Cyclotomic numerical semigroup polynomials with at most two irreducible factors
aa r X i v : . [ m a t h . A C ] J a n Cyclotomic numerical semigroup polynomials with fewirreducible factors
Alessio Borz`ı, Andr´es Herrera-Poyatos, Pieter MoreeJanuary 25, 2021
Abstract
A numerical semigroup S is cyclotomic if its semigroup polynomial P S is a product ofcyclotomic polynomials. The number of irreducible factors of P S (with multiplicity) is thepolynomial length ℓ ( S ) of S. We show that a cyclotomic numerical semigroup is completeintersection if ℓ ( S ) ≤
2. This establishes a particular case of a conjecture of Ciolan,Garc´ıa-S´anchez and Moree (2016) claiming that every cyclotomic numerical semigroupis complete intersection. In addition, we investigate the relation between ℓ ( S ) and theembedding dimension of S. The n -th cyclotomic polynomial is the minimal polynomial of any primitive n -th root of unityΦ n ( x ) = n Y j =1( j,n )=1 (cid:16) x − e πij/n (cid:17) = ϕ ( n ) X k =0 a n ( k ) x k , (1)where ϕ is Euler’s totient function. Its coefficients a n ( k ) are integers.Let N denote the non-negative integers. A numerical semigroup S is an additive submonoidof N with finite complement N \ S . The semigroup polynomial of S is defined by P S ( x ) =1 + ( x − P g ∈ N \ S x g . If p = q are primes and h p, q i is the numerical semigroup generated by p and q , then Φ pq = P h p,q i , (2)see for instance [10]. This identity can be used to reprove various properties of cyclotomicpolynomials, e.g., that a pq ( k ) ∈ { , , − } , a result due to Migotti [9]. More generally, if p and q are two coprime non-negative integers, then P h p,q i is a product of cyclotomic polynomials.There are various other interesting infinite families of numerical semigroups such that theirsemigroup polynomial has only cyclotomic factors. These facts led Ciolan, Garc´ıa-S´anchezand Moree [10] to define cyclotomic numerical semigroups as numerical semigroups whosesemigroup polynomial is a product of cyclotomic polynomials. For this family of numericalsemigroups it is easy to see that the following implications hold:complete intersection = ⇒ cyclotomic = ⇒ symmetric (3)(see Section 2.4). The converse of the second implication of (3) is far from true. In fact, forany odd integer F with F ≥
9, there is a numerical semigroup with Frobenius number F that1s symmetric and non-cyclotomic. This was proven independently by Garc´ıa-S´anchez (in theappendix of [7]), Herrera-Poyatos and Moree [7] and Sawhney and Stoner [13]. In the lattertwo papers it is shown (by quite different methods) that, for every k ≥
5, the polynomial1 − x + x k − x k − + x k , which is the semigroup polynomial of the symmetric numerical semigroup S k = h k, k +1 , . . . , k − i , is not a product of cyclotomic polynomials. Therefore, S k is symmetric, butnot cyclotomic. It was conjectured by Ciolan, Garc´ıa-S´anchez and Moree [4] that the converseof the first implication in (3) holds true, more precisely they made the following conjecture. Conjecture 1.1. [4, Conjecture 1].
A numerical semigroup S is cyclotomic if and only if itis complete intersection. Using the GAP package [5] the authors of [4] verified that Conjecture 1.1 holds true fornumerical semigroups with Frobenius number up to 70. Further, in the context of gradedalgebras, Borz`ı and D’Al`ı [2] prove a version of Conjecture 1.1 for Koszul algebras and forgraded algebras that have an irreducible h -polynomial.In this paper we classify all cyclotomic numerical semigroups such that their semigrouppolynomial has at most two irreducible polynomial factors. Theorem 1.2.
Let S be a cyclotomic numerical semigroup.1. If P S is irreducible, then S = h p, q i with p = q primes and P S = Φ pq .2. If P S is a product of two irreducible polynomials, then either(a) S = h p, q i with p, q distinct primes and P S = Φ pq Φ pq ; or(b) S = h p, q , qr i with p, q, r distinct primes such that p ∈ h q, r i and P S = Φ pq Φ q r . As a byproduct, since the numerical semigroups obtained in Theorem 1.2 are easily seento be complete intersections with the help of gluings (see Section 2), Conjecture 1.1 holds truein the cases studied in Theorem 1.2.
