.

Let $S$ be a numerical semigroup and let $x$ be one of its nonzero elements.
The _Apéry set of $S$ with respect to $x$_ is

$\operatorname{Ap}(S,x):=\\{y\in S\mid y-x\not\in S\\}.$

######  Proposition 39.

Let $S$ be a numerical semigroup such that
$\operatorname{msg}(S)=\\{a_{1},\ldots,a_{e}\\}$. If

(6) $\sum_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{e}a_{j}\in\operatorname{Ap}(S,a_{i})\ \text{for every}\
i\in\\{1,\ldots,e\\},$

then $S$ has full quotient rank.

###### Proof.

Suppose instead that $S$ does not have full quotient rank. Then, by [2,
Corollary 2.2], there exists $J\subseteq\\{1,\ldots,e\\}$ such that
$\sum_{j\in J}a_{j}=\sum_{j\in\\{1,\ldots,e\\}\setminus J}u_{j}a_{j}$ for some
$u_{j}\in\mathbb{N},\ j\in\\{1,\ldots,e\\}\setminus J,$ not all zero. Let
$i\in\\{1,\ldots,e\\}\setminus J$ be such that $u_{i}\neq 0$. Since

$\displaystyle\sum_{\begin{subarray}{c}j=1\\\ j\neq i\end{subarray}}^{e}a_{j}$
$\displaystyle=\sum_{j\in
J}a_{j}+\sum_{j\in\\{1,\ldots,e\\}\setminus(J\cup\\{i\\})}a_{j}=\sum_{j\in\\{1,\ldots,e\\}\setminus
J}u_{j}a_{j}+\sum_{j\in\\{1,\ldots,e\\}\setminus(J\cup\\{i\\})}a_{j}$
$\displaystyle=u_{i}a_{i}+\sum_{j\in\\{1,\ldots,e\\}\setminus(J\cup\\{i\\})}(u_{j}+1)a_{j},$

we have that

$\sum_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{e}a_{j}-a_{i}=(u_{i}-1)a_{i}+\sum_{j\in\\{1,\ldots,e\\}\setminus(J\cup\\{i\\})}(u_{j}+1)a_{j}\in
S,$

contradicting the hypothesis. ∎

It is worth asking whether there are numerical semigroups that satisfy the
condition (6). The answer is yes, as the following example shows.

######  Example 40.

Let $S=\langle 21,24,25,31\rangle$. Using the function

AperyListOfNumericalSemigroupWRTElement

included in the GAP ([6]) package NumericalSgps ([5]), we can easily check
that

$24+25+31=80\in\operatorname{Ap}(S,21),\quad
21+25+31=77\in\operatorname{Ap}(S,24),$
$21+24+31=76\in\operatorname{Ap}(S,25)\quad\text{and}\quad
21+24+25=70\in\operatorname{Ap}(S,31).$

Therefore, by Proposition 39, we conclude that $S$ has full quotient rank.

Note that the semigroup in the example above is not one of those shown in [2,
Theorem 3.1]. Let us introduce a new family of numerical semigroups with full
quotient rank.

In [8], _numerical semigroups having unique Betti element_ are introduced. By
[8, Theorem 12], we have that a numerical semigroup of embedding dimension
$e\geq 2$ has unique Betti number if and only if there exist relatively prime
integers $c_{1},\ldots,c_{e}$ greater than one such that

$\operatorname{msg}(S)=\left\\{\prod_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{e}c_{j}\mid i\in\\{1,\ldots,e\\}\right\\}.$

######  Corollary 41.

If $S$ is a numerical semigroup with a unique Betti element, then it has full
quotient rank.

###### Proof.

Let $\operatorname{msg}(S)=\\{a_{1},\ldots,a_{e}\\}$. Since $S$ is a numerical
semigroup having a unique Betti element, by [8, Theorem 12], there exist
relatively prime integers $c_{1},\ldots,c_{e}$ greater than one such that
$a_{i}=\prod_{\begin{subarray}{c}j=1\\\ j\neq i\end{subarray}}^{e}c_{j},\
i=1,\ldots,e$.

