Quantum integers and cyclotomy

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Journal of Number Theory 109 (2004) 120 135 www.elsevier.com/locate/jnt Quantum integers and cyclotomy Alexander Borisov a, Melvyn B.

1 Journal of Number Theory 109 (2004) Quantum integers and cyclotomy Alexander Borisov a, Melvyn B. Nathanson b,, Yang Wang c a Department of Mathematics, Pennsylvania State University, University Park, PA 16802, USA b Department of Computer Science andmathematics, Lehman College (CUNY), Bronx, NY 10468, USA c Department of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332, USA Received 6 October 2003; revised 5 May 2004 Communicated by A. Granville Abstract A sequence of functions F ={f n (q)} n=1 satisfies the functional equation for multiplication of quantum integers if f mn (q) = f m (q)f n (q m ) for all positive integers m and n. This paper describes the structure of all sequences of rational functions with coefficients in Q that satisfy this functional equation Elsevier Inc. All rights reserved. MSC: primary 39B05; 81R50; 11R18; 11T22; 11B13 Keywords: Quantum integers; Quantum polynomial; Polynomial functional equation; Cyclotomic polynomials 1. The functional equation for multiplication of quantum integers Let N ={1, 2, 3,...} denote the positive integers. For every n N, we define the polynomial [n] q = 1 + q + q 2 + +q n 1. Corresponding author. Fax: addresses: borisov@math.psu.edu (A. Borisov), melvyn.nathanson@lehman.cuny.edu (M.B. Nathanson), wang@math.gatech.edu (Y. Wang) X/$ - see front matter 2004 Elsevier Inc. All rights reserved. doi: /j.jnt

2 A. Borisov et al. / Journal of Number Theory 109 (2004) This polynomial is called the quantum integer n. The sequence of polynomials {[n] q } n=1 satisfies the following functional equation: f mn (q) = f m (q)f n (q m ) (1) for all positive integers m and n. Nathanson [1] asked for a classification of all sequences F ={f n (q)} n=1 of polynomials and of rational functions that satisfy the functional equation (1). The following statements are simple consequences of the functional equation. Proofs can be found in Nathanson [1]. Let F ={f n (q)} n=1 be any sequence of functions that satisfies (1). Then f 1(q) = f 1 (q) 2 = 0or1.Iff 1 (q) = 0, then f n (q) = f 1 (q)f n (q) = 0 for all n N, and F is a trivial solution of (1). In this paper, we consider only nontrivial solutions of the functional equation, that is, sequences F ={f n (q)} n=1 with f 1(q) = 1. Let P be a set of prime numbers, and let S(P) be the multiplicative semigroup of N generated by P. Then S(P) consists of all integers that can be represented as a product of powers of prime numbers belonging to P. Let F ={f n (q)} n=1 be a nontrivial solution of (1). We define the support supp(f) ={n N : f n (q) = 0}. There exists a unique set P of prime numbers such that supp(f) = S(P). Moreover, the sequence F is completely determined by the set {f p (q) : p P }. Conversely, if P is any set of prime numbers, and if {h p (q) : p P } is a set of functions such that h p1 (q)h p2 (q p 1 ) = h p2 (q)h p1 (q p 2 ) (2) for all p 1,p 2 P, then there exists a unique solution F ={f n (q)} n=1 of the functional equation (1) such that supp(f) = S(P) and f p (q) = h p (q) for all p P. For example, for the set P ={2, 5, 7}, the reciprocal polynomials h 2 (q) = 1 q + q 2, h 5 (q) = 1 q + q 3 q 4 + q 5 q 7 + q 8, h 7 (q) = 1 q + q 3 q 4 + q 6 q 8 + q 9 q 11 + q 12 satisfy the commutativity condition (2). Since h p (q) = [p] q 3 [p] q for p P,

