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The doubles of one half of a numerical semigroup
Journal of Number Theory 163 (2016) 375–384
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Journal of Number Theory www.elsevier.com/locate/jnt
The doubles of one half of a numerical semigroup Ze Gu ∗ , Xilin Tang Department of Mathematics, South China University of Technology, Guangzhou, Guangdong, 510640, PR China
a r t i c l e
i n f o
Article history: Received 22 April 2015 Received in revised form 11 November 2015 Accepted 11 November 2015 Available online 27 January 2016 Communicated by David Goss MSC: 20M14 11D07
a b s t r a c t Let F be the set of all numerical semigroups and d a positive integer. Define a mapping fd : F → F for every numerical semigroup S by fd (S) = Sd . In this paper, we show that the equivalence class (S)kerf d contains only finite maximal elements, but does not contain minimal element for every integer d ≥ 2. Moreover, we study the subset of (S)kerf 2 whose elements contain S and the subset of (S)kerf 2 whose elements are contained in S. In particular, we study the case of S has embedding dimension two. © 2016 Elsevier Inc. All rights reserved.
Keywords: Numerical semigroup Quotient Double Maximal element Minimal element
1. Introduction and preliminaries Let N be the set of nonnegative integers. A numerical semigroup is a subset S of N such that it is closed under addition, 0 ∈ S and N\S is finite. The elements of G(S) = N\S are the gaps of S and its cardinality, denoted by g(S), is called the genus of S. Another * Corresponding author. E-mail addresses: [email protected] (Z. Gu), [email protected] (X. Tang). http://dx.doi.org/10.1016/j.jnt.2015.11.026 0022-314X/© 2016 Elsevier Inc. All rights reserved.
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important invariant of S is the largest integer not belonging to S, known as the Frobenius number of S and denoted by F (S). If A ⊆ N is a nonempty set, we denote by A the submonoid of (N, +) generated by A, that is, A = {λ1 a1 + · · · + λn an | n ∈ N\{0}, a1 , · · · , an ∈ A, λ1 , · · · , λn ∈ N}. It is well known that A is a numerical semigroup if and only if gcd(A) = 1. If S is a numerical semigroup and S = A, then we say that A is a system of generators of S. In addition, if no proper subset of A generates S, then we say that A is a minimal system of generators of S. In [7] it is proved that every numerical semigroup S has a unique minimal system of generators and, moreover, such a system is finite. Denote by M (S) the minimal system of generators of S. Let M (S) = {n1 < n2 < · · · < np }. Then n1 is known as the multiplicity of S, denoted by m(S). The cardinality of M (S), p, is called the embedding dimension of S and is denoted by e(S). In particular, if e(S) = m(S) then S is a numerical semigroup with maximal embedding dimension. A partially ordered set (A, ≤) is a nonempty set A with an order relation ≤. For a, b ∈ A, if a ≤ b or b ≤ a, then a and b are comparable. Otherwise they are incomparable. A partially ordered set (A, ≤) is called a totally ordered set (also called a chain) if any two elements of A are comparable. An element a ∈ A is maximal in A if for every c ∈ A that is comparable to a, c ≤ a. An element b ∈ A is minimal in A if for every c ∈ A that is comparable to b, b ≤ c. Let S be a numerical semigroup and d a positive integer. Then in [7] it is proved that Sd = {x ∈ N | dx ∈ S} is also a numerical semigroup, called the quotient of S by d. The work on the numerical semigroups that are quotients of numerical semigroups of maximal embedding dimension and the numerical semigroups whose quotients are of maximal embedding dimension can be seen in [3,8]. When d = 2 we say that S2 is one half of the numerical semigroup S. Dually, we say that S is a double [4] of the numerical semigroup S2 . Denote by D(S) the set of all doubles of the numerical semigroup S, that is, D(S) = {T numerical semigroup | S = T2 }. In [4], A.M. Robles-Pérez et al. gave a representation for the elements of D(S) by a unique pair (m, H), where m is an odd integer of S and H is an m-upper subset of G(S). M. D’Anna and F. Strazzanti studied the set D(S) by the numerical duplication in [3]. Recently, F. Strazzanti gave a formula for the minimal genus of a multiple of a numerical semigroup S, a formula for the minimal genus of a symmetric double of S and the formula for the Frobenius number of the quotient of some families of numerical semigroups in [9]. Let S and T be two numerical semigroups such that S ⊆ T , and let d be a positive integer. T is called a d-extension [6] of S if dT = {dx | x ∈ T } ⊂ S. In [6], J.C. Rosales and M.B. Branco characterized the numerical semigroup that is a 2-extension of n1 , n2 by a subset of incomparable elements of B(n1 , n2 ), where B(n1 , n2 ) = {x ∈ G(n1 , n2 ) | 2x ∈ n1 , n2 }. In this paper, we want to study the set of numerical semigroups that have the same quotient by a positive integer. Let F be the set of all numerical semigroups and d a positive integer. Then F is a partially ordered set under the relation of set inclusion. We define a mapping fd : F → F for every numerical semigroup S by fd (S) = Sd .
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Then the inverse image of S under fd , denoted by fd−1 (S), is the set {T ∈ F | S = Td }. Moreover, the kernel of fd , denoted by kerf d , is the equivalence relation on F defined by {(S, T ) ∈ F × F | Sd = Td }. Denote the equivalence class containing S by (S)kerf d . In particular, D(S) = f2−1 (S) and (S)kerf 2 = D( S2 ), that is, the set of all doubles of S2 . In Section 2, we show that (S)kerf d has only finite maximal elements (Proposition 2.4), but no minimal element for every integer d ≥ 2 (Proposition 2.3). In Section 3, we study the subset {T ∈ F | S ⊆ T, S2 = T2 } of (S)kerf 2 , denoted by (S)⊆ kerf 2 . It is easy shown that every element of (S)⊆ kerf 2 is a 2-extension of S. Following the results given in [4], we give a definition, called an odd double subset of G(S) (Definition 3.2), and apply it to characterize the elements of the set (S)⊆ kerf 2 (Theorem 3.4). In particular, we precisely
describe the elements of the set (n1 , n2 )⊆ kerf 2 (Theorem 3.7). In Section 4, we study
the subset {T ∈ F | S ⊆ T, S2 = T2 } of (S)kerf 2 , denoted by (S)⊇ kerf 2 . According to the definitions and results in [2], we give a similar definition, called m-upper even subset of G(S) (Definition 4.1), and apply it to characterize the elements of the set (S)⊇ kerf 2 (Theorem 4.4). Finally, we give formulas for computing the genus and the Frobenius number of an element of the set (n1 , n2 )⊇ kerf 2 . Moreover, we find that the formulas match the results given by F. Strazzanti in [9]. 2. The maximal element and the minimal element of (S)kerf d Let S be a numerical semigroup and d be a positive integer. We denote the set of integers being divisible by d in S by Sd . Thus S2 means the set of even integers in S. However, to denote by So the set of odd integers in S corresponding to S2 , we use the notation Se instead of S2 . Moreover, from the definition of the quotient of S by d, we can deduce that d( Sd ) = Sd . In particular, 2( S2 ) = Se . Lemma 2.1. Let S, T be two numerical semigroups and let d be a positive integer. Then S T S T d = d if and only if Sd = Td . In particular, 2 = 2 if and only if Se = Te . Proof. It is straightforward. 2 Remark 2.2. (1) Let S be a numerical semigroup and n ∈ M (S). It is well known from [7] that S ∪ {F (S)} and S\{n} are also numerical semigroups. From Lemma 2.1, we have S∪{F (S)} S (resp. Sd = S\{n} d = d d ) if and only if F (S) (resp. n) is not divisible by d. (2) It is known that if S is a symmetric (resp. pseudo-symmetric) numerical semigroup then F (S) is odd (resp. even). Therefore, if S is symmetric (resp. pseudo-symmetric), then S2 = S∪{F2 (S)} (resp. S2 = S∪{F2 (S)} ). Proposition 2.3. Let S be a numerical semigroup and d ≥ 2 be a positive integer. Then (S)kerf d has no minimal element.