Theorem 1.3.
Suppose that the semigroup polynomial P S has at most two irreducible factors.Then S is cyclotomic if and only if it is complete intersection. Theorem 1.2 motivates the following definition. We define the polynomial length ℓ ( S ) of S as the number of irreducible factors of P S (with multiplicity). We study this quantity inSection 5.1.Our paper is organised as follows. In Section 2 we gather some preliminary material thatis partly expository and will be useful further on. In Section 3 we prove part 1 of Theorem1.2, and in Section 4 we prove part 2. Finally, in Section 5 we pose some conjectures involvingthe polynomial length of cyclotomic numerical semigroups.The computer algebra computations in this paper were done by using Macaulay2 [6], theGAP system [11] and, in particular, the NumericalSgps package [5]. For an introduction to numerical semigroups, see [12].2et S be a numerical semigroup. The embedding dimension e( S ) of S is the cardinality ofthe (unique) minimal generating system of S . The Frobenius number of S is F( S ) = max( N \ S ). For example, if S = h a, b i with b > a > S ) + 1 = ( a − b − T ⊆ S are two numerical semigroups, then clearly F( S ) ≤ F( T ), proves the following lemma. Lemma 2.1. If a, b ∈ S with gcd( a, b ) = 1 , then F( S ) + 1 ≤ ( a − b − . The
Hilbert series of S is H S ( x ) = X s ∈ S x s ∈ Z [[ x ]] , and the semigroup polynomial of S isP S ( x ) = (1 − x )H S ( x ) = 1 + ( x − X g ∈ N \ S x g . (4)Note that deg P S = F( S ) + 1. The second equality in (4) easily follows from H S ( x ) + P g ∈ N \ S x g = 1 / (1 − x ) , where here and in the sequel we work in Z [[ x ]] and use the shorthand1 / (1 − x ) for 1 + x + x + x + · · · . Some properties of a numerical semigroup, such as symmetry, can be captured in terms ofits semigroup polynomial. We need some notation in order to characterize symmetry in thisway. A numerical semigroup S is symmetric if for every integer n we have that n ∈ S if andonly if F( S ) − n / ∈ S . A polynomial f ( x ) = P di =0 α i x i ∈ Z [ x ] is palindromic (or self-reciprocal )if f ( x ) = x d f ( x − ), that is, its coefficients satisfy the relation α d − i = α i . Theorem 2.2. [10, Theorem 5].
A numerical semigroup S is symmetric if and only if P S ispalindromic. Let S , S and S be numerical semigroups, and let a ∈ S and a ∈ S be coprime integerssuch that they are not minimal generators of their respective semigroups. We say that S is the gluing of S and S at a a if S = a S + a S and we write S = a S + a a a S . Wewill keep this notation until the end of this section. By [12, Lemma 9.8] we have thate( S ) = e( S ) + e( S ) . (5)Let M be a submonoid of N , and let m ∈ M . The Ap´ery set of M at m is the setAp( M, m ) = { v ∈ M : v − m / ∈ M } . Gluings can be characterised in terms of Ap´ery sets as in Lemma 2.3.
Lemma 2.3. [12, Theorem 9.2].
