WHICH SETS ARE SETS OF LENGTHS IN ALL NUMERICAL MONOIDS?

arXiv:1802.09945v1 [] 26 Feb 2018

ALFRED GEROLDINGER AND WOLFGANG A. SCHMID
Abstract. We explicitly determine those sets of nonnegative integers which occur as sets of lengths in
all numerical monoids.

Dedicated to Jerzy Kaczorowski on the occasion of his 60th birthday.
1. Introduction and Main Result
Numerical monoids have been objects of interest ever since the work of Frobenius. Beyond number
theory, numerical monoids have close connections to various branches in commutative algebra. We
provide two examples. First, numerical semigroup rings and hence numerical monoids play a crucial role
in combinatorial commutative algebra (see [5, 4, 7]). Second, numerical monoids are the simplest cases of
finitely primary monoids which appear as localizations of non-principal orders in number fields at prime
ideals containing the conductor (see [16, Chapter 2.10]). Motivated by all these connections, the study
of the arithmetic of numerical monoids has found wide interest in the literature.
Factorization Theory originated from algebraic number theory before it branched out into various
subfields of algebra (see [16, 11, 3, 2, 8]). The goal is to understand from a qualitative and quantitative
point of view the various phenomena of non-uniqueness of factorizations into atoms (irreducible elements)
that can occur in non-factorial domains and monoids. We refer to Narkiewicz’s monograph [26] for a
presentation from a number theoretic point of view and to recent progress in the quantitative theory due
to Kaczorowski ([25]).
We briefly survey results on the arithmetic of numerical monoids with respect to sets of lengths and
then we discuss the more specific topics of the present paper.
By a numerical monoid we mean an additive submonoid of (N0 , +) whose complement in N0 is finite.
It is not difficult to show that finitely generated (commutative and cancellative) monoids (and numerical
monoids belong to this class) satisfy the main arithmetical finiteness properties studied in factorization
theory. To be able to be a little bit more precise, we fix notation and recall some basic definitions. Let
H be an additively written, commutative, and cancellative monoid. If a = u1 + . . . + uk , where k ∈ N
and u1 , . . . , uk are atoms of H, then k is called a factorization length of a and the set L(a) of all possible
factorization lengths is called the set of lengths of a. If a ∈ H is invertible, then we set L(a) = {0}, and
L(H) = {L(a) | a ∈ H} denotes the system of all sets of lengths. Suppose that H ⊂ (N0 , +) is numerical
and A(H) = {n1 , . . . , nt } is its set of atoms with t ∈ N and 1 ≤ n1 < . . . < nt . Then, clearly, t = 1 if
and only if n1 = 1 if and only if H = N0 . Suppose that t ≥ 2. Obviously, every nonzero element has a
factorization into atoms and max L(a) ≤ a/n1 for all a ∈ H. Furthermore, we have {n1 , n2 } ⊂ L(n1 n2 )
whence
{(N − i)n1 + in2 | i ∈ [0, N ]} ⊂ L(N n1 n2 ) for every N ∈ N .
Therefore, all sets of lengths are finite and there are arbitrarily large sets of lengths. For a finite set
L = {m1 , . . . , mℓ } ⊂ N0 with ℓ ∈ N0 and m1 < . . . < mℓ , we denote by ∆(L) = {mi − mi−1 | i ∈ [2, ℓ]}
2010 Mathematics Subject Classification. 20M13, 20M14.
Key words and phrases. numerical monoids, numerical semigroup algebras, sets of lengths, sets of distances.
This work was supported by the Austrian Science Fund FWF, Project Number P 28864-N35.
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ALFRED GEROLDINGER AND WOLFGANG A. SCHMID

the set of distances of L. The set
∆(H) =

[

∆(L)

L∈L(H)

is the set of distances of the monoid H (also called the delta set of H), and it is easy to see that
min ∆(H) = gcd ∆(H). Now we can formulate a main arithmetical finiteness result (the first statement
follows from [6, Proposition 2.9] and the second statement is a special case of [16, Theorem 4.3.6]).
Theorem A. Let H be a numerical monoid and A(H) = {n1 , . . . , nt } its set of atoms with t ∈ N≥2 and
1 < n1 < . . . < nt .
1. ∆(H) is finite and min ∆(H) = gcd(n2 − n1 , . . . , nt − nt−1 ).
2. There exists some M ∈ N0 such that every set of lengths L ∈ L(H) has the form
L = L′ ⊎ {y, y + d, . . . , y + ℓd} ⊎ L′′ ⊂ y + dZ ,
where y, ℓ ∈ N0 , L′ ⊂ y − [1, M ], L′′ ⊂ y + ℓd + [1, M ], and d = min ∆(H).
The following two questions ensue.
(1) Can the above structural results be improved or do realization theorems show that they are best
possible.
(2) Are there sets of lengths which are characteristic for a given numerical monoid (in the sense that
they do not occur as a sets of lengths in any other numerical monoid) and which sets of lengths
occur in any numerical monoid.
The focus of the present note is on question (2) for numerical monoids. But before considering
numerical monoids we summarize what is known for transfer Krull monoids over finite abelian groups.
They include principal orders in number fields, constitute the best investigated class of monoids in
factorization theory, but their arithmetic differs significantly from that of numerical monoids. Let H be a
transfer Krull monoid over a finite abelian group G. Then the system of sets of lengths of H just depends
on G. Indeed, we have