Let $\Bbbk$ be a field and consider the $\Bbbk-$algebra homomorphism

$\varphi:\Bbbk[x_{1},\ldots,x_{e}]\longrightarrow\Bbbk[S]:=\bigoplus_{s\in
S}\Bbbk\\{\chi^{s}\\}$

such that $\varphi(x_{i})=\chi^{a_{i}},\ i=1,\ldots,e$. By [8, Corollary 10],
a system of generators of $\ker\varphi$ is $\\{x_{i}^{c_{i}}-x_{j}^{c_{j}}\mid
1\leq i<j\leq e\\}$. Now, by [10, Theorem 6], we have that

$\operatorname{Ap}(S,a_{i})=\left\\{\sum_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{e}u_{j}a_{j}\mid 0\leq u_{j}\leq c_{j},\
j\in\\{1,\ldots,e\\}\setminus\\{i\\}\right\\},$

for every $i\in\\{1,\ldots,e\\}$. In particular,
$\sum_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{e}a_{j}\in\operatorname{Ap}(S,a_{i}),$ for every
$i\in\\{1,\ldots,e\\}$ and, by Proposition 39, we are done. ∎

Observe that if $S$ is numerical semigroup with a unique Betti element and
embedding dimension $e$, then it is the gluing of $\mathbb{N}$ and a numerical
semigroup with a unique Betti element and embedding dimension $e-1$ (see
Corollary 31); in particular, they are multiples of
$\operatorname{M}_{d}(S/d)-$embedding dimension one.

######  Remark 42.

Using [7, Proposition 3.10] and the same argument as that applied in the proof
of Corollary 41, one can prove that universally free numerical semigroups
(introduced in [7]) also have full quotient rank.

Finally, we delve into some questions related to Proposition 39. Of course,
the most immediate question is whether the condition (6) is also necessary. So
far we do not have sufficient evidence to formulate a conjecture, since all
examples of numerical semigroups having full rank occur as a by-product of the
results in [2].

Apart from this, an interesting consequence of (6) is the following.

######  Proposition 43.

Let $S$ be a numerical semigroup such that
$\operatorname{msg}(S)=\\{a_{1}<\ldots<a_{e}\\}$. If
$\sum_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{e}a_{j}\in\operatorname{Ap}(S,a_{i})$ for every
$i\in\\{1,\ldots,e\\}$, then $a_{1}\geq 2^{e-1}$.

###### Proof.

Clearly, $\sum_{j=2}^{e}a_{j}\in\operatorname{Ap}(S,a_{1})$ implies
$a_{J}:=\sum_{j\in J}a_{j}\in\operatorname{Ap}(S,a_{1})$ for every non-empty
$J\subseteq\\{2,\ldots,e\\}$. If there exist two different non-empty subsets
$J$ and $J^{\prime}$ of $\\{2,\ldots,e\\}$ with $a_{J}=a_{J^{\prime}}$, then
there exists $i\in J^{\prime}\setminus J$ such that

$\sum_{\begin{subarray}{c}j=1\\\ j\neq
i\end{subarray}}^{e}a_{j}=a_{J}+\sum_{j\in\\{1,\ldots,e\\}\setminus(J\cup\\{i\\})}a_{j}=a_{J^{\prime}}+\sum_{j\in\\{1,\ldots,e\\}\setminus(J\cup\\{i\\})}a_{j}\not\in\operatorname{Ap}(S,a_{i})$

which contradicts the hypothesis. Therefore, since
$0\in\operatorname{Ap}(S,a_{1})$, we conclude that there are at least
$2^{e-1}$ different elements in $\operatorname{Ap}(S,a_{1})$. Now, since the
cardinality of $\operatorname{Ap}(S,a_{1})$ is equal to $a_{1}$ (see, e.g.
[19, Lemma 2.4]), we are done. ∎

Recall that $\min(\operatorname{msg}(S))$ is called _multiplicity of $S$_,
denoted $\operatorname{m}(S)$. By [19, Proposition 2.10],
$\operatorname{e}(S)\leq\operatorname{m}(S)$. Therefore, if
$\operatorname{e}(S)>2$ and $S$ verifies the condition (6), then $S$ does not
have a maximum embedding dimension according to Lemma 43. In [2, Theorem 4.5]
it is shown that numerical semigroups with maximum embedding dimension do not
have full quotient rank. Therefore, if the converse of Proposition 39 is true,
Lemma 43 would provide an alternative proof of this fact.

### Data availability

Data sharing is not applicable to this article, as no datasets were generated
or analyzed during the present work.

