Almost Symmetric Numerical Semigroups Generated
by Four Elements
Takahiro NUMATA
（Accepted November 14, 2012）
ALMOST SYMMETRIC NUMERICAL SEMIGROUPS GENERATED BY FOUR ELEMENTS
TAKAHIRO NUMATA
Abstract. In this paper, we study almost symmetric numerical semigroups gen-erated by 4-elements. Rosales and Garc´ıa-S´anchez [RG2] proved that every al-most symmetric numerical semigroup can be constructed by removing some mini-mal generators from an irreducible numerical semigroup with the same Frobenius number. Using this result, we concretely construct almost symmetric numerical semigroups generated by 4-elements from an irreducible numerical semigroup gen-erated by 2 or 3-elements. In this situation, we prove that we have no almost symmetric numerical semigroups with embedding dimension 4 whose type is more than 3.
1. Introduction
In this paper, letN denote the set of nonnegative integers. A numerical semigroup
H is a subset of N that is closed under addition, contains the zero element and has
finite complement in N. Every numerical semigroup H is finitely generated and has the unique minimal system of generators a1, ..., ar ∈ N; that is H = ⟨a1, ..., ar⟩ =
{λ1a1 +· · · λrar | λ1, ..., λr ∈ N}. The condition #(N \ H) < ∞ is equivalent to
gcd(a1, ..., ar) = 1.
Let H =⟨a1, ..., ar⟩ be a numerical semigroup. We use the following notation:
• we set e(H) = min{a1, ..., ar}, emb(H) = r and call them the multiplicity
and embedding dimension of H, respectively.
• the maximum integer not belonging to H is the Frobenius number of H, denoted by F(H).
• the set PF(H) = {x ∈ Z \ H | x + h ∈ H for any 0 ̸= h ∈ H } is the set of pseudo-Frobenius numbers of H, and its cardinality is the type of H, denoted by t(H).
In general, it always holds emb(H)≤ e(H) and F(H) ∈ PF(H). If emb(H) = e(H), then H is said to have maximal embedding dimension. If H has maximal embedding dimension, then t(H) = emb(H)− 1 (see [RG]).
Let k be a field and t an indeterminate over k. The ring
k[H] := k[ta1, ..., tar]⊂ k[t]
is called the numerical semigroup ring of H = ⟨a1, ..., ar⟩ which naturally inherits a graded ring structure from H. The ring k[H] is a one-dimensional Cohen-Macaulay ring with homogeneous maximal ideal m = (ta1, ..., tar), and which is a natural
homomorphic image of a polynomial ring k[X1, ..., Xr]. The kernel p of this surjection is a binomial ideal, and it will be referred to as the defining ideal of k[H].
A numerical semigroup is irreducible if it cannot be expressed as an intersection of two numerical semigroups properly containing it. It is known that an irreducible numerical semigroup is either symmetric or pseudo-symmetric (see [RG] and Section 2 for definition of symmetric and pseudo-symmetric numerical semigroups).
The goal of this paper is to study almost symmetric numerical semigroups with embedding dimension 4. The notion of the almost symmetric numerical semigroups was introduced by Barucci and Fr¨oberg [BF]. Rosales and Garc´ıa-S´anchez [RG2] proved recently that every almost symmetric numerical semigroup is constructed by removing some minimal generators from an irreducible numerical semigroup with the same Frobenius number. Using this result, we can concretely construct all almost symmetric numerical semigroups. In particular, we want to construct those with embedding dimension 4. Why do we focus on the case of embedding dimension is 4 ? The first reason is that when the embedding dimension is 3, H is almost symmetric if and only if H is symmetric or pseudo-symmetric. In [NNW], the authors studied pseudo-symmetric numerical semigroup with embedding dimension 3. So the first interesting case is the case of the embedding dimension 4. The second reason is that we have a following question.
Question 1.1. If H is almost symmetric with emb(H) = 4, then do we have t(H)≤ 3
always ?
Indeed, we have no examples of almost symmetric numerical semigroups with embedding dimension 4 whose type is more than 3. So we expect that the question above is true.