(1.1)
L(H) = L B(G) ,
where B(G) is the (multiplicatively written) monoid of product-one sequences over G. The monoid H
is half-factorial (i.e., |L| = 1 for all L ∈ L(H)) if and only if |G| < 3. Suppose that |G| ≥ 3. Then
sets of lengths have a well-described structure (slightly more general than the one given in Theorem A.2)
and this description is known to be best possible ([28]). The set of distances ∆(H) is an interval with
min ∆(H) = 1 ([21]) whose maximum is unknown in general ([22]) (this is in contrast to the fact that in
finitely generated Krull monoids any finite set ∆ with min ∆ = gcd ∆ may occur as set of distances [18]).
The standing conjecture is that the system of sets of lengths is characteristic for the group (see [15]
for a survey, and [17, 23, 30, 29] for recent progress). This means that L(H) 6= L(H ′ ) for all Krull
monoids H ′ having prime divisors in all classes and class group G′ not being isomorphic to G (here we
need |G| ≥ 4). If true, this would yield another purely arithmetical characterization of the class group for
this class of monoids. Answering a question of Narkiewicz, Kaczorowski gave the first purely arithmetical
characterization of the class group (see [24]), and we refer to [16, Chapter 7] for further information on
such characterizations.
The question, which sets of nonnegative integers are sets of lengths in all non-half-factorial transfer
Krull monoids is completely answered.
Theorem B. We have
\

 (a) \
(b) 
(c) \
L B(G) ,
L(H) = y + 2k + [0, k] y, k ∈ N0 =
L B(G) =
(1)

(2)

where the intersection
• (1) is taken over all finite abelian groups G with |G| ≥ 3,

(3)

WHICH SETS ARE SETS OF LENGTHS IN ALL NUMERICAL MONOIDS?

3

• (2) is taken over all non-half-factorial transfer Krull monoids H over finite abelian groups, and
• (3) is taken over all finite groups with |G| ≥ 3.
We recall that Equation (a) easily follows from Equation (1.1), Equation (b) is proved in [19, Section
3], and Equation (c) can be found in [27, Proposition 4.1].
Getting back to numerical monoids, first recall that sets of lengths in numerical monoids and sets of
lengths in transfer Krull monoids are distinct. More precisely, if H is any numerical monoid distinct from
N0 and H ′ is any transfer Krull monoid (over any subset of any abelian group), then L(H) 6= L(H ′ ) by
[19, Theorem 5.5].
The standing conjecture on sets of distances of numerical monoids says that every finite set ∆ ⊂ N
with min ∆ = gcd ∆ occurs as the set of distances. However, this is far open and for partial results we
refer to [10]. Since every finite set L ⊂ N≥2 can be realized as a set of lengths in a numerical monoid
([20]), every finite set of positive integers is contained in the set of distances of some numerical monoid.
The maximum of the set of distances is unknown (in terms of the atoms) and this question seems to have
the same complexity as questions after the Frobenius number. For partial results and computational
approaches we refer to [9, 12, 13, 14].
There are numerical monoids containing no characteristic sets of lengths. Indeed, by [1], there are
distinct numerical monoids H1 and H2 such that L(H1 ) = L(H2 ). In our main result we determine all
sets of nonnegative integers which occur as sets of lengths in all numerical monoids. In particular, it
turns out these are only finitely many sets whereas the associated intersection for transfer Krull monoids
is infinite, as can be seen from Theorem B.
Theorem 1.1. We have

\


L(H) = {0}, {1}, {2} ,

where the intersection is taken over all numerical monoids H ( N0 . More precisely, for every t ∈ N≥6
we have
\