### Declarations

The authors are partially supported by Proyecto de Excelencia de la Junta de
Andalucía (ProyExcel_00868). The second author is partially supported by grant
PID2022-138906NB-C21 funded by MCIN/AEI/10.13039/501100011033 and by ERDF “A
way of making Europe”.

The authors have no relevant financial or non-financial interests to disclose.

All authors have contributed equally in the development of this work.

## References

  * [1] V. Barucci, D.E. Dobbs, M. Fontana, Maximality properties in numerical semigorups and applications to one-dimensional analytically irreducible local domain. Mem. Amer. Math. Soc. 125 (1997), no. 598.
  * [2] T. Bogart, C. O’Neill, K. Woods, When is a numerical semigroup a quotient?, Bull. Aust. Math. Soc. 109(1) (2024), 67–76.
  * [3] A. Brauer, J.E. Shockley, On a problem of Frobenius, J. Reine Angew. Math. 211 (1964), 215–220.
  * [4] M. Delgado, P.A. García Sánchez, J.C. Rosales, Numerical semigroups problem list, CIM Bulletin, 33 (2013), 15–26.
  * [5] M. Delgado, P.A. García Sánchez, J. Morais. NumericalSgps, A package for numerical semigroups, Version 1.3.1 (2022). Available at https://gap-packages.github.io/numericalsgps
  * [6] The GAP Group. GAP – Groups, Algorithms, and Programming, Version 4.12.2; 2022, https://www.gap-system.org.
  * [7] I. Gracía Marco, P.A. García Sánchez, I. Ojeda, C. Tatakis, Universally free numerical semigroups. J. Pure Appl. Algebra 228(5) (2024), 107551. https://doi.org/10.1016/j.jpaa.2023.107551
  * [8] P.A. García Sánchez, I. Ojeda, J.C. Rosales, Affine semigroups having a unique Betti element. J. Algebra Appl. 12(3) (2013), 250177, 11 pages.
  * [9] S.M. Johnson, A linear diophantine problem, Canad. J. Math. 12 (1960), 390–398.
  * [10] G. Márquez-Campos, I. Ojeda, J.M. Tornero On the computation of the Apéry set of numerical monoids and affine semigroups. Semigroup Forum 91(1) (2015), 139–158.
  * [11] M.A. Moreno, J. Nicola, E. Pardo, H. Thomas, Numerical semigroups that are not intersections of $d-$squashed semigroups. Canadian Math. Bull. 52 (2009), no. 4, 598–612.
  * [12] I. Ojeda and J.C. Rosales, The arithmetic extensions of a numerical semigroup. Comm. Algebra 48 (2020), no. 9, 3707–3715.
  * [13] J.L. Ramírez Alfonsín. The Diophantine Frobenius problem. Oxford Lecture Ser. Math. Appl., 30. Oxford University Press, Oxford, 2005.
  * [14] A.M. Robles-Pérez, J.C. Rosales, On sets related to integer partitions with quasi-required elements and disallowed elements. Aequat. Math. https://doi.org/10.1007/s00010-023-01005-5
  * [15] J.C. Rosales, M.B. Branco, Numerical semigroups that can be expressed as an intersection of symmetric numerical semigroups, J. Pure Appl. Algebra 171 (2002), 303–314.
  * [16] J.C. Rosales, P.A. García Sánchez, J.I. García-García, J.M. Urbano Blanco, Proportionally modular Diophantine inequalities. J. Number Theory 103 (2003), no. 2, 281–294.
  * [17] J.C. Rosales, J.M. Urbano Blanco, Proportionally modular Diophantine inequalities and full semigroups. Semigroup Forum 72 (2006), no. 3, 362–374.
  * [18] J.C. Rosales, P.A. García Sánchez, J.M. Urbano Blanco, The set of solutions of a proportionally modular Diophantine inequality. J. Number Theory 128 (2008), no. 3, 453–467.
  * [19] J.C. Rosales and P.A. García Sánchez, Numerical semigroups, Developments in Mathematics, Vol. 20, Springer-Verlag, New York, 2009.
  * [20] I. Swanson. Every numerical semigroup is one over $d$ of infinitely many symmetric numerical in Commutative algebra and its applications, 383–386. Walter de Gruyter GmbH & Co. KG, Berlin, 2009
  * [21] J.J. Sylvester, Mathematical questions with their solutions. Educational Times, vol. 41 (1884), p. 21.