3 122 A. Borisov et al. / Journal of Number Theory 109 (2004) it follows that f n (q) = [n] q 3 [n] q for all n S(P). (3) Moreover, f n (q) is a polynomial of degree 2(n 1) for all n S(P). Let F ={f n (q)} n=1 be a solution of the functional equation (1) with supp(f) = S(P) If P =, then supp(f) ={1}. It follows that f 1 (q) = 1 and f n (q) = 0 for all n 2. Also, for any prime p and any function h(q), there is a unique solution of the functional equation (1) with supp(f) = S({p}) and f p (q) = h(q). Thus, we only need to investigate solutions of (1) for card(p ) 2. If F ={f n (q)} n=1 and G ={g n(q)} n=1 are solutions of (1) with supp(f)=supp(g), then, for any integers d, e, r, and s, the sequence of functions H ={h n (q)} n=1, where h n (q) = f n (q r ) d g n (q s ) e, is also a solution of the functional equation (1) with supp(h) = supp(f). In particular, if F ={f n (q)} n=1 is a solution of (1), then H ={h n(q)} n=1 is another solution of (1), where h n (q) = The functional equation also implies that for all positive integers m and n, and { 1/fn (q) if n supp(f), 0 if n supp(f). f m (q)f n (q m ) = f n (q)f m (q n ) (4) k 1 f m k(q) = f m (q mi ). (5) i=0 Let F ={f n (q)} n=1 be a solution in rational functions of the functional equation (1) with supp(f) = S(P). Then there exist a completely multiplicative arithmetic function λ(n) with support S(P) and rational numbers t 0 and t 1 with t 0 (n 1) Z and t 1 (n 1) Z for all n S(P) such that, for every n S(P), we can write the rational function f n (q) uniquely in the form f n (q) = λ(n)q t 0(n 1) u n(q) v n (q), (6) where u n (q) and v n (q) are monic polynomials with nonzero constant terms, and

4 A. Borisov et al. / Journal of Number Theory 109 (2004) deg(u n (q)) deg(v n (q)) = t 1 (n 1) for all n supp(f). For example, let P be a set of prime numbers with card(p ) 2. Let λ(n) be a completely multiplicative arithmetic function with support S(P), and let t 0 be a rational number such that t 0 (n 1) Z for all n S(P). Let R be a finite set of positive integers and {t r } r R a set of integers. We construct a sequence F ={f n (q)} n=1 of rational functions as follows: For n S(P), we define f n (q) = λ(n)q t 0(n 1) r R[n] t r q r. (7) For n S(P) we set f n (q) = 0. Then r R [n]t r q r is a quotient of monic polynomials with coefficients in Q and nonzero constant terms. The sequence F ={f n (q)} n=1 satisfies the functional equation (1), and supp(f) = S(P). We shall prove that every solution of the functional equation (1) in rational functions with coefficients in Q is of the form (7). This provides an affirmative answer to Problem 6in[1] in the case of the field Q. 2. Roots of unity and solutions of the functional equation Let K be an algebraically closed field, and let K denote the multiplicative group of nonzero elements of K. Let Γ denote the group of roots of unity in K, that is, Γ ={ζ K : ζ n = 1 for some n N}. Since Γ is the torsion subgroup of K, every element in K \ Γ has infinite order. We define the logarithm group and the map by L(K) = K /Γ L : K L(K) L(a) = aγ for all a K. We write the group operation in L(K) additively L(a) + L(b) = aγ + bγ = abγ = L(ab).

5 124 A. Borisov et al. / Journal of Number Theory 109 (2004) Lemma 1. Let K be an algebraically closedfield, and L(K) its logarithm group. Then L(K) is a vector space over the field Q of rational numbers. Proof. Let a K and m/n Q. Since K is algebraically closed, there is an element b K such that b n = a m. We define Suppose m/n = r/s Q, and that m L(a) = L(b). n c s = a r for some c K. Since ms = nr, it follows that c ms = a mr = b nr = b ms, and so c/b Γ. Therefore, m n L(a) = L(b) = bγ = cγ = L(c) = r s L(a) and (m/n)l(a) is well defined. It is straightforward to check that L(K) is a Q-vector space. Lemma 2. Let P be a set of primes, card(p ) 2, andlet S(P) be the multiplicative semigroup generatedby P. For every integer m S(P) \{1} there is an integer n S(P) such that log m and log n are linearly independent over Q. Equivalently, for every integer m S(P)\{1} there is an integer n S(P) such that there exist integers r ands with m r = n s if andonly if r = s = 0. Proof. If m = p k is a prime power, let n be any prime in P \{p}. If m is divisible by more than one prime, let n be any prime in P. The result follows immediately from the fundamental theorem of arithmetic. Let K be a field. A function on K is a map f : K K { }. For example, f(q) could be a polynomial or a rational function with coefficients in K. We call f 1 (0) the set of zeros of f and f 1 ( ) the set of poles of f.