L(H) = {0}, {1}, {2} ,
|A(H)|=t

and for every t ∈ [2, 5] we have
\


L(H) = {0}, {1}, {2}, {3} ,

|A(H)|=t

where the intersections are taken over all numerical monoids H with the given properties.
2. Proof of the Main Theorem
Let H be a numerical monoid. Recall that L(0) = {0} by our convention and, by definition, for an
element u ∈ H we have L(u) = {1} if and only if u ∈ A(H). Thus {0} and {1} are elements of each of
the intersections. If A(H) = {n1 , . . . , nt }, where t ∈ N≥2 and 1 < n1 < . . . < nt , then L(2n1 ) = {2}.
Thus {2} is an element of each of the intersections as well.
For m ∈ N≥2 and d ∈ N, let Hm,d be the numerical monoid generated by {1 + (m − 1)d, 1 + md, . . . , 1 +
(2m − 2)d}; note that this is a numerical monoid because gcd(1 + (m − 1)d, . . . , 1 + (2m − 2)d) = 1, and
1 + (2m − 2)d < 2(1 + (m − 1)d) guarantees that each of the generating elements is an atom. By [6,
Theorem 3.9]
∆(Hm,d ) = {d}.
′
Thus, for distinct d and d , we get that L(Hm,d ) ∩ L(Hm,d′ ) cannot contain sets of cardinality greater
than 1, in other words this intersection is a subset of {{k} | k ∈ N0 }. This implies that each of the
intersections in the statement of our result is contained in {{k} | k ∈ N0 }.
To complete the proof of our result, it suffices to establish the following assertions.

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ALFRED GEROLDINGER AND WOLFGANG A. SCHMID

A1. For every m ≥ 2 and for every k ≥ 4, there is a numerical monoid H with |A(H)| = m such that
{k} ∈
/ L(H).
A2. For every m ≥ 6, there is a numerical monoid H with |A(H)| = m such that {3} ∈
/ L(H).
A3. If |A(H)| = 3, then {3} ∈ L(H).
A4. If |A(H)| = 4, then {3} ∈ L(H).
A5. If |A(H)| = 5, then {3} ∈ L(H).
Proof of A1. Let m ≥ 2 and let H be the numerical monoid generated by A = [m, 2m − 1]; note that
A(H) = A. First, we assert that it suffices to show that {4} ∈
/ L(H). Let k ≥ 5, and let a ∈ H with
k ∈ L(a), say, a = a1 +. . .+ak with ai ∈ A(H). Assuming {4} ∈
/ L(H), it follows that a′ = a1 +a2 +a3 +a4
has a factorization a′ = a′1 + · · · + a′l with a′i ∈ A(H) and l 6= 4. Then, a′1 + . . . + a′l + a5 + . . . + ak is a
factorization of lengths l + k − 4 of a, whence L(a) 6= {k}.
Now, let a ∈ H with 4 ∈ L(a). This means that a is in the 4-fold sumset of A, that is a ∈ 4A =
[4m, 8m − 4]. If a ≥ 5m, then a − m ∈ 4A and 4 ∈ L(a − m). Thus 5 ∈ 1 + L(a − m) ⊂ L(a), showing
that L(a) 6= {4}. If a ≤ 5m − 1, then a − (m + 1) ∈ [2m, 4m − 2] = 2A and 2 ∈ L(a − (m + 1)). Thus
3 ∈ 1 + L(a − (m + 1)) ⊂ L(a), and again L(a) 6= {4}.
[Proof of A1]
Proof of A2. Let m ≥ 6 and let H be the numerical monoid generated by
A = {m} ∪ [m + 3, 2m − 1] ∪ {2m + 1, 2m + 2} .
We note that A(H) = A. For the 2-fold, 3-fold, and 4-fold sumsets of A we obtain that
2A = {2m} ∪ [2m + 3, 4m + 4] ,
3A = {3m} ∪ [3m + 3, 6m + 6] ,