The numerical semigroup S is the gluing of S and S at m = a a , if and only if the map Ap( a S , m ) × Ap( a S , m ) → Ap(
S, m ) given by ( v, w ) v + w is bijective. If this is the case, then we have Ap(
S, m ) = Ap( a S , m )+Ap( a S , m ) . X w ∈ Ap(
S,m ) x w = (1 − x m ) H S ( x ) , (6)which follows from S = Ap( S, m ) + m N . Proposition 2.4 ([1, Corollary 4.4]) . The following statements are equivalent:1. S = a S + a a a S ;2. H S ( x ) = (1 − x a a )H S ( x a )H S ( x a ) ;3. P S ( x ) = P h a ,a i ( x )P S ( x a )P S ( x a ) .Proof. The equivalence of (2) and (3) is trivial, so we just need to prove the equivalence of(1) and (2). From Lemma 2.3 we infer that S = a S + a a a S if and only if X w ∈ Ap(
S,a a ) x w = X w ∈ Ap( a S ,a a ) X w ∈ Ap( a S ,a a ) x w + w == X w ∈ Ap( S ,a ) x a w X w ∈ Ap( S ,a ) x a w . Now apply (6) and divide both sides by 1 − x a a . Complete intersection numerical semigroups are usually introduced in the context of min-imal presentations [12, Chapter 8], where a numerical semigroup S is said to be completeintersection if the cardinality of a minimal presentation of S is equal to e( S ) −
1. In thispaper we will only need the recursive characterisation of complete intersection numericalsemigroups in terms of gluings.
Theorem 2.5. [12, Theorem 9.10].
A numerical semigroup is complete intersection if andonly if S is N or S is a gluing of two complete intersection numerical semigroups. Using this characterisation in terms of gluings along with Proposition 2.4 one can deter-mine the Hilbert series of S as follows. Corollary 2.6 ([1, Theorem 4.8]) . Let S = h n , . . . , n e i be a complete intersection numericalsemigroup. Then there are d , . . . , d e − ∈ S \ { n , . . . , n e } such that H S ( x ) = (1 − x d ) . . . (1 − x d e − )(1 − x n ) . . . (1 − x n e ) . The integers d i in the previous result are actually the Betti elements of S (with multiplic-ity). We refer to [12, Chapter 7] for a definition of Betti element, which will not be needed inthe remainder of this paper. If e( S ) = 2, so S = h a, b i = a N + ab b N , then Corollary 2.6 andthe fact that deg P S = F( S ) = ( a − b −
1) yieldP h a,b i ( x ) = (1 − x )(1 − x ab )(1 − x a )(1 − x b ) . (7)In the case when e( S ) = 3, we have the following result due to Herzog. Theorem 2.7. [8, Theorem 4.2.1].
Let S be a numerical semigroup. If e( S ) = 3 , then S iscomplete intersection if and only if it is symmetric. .3 Cyclotomic polynomials For an introduction to cyclotomic polynomials, see [15].From the definition (1) we have x n − Y d | n Φ d ( x ) . (8)This in combination with (7) yields for example thatP h a,b i = Y d | ab, d ∤ a, d ∤ b Φ d . (9)An important property of the cyclotomic polynomials is that they are irreducible over therationals, several famous mathematicians gave different proofs of this, cf. Weintraub [16].Hence, (8) gives the factorization of x n − M¨obius inversion formula (see [14, Proposition 3.7.1]) we infer from (8)that Φ n ( x ) = Y d | n ( x d − µ ( n/d ) , (10)where the M¨obius function µ is defined by µ ( n ) = n = 1;( − k n = p · · · p k with the p i distinct primes;0 otherwise . By taking degrees we obtain ϕ ( n ) = P d | n dµ ( n/d ). If n >
1, then P d | n µ ( n/d ) = 0, soequation (10) can be rewritten asΦ n ( x ) = Y d | n (1 − x d ) µ ( n/d ) . (11)Recall that a polynomial f of degree d is palindromic if f ( x ) = x d f ( x − ). As, for n > x ϕ ( n ) Φ n ( 1 x ) = x P d | n dµ ( n/d ) Y d | n (cid:0) − x d (cid:1) µ ( n/d ) = Y d | n ( x d − µ ( n/d ) = Φ n ( x ) , (12)we see that Φ n ( x ) is palindromic for n > n > n (0) = 1 andΦ n ( x ) ≡ − µ ( n ) x (mod x ) . (13)Let r be any natural number. It can also be deduced from (11) thatΦ nr ( x ) = Φ nr ( x r ) . (14)5 .4 Cyclotomic numerical semigroups A numerical semigroup S is cyclotomic if P S is a product of cyclotomic polynomials, that is,P S = Q d ∈D Φ f d d , for some finite set D , with f d ≥
1. As P S (1) = 1 , Φ does not appear in thisproduct.Corollary 2.6 gives rise to the following result of Ciolan et al. [4]. Corollary 2.8.