6 A. Borisov et al. / Journal of Number Theory 109 (2004) Theorem 1. Let K be an algebraically closedfield. Let F ={f n (q)} n=1 be a sequence of functions on K that satisfies the functional equation (1). Let P be the set of primes such that supp(f) = S(P). If card(p ) 2 andif, for every n supp(f), the function f n (q) has only finitely many zeros andonly finitely many poles, then every zero and pole of f n (q) is either 0 or a root of unity. Proof. The proof is by contradiction. Let Γ be the group of roots of unity in K. Suppose that f n (a) = 0 for some n supp(f) and a K \ Γ. By Lemma 2, there is an integer m S(P) such that log m and log n are linearly independent over Q. Since a has infinite order in the multiplicative group K and fn 1(0) is finite, there are positive integers k and M = mk such that a M is not a zero of the function f n (q). By (4), we have f M (q)f n (q M ) = f n (q)f M (q n ). Therefore, f M (a)f n (a M ) = f n (a)f M (a n ) = 0. Since f n (a M ) = 0, it follows from (5) that k 1 0 = f M (a) = f m k(a) = f m (a mi ) and so i=0 f m (a mi ) = 0 for some i such that 0 i k 1. Let b = a mi. Then f m (b) = 0, b K \ Γ,

7 126 A. Borisov et al. / Journal of Number Theory 109 (2004) and L(b) = m i L(a) (8) Since fm 1(0) is finite, there are positive integers l and N = nl such that z N is not a zero of f m (q) for every z fm 1(0) with z K \ Γ. Since K is algebraically closed, we can choose c K such that c N = b. Then f m (c) = 0, c K \ Γ and NL(c) = L(b). (9) Again applying (4), we have f m (q)f N (q m ) = f N (q)f m (q N ) and so f m (c)f N (c m ) = f N (c)f m (c N ) = f N (c)f m (b) = 0. It follows that l 1 0 = f N (c m ) = f n l(c m ) = f n (c mnj ) and so f n (c mnj ) = 0 for some j such that 0 j l 1. Let a = c mnj. j=0

8 A. Borisov et al. / Journal of Number Theory 109 (2004) Then f n (a ) = 0, a K \ Γ, and L(a ) = mn j L(c) (10) Combining (8) (10), we obtain that is, L(a ) = mnj N mi+1 L(b) = L(a) nl j L(a ) = mi n j L(a), where 1 i k and 1 j l. (11) What we have accomplished is the following: Given an element a fn 1 (0) that is neither 0 nor a root of unity, we have constructed another element a fn 1 (0) that is also neither 0 nor a root of unity, and that satisfies (11). Iterating this process, we obtain an infinite sequence of such elements. However, the number of zeros of f n (q) is finite, and so the elements in this sequence cannot be pairwise distinct. It follows that there is an element such that a f 1 (0) \ (Γ {0}) n L(a) = mr n s L(a), where r and s are positive integers. Then ( ) ( ) a ns Γ = L a ns = n s L(a) = m r L(a) = L a mr = a mr Γ. Since a is not a root of unity, it follows that m r = n s,

9 128 A. Borisov et al. / Journal of Number Theory 109 (2004) which contradicts the linear independence of log m and log n over Q. Therefore, the zeros of the functions f n (q) belong to Γ {0} for all n supp(f). Replacing the sequence F ={f n (q)} n supp(f) with F ={1/f n (q)} n supp(f), we conclude that the poles of the functions f n (q) also belong to Γ {0} for all n supp(f). This completes the proof. 3. Rational solutions of the functional equation In this section, we shall completely classify sequences of rational functions with rational coefficients that satisfy the functional equation for quantum multiplication. For k 1, let Φ k (q) denote the kth cyclotomic polynomial. Then F k (q) = q k 1 = d k Φ d (q) and Φ k (q) = d k F d (q) μ(k/d), (12) where μ(k) is the Möbius function. Let ζ be a primitive dth root of unity. Then F k (ζ) = 0 if and only if d is a divisor of k. We define Note that F 0 (q) = Φ 0 (q) = 1. F k (q) = q k 1 = (q 1)(1 + q + +q k 1 ) = F 1 (q)[k] q (13) for all k 1. A multiset U = (U 0, δ) consists of a finite set U 0 of positive integers and a function δ : U 0 N. The positive integer δ(u) is called the multiplicity of u. Multisets U = (U 0, δ) and U = (U 0, δ ) are equal if U 0 = U 0 and δ(u) = δ (u) for all u U 0. Similarly, U U if U 0 U 0 and δ(u) δ (u) for all u U 0. The multisets U and U are disjoint if U 0 U 0 =. We define f u (q) = f u (q) δ(u) u U u U 0 and max(u) = max(u 0 ). If U 0 =, then we set max(u) = 0 and u U f u(q) = 1.