The main part of this paper is Section 3. We construct almost numerical semi-groups with embedding dimension 4 from an irreducible numerical semigroup with embedding dimension 2 or 3. Then we show that our Question 1.1 is true in those cases.
2. Preliminaries
We recall the definition of symmetric and pseudo-symmetric numerical semi-groups.
Definition 2.1. Let H be a numerical semigroup.
(1) H is symmetric if for every a∈ Z, either a ∈ H or F(H) − a ∈ H.
(2) H is pseudo-symmetric if F(H) is even and for every a ∈ Z, a ̸= F(H)/2, either a∈ H or F(H) − a ∈ H.
By definition, if H is symmetric, then F(H) is odd. It is well known that H is symmetric if and only if k[H] is Gorenstein. Symmetric and pseudo-symmetric numerical semigroups are characterized by pseudo-Frobenius numbers (see [RG]). Proposition 2.2. Let H be a numerical semigroup.
(1) H is symmetric if and only if t(H) = 1.
Next, we introduce almost symmetric numerical semigroups from [BF]. We set L(H) :={x ∈ Z \ H | F(H) − x /∈ H}.
Definition 2.3. [BF] A numerical semigroup H is almost symmetric if L(H) ⊂ PF(H).
Nari [N] proved that almost symmetric numerical semigroups are characterized by symmetry in PF(H). The result is an useful criterion for H is to be almost symmetric.
Theorem 2.4. [N] Let H be a numerical semigroup with t(H) = t. Set PF(H) =
{f1 <· · · < ft= F(H)}. Then H is almost symmetric if and only if fi+ft−i = F(H)
for i = 1, ..., t− 1.
Remark 2.5. For a numerical semigroup H, H is pseudo-symmetric if and only if H is almost symmetric with t(H) = 2 (see [BF]).
It is known that if emb(H) = 3, then t(H)≤ 2 (see [FGH]). Hence, by Proposition 2.2 and Theorem 2.4, when H is not symmetric with emb(H) = 3, H is almost symmetric if and only if H is pseudo-symmetric.
The following proposition is a basic correspondence of their invariants between H and k[H] (see [GW]).
Proposition 2.6. Let H be a numerical semigroup and R = k[H] its semigroup ring.
(1) t(H) = r(R), (2) F(H) = a(R),
where r(R) and a(R) are the Cohen-Macaulay type and a-invariant of R, respectively.
3. Construction of almost symmetric numerical
semigroups with embedding dimension four
The following is the key theorem to achieve our goal.
Theorem 3.1. [RG2] Let T be a numerical semigroup. Then T is almost symmetric if and only if there exists an irreducible numerical semigroup H with F(H) = F(T )
such that T = H\ A, where A is a set of minimal generators of H such that
(∗) A⊂ [F(H)/2, F(H)] and x + y − F(H) /∈ T for any x, y ∈ A.
When this is the case, t(T ) = 2· #(A) + t(H).
Using Theorem 3.1, we can construct all almost symmetric numerical semigroups from irreducible numerical semigroups. Note that an irreducible numerical semi-group is either symmetric or pseudo-symmetric (see [RG]). So we consider symmet-ric and pseudo-symmetsymmet-ric case, respectively.
Remark 3.2. If we do not assume that A ⊂ [F(H)/2, F(H)], then there may be two
irreducible numerical semigroups H which satisfy the condition of Theorem 3.1. For example, let H = ⟨5, 7, 8, 9⟩ and H′ = ⟨6, 7, 8, 9, 10⟩, where F(H) = F(H′) = 11. Although F(H)/2 > 5, T = H \ {5} = H′\ {6} is almost symmetric.
In the following, we always assume that H is an irreducible numerical semigroup and T = H \ A, where A is a set of minimal generators of H which satisfies the condition of Theorem 3.1. In this situation, we say the set A satisfies Condition (∗). 3.1. The case of emb(H) = 2.
Let H = ⟨a, b⟩ be a numerical semigroup and H1 = H \ {b}. Then since
H1 = ⟨a, a + b, 2b, 3b⟩, we have that emb(H1) ≤ 4 and a, a + b are always
mini-mal generators of H1.