Every complete intersection numerical semigroup is cyclotomic.Proof.
By Corollary 2.6 and (8) we have P S = (1 − x ) H S ( x ) = Q Φ f n n , with possibly f n < . However, this would imply that P S ( x ) has a pole at x = e πi/n , contradicting the fact thatP S is a polynomial.Observe that the product of two palindromic polynomials is palindromic. Hence, byapplying Theorem 2.2 and recalling that Φ n is palindromic for n > Corollary 2.9 ([4, Theorem 1]) . Every cyclotomic numerical semigroup is symmetric.
Corollary 2.10 ([4, Lemma 7]) . Conjecture 1.1 holds true for those S with e( S ) ≤ .Proof. This follows from Corollary 2.9 and Theorem 2.7.To conclude this section, we note that the cyclotomicity of numerical semigroups is pre-served under gluing.
Corollary 2.11. If S is the gluing of S and S at a a , then S is cyclotomic if and only ifboth S and S are cyclotomic.Proof. This follows from Proposition 2.4. In this section we classify cyclotomic numerical semigroups having irreducible semigroup poly-nomial. They are given in Corollary 3.3.
Lemma 3.1.
Let S be a numerical semigroup such that Φ n divides P S for some n . If p, q ∈ S are two different primes dividing n , then S = h p, q i and P S = Φ pq .Proof. Recall that deg Φ n = ϕ ( n ), see (1), and deg P S = F( S ) + 1. Hence,( p − q − ≤ ϕ ( n ) ≤ deg P S = F( S ) + 1 . By Lemma 2.1 with a = p and b = q it follows that( p − q − ≤ ϕ ( n ) ≤ F( S ) + 1 ≤ ( p − q − , and we conclude that P S = Φ n and ϕ ( n ) = ϕ ( pq ). Writing n = kpq we have ϕ ( pq ) = ϕ ( n ) = ϕ ( kpq ) ≥ ϕ ( k ) ϕ ( pq ) and hence ϕ ( k ) = 1 , implying k = 1 or k = 2. Consequently, n = pq or n = 2 pq . From (13) and the equality P S ( x ) ≡ Φ n ( x ) (mod x ) , we obtain µ ( n ) = 1 and so n = pq and hence S = h p, q i . Theorem 3.2.
Let S be a numerical semigroup such that P S ( x ) = Φ n ( x j ) h . Then S = h p, q i with p = q primes, n = pq and j = h = 1 . roof. On the one hand we have P S ( x ) ≡ − x (mod x ), and on the other Φ n ( x j ) k ≡ − µ ( n ) hx j (mod x j ) . We conclude that µ ( n ) = j = h = 1, so P S ( x ) = Φ n ( x ). From1 = P S (1) = Φ n (1) we infer n ≥
2. Thus n is a product of an even number of distinctprimes p < p < . . . < p k , that is n = p · · · p k . From (11) and the fact that if d | n , then µ ( n/d ) = µ ( d ), we obtain Φ n ( x ) Y d | nµ ( d )= − (1 − x d ) = Y d | nµ ( d )=1 (1 − x d ) . (15)Recall that Φ n ( x ) = P S ( x ) = (1 − x )H S ( x ). On dividing both sides of (15) by 1 − x andreducing the resulting identity modulo x p +1 , we find that(1 − x p )(1 − x p )H S ( x ) ≡ x p +1 ) , which can be rewritten asH S ( x ) ≡ x p H S ( x ) + x p H S ( x ) (mod x p +1 ) . We deduce that both p and p are in S . Consequently, we obtain S = h p , p i by Lemma3.1. Corollary 3.3 (Part 1 of Theorem 1.2) . A cyclotomic numerical semigroup S has irreduciblesemigroup polynomial if and only if S = h p, q i for some distinct primes p and q . If this is thecase, then P S = Φ pq . In this section we classify the cyclotomic numerical semigroups with polynomial length 2, asit was announced in part 2 of Theorem 1.2. As a consequence of this result, it follows thatevery cyclotomic numerical semigroup with polynomial length 2 is complete intersection. Ourproof of part 2 of Theorem 1.2 uses the following three lemmas.