and

4A = {4m} ∪ [4m + 3, 8m + 8] .
which implies that 3A ⊂ 2A ∪ 4A. Thus for every a ∈ H with 3 ∈ L(a) it follows that L(a) ∩ {2, 4} =
6
∅.
[Proof of A2]
Proof of A3. Assume to the contrary that there exists a numerical monoid H with three atoms, say
A(H) = {n1 , n2 , n3 } with 1 < n1 < n2 < n3 , such that {3} 6∈ L(H). Since 3 ∈ L(2n1 + n2 ), the
element 2n1 + n2 must have a further factorization length. Since 2n1 + n2 cannot be a multiple of n1 , it
follows that max L(2n1 + n2 ) = 3. Thus, 2 ∈ L(2n1 + n2 ) and it follows that 2n1 + n2 = 2n3 . Similarly,
we infer that 3n1 must have a factorization of length 2. Since 3n1 < 2n1 + n2 = 2n3 , it follows that
3n1 ∈ {2n2 , n2 + n3 }.
Suppose that 3n1 = n2 + n3 . Then, using the just established equalities, n2 − n1 = (2n1 + n2 ) −
3n1 = 2n3 − (n2 + n3 ) = n3 − n2 =: d. Thus n2 = n1 + d and n3 = n1 + 2d which implies that
3n1 = n2 + n3 = 2n1 + 3d whence n1 = 3d. Since gcd(n1 , n2 , n3 ) = 1, it follows that d = 1 whence
(n1 , n2 , n3 ) = (3, 4, 5). However, since L(11) = {3}, we obtain a contradiction.
Suppose that 3n1 = 2n2 . Then n2 − n1 = 2(n3 − n2 ), say n3 − n2 = d. Then n2 = n1 + 2d,
3n1 = 2n1 + 4d whence n1 = 4d, n2 = 6d, and n3 = 7d. Since gcd(n1 , n2 , n3 ) = 1, it follows that d = 1
whence (n1 , n2 , n3 ) = (4, 6, 7). However, since L(15) = {3}, we obtain a contradiction. [Proof of A3]
Proof of A4. Assume to the contrary that there exists a numerical monoid H with four atoms, say
A(H) = {n1 , n2 , n3 , n4 } with 1 < n1 < n2 < n3 < n4 , such that {3} 6∈ L(H). Then as in A3 we obtain
2 ∈ L(3n1 ) and 2 ∈ L(2n1 + n2 ) which implies
(2.1)

2n1 + n2 ≥ 2n3 .

If 2n1 + n3 would have a factorization of length at least four, then
(2.1)

2n1 + n3 ≥ n1 + 3n2 > (2n1 + n2 ) + n2 ≥ 2n3 + n2 ,
a contradiction. Thus 2 ∈ L(2n1 + n3 ) which implies 2n1 + n3 ∈ {n2 + n4 , 2n4 } and hence
(2.2)

2n1 + n3 ≥ n2 + n4 .

WHICH SETS ARE SETS OF LENGTHS IN ALL NUMERICAL MONOIDS?

5

If 2n1 + n4 would have a factorization of length two, then 2n1 + n4 ≤ 2n3 but
(2.1)

2n1 + n4 > 2n1 + n2 ≥ 2n3 ,

a contradiction.

Therefore, 2n1 + n4 has a factorization of length at least four which implies that
2n1 + n4 ≥ n1 + 3n2 = (2n1 + n2 ) + n2 + (n2 − n1 )
(2.1)

(2.3)

≥ 2n3 + n2 + (n2 − n1 ) = (2n1 + n3 ) + (n3 − n1 ) + 2(n2 − n1 )

(2.2)

≥ n2 + n4 + (n3 − n1 ) + 2(n2 − n1 ) .

Consequently, we infer that
2n1 ≥ n2 + (n3 − n1 ) + 2(n2 − n1 ) whence

3n1 ≥ n2 + n3 + 2(n2 − n1 ) > n2 + n3

which implies that 3n1 ∈ {2n3 , n2 + n4 , n3 + n4 , 2n4 }. If 3n1 = 2n3 , then 2n1 + n2 ≥ n3 + n4 whence
2n1 + n3 = 2n4 and if 3n1 ≥ n2 + n4 , then 2n1 + n3 > n2 + n4 whence 2n1 + n3 = 2n4 . Thus in any case
we have 2n1 + n3 = 2n4 and we can improve the last inequality in (2.3) whence
2n1 + n4 ≥ 2n4 + (n3 − n1 ) + 2(n2 − n1 ) .
Therefore, 2n1 ≥ n4 + (n3 − n1 ) + 2(n2 − n1 ) and adding n1 we obtain that 3n1 ≥ n4 + n3 + 2(n2 − n1 ) >
n3 + n4 . This implies that 3n1 = 2n4 , a contradiction to 2 ∈ L(2n1 + n2 ).
[Proof of A4]
Proof of A5. Again, assume to the contrary that there exists a numerical monoid H with five atoms,
say A(H) = {n1 , n2 , n3 , n4 , n5 } with 1 < n1 < n2 < n3 < n4 < n5 , such that {3} 6∈ L(H). Then as in
A4 we obtain that 2 is an element of L(3n1 ), of L(2n1 + n2 ), and of L(2n1 + n3 ). Moreover,
(2.4)

2n1 + n2 ≥ 2n3 and 2n1 + n3 ≥ n2 + n4 .