Lemma 3.3. Let H = ⟨a, b⟩ be a numerical semigroup and H1 = H \ {b}. Then
emb(H1) = 4 if and only if a≥ 4. In this case, H1 =⟨a, a + b, 2b, 3b⟩.
Proof. It is easily seen that if emb(H1) = 4, then H1 = ⟨a, a + b, 2b, 3b⟩. If H1 =
⟨a, a + b, 2b, 3b⟩, then 2b, 3b /∈ ⟨a⟩ and hence a ≥ 4. Conversely, if a ≥ 4, then
2b, 3b /∈ ⟨a⟩. This implies 2b /∈ ⟨a, a + b, 3b⟩ and 3b /∈ ⟨a, a + b, 2b⟩. Hence H1 =
⟨a, a + b, 2b, 3b⟩ and emb(H1) = 4. □
Proposition 3.4. Let H = ⟨a, b⟩ be a numerical semigroup and A ⊂ {a, b} with #A = 1. If emb(H \ A) = 4 and the set A satisfies Condition (∗), then H = ⟨2, 5⟩
and A ={2}, or H = ⟨3, 4⟩ and A = {3}.
Proof. We may assume that A ={b}. Since F(H) = ab − a − b, we have (a − 3)b < a
from F(H)/2 < b. Since a ≥ 4 by Lemma 3.3, we get a = 5 and b = 2, or a = 4 and b = 3. Hence H = ⟨2, 5⟩ or H = ⟨3, 4⟩. Then A = {b} satisfies Condition (∗),
respectively. □
Next, we consider the case of removing 2-generators from H =⟨a, b⟩.
Proposition 3.5. Let H = ⟨a, b⟩ be a numerical semigroup and A = {a, b}. If the
set A satisfies Condition (∗), then H = ⟨3, 4⟩ and H \ A = ⟨6, 7, 8, 9, 10⟩.
Proof. We may assume that 2 ≤ a < b. Since F(H) = ab − a − b, Condition (∗)
implies that
ab− a − b
2 < a < b < ab− a − b.
Then we have that ab−3a−b < 0 and ab−a−2b > 0. Hence 2(b−3)a < 2b < (b−1)a.
This yields that b≤ 4 and thus (a, b) = (3, 4). □
Theorem 3.6. Almost symmetric numerical semigroups with embedding dimension 4 that are constructed from H =⟨a, b⟩ under Condition (∗) are
⟨4, 5, 6, 7⟩ and ⟨4, 6, 7, 9⟩ .
Proof. By Proposition 3.4 and Proposition 3.5, the pairs of H =⟨a, b⟩ and A ⊂ {a, b}
that satisfy Condition (∗) are H = ⟨2, 5⟩ and A = {2}, or H = ⟨3, 4⟩ and A = {3}. Then H \ A = ⟨4, 5, 6, 7⟩ or ⟨4, 6, 7, 9⟩, which are almost symmetric with t(H) = 3
by Theorem 3.1. □
By Theorem 3.6, we conclude that we cannot construct almost symmetric numer-ical semigroups T with t(T ) > 3 from H =⟨a, b⟩.
3.2. The case of emb(H) = 3.
Next, we consider the case of H = ⟨a, b, c⟩. Then the embedding dimension of
H1 = H\ {b} is at most 6, i.e. H1 =⟨a, c, a + b, b + c, 2b, 3b⟩.
Lemma 3.7. Let H = ⟨a, b, c⟩ be a numerical semigroup and H1 = H \ {b}. If
emb(H1) = 4, then one of the following conditions holds:
(a) 2b∈ ⟨a, c⟩, a + b /∈ ⟨c⟩ and b + c /∈ ⟨a⟩, H1 =⟨a, c, a + b, b + c⟩.
(b) 2b /∈ ⟨a, c⟩, 3b ∈ ⟨a, c⟩, a + b ∈ ⟨c⟩ and b + c /∈ ⟨a⟩, H1 =⟨a, c, b + c, 2b⟩.
(c) 2b /∈ ⟨a, c⟩, 3b ∈ ⟨a, c⟩, a + b /∈ ⟨c⟩ and b + c ∈ ⟨a⟩, H1 = ⟨a, c, a + b, 2b⟩ (If
we change a and c in (b), then we get case (c)).