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Proof. For any numerical semigroup T , there exists n ∈ M (T ) such that n is not divisible by d since gcd(M (T )) = 1. According to Remark 2.2(1), we have Td = T \{n} . Therefore, d if T ∈ (S)kerf d , then there exists T = T \{n} ⊂ T for some n ∈ M (T ) such that T ∈ (S)kerf d . It follows that there is no minimal element in (S)kerf d . 2 Let S be a numerical semigroup. A numerical semigroup T that contains S is called an oversemigroup of S. Given two sets X and Y of integers, we denote X + Y = {x + y | x ∈ X and y ∈ Y }. Proposition 2.4. Let S be a numerical semigroup and d be a positive integer. Then (S)kerf d has finite maximal elements. Proof. Let T ∈ (S)kerf d and f be the Frobenius number of Sd . Then d({f } + N+ ) ⊆ Td . If F (T ) > df , then d F (T ), since F (T ) ∈ / d({f } + N+ ). From Remark 2.2(1), we have S∪{F (T )} S T = = . Hence T ∪ {F (T )} ∈ (S)kerf d and so T is not maximal in (S)kerf d . d d d Thus the maximal elements in (S)kerf d are contained in the set of oversemigroups of {0, df, →} (the symbol → means that every integer greater than df belongs to the set) that are in (S)kerf d . Obviously, this set is not empty and finite. 2 Remark 2.5. (1) Clearly, ((S)kerf d , ⊆) is a partially ordered set. However, it is not necessarily a totally ordered set. For example, let S = 3, 5 and T = 6, 7, 8, 9, 10, 11. Then Se = Te = {2n | n ≥ 3, n ∈ N}. Therefore, T ∈ (S)kerf 2 . But S and T are incomparable since 5 ∈ S\T and 7 ∈ T \S. (2) In particular, if d ∈ S, then Sd = N. Therefore, N ∈ (S)kerf d is the maximum element of (S)kerf d . In the remaining of this paper, we study the elements that are comparable to S in (S)kerf 2 in two cases. 3. The doubles of
S 2
containing S
Firstly, we give some notations used in the remainder of this paper. Let S be a numerical semigroup. We denote the set of odd (resp. even) integers in G(S) by Go (S) (resp. Ge (S)). Moreover, we denote B(S) = {x ∈ G(S) | 2x ∈ S} and denote the set of all odd (resp. even) integers in B(S) by Bo (S) (resp. Be (S)), that is, Bo (S) = {x ∈ Go (S) | 2x ∈ S} (resp. Be (S) = {x ∈ Ge (S) | 2x ∈ S}). Denote the set of odd (resp. even) integers in M (S) by Mo (S) (resp. Me (S)). Lemma 3.1. Let S be a numerical semigroup and H be a subset of Go (S). Then H + So ⊆ Se if and only if H + Mo (S) ⊆ Se . Proof. We need only to prove the sufficiency. For any x ∈ So , there exist x1 ∈ So and x2 ∈ Se such that x = x1 + x2 . If x1 ∈ Mo (S), then we obtain the conclusion. Otherwise,
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we have x1 = x11 + x12 for some x11 ∈ So and x12 ∈ Se \{0}. By recurrence, we can show that x ∈ Mo (S) + Se . Thus we have our conclusion. 2 If T ∈ (S)⊆ kerf 2 , then we can show that T is a 2-extension of S. Inspired by the representations of the 2-extensions of the numerical semigroup with embedding dimension two in [6], we give the following definition for characterizing the elements in (S)⊆ kerf 2 . Definition 3.2. Let S be a numerical semigroup and H be a subset of G(S). H is called an odd double subset of G(S) if it satisfies the following conditions: (1) H ⊆ Bo (S); (2) H + H ⊆ Se ; (3) H + Mo (S) ⊆ Se . Definition 3.3. Let S be a numerical semigroup. We define an order relation, ≤S , on Z as follows: x ≤S y if and only if y − x ∈ S. Theorem 3.4. Let S be a numerical semigroup. Then (S)⊆ kerf 2 = {S + (H ∪ {0}) | H is an odd double subset of incomparable elements of G(S)}. Moreover, let A and B be two odd double subsets of incomparable elements of G(S). Then S + (A ∪ {0}) = S + (B ∪ {0}) if and only if A = B. Proof. (1) Let T = S + (H ∪ {0}) for some odd double subset H of G(S). Then S ⊆ T . Since H + H ⊆ Se , we have T + T ⊆ T . Thus T is a numerical semigroup containing S. Moreover, it is easy to show that Te = Se from the conditions satisfied by H and Lemma 3.1. Consequently, T ∈ (S)⊆ kerf 2 .
Next let T ∈ (S)⊆ kerf 2 and let H = T \S. Then T = S + (H ∪ {0}) and H ⊆ Bo (S), H + H ⊆ Te = Se , H + So ⊆ Te = Se . Therefore, H is an odd double subset of G(S) from Lemma 3.1. According to Definition 3.3, we obtain S + (H ∪ {0}) = S + (Minimals≤S (H) ∪ {0}). It is obvious that Minimals≤S (H) is a set of incomparable elements. Thus we have our conclusion. (2) Let A and B be odd double subsets of incomparable elements of G(S) such that S + (A ∪ {0}) = S + (B ∪ {0}). If a ∈ A, then a ∈ S + (B ∪ {0}) and a ∈ / S. Thus there exists b ∈ B such that a ∈ S + {b} and so b ≤S a. Also b ∈ S + (A ∪ {0}) and b ∈ / S, then there exists a ∈ A such that b ∈ S + {a } and thus a ≤S b. Therefore, a ≤S b ≤S a. Since A is a set of incomparable elements, we have a = b = a. Thus a ∈ B and so A ⊆ B. In the same way we show that B ⊆ A. Therefore, A = B. 2
In the last of this section, we describe the doubles of one half of a numerical semigroup with embedding dimension two. Let n1 and n2 be two integers greater than or equal to 2 such that gcd(n1 , n2 ) = 1. Given a rational number q we denote by q the integer max{z ∈ Z | z ≤ q}.
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Lemma 3.5. (See [5, Proposition 4].) Let x ∈ N. Then x ∈ B(n1 , n2 ) if and only if x = n1 n2 − an1 − bn2 for some integers a and b verifying that 1 ≤ a ≤ n22 and 1 ≤ b ≤ n21 . Lemma 3.6. (See [6, Lemma 2].) B(n1 , n2 ) + B(n1 , n2 ) ⊆ n1 , n2 . Since gcd(n1 , n2 ) = 1, we have at least one of n1 and n2 is an odd integer. Without loss of generality, we suppose that n1 is odd. Theorem 3.7. Let n1 and n2 be two integers greater than or equal to 2 such that gcd(n1 , n2 ) = 1. (1) If n1 and n2 are odd, then (n1 , n2 )⊆ kerf 2 = {n1 , n2 , n1 , n2 ∪ {n1 n2 − n1 − n2 }};
(2) If n1 is odd and n2 is even, then (n1 , n2 )⊆ kerf 2 = {n1 , n2 } ∪ {n1 , n2 ∪ {n1 n2 − n1 − n2 , n1 n2 − n1 − 2n2 , . . . , n1 n2 − n1 − bn2 } | b ∈ {1, 2, . . . , n12−1 }}.