Lemma 4.1.
Let S be a cyclotomic numerical semigroup. Hence P S ( x ) = Y d ∈D Φ d ( x ) f d for some finite set of positive integers D and positive integers f d . Then we have X d ∈D f d µ ( d ) = 1 . In particular, there exists an integer d > such that µ ( d ) = 1 and Φ d | P S .Proof. Since P S (1) = 1, we have 1
6∈ D . In view of (13), we obtainP S ( x ) ≡ − x X d ∈D f d µ ( d ) (mod x ) . Recalling that P S ( x ) ≡ − x (mod x ), it follows that 1 = P d ∈D f d µ ( d ), and also that theremust be some integer d > D with µ ( d ) = 1.7 emma 4.2. Let S be a numerical semigroup such that P S ( x ) = Φ n ( x ) f ( x q ) (16) for some integers n, q > such that µ ( n ) = 1 and f ( x ) ∈ Z [ x ] is of positive degree. Then q isa prime number and n = pq for some other prime p ∈ S .Proof. Since µ ( n ) = 1 by assumption, we can write n = p · · · p k with p < p < . . . < p k primes. Using (11) and reducing the resulting expression modulo x p +1 we obtain from (16)H S ( x )(1 − x p )(1 − x p ) ≡ f ( x q ) (mod x p +1 ) , which can be rewritten asH S ( x ) ≡ f ( x q ) + x p H S ( x ) + x p H S ( x ) (mod x p +1 ) . (17)Since p and p can not belong to S at the same time by Lemma 3.1, we see that f ( x q ) containsa monomial with exponent p or p . Consequently, q divides p or p , that is, q ∈ { p , p } .Furthermore, if p ∈ { p , p } \ { q } , then q does not divide p and we find that p ∈ S by (17).Note that { p , p } = { p, q } . It remains to show that n = p p = pq . In order to obtain a contradiction we assume that k >
1. Using (11) and reducing the resulting expression modulo x p +1 , we obtain from (16)H S ( x )(1 − x p )(1 − x q )(1 − x p ) ≡ f ( x q )(1 − x pq ) (mod x p +1 ) , which can be simplified toH S ( x ) ≡ x p H S ( x ) + x p H S ( x ) + f ( x q ) (1 − x pq )(1 − x q ) (mod x p +1 ) . (18)Note that f ( x q ) (1 − x pq )(1 − x q ) = g ( x q ) , for some g ( x ) ∈ Z [ x ] . Since q does not divide p , we must have p ∈ S in view of (18). Since p ∈ S , we conclude by Lemma 3.1 with a = p and b = p that S = h p, p i and hence P S = Φ pp is irreducible, whereas by assumption it has at least two irreducible factors. Lemma 4.3.