Proof. It is easy to check that H1 is either one of the above forms since a and c are
always minimal generators of H1. □
First, we consider the symmetric case. The following result is a characterization of symmetric numerical semigroups with embedding dimension 3.
Theorem 3.8. [H],[W] Let H =⟨a, b, c⟩ be a symmetric numerical semigroup. (1) Changing order of a, b and c if necessary, we can write a = a′d, b = b′d where
gcd(a, b) = d > 1 and c ∈ ⟨a′, b′⟩, c ̸= a′, b′. In this case, we denote by
H =⟨d ⟨a′, b′⟩ , c⟩.
(2) If H =⟨d ⟨a′, b′⟩ , c⟩, then F(H) = d(a′b′ − a′− b′) + c(d− 1).
We can classify symmetric numerical semigroups H with emb(H1) = 4 in term of
Theorem 3.8.
Theorem 3.9. Let H = ⟨a, b, c⟩ be a symmetric numerical semigroup and H1 =
H \ {b}. If emb(H1) = 4, then, changing a and c if necessary, H is one of the
following form:
(1) ⟨d ⟨2, b′⟩ , c⟩, where a = 2d, b = b′d and c̸= 2 + b′.
(2) ⟨2 ⟨a′, c′⟩ , b⟩, where a = 2a′ and c = 2c′.
(3) ⟨d ⟨3, b′⟩ , c⟩, where a = 3d, b = b′d and c = 3 + b′.
Proof. By Lemma 3.7, H must satisfy one of condtions (a), (b) or (c).
First, we assume that H satisfies the condition (a). Then one of the following two cases occurs:
(i) If 2b∈ ⟨a⟩, then we see that H = ⟨d ⟨2, b′⟩ , c⟩, where a = 2d, b = b′d. In this situation, it holds that b + c /∈ ⟨a⟩ since otherwise, a, b and c are dividisible by d. Since c ∈ ⟨2, b′⟩, it follows that a + b /∈ ⟨c⟩ if and only if c ̸= 2 + b′. Hence we see that H is as in (1).
(ii) If 2b∈ ⟨a, c⟩ and 2b /∈ ⟨a⟩ , ⟨c⟩, then H = ⟨2 ⟨a′, c′⟩ , b⟩, where a = 2a′, c = 2c′. Then a + b /∈ ⟨c⟩ and b + c /∈ ⟨a⟩. Hence in this case, we see that H is as in (2).
Next, we assume that H satisfies the condition (b) (or (c) if changing a and c). Then we should also consider following two cases:
(i) If 3b ∈ ⟨a⟩, we know that H = ⟨d ⟨3, b′⟩ , c⟩, where a = 3d, b = b′d. In this case, b + c /∈ ⟨a⟩ since otherwise, a, b, c are divisible by d. It is also easily seen that a + b∈ ⟨c⟩ if and only if c = 3 + b′. Hence H is as in (3).
(ii) If 3b ∈ ⟨a, c⟩ and 3b /∈ ⟨a⟩ , ⟨c⟩, we guess that H = ⟨3 ⟨a′, c′⟩ , b⟩, where
a = 3a′, c = 3c′. But in this case, a + b /∈ ⟨c⟩ and b + c /∈ ⟨a⟩ since
otherwise, a, b, c are divisible by 3. So we see that emb(H1) > 4, which is a
contradiction.
Thus, we conclude that H is the one of forms of (1), (2) or (3) if emb(H1) = 4. □
In each case of Theorem 3.9, there are examples that the set A = {b} satisfies Condition (∗), See Example 3.14.
Now, we consider the case of removing 2 or 3 generators from H =⟨a, b, c⟩. Lemma 3.10. Let H =⟨a, b, c⟩ be a symmetric numerical semigroup. Assume that
H =⟨d ⟨a′, b′⟩ , c⟩, where a = a′d and b = b′d. Then
(1) if F(H)/2 < b, then a + b > c(d− 1) if a′ = 2, or a > c(d− 1) if a′ ≥ 3. (2) if F(H)/2 < c, then d = 2.