10 A. Borisov et al. / Journal of Number Theory 109 (2004) Lemma 3. Let U and U be multisets of positive integers. Then if andonly if U = U. = u U u U F u (q), (14) Proof. Let k = max(u U ). Let ζ be a primitive kth root of unity. If k U, then F u (ζ) = u U u U F u (ζ) = 0, and so k U. Dividing (14) byf k (q), reducing the multiplicity of k in the multisets U and U by 1, and continuing inductively, we obtain U = U. Let F ={f n (q)} n=1 be a nontrivial solution of the functional equation (1), where f n (q) is a rational function with rational coefficients for all n supp(f). Because of the standard representation (6), we can assume that f n (q) = u n(q) v n (q), where u n (q) and v n (q) are monic polynomials with nonzero constant terms. By Theorem 1, the zeros of the polynomials u n (q) and v n (q) are roots of unity, and so we can write f n (q) = u U n Φ u(q) v V n Φ v(q), where U n and V n are disjoint multisets of positive integers. Applying (12), we replace each cyclotomic polynomial in this expression with a quotient of polynomials of the form F k (q). Then f n (q) = u U n v V n, (15) where U n and V n are disjoint multisets of positive integers. Let f n (q) = u U n v V n = u U n F u (q) v V n F v (q),

11 130 A. Borisov et al. / Journal of Number Theory 109 (2004) where U n and V n are disjoint multisets of positive integers and U n and V n multisets of positive integers. Then are disjoint =. u U n V n v U n V n By Lemma 3, we have the multiset identity U n V n = U n V n. Since U n V n =, it follows that U n U n and so U n = U n. Similarly, V n = V n. Thus, the representation (15) is unique. We introduce the following notation for the dilation of a set: For any integer d and any set S of integers, d S ={ds : s S}. Lemma 4. Let F ={f n (q)} n=1 be a nontrivial solution of the functional equation (1) with supp(f) = S(P), where card(p ) 2. Let f n (q) = u U n v V n and U n and V n are disjoint multisets of positive integers. For every prime p P, let m p = max(u p V p ). There exists an integer r such that m p = rp for every p P. Moreover, either m p U p for all p P or m p V p for all p P. Proof. Let p 1 and p 2 be prime numbers in P, and let Equivalently, m p1 p 1 m p 2 p 2. p 2 m p1 p 1 m p2.

12 A. Borisov et al. / Journal of Number Theory 109 (2004) Applying functional equation (4) with m = p 1 and n = p 2, we obtain where u U p1 v V p1 u U p2 F u (q p 1) v V p2 F v (q p 1) = u U p2 v V p2 u U p1 F u (q p 2) v V p1 F v (q p 2), U p1 V p1 = U p2 V p2 =. The identity F n (q m ) = ( q m) n 1 = q mn 1 = F mn (q), implies that u U p1 p 1 U p2 v V p1 p 1 V p2 = = = u U p1 v V p1 u U p2 v V p2 u U p2 p 2 U p1 v V p2 p 2 V p1. u p 1 U p2 v p 1 V p2 s p 2 U p1 t p 2 V p1 By the uniqueness of the representation (15), it follows that U p1 (p 1 U p2 ) V p2 (p 2 V p1 ) = U p2 (p 2 U p1 ) V p1 (p 1 V p2 ). Recall that m p1 = max ( U p1 V p1 ). If m p1 U p1, then p 2 m p1 p 2 U p1

13 132 A. Borisov et al. / Journal of Number Theory 109 (2004) and so p 2 m p1 U p1 (p 1 U p2 ) V p2 (p 2 V p1 ). However, (i) p 2 m p1 U p1 since p 2 m p1 >m p1 = max ( ) U p1 V p1, (ii) p 2 m p1 p 2 V p1 since m p1 U p1 and U p1 V p1 =, (iii) p 2 m p1 V p2 since p 2 m p1 p 1 m p2 >m p2 = max ( ) U p2 V p2. If p 2 m p1 >p 1 m p2 = max ( ) p 1 U p2, then p2 m p1 p 1 U p2. This is impossible, and so and p 2 m p1 = p 1 m p2 p 1 U p2, m p2 U p2, m p1 p 1 = m p 2 p 2 = r for all p 1,p 2 P. Similarly, if m p1 V p1 for some p 1 P, then m p2 V p2 for all p 2 P. This completes the proof. Theorem 2. Let F ={f n (q)} n=1 be a sequence of rational functions with coefficients in Q that satisfies the functional equation (1). If supp(f) = S(P), where P is a set of prime numbers andcard(p ) 2, then there are (i) a completely multiplicative arithmetic function λ(n) with support S(P), (ii) a rational number t 0 such that t 0 (n 1) is an integer for all n S(P), (iii) a finite set R of positive integers anda set {t r } r R of integers such that f n (q) = λ(n)q t 0(n 1) r R[n] t r q r for all n supp(f). (16) Proof. It suffices to prove (16) for all p P. Recalling the representation (6), we only need to investigate the case f p (q) = u U p v V p,