Let S be a numerical semigroup such that P S ( x ) = Φ pq ( x )Φ l ( x q ) for some distinct prime numbers p and q , and l a multiple of q . Then either1. S = h p, q i and l = pq ; or2. S = h p, q , qr i for some prime number r such that r
6∈ { p, q } and p ∈ h q, r i , and l = qr .Proof. We can write H S ( x ) = Φ l ( x q ) 1 − x pq (1 − x q )(1 − x p ) = Φ l ( x q )1 − x q q − X i =0 x ip . (19)8ote that Φ l ( x ) / (1 − x ) ∈ Z [[ x ]] . We write it as P ∞ j =0 a j x j . We consider S ′ = { j ∈ N : a j = 0 } .Since Φ l (0) = 1, it follows that a = 1 and hence 0 ∈ S ′ . We are going to prove that S ′ is anumerical semigroup with P S ′ = Φ l . Equation (19) can be rewritten in Z [[ x ]] asH S ( x ) = ∞ X j =0 a j x jq q − X i =0 x ip ! = ∞ X j =0 a j q − X i =0 x jq + ip . (20)Since p and q are prime numbers, if jq + ip = j ′ q + i ′ p with j, j ′ ∈ N and i, i ′ ∈ { , , . . . , q − } ,then i = i ′ and j = j ′ . Consequently, a j is the qj -th coefficient of H S and hence a j ∈ { , } .Note that j ∈ S ′ if and only if qj ∈ S . Since S is a numerical semigroup, we see that S ′ isclosed under addition. Furthermore, we have (1 − x )H S ′ ( x ) = Φ l ( x ). As a consequence, S ′ is anumerical semigroup with polynomial Φ l . By Theorem 3.2 and the assumption q | l, we obtain l = qr , where r is a prime number different from q , and S ′ = h q, r i . Summarizing, we haveP S ( x ) = Φ pq ( x )Φ qr ( x q ) = Φ pq ( x )Φ q r ( x ), where we used (14), and H S ( x ) = H S ′ ( x q ) P q − i =0 x ip .From the latter equality, we obtain S = qS ′ + p N . There are two possibilities: • r = p . Then P S = Φ pq Φ pq = P h p,q i , where in the latter equality we used (9). That is, S = h p, q i . • r = p . Then S = q h r, q i + qp p N is a gluing and, thus, S = h p, q , qr i . Proof of part 2 of Theorem 1.2.
By Lemma 4.1, our assumption ℓ ( S ) = 2 implies that we canwrite P S = Φ n Φ m with µ ( n ) = 1 and µ ( m ) = 0 . Let q be the smallest prime such that q | m and put l = m/q. On applying (14) we findP S ( x ) = Φ n ( x )Φ l ( x q ) . (21)Note that n, l > . By Lemma 4.2 it now follows that P S ( x ) = Φ pq ( x )Φ l ( x q ) for some prime p = q. The proof is completed on invoking Lemma 4.3 (note that q | l ). Let S be a numerical semigroup. Recall that we define the polynomial length ℓ ( S ) of S asthe number of irreducible factors of P S (with multiplicity). If q > p are two primes we haveP h p,q i ( x ) = Φ pq ( x ) by (2) and hence the polynomial length of h p, q i is 1. This observation isgeneralized in Example 5.1. This example involves some more notation that we now introduce.Let d ( n ) = P d | n n . We have d ( ab ) ≤ d ( a ) d ( b ) withequality if a and b are coprime. By ir( f ) we denote the number of irreducible prime factorsof f, and so ℓ ( S ) = ir( P S ) . A fundamental observation we will use is thatir( x n −
1) = d ( n ) (22)(this is a consequence of (8) and the irreducibility of cyclotomic polynomials). Using this and(10) we obtain the (known) identityir(Φ n ) = 1 = X δ | n d ( δ ) µ ( n/δ ) . S is complete intersection with minimal generators n , . . . , n e , and Betti elements b , . . . , b e − (with multiplicity), then from Corollary 2.6 and (22) we have ℓ ( S ) = e − X j =1 d ( b j ) − e X j =1 d ( n j ) + 1 . (23) Example 5.1.
Let b > a > h a, b i is ab , see(7), so we have ir( P h a,b i ( x )) = d ( ab ) − d ( a ) − d ( b ) + 1 . From (7) and the multiplicativity of the sum of divisors function d we find that l ( h a, b i ) = ( d ( a ) − d ( b ) − . If S = a S + a a a S , then from Proposition 2.4 and the latter example, we obtain theinequality ℓ ( S ) ≥ ℓ ( S ) + ℓ ( S ) + ( d ( a ) − d ( a ) − . (24) Remark 5.2.
This lower bound is sharp. Let S = h , i and S = h , i . Recall that S and S have length 1 (Corollary 3.3). We consider the following gluing of S and S , S = 12 h , i + 5 h , i . Then the polynomial of S isP S ( x ) = P h , i ( x )Φ ( x )Φ ( x ) = P h , i ( x )Φ ( x )Φ ( x ) , and ℓ ( S ) = ( d (12) − d (5) −
1) + ℓ ( S ) + ℓ ( S ) = 7. Proposition 5.3. If S is a complete intersection numerical semigroup, then we have e( S ) ≤ ℓ ( S ) + 1 .Proof. We proceed by induction on e( S ). If e( S ) = 2, then the result is trivial. Now letus assume that the result is true for every numerical semigroup with embedding dimensionsmaller than e( S ) >
2. The numerical semigroup S is a gluing of two complete intersectionnumerical semigroups S and S by Theorem 2.5. From (5), our induction hypothesis, and(24) we have e( S ) = e( S ) + e( S ) ≤ ℓ ( S ) + ℓ ( S ) + 2 ≤ ℓ ( S ) + 1 . The next result shows that the inequality in Proposition 5.3 is sharp.