We proceed to show that 2 ∈ L(2n1 + n4 ). Assume not. Then, L(2n1 + n4 ) contains an element greater
than or equal to 4 and it follows that 2n1 + n4 ≥ n1 + 3n2 . Similarly to A4 we get that, using (2.4),
2n1 + n4 ≥ n1 + 3n2
= (2n1 + n2 ) + n2 + (n2 − n1 )
≥ 2n3 + n2 + (n2 − n1 )
whence n4 ≥ n3 +(n3 −n1 )+2(n2 −n1 ). In combination with 2n1 +n3 ≥ n2 +n4 , that is, n3 ≥ n4 +n2 −2n1 ,
we get that n4 ≥ n4 + (n3 − n1 ) + 3(n2 − n1 ) − n1 . Equivalently, n1 ≥ (n3 − n1 ) + 3(n2 − n1 ) and
5n1 ≥ n3 + 3n2 . This yields 3n1 > n2 + n3 . Moreover, 2n1 + n4 ≥ n1 + 3n2 , means n1 + n4 ≥ 3n2 , and
this implies 2n1 +n2 < 3n2 ≤ n1 +n4 . Thus, 2n1 +n2 ∈ {n2 +n3 , 2n3 }. Yet, since 2n1 +n2 > 3n1 > n2 +n3
this is a contradiction, both 2n1 + n2 and 3n1 would need to equal 2n3 . This contradiction shows that
max L(2n1 + n4 ) < 4, and whence 2 ∈ L(2n1 + n4 ).
We consider the possible factorizations of 2n1 + n4 of length 2. The factorization must not contain
n1 or n4 . Moreover, 2n1 + n4 is strictly greater than 2n1 + n2 ≥ 2n3 and 2n1 + n3 ≥ n2 + n4 . Thus,
2n1 + n4 ∈ {n2 + n5 , n3 + n5 , 2n5 } and we distinguish these three cases.
CASE 1. 2n1 + n4 = n2 + n5 . Since 2n1 + n3 < 2n1 + n4 = n2 + n5 and since by (2.4) we have
2n1 + n3 ≥ n2 + n4 , it follows that 2n1 + n3 ∈ {n2 + n4 , 2n4 }. We distinguish the two cases.
Case 1.1. 2n1 + n3 = n2 + n4 . Since 2n3 ≤ 2n1 + n2 < 2n1 + n3 = n2 + n4 , we get that 2n1 + n2 = 2n3 .
Considering differences we get that n4 − n3 = n5 − n4 and moreover n3 − n2 = (n4 − n3 ) + (n2 − n3 ).
Thus, n4 − n3 = 2(n3 − n2 ). We set d = n3 − n2 . We have that 3n1 ∈ {2n2 , n2 + n3 }. We distinguish the
two cases.
Case 1.1.1. 3n1 = 2n2 . Considering differences we get n2 − n1 = 2(n3 − n2 ) = 2d. Consequently,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 3d, n1 + 5d, n1 + 7d). From 3n1 = 2n2 we infer that n1 = 4d. We
get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (4, 6, 7, 9, 11). Thus, n1 + n3 = n5 , a contradiction.

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ALFRED GEROLDINGER AND WOLFGANG A. SCHMID