Proof. (1) If n1 , n2 are odd, then Mo (n1 , n2 ) = {n1 , n2 }. Let T ∈ (n1 , n2 )⊆ kerf 2 . Then from Theorem 3.4, we have T = n1 , n2 + (H ∪ {0}) for some odd double subset H of incomparable elements of G(n1 , n2 ). If H = ∅, then we suppose that n1 n2 − an1 − bn2 ∈ H for some integers a and b. Since H + Mo (n1 , n2 ) ⊆ n1 , n2 , we have n1 n2 − an1 − bn2 + n1 = n1 n2 − (a − 1)n1 − bn2 ∈ n1 , n2 and n1 n2 − an1 − bn2 + n2 = n1 n2 −an1 −(b−1)n2 ∈ n1 , n2 . Thus by Lemma 3.5, we have a = 1 and b = 1. Therefore, H = {n1 n2 − n1 − n2 }. Moreover, such H is an odd double subset of G(n1 , n2 ) from Lemma 3.6. Consequently, (n1 , n2 )⊆ kerf 2 = {n1 , n2 , n1 , n2 ∪ {n1 n2 − n1 − n2 }}. (2) If n1 is odd and n2 is even, then Mo (n1 , n2 ) = {n1 }. Let T = n1 , n2 +(H∪{0}) ∈ (n1 , n2 )⊆ kerf 2 for some odd double subset H of incomparable elements of G(n1 , n2 ). If H = ∅, then there exist integers a and b such that n1 n2 − an1 − bn2 ∈ H. Since H + Mo (n1 , n2 ) ⊆ n1 , n2 , we have n1 n2 − an1 − bn2 + n1 = n1 n2 − (a − 1)n1 − bn2 ∈ n1 , n2 . Thus a = 1 from Lemma 3.5. Moreover, as H is a set of incomparable elements, we have H = {n1 n2 − n1 − bn2 } for some integer b ∈ {1, 2, . . . , n12−1 }. Also, such H satisfies the conditions in Definition 3.2. Consequently, we have (n1 , n2 )⊆ kerf 2 = n1 −1 {n1 , n2 }∪{n1 , n2 +{0, n1 n2 −n1 −bn2 } | b ∈ {1, 2, . . . , 2 }} = {n1 , n2 }∪{n1 , n2 ∪ {n1 n2 − n1 − n2 , n1 n2 − n1 − 2n2 , . . . , n1 n2 − n1 − bn2 } | b ∈ {1, 2, . . . , n12−1 }}. 2 4. The doubles of
S 2
contained in S
In this section, we mainly represent the elements of the set (S)⊇ kerf 2 = {T ∈ F | T ⊆ S,
= T2 }. If T ∈ (S)⊇ kerf 2 , then from Lemma 2.1 we know Se ⊂ T and T \Se has only odd integers. In [4], A.M. Robles-Pérez et al. characterized the elements of D(S) by the m-upper subset of G(S), where m ∈ So . Combining the properties of the elements in (S)⊇ kerf 2 and the idea given in [4], we give the following definition for representing the S 2
elements of the set (S)⊇ kerf 2 .
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Definition 4.1. Let S be a numerical semigroup, m ∈ So and I be a subset of Ge (S). We say that I is an m-upper even subset of G(S) if it satisfies the following conditions: (1) I + {m} ⊆ So ; (2) If i ∈ I, then {j ∈ Ge (S) | i ≤S j} ⊆ I. Lemma 4.2. Let S be a numerical semigroup, m ∈ So and I be a subset of Ge (S). Denote T = Se ∪ (Se + {m}) ∪ (I + {m}). Then T ∈ (S)⊇ kerf 2 if and only if I is an m-upper even subset of G(S). Proof. (⇒) Since T ⊆ S, I ⊆ Ge (S) and m ∈ So , we can deduce that I satisfies the condition (1) in Definition 4.1. Let i ∈ I and j ∈ Ge (S). If i ≤S j, then j = i + s for some s ∈ Se . Thus j + m = (i + m) + s ∈ (I + {m}) + Se ⊆ T . Therefore, j ∈ I. (⇐) We first show that T is a numerical semigroup. Let s be an odd positive integer. Then m + 2F (S) + s = 2(F (S) + k) ∈ Se for some k ∈ N\{0}. If t is an even positive integer, then m + 2F (S) + t = m + 2(F (S) + l) ∈ Se + {m} for some l ∈ N\{0}. Therefore, for any integer x ≥ m + 2F (S) + 1, we obtain x ∈ T and so N\T is finite. 0 ∈ T is obvious. Hence, it suffices to prove that T is closed under addition. We consider the following cases: (1) If x, y ∈ Se ∪ (Se + {m}), then it is clear that x + y ∈ Se ∪ (Se + {m}). (2) If x ∈ Se + {m} and y ∈ I + {m}, then x = s + m and y = i + m for some s ∈ Se and some i ∈ I. Thus x + y = s + [(i + m) + m] ∈ Se from I + {m} ⊆ So . (3) If x, y ∈ I + {m}, then there exist i, j ∈ I such that x = i + m and y = j + m. Thus x + y = (i + m) + (j + m) ∈ So + So ⊆ Se . (4) If x ∈ Se and y ∈ I + {m}, then x + y = (x + i) + m for some i ∈ I. If x + i ∈ S, then x + y ∈ Se + {m}. If x + i ∈ G(S), then x + i ∈ I since i ≤S x + i. Therefore, x + y ∈ I + {m}. It is clear that T ⊆ S and Te = Se . Thus
S 2
=
T 2
by Lemma 2.1.
2
Let S be a numerical semigroup and let n be one of its nonzero elements. The Apéry set (see [1]) of n in S is Ap(S, n) = {s ∈ S | s − n ∈ / S}. For a set A, we denote by |A| its cardinality. It is well known (see [7]) that |Ap(S, n)| = n. Moreover, Ap(S, n) = {w(0) = 0, w(1), . . . , w(n − 1)}, where w(i) is the least element of S congruent with i modulo n. We denote by Apo (S, n) (resp. Ape (S, n)) the set of all odd (resp. even) integers in Ap(S, n). Since 2( S2 ) = Se , from Lemma 5 in [4], we have S = Se ∪ (Se + {m}) ∪ Apo (S, m) where m ∈ So .