Proposition 5.4.
Let e ≥ be an integer. For every l ≥ e − there exists a completeintersection numerical semigroup S such that ℓ ( S ) = l and e( S ) = e .Proof. For every k ≥
1, we inductively construct a family of numerical semigroups S ( e ) k , suchthat e ( S ( e ) k ) = e and ℓ ( S ( e ) k ) = e + k −
2, as follows: • S (2) k = h p k , p i for some distinct primes p and p ; • S ( e +1) k = p S ( e ) k + p e +1 p p e +1 N for some prime p e +1 ∈ S ( e ) k that is not a minimal generator.By (5) we conclude that e ( S ( e ) k ) = e ( S ( e − k ) + 1 = · · · = e ( S (2) k ) + e − e. P S ( e ) k ( x ) = P S ( e − k ( x p )Φ p p e ( x ). Applying this formula recur-sively we obtain P S ( e ) k ( x ) = P S (2) k ( x p e − ) e Y i =3 Φ p p i ( x p e − i ) . Now, inserting P S (2) k ( x ) = Q kj =1 Φ p j p ( x ) , which follows by (9), and applying (14), we inferthat P S ( e ) k ( x ) = e Y i =3 Φ p e − i +11 p i ( x ) k Y j =1 Φ p e + j − p ( x ) , and hence ℓ ( S ( e ) k ) = e + k − Conjecture 5.5.
Let S be a cyclotomic numerical semigroup. Then e( S ) ≤ ℓ ( S ) + 1 . Using Proposition 5.3 we see thatConjecture 1 . ⇒ Conjecture 5 . . Assuming Conjecture 5.5 holds true, the proof of Theorem 1.2 can be greatly simplified.Namely, by assumption ℓ ( S ) ≤
2, so we would have e( S ) ≤ ℓ ( S ) + 1 ≤ S wouldbe complete intersection by Corollary 2.10.To conclude this section, we have computed all the cyclotomic numerical semigroups withFrobenius number at most 70, and classified them in terms of their polynomial length. Thesecyclotomic numerical semigroups are complete intersections, as it was computationally checkedin [4], and there are 835 of them. The results are displayed in Table 1. The largest polynomiallength found among these semigroups is 8. Recall that Sections 3 and 4 of the present paperstudy cyclotomic numerical semigroups of polynomial length at most 2, which add up to 138semigroups out of the 835 found.Table 1: Number of cyclotomic numerical semigroups with Frobenius num-ber at most 70, grouped by their polynomial length.Length 1 2 3 4 5 6 7 8Number of semigroups 33 105 224 196 165 74 34 4 Let S be a numerical semigroup. Since P S (0) = 1, there exist unique integers e j such thatthe formal identity P S ( x ) = ∞ Y j =1 (cid:0) − x j (cid:1) e j (25)holds [3, Lemma 3.1]. The sequence e = { e j } j ∈ N is known as the cyclotomic exponent sequence of S . This sequence was introduced in [4, Section 6] and later studied in [3]. From (25) andthe uniqueness of the exponents e , one can show that S is a cyclotomic numerical semigroup11f and only if e has only a finite number of non-zero elements, see [3, Proposition 2.4] fordetails. If this is the case, then, by (22), we obtain ℓ ( S ) = ∞ X j =1 e j d ( j ) . This equality generalizes equation (23), which gives the length of a complete intersectionnumerical semigroup.One of the main results of [3] is the following.