14 A. Borisov et al. / Journal of Number Theory 109 (2004) where U p and V p are disjoint multisets of positive integers. Let m p = max(u p V p ). By Lemma 4, there is a nonnegative integer m such that m p = mp for all p P. We can assume that m p U p for all p P. The proof is by induction on m. Ifm = 0, then U p = V p = and f p (q) = 1 for all p P, hence (16) holds with R =. Let m = 1, and suppose that m p = p U p for all p P. Then f p (q) = u U p v V p = (q p 1) u U p F u(q). v V p Since q p 1 = F 1 (q)[p] q, we have g p (q) = f p(q) [p] q = (qp 1) u U p \{p} F u(q) [p] q v V p = F 1(q) u U p \{p} F u(q) v V p u U p = F u(q) v V p F v(q), where U p V p =. The sequence of rational functions G ={g n(q)} n=1 is also a solution of the functional equation (1), and either max(u p V p ) = 0 for all p P or max(u p V p ) = p for all p P. If max(u p V p ) = p for all p P, then we construct the sequence H ={h n(q)} n=1 of rational functions h n (q) = g n(q) [n] q = f n(q) [n] 2. q Continuing inductively, we obtain a positive integer t such that f n (q) =[n] t q for all n supp(f). Thus, (16) holds in the case m = 1. Let m be an integer such that the Theorem holds whenever m p <mp for all p P, and let F ={f n (q)} n=1 be a solution of the functional equation (1) with supp(f)=s(p)

15 134 A. Borisov et al. / Journal of Number Theory 109 (2004) and m p = mp and m p U p for all p P. The sequence G ={g n (q)} n=1 with g n (q) = f n(q) [n] q r is a solution of the functional equation (1). Since F rp (q) = q rp 1 = ( q r 1 ) ( 1 + q r + +q r(p 1)) = F r (q)[p] q r, it follows that g p (q) = (qm p 1) u U p \{m p } F u(q) [p] q r v V p = (qmp 1) u U p \{mp} F u(q) [p] q r v V p = F r(q) u U p \{mp} F u(q) v V p u U p = F u(q) v V p F v(q), where U p V p =, and max(u p V p ) mp. If max(u p V p ) = mp, then mp U p. We repeat the construction with h n (q) = g n(q) [n] q r = f n(q) [n] 2 q r. Continuing this process, we eventually obtain a positive integer t r such that the sequence of rational functions satisfies the functional equation (1), and { f n (q) [n] t r q r f p (q) [p] t = r q r } n=1 u U p F u(q) v V p F v(q),

16 A. Borisov et al. / Journal of Number Theory 109 (2004) where U p V p = and max(u p V p )<mp.it follows from the induction hypothesis there is a finite set R of positive integers and a set {t r } r R of integers such that f n (q) = r R[n] t r q r for all n supp(f). This completes the proof. There remain two related open problems. First, we would like to have a simple criterion to determine when a sequence of rational functions satisfying the functional equation (1) is actually a sequence of polynomials. It is sufficient that all of the integers t r in the representation (16) be nonnegative, but the example in (3) shows that this condition is not necessary. Second, we would like to have a structure theorem for rational function solutions and polynomial solutions to the functional equation (1) with coefficients in an arbitrary field, not just the field of rational numbers. Acknowledgments Nathanson s work was supported in part by Grants from the NSA Mathematical Sciences Program and the PSC-CUNY Research Award Program. Yang s work was supported in part by National Science Foundation Grants DMS and DMS Nathanson and Wang began their collaboration at a conference on Combinatorial and Number Theoretic Methods in Harmonic Analysis at the Schrödinger Institute in Vienna in February, References [1] M.B. Nathanson, A functional equation arising from multiplication of quantum integers, J. Number Theory 103 (2003)

LINEAR QUANTUM ADDITION RULES. Melvyn B. Nathanson 1 Department of Mathematics, Lehman College (CUNY), Bronx, New York 10468, USA

LINEAR QUANTUM ADDITION RULES. Melvyn B. Nathanson 1 Department of Mathematics, Lehman College (CUNY), Bronx, New York 10468, USA INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7(2) (2007), #A27 LINEAR QUANTUM ADDITION RULES Melvyn B. Nathanson 1 Department of Mathematics, Lehman College (CUNY), Bronx, New York 10468,