Proof. (1) Since F(H) = d(a′b′− a′− b′) + c(d− 1) = a′b− a − b + c(d − 1) < 2b, we
get a− c(d − 1) > (a′− 3)b. Hence the assertion follows from this inequality. (2) From F(H) < 2c , we have d F(H′) < (3− d)c, where H′ =⟨a′, b′⟩. So we get
d = 2. □
Lemma 3.11. Let H =⟨a, b, c⟩ be a symmetric numerical semigroup. If F(H)/2 <
x and F(H)/2 < y for some x, y ∈ {a, b, c}, x ̸= y, then H = ⟨4, b, c⟩ = ⟨2 ⟨2, b′⟩ , c⟩,
where b− 4 < c.
Proof. We may assume that H = ⟨d ⟨a′, b′⟩ , c⟩, where a = a′d, b = b′d and a′ < b′.
First, we assume that F(H)/2 < a, b.
(i) If a′ ≥ 3 and b′ ≥ 3, then c < a/(d − 1) = and c < b/(d − 1) by Lemma 3.10. Since a = a′d, b = b′d and d > 1, we get the following inequalities;
(3.1) c < a d− 1 = d d− 1a ′ ≤ 2a′, c < b d− 1 = d d− 1b ′ ≤ 2b′.
Then we see that c = a′+ b′ since c∈ ⟨a′, b′⟩. But it is a contradiction since we get b′ < a′ and a′ < b′ by (3.1).
(ii) If a′ = 2, then F(H) = d(b′− 2) − c + dc. Since F(H)/2 < a, we get
(3.2) c < 3a− b
d− 1 =
d
d− 1· (6 − b
′)≤ 2(6 − b′).
So we have b′ = 3 since c > 3, and hence c = 4 or 5.
If c = 4, then d = 2 or 3 by (3.2). When d = 2, it is a contradiction since
c = 4. If d = 3, then we see that H = ⟨3 ⟨2, 3⟩ , 4⟩ = ⟨6, 9, 4⟩. Note that we
can also write as H =⟨4, 6, 9⟩ = ⟨2 ⟨2, 3⟩ , 9⟩.
If c = 5, then we get d = 2 by (3.2). Hence we see that H =⟨2 ⟨2, 3⟩ , 5⟩ = ⟨4, 6, 5⟩.
Next we assume that F(H)/2 < b and F(H)/2 < c. Then we may also assume that b′ ≥ 3 and we have d = 2 by Lemma 3.10. If a′ > 2, then we get c < 6 since F(H) = 2(a′b′−a′−b′)+c≥ 2(3b′−3−b′)+c. This is a contradiction since c∈ ⟨a′, b′⟩ and a′, b′ > 2. Hence we have a′ = 2. Then b−4 < c since F(H) = 2(b′−2)+c < 2c. So in this case, we see that H =⟨2 ⟨2, b′⟩ , c⟩, where b − 4 < c.
Hence, we conclude that H = ⟨2 ⟨2, b′⟩ , c⟩, where b − 4 < c if F(H)/2 < x and
F(H)/2 < y for some x, y ∈ {a, b, c}, x ̸= y. □
Lemma 3.12. Let H =⟨a, b, c⟩ be a symmetric numerical semigroup. If F(H)/2 < z
for any z ∈ {a, b, c}, then H = ⟨4, 5, 6⟩.
Proof. We may assume that H = ⟨d ⟨a′, b′⟩ , c⟩, where a = a′d, b = b′d and a′ < b′.
Then we have d = 2 by Lemma 3.10.
(i) If a′ = 2 and b′ ≥ 3, then it follows that b = 6 and c = 5 since F(H) =
b− 4 + c < 2a = 8. Hence H = ⟨4, 6, 5⟩.
(ii) If a′, b′ ≥ 3, then c < a and c < b by Lemma 3.10. Since c ∈ ⟨a′, b′⟩, we write
c = λ1a′ + λ2b′, where λ1, λ2 ≥ 0. So we have that λ2b′ < (2− λ1)a′ and
λ1a′ < (2− λ2)b′. Then it must be λ1 = λ2 = 1 since c ̸= a′, b′. Hence we
get b′ < a′ and a′ < b′, which is a contradiction.