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Lemma 4.3. Let S be a numerical semigroup and m be the least odd integer of S. Then I = {w − m | w ∈ Apo (S, m)} is an m-upper even subset of G(S). Proof. If w ∈ Apo (S, m), then w − m ∈ / S and w − m is an even integer. Thus w − m ∈ Ge (S) and so I ⊆ Ge (S). Next we show that I satisfies the conditions for being an m-upper even subset of G(S). (1) If i ∈ I, then there exists w ∈ Apo (S, m) such that i = w − m. Thus i + m = w − m + m = w ∈ Apo (S, m) ⊆ So . Therefore, I + {m} ⊆ So . (2) Let i ∈ I and j ∈ Ge (S) be such that i ≤S j. Then i = w − m for some w ∈ Apo (S, m) and j − i ∈ Se . Thus j + m − w = j − i ∈ Se . Therefore, j + m = (j + m − w) + w ∈ So . Moreover, since j ∈ Ge (S), we have j + m ∈ Apo (S, m). Consequently, j = (j + m) − m ∈ I. 2 Theorem 4.4. Let S be a numerical semigroup. Then (S)⊇ kerf 2 = {Se ∪ (Se + {m}) ∪ (I + {m}) | m ∈ So and I is an m-upper even subset of G(S)}. Moreover, let m1 , m2 ∈ So and I1 , I2 be two m-upper even subsets of G(S). Then Se ∪ (Se + {m1 }) ∪ (I1 + {m1 }) = Se ∪ (Se + {m2 }) ∪ (I2 + {m2 }) if and only if m1 = m2 and I1 = I2 . Proof. From Lemma 4.2, we obtain that if m ∈ So and I is an m-upper even subset ⊇ of G(S), then Se ∪ (Se + {m}) ∪ (I + {m}) ∈ (S)⊇ kerf 2 . Now let T ∈ (S)kerf 2 and m be the least odd integer of T . Then Se = Te and so T = Te ∪ (Te + {m}) ∪ Apo (T, m) = Se ∪ (Se + {m}) ∪ Apo (T, m). By Lemma 4.3, we have I = {w − m | w ∈ Apo (T, m)} is an m-upper even subset of G(T ). Since Se = Te , Ge (S) = Ge (T ). Therefore, I ⊆ Ge (S). Also, if i ∈ I and j ∈ Ge (S) is such that i ≤S j, then j ∈ Ge (T ) and i ≤T j. Thus, {j ∈ Ge (S) | i ≤S j} = {j ∈ Ge (T ) | i ≤T j} ⊆ I. Moreover, as T ⊆ S, we have m ∈ So and I +{m} ⊆ To ⊆ So . Consequently, I is also an m-upper even subset of G(S). Observe that Apo (T, m) = I + {m}. Thus, T = Se ∪ (Se + {m}) ∪ (I + {m}). Next let T1 = Se ∪ (Se + {m1 }) ∪ (I1 + {m1 }) and T2 = Se ∪ (Se + {m2 }) ∪ (I2 + {m2 }) such that T1 = T2 . Then m1 and m2 are the least odd integers of T1 and T2 respectively and thus m1 = m2 . Besides, I1 + {m1 } = I2 + {m2 }. Therefore, I1 = I2 . 2 Let S be a numerical semigroup. We denote go (S) = |Go (S)| and ge (S) = |Ge (S)|. Moreover, denote the largest odd (resp. even) integer not belonging to S by Fo (S) (resp. Fe (S)). It is clear that Fe (S) = 2F ( S2 ). If T = Se ∪ (Se + {m}) ∪ (I + {m}) for some m ∈ So and some m-upper even subset I of G(S), then G(T ) = Ge (S) ∪ (Ge (S)\I + {m}) ∪ {1, 3, . . . , m − 2}. Therefore, we have the following. Proposition 4.5. Let S be a numerical semigroup and let T = Se ∪ (Se + {m}) ∪ (I + {m}) for some m ∈ So and some m-upper even subset I of G(S). Then (1) g(T ) = 2ge (S) +
m−1 2
− |I|;
Z. Gu, X. Tang / Journal of Number Theory 163 (2016) 375–384
(2) F (T ) =
max{Fe (S), m − 2}
383
if I = Ge (S);
max{Fe (S), max(Ge (S)\I) + m}
if I = Ge (S).
Remark 4.6. (1) Let F (S) be odd in Proposition 4.5. Then Fe (S) < F (S). Since T ⊆ S, we have F (T ) ≥ F (S). Thus, F (T ) = m −2 if I = Ge (S) and F (T ) = max(Ge (S)\I) +m if I = Ge (S). Therefore, F (T ) is odd. (2) Let T be symmetric in Proposition 4.5. Since Fe (S) ∈ / T and F (T ) is odd, we have F (T ) − Fe (S) ∈ To . Thus F (T ) − Fe (S) = x + m for some x ∈ Se or x ∈ I. Hence F (T ) = Fe (S)+m+x > m−2 and so F (T ) = max(Ge (S)\I)+m. Moreover, max Ge (S) = Fe (S). Consequently, x = 0 and thus F (T ) = Fe (S) + m. From Corollary 4.6 given by F. Strazzanti in [9], we know that F ( T2 ) = F (T2)−m . It follows that F (T ) = 2F ( T2 ) + m = 2F ( S2 ) + m = Fe (S) + m. Hence our result completely matches the conclusion in [9]. Let n1 and n2 be two integers defined as in Section 3. It is well known that 2 −1) F (n1 , n2 ) = n1 n2 − n1 − n2 and g(n1 , n2 ) = (n1 −1)(n . Therefore, combining 2 Lemma 15 in [3] and Proposition 7, Lemma 16 in [4], we have the following result. Proposition 4.7. Let n1 and n2 be two integers greater than or equal to 2 such that gcd(n1 , n2 ) = 1. (1) If n1 , n2 are odd, then Fe (n1 , n2 ) = n1 n2 −n1 −n2 −min{n1 , n2 } and ge (n1 , n2 ) = (n1 −1)(n2 −1) ; 4 (2) If n1 is odd and n2 is even, then Fe (n1 , n2 ) = n1 n2 − 2n1 − n2 and ge (n1 , n2 ) = (n1 −1)(n2 −2) . 4 Since F (n1 , n2 ) is odd, we have the following result from Remark 4.6(1). Proposition 4.8. Let T = n1 , n2 e ∪ (n1 , n2 e + {m}) ∪ (I + {m}) for some m ∈ n1 , n2 o and some m-upper even subset of G(n1 , n2 ). (1) If n1 , n2 are odd, then (i) g(T ) =
(n1 −1)(n2 −1) 2
(ii) F (T ) =
+
m−1 2
− |I|;
m−2
if I = Ge (n1 , n2 );
max(Ge (n1 , n2 )\I) + m
if I = Ge (n1 , n2 ).
(2) If n1 is odd and n2 is even, then (iii) g(T ) = (iv) F (T ) =
(n1 −1)(n2 −2) 2
+
m−1 2
− |I|;
m−2
if I = Ge (n1 , n2 );
max(Ge (n1 , n2 )\I) + m
if I = Ge (n1 , n2 ).
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Remark 4.9. Let T be symmetric in Proposition 4.8. From Remark 4.6(2), we have that F (T ) = Fe (n1 , n2 ) + m. Moreover, g(T ) = F (T2)+1 . Therefore: (1) If n1 , n2 are odd, then F (T ) = n1 n2 − n1 − n2 − min{n1 , n2 } + m and g(T ) = n1 n2 −n1 −n2 −min{n1 ,n2 }+m+1 ; 2 (2) If n1 is odd and n2 is even, then F (T ) = n1 n2 − 2n1 − n2 + m and g(T ) = n1 n2 −2n1 −n2 +m+1 . 2 Acknowledgment We thank the referee whose comments led to significant improvements to this paper. References [1] R. Apéry, Sur les branches superlinéaires des courbes algébriques, C. R. Acad. Sci. Paris 222 (1946) 1198–1200. [2] M. D’Anna, F. Strazzanti, The numerical duplication of a numerical semigroup, Semigroup Forum 87 (2013) 149–160. [3] D.E. Dobbs, H.J. Smith, Numerical semigroups whose fractions are of maximal embedding dimension, Semigroup Forum 82 (2011) 412–422. [4] A.M. Robles-Pérez, J.C. Rosales, P. Vasco, The doubles of a numerical semigroup, J. Pure Appl. Algebra 213 (2009) 387–396. [5] J.C. Rosales, Fundamental gaps of numerical semigroups generated by two elements, Linear Algebra Appl. 405 (2005) 200–208. [6] J.C. Rosales, M.B. Branco, Two-extension of a numerical semigroup with embedding dimension two, Comm. Algebra 42 (2014) 1690–1697. [7] J.C. Rosales, P.A. García-Sánchez, Numerical Semigroups, Springer, New York, 2009. [8] H.J. Smith, Numerical semigroups that are fractions of numerical semigroups of maximal embedding dimension, JP J. Algebra Number Theory Appl. 17 (2010) 69–96. [9] F. Strazzanti, Minimal genus of a multiple and Frobenius number of a quotient of a numerical semigroup, Internat. J. Algebra Comput. 25 (06) (2015) 1043–1053.