Case 1.1.2. 3n1 = n2 + n3 . Considering differences we get that n2 − n1 = n3 − n2 = d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 4d, n1 + 6d). From 3n1 = n2 + n3 we infer that n1 = 3d
whence n1 + n2 = n4 , a contradiction.
Case 1.2. 2n1 + n3 = 2n4 . We get that 2n1 + n2 ∈ {2n3 , n3 + n4 }. We distinguish the two cases.
Case 1.2.1. 2n1 + n2 = 2n3 . Considering differences we get that n3 − n2 = 2(n4 − n3 ). Moreover,
n4 − n3 = (n5 − n4 ) + (n2 − n4 ). Thus, setting d = n4 − n3 we have n3 − n2 = 2d and n5 − n4 = 4d. We
have that 3n1 ∈ {2n2 , n2 + n3 } and distinguish cases.
Case 1.2.1.1. 3n1 = 2n2 . Considering differences we get that n2 − n1 = 2(n3 − n2 ) = 4d. It follows that
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 4d, n1 + 6d, n1 + 7d, n1 + 11d). From 3n1 = 2n2 we infer that n1 = 8d. We
get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (8, 12, 14, 15, 19). We check that L(35) = {3}, a contradiction.
Case 1.2.1.2. 3n1 = n2 + n3 . Considering differences we get that n2 − n1 = n3 − n2 . It follows that
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 4d, n1 + 5d, n1 + 9d). From 3n1 = n2 + n3 we infer that n1 = 6d.
We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (6, 8, 10, 11, 15). We check that L(27) = {3}, a contradiction.
Case 1.2.2. 2n1 + n2 = n3 + n4 . It follows that 3n1 ∈ {2n2 , n2 + n3 , n2 + n4 , 2n3 }. We distinguish cases.
Case 1.2.2.1. 3n1 = 2n2 . Considering differences we get that n3 − n2 = n4 − n3 =: d. Moreover,
n2 − n1 = (n4 − n2 ) + (n3 − n2 ) = 3d and n4 − n1 = n5 − n2 . Thus, (n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 3d, n1 +
4d, n1 + 5d, n1 + 8d). From 3n1 = 2n2 we infer that n1 = 6d. We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) =
(6, 9, 10, 11, 14). We check that L(26) = {3}, a contradiction.
Case 1.2.2.2. 3n1 = n2 + n3 . Considering differences we get that n3 − n2 = n4 − n3 =: d and n2 − n1 =
n4 −n2 = 2d. Moreover, n4 −n1 = n5 −n3 . Thus, (n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 +2d, n1 +3d, n1 +4d, n1 +7d).
From 3n1 = n2 + n3 we infer that n1 = 5d whence n1 + n2 = n5 , a contradiction.
Case 1.2.2.3. 3n1 = n2 +n4 . Considering differences we get that n2 −n1 = n3 −n2 =: d and n3 −n2 = n4 −
n3 = d. Moreover, n4 −n1 = n5 −n4 = 3d. Thus, (n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 +d, n1 +2d, n1 +3d, n1 +6d).
From 3n1 = n2 + n4 we infer that n1 = 4d whence n1 + n3 = n5 , a contradiction.
Case 1.2.2.4. 3n1 = 2n3 . Considering differences we get that n2 − n1 = n4 − n3 =: d and n3 − n2 =
n4 − n3 = d. Moreover, n4 − n1 = n5 − n3 + (n2 − n3 ) and thus n5 − n3 = (n4 − n1 ) + (n3 − n2 ) = 4d.
Thus, (n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 3d, n1 + 6d). From 3n1 = 2n3 we infer that n1 = 4d.
We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (4, 5, 6, 7, 10). Thus, n1 + n3 = n5 , a contradiction.
CASE 2. 2n1 + n4 = n3 + n5 . Since 2n1 + n3 < 2n1 + n4 = n3 + n5 , it follows that 2n1 + n3 ∈
{n2 + n4 , 2n4 , n2 + n5 }. We distinguish the three cases.
Case 2.1. 2n1 + n3 = n2 + n4 . We get that 2n1 + n2 = 2n3 . Considering differences we get that
n4 − n3 = (n3 − n2 ) + (n5 − n4 ) and n3 − n2 = (n4 − n3 ) + (n2 − n3 ). It follows that n4 − n3 = 2(n3 − n2 )
and n5 − n4 = (n4 − n3 ) − (n3 − n2 ). We set d = n3 − n2 to get n4 − n3 = 2d and n5 − n4 = d. We infer
that 3n1 ∈ {2n2 , n2 + n3 } and distinguish the two cases.
Case 2.1.1. 3n1 = 2n2 . Considering differences we get that n2 − n1 = 2(n3 − n2 ) = 2d. It follows that
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 3d, n1 + 5d, n1 + 6d). From 3n1 = 2n2 we infer that n1 = 4d whence
n1 + n2 = n5 , a contradiction.
Case 2.1.2. 3n1 = n2 + n3 . Considering differences we get that n2 − n1 = n3 − n2 = d. It follows that
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 4d, n1 + 5d). From 3n1 = n2 + n3 we infer that n1 = 3d
whence n1 + n2 = n4 , a contradiction.
Case 2.2. 2n1 + n3 = 2n4 . We infer that 2n1 + n2 ∈ {2n3 , n3 + n4 } and distinguish the two cases.
Case 2.2.1. 2n1 + n2 = 2n3 . Considering differences we get that n3 − n2 = 2(n4 − n3 ). Moreover,
n4 − n3 = n5 − n4 + (n3 − n4 ) and thus n5 − n4 = 2(n4 − n3 ). We infer that 3n1 ∈ {2n2 , n2 + n3 , n2 + n4 }
and distinguish three cases. Set d = n4 − n3 .
Case 2.2.1.1. 3n1 = 2n2 . Considering differences we get that n2 − n1 = 2(n3 − n2 ) = 4d. Thus, it follows
that (n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 4d, n1 + 6d, n1 + 7d, n1 + 9d). From 3n1 = 2n2 we infer that n1 = 8d.
We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (8, 12, 14, 15, 17). We check that L(33) = {3}, a contradiction.