Hence we conclude that H =⟨4, 5, 6⟩. □
Theorem 3.13. Let H = ⟨a, b, c⟩ be a symmetric numerical semigroup and A ⊂ {a, b, c} with #A ≥ 2. If H ̸= ⟨4, 5, 6⟩, then the set A never satisfies Condition (∗). Proof. If #A = 3, then it follows by Lemma 3.12.
If #A = 2, then H = ⟨4, b, c⟩ = ⟨2 ⟨2, b′⟩ , c⟩ by Lemma 3.11. Then we see that F(H)/2 < b, c by the proof of Lemma 3.12. Since F(H) = b + c − 4, it follows that b + c− F(H) = 4 ∈ H \ {b, c}, which implies that A = {b, c} does not satisfy
Condition (∗). □
Theorem 3.13 implies that⟨4, 5, 6⟩ is the only symmetric numerical semigroup with embedding dimension 3 which constructs almost symmetric numerical semigroups by removing 2 or 3 elements under Condition (∗). Thus, the following example (1) shows that if H =⟨a, b, c⟩ is symmetric and emb(H \ A) = 4, then t(H \ A) = 3. Example 3.14. (1) Let H = ⟨4, 5, 6⟩. Then all almost symmetric numerical semigroups constructed from H are H \ {5} = ⟨4, 6, 9, 11⟩, H \ {4} = ⟨5, 6, 8, 9⟩,
H\ {4, 5} = ⟨6, 8, 9, 10, 11, 13⟩ and H \ {4, 5, 6} = ⟨8, 9, 10, 11, 12, 13, 14, 15⟩.
(2) Let H =⟨5 ⟨2, 5⟩ , 8⟩. Then H \ {25} = ⟨8, 10, 33, 35⟩ is almost symmetric and
PF(H\ {25}) = {22, 25, 47}.
(3) Let H = ⟨6, 8, 11⟩ = ⟨2 ⟨3, 4⟩ , 11⟩. Then H \ {11} = ⟨6, 8, 17, 19⟩ is almost symmetric with PF(H \ {11}) = {10, 11, 21}.
Now, we consider the case of H =⟨a, b, c⟩ is pseudo-symmetric. Let H1 = H\{b}.
Then we prove that if emb(H1) = 4, then H1 has maximal embedding dimension.
This means that A ={b} never satisfies Condition (∗). Let us begin by introducing the characterization of pseudo-symmetric numerical semigroups with emb(H) = 3 from [NNW].
Let H = ⟨a, b, c⟩ be a numerical semigroup which is not symmetric and R =
k[H] ∼= k[X, Y, Z]/p be its semigroup ring. Then it is known that the defining ideal
p is generated by 2× 2 maximal minors of the matrix (3.3) ( Xα Yβ Zγ Yβ′ Zγ′ Xα′ )
where α, β, γ, α′, β′ and γ′ are positive integers (see [H]). Since a = dim
kk[H]/(ta),
and likewise for b and c, we get the following equations:
a = βγ + β′γ + β′γ′,
b = γα + γ′α + γ′α′,
c = αβ + α′β + α′β′.
(3.4)
Theorem 3.15. [NNW] Let H = ⟨a, b, c⟩ be a numerical semigroup which is not symmetric. Then H is pseudo-symmetric if and only if
(1) α = β = γ = 1 if β′b > αa, (2) α′ = β′ = γ′ = 1 if β′b < αa,
in (3.3). When the case (1) (resp. (2)) occurs, F(H) = 2α′β′γ′ − 2 (resp. F(H) =
2αβγ− 2).
By Theorem 3.15 and (3.4), we have either (3.5) or (3.6):
a = β′γ′+ β′+ 1 b = γ′α′+ γ′ + 1, c = α′β′ + α′+ 1. (3.5) a = βγ + γ + 1 b = γα + α + 1, c = αβ + β + 1. (3.6)
Lemma 3.16. Let H =⟨a, b, c⟩ be a numerical semigroup. If H is pseudo-symmetric, the following holds:
(1) If 2b∈ ⟨a, c⟩, then b + c ∈ ⟨a⟩ or a + b ∈ ⟨c⟩. (2) If a + b∈ ⟨c⟩, then 2a ∈ ⟨b, c⟩ or 2b ∈ ⟨a, c⟩. (3) If b + c∈ ⟨a⟩, then 2c ∈ ⟨a, b⟩ or 2b ∈ ⟨a, c⟩.