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Journal of Number Theory www.elsevier.com/locate/jnt
The doubles of one half of a numerical semigroup Ze Gu ∗ , Xilin Tang Department of Mathematics, South China University of Technology, Guangzhou, Guangdong, 510640, PR China
a r t i c l e
i n f o
Article history: Received 22 April 2015 Received in revised form 11 November 2015 Accepted 11 November 2015 Available online 27 January 2016 Communicated by David Goss MSC: 20M14 11D07
a b s t r a c t Let F be the set of all numerical semigroups and d a positive integer. Define a mapping fd : F → F for every numerical semigroup S by fd (S) = Sd . In this paper, we show that the equivalence class (S)kerf d contains only finite maximal elements, but does not contain minimal element for every integer d ≥ 2. Moreover, we study the subset of (S)kerf 2 whose elements contain S and the subset of (S)kerf 2 whose elements are contained in S. In particular, we study the case of S has embedding dimension two. © 2016 Elsevier Inc. All rights reserved.
Keywords: Numerical semigroup Quotient Double Maximal element Minimal element
1. Introduction and preliminaries Let N be the set of nonnegative integers. A numerical semigroup is a subset S of N such that it is closed under addition, 0 ∈ S and N\S is finite. The elements of G(S) = N\S are the gaps of S and its cardinality, denoted by g(S), is called the genus of S. Another * Corresponding author. E-mail addresses: [email protected] (Z. Gu), [email protected] (X. Tang). http://dx.doi.org/10.1016/j.jnt.2015.11.026 0022-314X/© 2016 Elsevier Inc. All rights reserved.
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important invariant of S is the largest integer not belonging to S, known as the Frobenius number of S and denoted by F (S). If A ⊆ N is a nonempty set, we denote by A the submonoid of (N, +) generated by A, that is, A = {λ1 a1 + · · · + λn an | n ∈ N\{0}, a1 , · · · , an ∈ A, λ1 , · · · , λn ∈ N}. It is well known that A is a numerical semigroup if and only if gcd(A) = 1. If S is a numerical semigroup and S = A, then we say that A is a system of generators of S. In addition, if no proper subset of A generates S, then we say that A is a minimal system of generators of S. In [7] it is proved that every numerical semigroup S has a unique minimal system of generators and, moreover, such a system is finite. Denote by M (S) the minimal system of generators of S. Let M (S) = {n1 < n2 < · · · < np }. Then n1 is known as the multiplicity of S, denoted by m(S). The cardinality of M (S), p, is called the embedding dimension of S and is denoted by e(S). In particular, if e(S) = m(S) then S is a numerical semigroup with maximal embedding dimension. A partially ordered set (A, ≤) is a nonempty set A with an order relation ≤. For a, b ∈ A, if a ≤ b or b ≤ a, then a and b are comparable. Otherwise they are incomparable. A partially ordered set (A, ≤) is called a totally ordered set (also called a chain) if any two elements of A are comparable. An element a ∈ A is maximal in A if for every c ∈ A that is comparable to a, c ≤ a. An element b ∈ A is minimal in A if for every c ∈ A that is comparable to b, b ≤ c. Let S be a numerical semigroup and d a positive integer. Then in [7] it is proved that Sd = {x ∈ N | dx ∈ S} is also a numerical semigroup, called the quotient of S by d. The work on the numerical semigroups that are quotients of numerical semigroups of maximal embedding dimension and the numerical semigroups whose quotients are of maximal embedding dimension can be seen in [3,8]. When d = 2 we say that S2 is one half of the numerical semigroup S. Dually, we say that S is a double [4] of the numerical semigroup S2 . Denote by D(S) the set of all doubles of the numerical semigroup S, that is, D(S) = {T numerical semigroup | S = T2 }. In [4], A.M. Robles-Pérez et al. gave a representation for the elements of D(S) by a unique pair (m, H), where m is an odd integer of S and H is an m-upper subset of G(S). M. D’Anna and F. Strazzanti studied the set D(S) by the numerical duplication in [3]. Recently, F. Strazzanti gave a formula for the minimal genus of a multiple of a numerical semigroup S, a formula for the minimal genus of a symmetric double of S and the formula for the Frobenius number of the quotient of some families of numerical semigroups in [9]. Let S and T be two numerical semigroups such that S ⊆ T , and let d be a positive integer. T is called a d-extension [6] of S if dT = {dx | x ∈ T } ⊂ S. In [6], J.C. Rosales and M.B. Branco characterized the numerical semigroup that is a 2-extension of n1 , n2 by a subset of incomparable elements of B(n1 , n2 ), where B(n1 , n2 ) = {x ∈ G(n1 , n2 ) | 2x ∈ n1 , n2 }. In this paper, we want to study the set of numerical semigroups that have the same quotient by a positive integer. Let F be the set of all numerical semigroups and d a positive integer. Then F is a partially ordered set under the relation of set inclusion. We define a mapping fd : F → F for every numerical semigroup S by fd (S) = Sd .
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Then the inverse image of S under fd , denoted by fd−1 (S), is the set {T ∈ F | S = Td }. Moreover, the kernel of fd , denoted by kerf d , is the equivalence relation on F defined by {(S, T ) ∈ F × F | Sd = Td }. Denote the equivalence class containing S by (S)kerf d . In particular, D(S) = f2−1 (S) and (S)kerf 2 = D( S2 ), that is, the set of all doubles of S2 . In Section 2, we show that (S)kerf d has only finite maximal elements (Proposition 2.4), but no minimal element for every integer d ≥ 2 (Proposition 2.3). In Section 3, we study the subset {T ∈ F | S ⊆ T, S2 = T2 } of (S)kerf 2 , denoted by (S)⊆ kerf 2 . It is easy shown that every element of (S)⊆ kerf 2 is a 2-extension of S. Following the results given in [4], we give a definition, called an odd double subset of G(S) (Definition 3.2), and apply it to characterize the elements of the set (S)⊆ kerf 2 (Theorem 3.4). In particular, we precisely
describe the elements of the set (n1 , n2 )⊆ kerf 2 (Theorem 3.7). In Section 4, we study
the subset {T ∈ F | S ⊆ T, S2 = T2 } of (S)kerf 2 , denoted by (S)⊇ kerf 2 . According to the definitions and results in [2], we give a similar definition, called m-upper even subset of G(S) (Definition 4.1), and apply it to characterize the elements of the set (S)⊇ kerf 2 (Theorem 4.4). Finally, we give formulas for computing the genus and the Frobenius number of an element of the set (n1 , n2 )⊇ kerf 2 . Moreover, we find that the formulas match the results given by F. Strazzanti in [9]. 2. The maximal element and the minimal element of (S)kerf d Let S be a numerical semigroup and d be a positive integer. We denote the set of integers being divisible by d in S by Sd . Thus S2 means the set of even integers in S. However, to denote by So the set of odd integers in S corresponding to S2 , we use the notation Se instead of S2 . Moreover, from the definition of the quotient of S by d, we can deduce that d( Sd ) = Sd . In particular, 2( S2 ) = Se . Lemma 2.1. Let S, T be two numerical semigroups and let d be a positive integer. Then S T S T d = d if and only if Sd = Td . In particular, 2 = 2 if and only if Se = Te . Proof. It is straightforward. 2 Remark 2.2. (1) Let S be a numerical semigroup and n ∈ M (S). It is well known from [7] that S ∪ {F (S)} and S\{n} are also numerical semigroups. From Lemma 2.1, we have S∪{F (S)} S (resp. Sd = S\{n} d = d d ) if and only if F (S) (resp. n) is not divisible by d. (2) It is known that if S is a symmetric (resp. pseudo-symmetric) numerical semigroup then F (S) is odd (resp. even). Therefore, if S is symmetric (resp. pseudo-symmetric), then S2 = S∪{F2 (S)} (resp. S2 = S∪{F2 (S)} ). Proposition 2.3. Let S be a numerical semigroup and d ≥ 2 be a positive integer. Then (S)kerf d has no minimal element.