WHICH SETS ARE SETS OF LENGTHS IN ALL NUMERICAL MONOIDS?

7

Case 2.2.1.2. 3n1 = n2 + n3 . Considering differences we get that n2 − n1 = n3 − n2 = 2d. Thus, it
follows that (n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 4d, n1 + 5d, n1 + 7d). From 3n1 = n2 + n3 we infer
that n1 = 6d. We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (6, 8, 10, 11, 13). We check that L(25) = {3}, a
contradiction.
Case 2.2.1.3. 3n1 = n2 + n4 . Considering differences we get that n2 − n1 = (n3 − n2 ) + (n3 − n4 ) = d.
Thus, it follows that (n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 3d, n1 + 4d, n1 + 6d). From 3n1 = n2 + n4 we
infer that n1 = 5d whence n1 + n2 = n5 , a contradiction.
Case 2.2.2. 2n1 + n2 = n3 + n4 . Considering differences we get that n3 − n2 = n4 − n3 =: d. Moreover,
n4 − n3 = n5 − n4 + (n3 − n4 ) and thus n5 − n4 = 2(n4 − n3 ) = 2d. We observe that 3n1 ∈ {2n2 , n2 +
n3 , n2 + n4 , 2n3 } and distinguish cases.
Case 2.2.2.1. 3n1 = 2n2 . Considering differences we get that n2 − n1 = n4 − n2 + (n3 − n2 ) = 3d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 3d, n1 + 4d, n1 + 5d, n1 + 7d). From 3n1 = 2n2 we infer that n1 = 6d. We
get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (6, 9, 10, 11, 13). We check that L(25) = {3}, a contradiction.
Case 2.2.2.2. 3n1 = n2 + n3 . Considering differences we get that n2 − n1 = n4 − n2 = 2d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 3d, n1 + 4d, n1 + 6d). From 3n1 = n2 + n3 we infer that n1 = 5d.
We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (5, 7, 8, 9, 11). We check that L(21) = {3}, a contradiction.
Case 2.2.2.3. 3n1 = n2 + n4 . Considering differences we get that n2 − n1 = n3 − n2 = d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 3d, n1 + 5d). From 3n1 = n2 + n4 we infer that n1 = 4d
whence n1 + n2 = n5 , a contradiction.
Case 2.2.2.4. 3n1 = 2n3 . Considering differences we get that n2 − n1 = n4 − n3 = d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 3d, n1 + 5d). From 3n1 = 2n3 we infer that n1 = 4d
whence n1 + n2 = n5 , a contradiction.
Case 2.3. 2n1 + n3 = n2 + n5 . We obtain that 2n1 + n2 ∈ {2n3 , n3 + n4 , 2n4 } and distinguish the three
cases.
Case 2.3.1. 2n1 + n2 = 2n3 . Considering differences we get n4 − n3 = n3 − n2 =: d. Moreover,
n4 − n2 = n5 − n3 = 2d and therefore n5 − n4 = d. We infer that 3n1 ∈ {2n2 , n2 + n3 , n2 + n4 } and
distinguish the three cases.
Case 2.3.1.1. 3n1 = 2n2 . Considering differences we get that n2 − n1 = 2(n3 − n2 ) = 2d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 3d, n1 + 4d, n1 + 5d). From 3n1 = 2n2 we infer that n1 = 4d
whence 2n1 = n4 , a contradiction.
Case 2.3.1.2. 3n1 = n2 + n3 . Considering differences we get that n2 − n1 = n3 − n2 = d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 3d, n1 + 4d). From 3n1 = n2 + n3 we infer that n1 = 3d
whence 2n1 = n4 , a contradiction.
Case 2.3.1.3. 3n1 = n2 + n4 . Considering differences we get that n2 − n1 = n3 − n2 − (n4 − n3 ) = 0, a
contradiction.
Case 2.3.2. 2n1 + n2 = n3 + n4 . Considering differences we get that n4 − n3 = n3 − n2 =: d and
n5 − n4 = n4 − n2 = 2d. We have 3n1 ∈ {2n2 , n2 + n3 , n2 + n4 , 2n3 }. We distinguish the four cases.
Case 2.3.2.1. 3n1 = 2n2 . Considering differences we get that n2 − n1 = n4 − n2 + (n3 − n2 ) = 3d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 3d, n1 + 4d, n1 + 5d, n1 + 7d). From 3n1 = 2n2 we infer that n1 = 6d. We
get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (6, 9, 10, 11, 13). We check that L(25) = {3}, a contradiction.
Case 2.3.2.2. 3n1 = n2 + n3 . Considering differences we get that n2 − n1 = n4 − n2 = 2d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 3d, n1 + 4d, n1 + 6d). From 3n1 = n2 + n3 we infer that n1 = 5d.
We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (5, 7, 8, 9, 11). We check that L(21) = {3}, a contradiction.
Case 2.3.2.3. 3n1 = n2 + n4 . Considering differences we get that n2 − n1 = n3 − n2 = d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 3d, n1 + 5d). From 3n1 = n2 + n4 we infer that n1 = 4d
whence n1 + n2 = n5 , a contradiction.
Case 2.3.2.4. 3n1 = 2n3 . Considering differences we get that n2 − n1 = n4 − n3 = d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 3d, n1 + 5d). From 3n1 = 2n3 we infer that n1 = 4d
whence n1 + n2 = n5 , a contradiction.