Proof. Since H is pseudo-symmetric, α = β = γ = 1 or α′ = β′ = γ′ = 1 in (3.3) by
Theorem 3.15.
(1) First, assume that α = β = γ = 1. Since 2b ∈ ⟨a, c⟩, it must be β′ = 1 and hence b + c ∈ ⟨a⟩. Next, we assume that α′ = β′ = γ′ = 1. Then we have β = 1 since 2b∈ ⟨a, c⟩. This implies that a + b ∈ ⟨c⟩.
(2) If α = β = γ = 1, then α′ = 1 since a + b ∈ ⟨c⟩, which implies 2a ∈ ⟨b, c⟩. If α′ = β′ = γ′ = 1, then we have β = 1 since a + b∈ ⟨c⟩, and hence 2b ∈ ⟨a, c⟩.
(3) Changing a and c in (2), we get the assertion. □
Remark 3.17. In Lemma 3.16, we need to assume pseudo-symmetric for H. For example, let H = ⟨5, 8, 6⟩ , H′ = ⟨5, 7, 6⟩, both of which are not pseudo-symmetric and do not satisfy (1), (2), respectively.
Theorem 3.18. Let H =⟨a, b, c⟩ be a pseudo-symmetric numerical semigroup and
A⊂ {a, b, c} with #A = 1. If emb(H \ A) = 4, then H \ A has maximal embedding
Proof. We may assume that A = {b}. We use the classification of Lemma 3.7. First, assume that H1 = H \ A is as in the case of (a) in Lemma 3.7. Then
b + c ∈ ⟨a⟩ or a + b ∈ ⟨c⟩ by Lemma 3.16 since 2b ∈ ⟨a, c⟩. Hence emb(H1) < 4,
which is a contradiction.
Next, we consider the case H1 is as in the case (b).
(i) If α = β = γ = 1, we have α′ = 1 since a + b ∈ ⟨c⟩, and β′ = 2 since
2b /∈ ⟨a, c⟩ and 3b ∈ ⟨a, c⟩. Then it necessary holds that 2a ∈ ⟨b, c⟩ by
Lemma 3.16. By (3.5), we get a = 2γ′ + 3, b = 2γ′ + 1 and c = 4. Hence
H1 =⟨4, 2γ′+ 3, 2γ′+ 5, 4γ′+ 2⟩ has maximal embedding dimension.
(ii) If α′ = β′ = γ′ = 1, then β = 2 since 2b /∈ ⟨a, c⟩ and 3b ∈ ⟨a, c⟩. Then
a + b /∈ ⟨c⟩, which contradicts the condition of (b).
Lastly, we consider H1 is as in the case (c).
(i) If α = β = γ = 1, then β′ = 2 from the condition of (c). In this case,
b + c /∈ ⟨a⟩, which contradicts the condition of (c).
(ii) If α′ = β′ = γ′ = 1, we have β = 2 and γ = 1 by the condition of (c) and Lemma 3.16. Then we see that a = 4, b = 2α + 1 and c = 2α + 3 from (3.6). Hence H1 =⟨4, 2α + 3, 2α + 5, 4α + 2⟩ has maximal embedding dimension.
In conclusion, H1 has maximal embedding dimension if emb(H1) = 4. Then
t(H1) = 3 and hence A does not satisfy Condition (∗). □
Next, we consider the case of removing 2 or 3 elements.
Lemma 3.19. Let H = ⟨a, b, c⟩ be a pseudo-symmetric numerical semigroup and
A ⊂ {a, b, c} with #A = 2. If F(H)/2 < x and F(H)/2 < y for x, y ∈ A, x ̸= y,
then H is either ⟨ 3,f 2 + 3, f + 3 ⟩ , or ⟨ 4,f 2 + 2, f 2 + 4 ⟩ , where F(H) = f .