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Proof. For any numerical semigroup T , there exists n ∈ M (T ) such that n is not divisible by d since gcd(M (T )) = 1. According to Remark 2.2(1), we have Td = T \{n} . Therefore, d if T ∈ (S)kerf d , then there exists T = T \{n} ⊂ T for some n ∈ M (T ) such that T ∈ (S)kerf d . It follows that there is no minimal element in (S)kerf d . 2 Let S be a numerical semigroup. A numerical semigroup T that contains S is called an oversemigroup of S. Given two sets X and Y of integers, we denote X + Y = {x + y | x ∈ X and y ∈ Y }. Proposition 2.4. Let S be a numerical semigroup and d be a positive integer. Then (S)kerf d has finite maximal elements. Proof. Let T ∈ (S)kerf d and f be the Frobenius number of Sd . Then d({f } + N+ ) ⊆ Td . If F (T ) > df , then d F (T ), since F (T ) ∈ / d({f } + N+ ). From Remark 2.2(1), we have S∪{F (T )} S T = = . Hence T ∪ {F (T )} ∈ (S)kerf d and so T is not maximal in (S)kerf d . d d d Thus the maximal elements in (S)kerf d are contained in the set of oversemigroups of {0, df, →} (the symbol → means that every integer greater than df belongs to the set) that are in (S)kerf d . Obviously, this set is not empty and finite. 2 Remark 2.5. (1) Clearly, ((S)kerf d , ⊆) is a partially ordered set. However, it is not necessarily a totally ordered set. For example, let S = 3, 5 and T = 6, 7, 8, 9, 10, 11. Then Se = Te = {2n | n ≥ 3, n ∈ N}. Therefore, T ∈ (S)kerf 2 . But S and T are incomparable since 5 ∈ S\T and 7 ∈ T \S. (2) In particular, if d ∈ S, then Sd = N. Therefore, N ∈ (S)kerf d is the maximum element of (S)kerf d . In the remaining of this paper, we study the elements that are comparable to S in (S)kerf 2 in two cases. 3. The doubles of
S 2
containing S
Firstly, we give some notations used in the remainder of this paper. Let S be a numerical semigroup. We denote the set of odd (resp. even) integers in G(S) by Go (S) (resp. Ge (S)). Moreover, we denote B(S) = {x ∈ G(S) | 2x ∈ S} and denote the set of all odd (resp. even) integers in B(S) by Bo (S) (resp. Be (S)), that is, Bo (S) = {x ∈ Go (S) | 2x ∈ S} (resp. Be (S) = {x ∈ Ge (S) | 2x ∈ S}). Denote the set of odd (resp. even) integers in M (S) by Mo (S) (resp. Me (S)). Lemma 3.1. Let S be a numerical semigroup and H be a subset of Go (S). Then H + So ⊆ Se if and only if H + Mo (S) ⊆ Se . Proof. We need only to prove the sufficiency. For any x ∈ So , there exist x1 ∈ So and x2 ∈ Se such that x = x1 + x2 . If x1 ∈ Mo (S), then we obtain the conclusion. Otherwise,
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we have x1 = x11 + x12 for some x11 ∈ So and x12 ∈ Se \{0}. By recurrence, we can show that x ∈ Mo (S) + Se . Thus we have our conclusion. 2 If T ∈ (S)⊆ kerf 2 , then we can show that T is a 2-extension of S. Inspired by the representations of the 2-extensions of the numerical semigroup with embedding dimension two in [6], we give the following definition for characterizing the elements in (S)⊆ kerf 2 . Definition 3.2. Let S be a numerical semigroup and H be a subset of G(S). H is called an odd double subset of G(S) if it satisfies the following conditions: (1) H ⊆ Bo (S); (2) H + H ⊆ Se ; (3) H + Mo (S) ⊆ Se . Definition 3.3. Let S be a numerical semigroup. We define an order relation, ≤S , on Z as follows: x ≤S y if and only if y − x ∈ S. Theorem 3.4. Let S be a numerical semigroup. Then (S)⊆ kerf 2 = {S + (H ∪ {0}) | H is an odd double subset of incomparable elements of G(S)}. Moreover, let A and B be two odd double subsets of incomparable elements of G(S). Then S + (A ∪ {0}) = S + (B ∪ {0}) if and only if A = B. Proof. (1) Let T = S + (H ∪ {0}) for some odd double subset H of G(S). Then S ⊆ T . Since H + H ⊆ Se , we have T + T ⊆ T . Thus T is a numerical semigroup containing S. Moreover, it is easy to show that Te = Se from the conditions satisfied by H and Lemma 3.1. Consequently, T ∈ (S)⊆ kerf 2 .
Next let T ∈ (S)⊆ kerf 2 and let H = T \S. Then T = S + (H ∪ {0}) and H ⊆ Bo (S), H + H ⊆ Te = Se , H + So ⊆ Te = Se . Therefore, H is an odd double subset of G(S) from Lemma 3.1. According to Definition 3.3, we obtain S + (H ∪ {0}) = S + (Minimals≤S (H) ∪ {0}). It is obvious that Minimals≤S (H) is a set of incomparable elements. Thus we have our conclusion. (2) Let A and B be odd double subsets of incomparable elements of G(S) such that S + (A ∪ {0}) = S + (B ∪ {0}). If a ∈ A, then a ∈ S + (B ∪ {0}) and a ∈ / S. Thus there exists b ∈ B such that a ∈ S + {b} and so b ≤S a. Also b ∈ S + (A ∪ {0}) and b ∈ / S, then there exists a ∈ A such that b ∈ S + {a } and thus a ≤S b. Therefore, a ≤S b ≤S a. Since A is a set of incomparable elements, we have a = b = a. Thus a ∈ B and so A ⊆ B. In the same way we show that B ⊆ A. Therefore, A = B. 2
In the last of this section, we describe the doubles of one half of a numerical semigroup with embedding dimension two. Let n1 and n2 be two integers greater than or equal to 2 such that gcd(n1 , n2 ) = 1. Given a rational number q we denote by q the integer max{z ∈ Z | z ≤ q}.
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Lemma 3.5. (See [5, Proposition 4].) Let x ∈ N. Then x ∈ B(n1 , n2 ) if and only if x = n1 n2 − an1 − bn2 for some integers a and b verifying that 1 ≤ a ≤ n22 and 1 ≤ b ≤ n21 . Lemma 3.6. (See [6, Lemma 2].) B(n1 , n2 ) + B(n1 , n2 ) ⊆ n1 , n2 . Since gcd(n1 , n2 ) = 1, we have at least one of n1 and n2 is an odd integer. Without loss of generality, we suppose that n1 is odd. Theorem 3.7. Let n1 and n2 be two integers greater than or equal to 2 such that gcd(n1 , n2 ) = 1. (1) If n1 and n2 are odd, then (n1 , n2 )⊆ kerf 2 = {n1 , n2 , n1 , n2 ∪ {n1 n2 − n1 − n2 }};
(2) If n1 is odd and n2 is even, then (n1 , n2 )⊆ kerf 2 = {n1 , n2 } ∪ {n1 , n2 ∪ {n1 n2 − n1 − n2 , n1 n2 − n1 − 2n2 , . . . , n1 n2 − n1 − bn2 } | b ∈ {1, 2, . . . , n12−1 }}.
