tailieunhanh - Arf numerical semigroups

The aim of this work is to exhibit the relationship between the Arf closure of a numerical semigroup S and its Lipman semigroup L(S). This relationship is then used to give direct proofs of some characterizations of Arf numerical semigroups through their Lipman sequences of semigroups. | Turk J Math (2017) 41: 1448 –1457 ¨ ITAK ˙ c TUB ⃝ Turkish Journal of Mathematics doi: Research Article Arf numerical semigroups 1 1,∗ ˙ ˙ Sedat ILHAN , Halil Ibrahim KARAKAS ¸2 Department of Mathematics, Faculty of Science, Dicle University, Diyarbakır, Turkey 2 Faculty of Commercial Sciences, Ba¸skent University, Ankara, Turkey Received: • Accepted/Published Online: • Final Version: Abstract: The aim of this work is to exhibit the relationship between the Arf closure of a numerical semigroup S and its Lipman semigroup L(S). This relationship is then used to give direct proofs of some characterizations of Arf numerical semigroups through their Lipman sequences of semigroups. We also give an algorithmic construction of the Arf closure of a numerical semigroup via its Lipman sequence of semigroups. Key words: Numerical semigroups, Arf numerical semigroups, Arf closure 1. Introduction Let N0 denote the set of nonnegative integers. A subset S ⊆ N0 satisfying (i) 0 ∈ S (ii) x, y ∈ S ⇒ x + y ∈ S (iii) |N0 \ S| be a numerical semigroup with minimal system of generators a1 = m . Proof The inclusion ⊆ L(S) is clear by Lemma . Now let x ∈ L(S). Then x = z − km, z ∈ kM, k ≥ 1. We write z = b1 + b2 + · · · + bk with bi ∈ M, 1 ≤ i ≤ k . For each i = 1, . . . , k , ∑e we can write bi = ci1 m + j=2 cij aj with cij ∈ N0 and at least one of the cij ̸= 0 for 1 ≤ j ≤ e. Hence, x= k e ∑ ∑ (ci1 m + cij aj ) − km. i=1 j=2 This sum can be arranged as x=( k ∑ e ∑ i=1 j=1 where ∑k i=1 ∑e j=1 cij cij − k)m + k ∑ e ∑ cij (aj − m), i=1 j=2 − k ≥ 0, proving that x ∈. 2 Lemma Let S be a numerical semigroup with the maximal ideal M and the multiplicity m . The following are equivalent for any positive integer h : (i) (h + 1)M = m + hM , (ii) L(S) = hM − hM , (iii) L(S) = −hm + hM , (iv) |hM \ (h + 1)M | = m. Proof (i) ⇒ (ii) As in the proof of Lemma , this gives (h