Proof. By Theorem 3.15, we may assume that α′ = β′ = γ′ = 1. Then F(H) =
2αβγ− 2. We may also assume that F(H)/2 < a and F(H)/2 < b by considering of
changing order of a, b, c. Then we get the following inequalities by (3.6). αβγ < βγ + γ + 2,
αβγ < γα + α + 2. (3.7)
The pairs (α, β, γ) which satisfy (3.7) are, changing order of α, β, γ if necessary, (1, 1, γ) and (2, 1, γ), where γ is any positive integer. Then we see that H is the
above form by (3.6). □
Lemma 3.20. Let H = ⟨a, b, c⟩ be a pseudo-symmetric numerical semigroup and
A ={a, b, c}. If F(H)/2 < z for any z ∈ A, then H = ⟨3, 4, 5⟩, ⟨3, 5, 7⟩, or ⟨4, 5, 7⟩.
Proof. We may assume that α′ = β′ = γ′ = 1. By the proof of Lemma 3.19,
we may also assume that (α, β) = (1, 1) or (2, 1). In both cases, we have γ < 3 from F(H)/2 < c and (3.6). Then the pairs (α, β, γ) which H is to be a numerical semigroup are (1, 1, 2), (1, 1, 3) and (1, 2, 2). By (3.6), we see that H = ⟨3, 4, 5⟩,
Note that ⟨3, 4, 5⟩, ⟨3, 5, 7⟩ and ⟨4, 5, 7⟩ do not satisfy Condition (∗) since a min-imal generator of those numerical semigroups is more than F(H). Hence we have the following proposition.
Proposition 3.21. Let H = ⟨a, b, c⟩ be a pseudo-symmetric numerical semigroup
and A ={a, b, c}. Then the set A never satisfies Condition (∗).
Now we can prove our main theorem.
Theorem 3.22. Let H =⟨a, b, c⟩ be a pseudo-symmetric numerical semigroup and
A⊂ {a, b, c} with #A ≥ 2. If H ̸= ⟨4, 5, 7⟩, then the set A never satisfies Condition
(∗).
Proof. If #A = 3, then the assertion follows from Proposition 3.21.
If #A = 2, then H =⟨3, f/2 + 3, f + 3⟩ or ⟨4, f/2 + 2, f/2 + 4⟩ by Lemma 3.19. First, assume that H = ⟨3, f/2 + 3, f + 3⟩. We may assume that f ≥ 8 and A = {f/2 + 3, f + 3}. But, then f + 3 > f = F(H), which implies the set A does not satisfy Condition (∗).
Next we assume that H = ⟨4, f/2 + 2, f/2 + 4⟩. Since H ̸= ⟨4, 5, 7⟩, we may also assume that f ≥ 10. Then we guess that A = {f/2 + 2, f/2 + 4}. But,
2b− F(H) = 4α + 2 − 4α + 2 = 4 ∈ H \ A, which implies that A does not satisfy
Condition (∗). □
By Theorem 3.22, we conclude that if H = ⟨a, b, c⟩ is pseudo-symmetric and if the set A satisfies Condition (∗), then H = ⟨4, 5, 7⟩ and A = {4, 5}. Since H \ A = ⟨7, 8, 9, 10, 11, 12, 13⟩ in this case, we cannot construct almost symmetric numerical semigroups whose embedding dimension is 4 from pseudo-symmetric semigroup of embedding dimension 3.
Also, we conclude that all almost symmetric semigroups of embedding dimension 4 obtained from irreducible semigroups of embedding dimension≤ 3 by the method of [RG2] have type ≤ 3.
Acknowledgment. I would like to thank Professor Kei-ichi Watanabe for valuable comments and suggestions. I am also grateful to Professor Pedro A. Garc´ıa-S´anchez for helpful remarks and suggestions and also for sending us the paper [RG2]. Finally, I would like to thank the refree for many valuable comments and, in particular improving the proof of Proposition 3.5.
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Graduate School of Integrated Basic Sciences, Nihon University, Setagaya-ku, Tokyo, 156-0045, JAPAN