Theorem 5.6 ([3, Theorem 1.1]) . Let S = N be a numerical semigroup and let e be itscyclotomic exponent sequence. Then1. e = 1; e j = 0 for every j ≥ not in S ; e j = − for every minimal generator j of S ;4. e j = 0 for every j in S that has only one factorization and is not a minimal generator. As a consequence of this theorem, the set { n ∈ N : e n < } is a system of generatorsof S . In the case of cyclotomic numerical semigroups, the authors of [3] made the followingconjecture. Conjecture 5.7 ([3, Conjecture 7.1]) . Let S be a cyclotomic numerical semigroup and let e be its cyclotomic exponent sequence. Then n ∈ N is a minimal generator of S if and only if e n < . The cyclotomic exponent sequence of a complete intersection numerical semigroup can beeasily obtained from Corollary 2.6. Note that for these semigroups the only integers with e n < . ⇒ Conjecture 5 . . In the rest of this section we relate Conjectures 5.5 and 5.7. In order to do so, we formulatea further conjecture.
Conjecture 5.8.
Let S be a cyclotomic numerical semigroup. Hence P S ( x ) = Y d ∈D Φ d ( x ) f d for some finite set of positive integers D and positive integers f d . Let e the cyclotomic exponentsequence of S . Then { d ≥ e d > } ⊆ D , and e d ≤ f d for every d ≥ with e d > . In addition, we will need the following proposition, the proof of which we include forcompleteness.
Proposition 5.9 ([3, Proposition 2.3]) . Let S be a numerical semigroup and let e be itscyclotomic exponent sequence. If S is cyclotomic, then P j ≥ e j = 0 . roof. Let N be the largest index j such that e j = 0 . Then we haveP S ( x ) = (1 − x ) P j ≤ N e j G S ( x ) , for some rational function G S ( x ) satisfying G S (1)
6∈ { , ∞} (in fact G S (1) = Q j ≤ N j e j ). SinceP S (1) = 1, it follows that P j ≥ e j = 0. Proposition 5.10.
The following implications hold:Conjecture 1.1 = ⇒ Conjecture 5.7 = ⇒ Conjecture 5.8 = ⇒ Conjecture 5.5.Proof.
The first implication follow from Corollary 2.6. Let S be a cyclotomic numerical semi-group and let n , . . . , n e be its minimal generators. Let us assume that S satisfies Conjecture5.7. Then we have e Y i =1 (1 − x n i )P S ( x ) = Y d ∈ N ; e d > (1 − x d ) e d Let d ∈ N with e d > d ≥
2. We are going to prove that Φ e d d divides P S . Recall that Φ e d d divides (1 − x d ) e d exactly. Now we argue that Φ d does not divide Q ei =1 (1 − x n i ). Note thatΦ d divides Q ei =1 (1 − x n i ) if and only if there exists i ∈ { , , . . . , e } such that d divides n i . ByTheorem 5.6 it follows that d = n i and d ∈ S. Since by assumption n i is a minimal generatorof S , we conclude that d does not divide n i . This along with the fact that Φ d is irreducibleallows us to conclude that Φ e d d divides P S and, hence, e d ≤ f d .Now let us assume that S satisfies Conjecture 5.8. Since { d ∈ N : e d < } is a (finite)system of generators of S by Theorem 5.6, we find thate( S ) ≤ X d ∈ N ; e d < ( − e d ) . From Proposition 5.9, we obtain X d ∈ N ; e d < ( − e d ) = X d ∈ N ; e d > e d . Finally, because we are assuming that S satisfies Conjecture 5.8, we have X d ∈ N ; e d > e d ≤ e + X d ∈D f d = ℓ ( S ) + 1 . We conclude that e( S ) ≤ ℓ ( S ) + 1. Acknowledgement.
The authors thank Pedro A. Garc´ıa-S´anchez for putting the authors incontact with each other and for helpful conversations on this work.
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Alessio Borz`ı
Mathematics Institute, University of Warwick, Coventry CV4 7AL, UnitedKingdom.Andr´es Herrera-Poyatos [email protected]
Department of Computer Science, University of Oxford, Wolfson Building,Parks Road, Oxford, OX1 3QD, United Kingdom.Pieter Moree [email protected]