Proof. (1) If n1 , n2 are odd, then Mo (n1 , n2 ) = {n1 , n2 }. Let T ∈ (n1 , n2 )⊆ kerf 2 . Then from Theorem 3.4, we have T = n1 , n2 + (H ∪ {0}) for some odd double subset H of incomparable elements of G(n1 , n2 ). If H = ∅, then we suppose that n1 n2 − an1 − bn2 ∈ H for some integers a and b. Since H + Mo (n1 , n2 ) ⊆ n1 , n2 , we have n1 n2 − an1 − bn2 + n1 = n1 n2 − (a − 1)n1 − bn2 ∈ n1 , n2 and n1 n2 − an1 − bn2 + n2 = n1 n2 −an1 −(b−1)n2 ∈ n1 , n2 . Thus by Lemma 3.5, we have a = 1 and b = 1. Therefore, H = {n1 n2 − n1 − n2 }. Moreover, such H is an odd double subset of G(n1 , n2 ) from Lemma 3.6. Consequently, (n1 , n2 )⊆ kerf 2 = {n1 , n2 , n1 , n2 ∪ {n1 n2 − n1 − n2 }}. (2) If n1 is odd and n2 is even, then Mo (n1 , n2 ) = {n1 }. Let T = n1 , n2 +(H∪{0}) ∈ (n1 , n2 )⊆ kerf 2 for some odd double subset H of incomparable elements of G(n1 , n2 ). If H = ∅, then there exist integers a and b such that n1 n2 − an1 − bn2 ∈ H. Since H + Mo (n1 , n2 ) ⊆ n1 , n2 , we have n1 n2 − an1 − bn2 + n1 = n1 n2 − (a − 1)n1 − bn2 ∈ n1 , n2 . Thus a = 1 from Lemma 3.5. Moreover, as H is a set of incomparable elements, we have H = {n1 n2 − n1 − bn2 } for some integer b ∈ {1, 2, . . . , n12−1 }. Also, such H satisfies the conditions in Definition 3.2. Consequently, we have (n1 , n2 )⊆ kerf 2 = n1 −1 {n1 , n2 }∪{n1 , n2 +{0, n1 n2 −n1 −bn2 } | b ∈ {1, 2, . . . , 2 }} = {n1 , n2 }∪{n1 , n2 ∪ {n1 n2 − n1 − n2 , n1 n2 − n1 − 2n2 , . . . , n1 n2 − n1 − bn2 } | b ∈ {1, 2, . . . , n12−1 }}. 2 4. The doubles of
S 2
contained in S
In this section, we mainly represent the elements of the set (S)⊇ kerf 2 = {T ∈ F | T ⊆ S,
= T2 }. If T ∈ (S)⊇ kerf 2 , then from Lemma 2.1 we know Se ⊂ T and T \Se has only odd integers. In [4], A.M. Robles-Pérez et al. characterized the elements of D(S) by the m-upper subset of G(S), where m ∈ So . Combining the properties of the elements in (S)⊇ kerf 2 and the idea given in [4], we give the following definition for representing the S 2
elements of the set (S)⊇ kerf 2 .
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Definition 4.1. Let S be a numerical semigroup, m ∈ So and I be a subset of Ge (S). We say that I is an m-upper even subset of G(S) if it satisfies the following conditions: (1) I + {m} ⊆ So ; (2) If i ∈ I, then {j ∈ Ge (S) | i ≤S j} ⊆ I. Lemma 4.2. Let S be a numerical semigroup, m ∈ So and I be a subset of Ge (S). Denote T = Se ∪ (Se + {m}) ∪ (I + {m}). Then T ∈ (S)⊇ kerf 2 if and only if I is an m-upper even subset of G(S). Proof. (⇒) Since T ⊆ S, I ⊆ Ge (S) and m ∈ So , we can deduce that I satisfies the condition (1) in Definition 4.1. Let i ∈ I and j ∈ Ge (S). If i ≤S j, then j = i + s for some s ∈ Se . Thus j + m = (i + m) + s ∈ (I + {m}) + Se ⊆ T . Therefore, j ∈ I. (⇐) We first show that T is a numerical semigroup. Let s be an odd positive integer. Then m + 2F (S) + s = 2(F (S) + k) ∈ Se for some k ∈ N\{0}. If t is an even positive integer, then m + 2F (S) + t = m + 2(F (S) + l) ∈ Se + {m} for some l ∈ N\{0}. Therefore, for any integer x ≥ m + 2F (S) + 1, we obtain x ∈ T and so N\T is finite. 0 ∈ T is obvious. Hence, it suffices to prove that T is closed under addition. We consider the following cases: (1) If x, y ∈ Se ∪ (Se + {m}), then it is clear that x + y ∈ Se ∪ (Se + {m}). (2) If x ∈ Se + {m} and y ∈ I + {m}, then x = s + m and y = i + m for some s ∈ Se and some i ∈ I. Thus x + y = s + [(i + m) + m] ∈ Se from I + {m} ⊆ So . (3) If x, y ∈ I + {m}, then there exist i, j ∈ I such that x = i + m and y = j + m. Thus x + y = (i + m) + (j + m) ∈ So + So ⊆ Se . (4) If x ∈ Se and y ∈ I + {m}, then x + y = (x + i) + m for some i ∈ I. If x + i ∈ S, then x + y ∈ Se + {m}. If x + i ∈ G(S), then x + i ∈ I since i ≤S x + i. Therefore, x + y ∈ I + {m}. It is clear that T ⊆ S and Te = Se . Thus
S 2
=
T 2
by Lemma 2.1.
2
Let S be a numerical semigroup and let n be one of its nonzero elements. The Apéry set (see [1]) of n in S is Ap(S, n) = {s ∈ S | s − n ∈ / S}. For a set A, we denote by |A| its cardinality. It is well known (see [7]) that |Ap(S, n)| = n. Moreover, Ap(S, n) = {w(0) = 0, w(1), . . . , w(n − 1)}, where w(i) is the least element of S congruent with i modulo n. We denote by Apo (S, n) (resp. Ape (S, n)) the set of all odd (resp. even) integers in Ap(S, n). Since 2( S2 ) = Se , from Lemma 5 in [4], we have S = Se ∪ (Se + {m}) ∪ Apo (S, m) where m ∈ So .