8

ALFRED GEROLDINGER AND WOLFGANG A. SCHMID

Case 2.3.3. 2n1 + n2 = 2n4 . Considering differences we get that n4 − n3 = n3 − n2 =: d and n3 − n2 =
(n5 − n4 ) + (n2 − n4 ), that is, n5 − n4 = n4 − n2 + (n3 − n2 ) = 3d. We have 3n1 ∈ {2n2 , n2 + n3 , n2 +
n4 , 2n3 , n3 + n4 }. We distinguish the five cases.
Case 2.3.3.1. 3n1 = 2n2 . Considering differences we get that n2 − n1 = 2(n4 − n2 ) = 4d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 4d, n1 + 5d, n1 + 6d, n1 + 9d). From 3n1 = 2n2 we infer that n1 = 8d. We
get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (8, 12, 13, 14, 17). We check that L(31) = {3}, a contradiction.
Case 2.3.3.2. 3n1 = n2 +n3 . Considering differences we get that n2 −n1 = n4 −n2 +(n4 −n3 ) = 3d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 3d, n1 + 4d, n1 + 5d, n1 + 8d). From 3n1 = n2 + n3 we infer that n1 = 7d.
We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (7, 10, 11, 12, 15). We check that L(27) = {3}, a contradiction.
Case 2.3.3.3. 3n1 = n2 + n4 . Considering differences we get that n2 − n1 = n4 − n2 = 2d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 3d, n1 + 4d, n1 + 7d). From 3n1 = n2 + n4 we infer that n1 = 6d.
We get d = 1, and (n1 , n2 , n3 , n4 , n5 ) = (6, 8, 9, 10, 13). We check that L(23) = {3}, a contradiction.
Case 2.3.3.4. 3n1 = 2n3 . Considering differences we get that n2 − n1 = 2(n4 − n3 ) = 2d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + 2d, n1 + 3d, n1 + 4d, n1 + 7d). From 3n1 = 2n3 we infer that n1 = 6d, and
we conclude as in the preceding case.
Case 2.3.3.5. 3n1 = n3 + n4 . Considering differences we get that n2 − n1 = n4 − n3 = d. Thus,
(n1 , n2 , n3 , n4 , n5 ) = (n1 , n1 + d, n1 + 2d, n1 + 3d, n1 + 6d). From 3n1 = n3 + n4 we infer that n1 = 5d
whence n1 + n2 = n5 , a contradiction.
CASE 3. 2n1 + n4 = 2n5 . It follows that 2n1 > n5 . We consider 2n1 + n5 . Since 2n5 < 2n1 + n5 < 4n1 .
The first inequality shows that L(2n1 + n5 ) cannot contain 2, the second shows that L(2n1 + n5 ) cannot
contain 4 or any larger element. Thus, L(2n1 + n5 ) = {3}.
[Proof of A5]
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Institute for Mathematics and Scientific Computing, University of Graz, NAWI Graz, Heinrichstraße 36,
8010 Graz, Austria
E-mail address: alfred.geroldinger@uni-graz.at
URL: http://imsc.uni-graz.at/geroldinger
Université Paris 13, Sorbonne Paris Cité, LAGA, CNRS, UMR 7539, Université Paris 8, F-93430, Villetaneuse,
France, and, Laboratoire Analyse, Géométrie et Applications (LAGA, UMR 7539), COMUE Université Paris
Lumières, Université Paris 8, CNRS, 93526 Saint-Denis cedex, France
E-mail address: schmid@math.univ-paris13.fr