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Lemma 4.3. Let S be a numerical semigroup and m be the least odd integer of S. Then I = {w − m | w ∈ Apo (S, m)} is an m-upper even subset of G(S). Proof. If w ∈ Apo (S, m), then w − m ∈ / S and w − m is an even integer. Thus w − m ∈ Ge (S) and so I ⊆ Ge (S). Next we show that I satisfies the conditions for being an m-upper even subset of G(S). (1) If i ∈ I, then there exists w ∈ Apo (S, m) such that i = w − m. Thus i + m = w − m + m = w ∈ Apo (S, m) ⊆ So . Therefore, I + {m} ⊆ So . (2) Let i ∈ I and j ∈ Ge (S) be such that i ≤S j. Then i = w − m for some w ∈ Apo (S, m) and j − i ∈ Se . Thus j + m − w = j − i ∈ Se . Therefore, j + m = (j + m − w) + w ∈ So . Moreover, since j ∈ Ge (S), we have j + m ∈ Apo (S, m). Consequently, j = (j + m) − m ∈ I. 2 Theorem 4.4. Let S be a numerical semigroup. Then (S)⊇ kerf 2 = {Se ∪ (Se + {m}) ∪ (I + {m}) | m ∈ So and I is an m-upper even subset of G(S)}. Moreover, let m1 , m2 ∈ So and I1 , I2 be two m-upper even subsets of G(S). Then Se ∪ (Se + {m1 }) ∪ (I1 + {m1 }) = Se ∪ (Se + {m2 }) ∪ (I2 + {m2 }) if and only if m1 = m2 and I1 = I2 . Proof. From Lemma 4.2, we obtain that if m ∈ So and I is an m-upper even subset ⊇ of G(S), then Se ∪ (Se + {m}) ∪ (I + {m}) ∈ (S)⊇ kerf 2 . Now let T ∈ (S)kerf 2 and m be the least odd integer of T . Then Se = Te and so T = Te ∪ (Te + {m}) ∪ Apo (T, m) = Se ∪ (Se + {m}) ∪ Apo (T, m). By Lemma 4.3, we have I = {w − m | w ∈ Apo (T, m)} is an m-upper even subset of G(T ). Since Se = Te , Ge (S) = Ge (T ). Therefore, I ⊆ Ge (S). Also, if i ∈ I and j ∈ Ge (S) is such that i ≤S j, then j ∈ Ge (T ) and i ≤T j. Thus, {j ∈ Ge (S) | i ≤S j} = {j ∈ Ge (T ) | i ≤T j} ⊆ I. Moreover, as T ⊆ S, we have m ∈ So and I +{m} ⊆ To ⊆ So . Consequently, I is also an m-upper even subset of G(S). Observe that Apo (T, m) = I + {m}. Thus, T = Se ∪ (Se + {m}) ∪ (I + {m}). Next let T1 = Se ∪ (Se + {m1 }) ∪ (I1 + {m1 }) and T2 = Se ∪ (Se + {m2 }) ∪ (I2 + {m2 }) such that T1 = T2 . Then m1 and m2 are the least odd integers of T1 and T2 respectively and thus m1 = m2 . Besides, I1 + {m1 } = I2 + {m2 }. Therefore, I1 = I2 . 2 Let S be a numerical semigroup. We denote go (S) = |Go (S)| and ge (S) = |Ge (S)|. Moreover, denote the largest odd (resp. even) integer not belonging to S by Fo (S) (resp. Fe (S)). It is clear that Fe (S) = 2F ( S2 ). If T = Se ∪ (Se + {m}) ∪ (I + {m}) for some m ∈ So and some m-upper even subset I of G(S), then G(T ) = Ge (S) ∪ (Ge (S)\I + {m}) ∪ {1, 3, . . . , m − 2}. Therefore, we have the following. Proposition 4.5. Let S be a numerical semigroup and let T = Se ∪ (Se + {m}) ∪ (I + {m}) for some m ∈ So and some m-upper even subset I of G(S). Then (1) g(T ) = 2ge (S) +
m−1 2
− |I|;
Z. Gu, X. Tang / Journal of Number Theory 163 (2016) 375–384
(2) F (T ) =
max{Fe (S), m − 2}
383
if I = Ge (S);
max{Fe (S), max(Ge (S)\I) + m}
if I = Ge (S).
Remark 4.6. (1) Let F (S) be odd in Proposition 4.5. Then Fe (S) < F (S). Since T ⊆ S, we have F (T ) ≥ F (S). Thus, F (T ) = m −2 if I = Ge (S) and F (T ) = max(Ge (S)\I) +m if I = Ge (S). Therefore, F (T ) is odd. (2) Let T be symmetric in Proposition 4.5. Since Fe (S) ∈ / T and F (T ) is odd, we have F (T ) − Fe (S) ∈ To . Thus F (T ) − Fe (S) = x + m for some x ∈ Se or x ∈ I. Hence F (T ) = Fe (S)+m+x > m−2 and so F (T ) = max(Ge (S)\I)+m. Moreover, max Ge (S) = Fe (S). Consequently, x = 0 and thus F (T ) = Fe (S) + m. From Corollary 4.6 given by F. Strazzanti in [9], we know that F ( T2 ) = F (T2)−m . It follows that F (T ) = 2F ( T2 ) + m = 2F ( S2 ) + m = Fe (S) + m. Hence our result completely matches the conclusion in [9]. Let n1 and n2 be two integers defined as in Section 3. It is well known that 2 −1) F (n1 , n2 ) = n1 n2 − n1 − n2 and g(n1 , n2 ) = (n1 −1)(n . Therefore, combining 2 Lemma 15 in [3] and Proposition 7, Lemma 16 in [4], we have the following result. Proposition 4.7. Let n1 and n2 be two integers greater than or equal to 2 such that gcd(n1 , n2 ) = 1. (1) If n1 , n2 are odd, then Fe (n1 , n2 ) = n1 n2 −n1 −n2 −min{n1 , n2 } and ge (n1 , n2 ) = (n1 −1)(n2 −1) ; 4 (2) If n1 is odd and n2 is even, then Fe (n1 , n2 ) = n1 n2 − 2n1 − n2 and ge (n1 , n2 ) = (n1 −1)(n2 −2) . 4 Since F (n1 , n2 ) is odd, we have the following result from Remark 4.6(1). Proposition 4.8. Let T = n1 , n2 e ∪ (n1 , n2 e + {m}) ∪ (I + {m}) for some m ∈ n1 , n2 o and some m-upper even subset of G(n1 , n2 ). (1) If n1 , n2 are odd, then (i) g(T ) =
(n1 −1)(n2 −1) 2
(ii) F (T ) =
+
m−1 2
− |I|;
m−2
if I = Ge (n1 , n2 );
max(Ge (n1 , n2 )\I) + m
if I = Ge (n1 , n2 ).
(2) If n1 is odd and n2 is even, then (iii) g(T ) = (iv) F (T ) =
(n1 −1)(n2 −2) 2
+
m−1 2
− |I|;
m−2
if I = Ge (n1 , n2 );
max(Ge (n1 , n2 )\I) + m
if I = Ge (n1 , n2 ).
384
Z. Gu, X. Tang / Journal of Number Theory 163 (2016) 375–384
Remark 4.9. Let T be symmetric in Proposition 4.8. From Remark 4.6(2), we have that F (T ) = Fe (n1 , n2 ) + m. Moreover, g(T ) = F (T2)+1 . Therefore: (1) If n1 , n2 are odd, then F (T ) = n1 n2 − n1 − n2 − min{n1 , n2 } + m and g(T ) = n1 n2 −n1 −n2 −min{n1 ,n2 }+m+1 ; 2 (2) If n1 is odd and n2 is even, then F (T ) = n1 n2 − 2n1 − n2 + m and g(T ) = n1 n2 −2n1 −n2 +m+1 . 2 Acknowledgment We thank the referee whose comments led to significant improvements to this paper. References [1] R. Apéry, Sur les branches superlinéaires des courbes algébriques, C. R. Acad. Sci. Paris 222 (1946) 1198–1200. [2] M. D’Anna, F. Strazzanti, The numerical duplication of a numerical semigroup, Semigroup Forum 87 (2013) 149–160. [3] D.E. Dobbs, H.J. Smith, Numerical semigroups whose fractions are of maximal embedding dimension, Semigroup Forum 82 (2011) 412–422. [4] A.M. Robles-Pérez, J.C. Rosales, P. Vasco, The doubles of a numerical semigroup, J. Pure Appl. Algebra 213 (2009) 387–396. [5] J.C. Rosales, Fundamental gaps of numerical semigroups generated by two elements, Linear Algebra Appl. 405 (2005) 200–208. [6] J.C. Rosales, M.B. Branco, Two-extension of a numerical semigroup with embedding dimension two, Comm. Algebra 42 (2014) 1690–1697. [7] J.C. Rosales, P.A. García-Sánchez, Numerical Semigroups, Springer, New York, 2009. [8] H.J. Smith, Numerical semigroups that are fractions of numerical semigroups of maximal embedding dimension, JP J. Algebra Number Theory Appl. 17 (2010) 69–96. [9] F. Strazzanti, Minimal genus of a multiple and Frobenius number of a quotient of a numerical semigroup, Internat. J. Algebra Comput. 25 (06) (2015) 1043–1053.