On sum-product representations in Z q

Transcription

1 J. Eur. Math. Soc. 8, c European Mathematical Society 2006 Mei-Chu Chang On sum-product representations in Z q Received April 15, 2004 and in revised form July 14, 2005 Abstract. The purpose of this paper is to investigate efficient representations of the residue classes modulo q, by performing sum and product set operations starting from a given subset A of Z q. We consider the case of very small sets A and composite q for which not much seemed known (nontrivial results were recently obtained when q is prime or when log A log q). Roughly speaking we show that all residue classes are obtained from a k-fold sum of an r-fold product set of A, where r log q and log k log q, provided the residue sets π q (A) are large for all large divisors q of q. Even in the special case of prime modulus q, some results are new, when considering large but bounded sets A. It follows for instance from our estimates that one can obtain r as small as r log q/ log A with similar restriction on k, something not covered by earlier work of Konyagin and Shparlinski. On the technical side, essential use is made of Freiman s structural theorem on sets with small doubling constant. Taking for A = H a possibly very small multiplicative subgroup, bounds on exponential sums and lower bounds on min a Z q max x H ax/q are obtained. This is an extension to the results obtained by Konyagin, Shparlinski and Robinson on the distribution of solutions of x m = a (mod q) to composite modulus q. 0. Introduction In this paper, we consider the following problem. Consider a subset H Z q (q N arbitrary) such that π p (H ) > 1 for all prime divisors p q, where π p denotes the quotient map mod p. Let kh be the k-fold sum set, and H r the r-fold product set of H. Then kh = Z q for some k N. One may for instance take k = q 3 (see proof of Theorem 2). Assume now we allow both addition and multiplication and seek for a representation Z q = kh r ; how small may we take k and r? In this context, we show the following: Theorem A. There is a function κ = κ (κ, M) such that κ 0 if κ 0 and M, with the following property. Let q N be odd and H Z q such that Then (This will be proved in 6.) π p (H ) > 1 for all prime divisors p of q, (0.1) π q (H ) > M for all divisors q q with q > q κ. (0.2) Z q = kh r with k < q κ and r < κ log q. (0.3) M. C. Chang: Mathematics Department, University of California, Riverside, CA 92507, USA; mcc@math.ucr.edu

2 436 Mei-Chu Chang The main motivation for this work comes from a recent line of reseach in combinatorial number theory and its applications to exponential sums in finite fields and residue classes (cf. [BKT], [BGK], [BC], [B]). If we consider in particular a subset A F p, p prime, such that A > p ε for some fixed (and arbitrary) ε > 0, then ka k = F p provided k > k(ε) and also max e p (ax 1 x k ) < p δ(ε) A k (a,p)=1 x 1,...,x k A for some δ(ε) > 0. This and related estimates had very significant applications to the theory of Gauss sums and various issues related to pseudo-randomness (see [B], [BKSSW], [BIW] for instance). One of the main shortcomings of the results that are presently available is the break-down of the method, starting from the sum-product theorem in [BKT], if we let ε = ε(p) be small. The boundary of the assumption here is ε 1/log log p, which is likely much stronger than necessary for such results to hold. More precisely, letting H < Fp, one could expect an equidistribution result of the form max e p (ax) < o( H ) ( ) (a,p)=1 x H to hold whenever log H log log p, which at this stage we can only establish if log H > log p/(log log p) ε (see [BGK]). It became apparently clear that the underlying ideas as developed in [BKT], [BGK] are insufficient to reach this goal (in particular they seem unable to produce a result such as the theorem stated above). Our purpose here is to explore the use of Freiman s theorem in sum-product problems (which was not used in [BKT]). Freiman s theorem (see [N] for instance) is one of the deepest result in additive number theory, providing a very specific description of subsets A of a torsion-free Abelian group with small sumset, i.e. 2A = A + A < K A, with K not too large. The results of this paper are new and based on a new approach. They do not provide the answers to the primary questions we are interested in, such as understanding when ( ) holds, but bring new techniques into play through related and more modest aims. Our bound in (0.3) is essentially optimal. Consider a composite q = p 1 p 2, where p 1 and p 2 are prime. Let p qκ. Define H = {1, θ} + p 1 {0, 1,..., p 2 1} where θ is of multiplicative order 2 (mod p 1 ). Hence H Z q. Obviously (0.1), (0.2) hold. Since kh r = kh {x + yθ : x, y N, x + y = k} + p 1 {0, 1,..., p 2 1}, (0.3) requires k p qκ, hence κ κ. The argument used to prove Theorem A has the following interesting consequence for subsets A Z p, p prime.

3 Sum-product representations in Z q 437 Theorem B. Given K > 1, there is K = K (K) as K such that the following holds. Let θ Z p be such that θ is not a root of any polynomial in Z p [x] of degree at most K and coefficients bounded by K (as integers). Then, if A Z p is an arbitrary set and K < A < p/k, we have A + θa > K A. Remark. For a similar result over characteristic 0, by Konyagin and Łaba, see [KL]. Returning to exponential sums with prime modulus (see (6.20)), we do obtain the following extension for composite modulus: Theorem C. Let H < Z q (q arbitrary) and assume H M > 1. Then max e q (ax) < H cq δ(m) (0.4) (a,q)=1 x H where δ(m) 0 as M (independently of q). This theorem will be proved in 8. In the proof, two cases are distinguished. If H contains an element θ of large multiplicative order, it turns out that one may proceed by a slight modification of the proof of (6.20) (Theorem 4.2 in [KS] for q prime). If all elements of H are of low order, we use the sum-product type results developed earlier in the paper. In the case q is a prime, Theorems A and C may be obtained by combining a theorem by Konyagin and a theorem in the book by Konyagin and Shparlinski [KS], except that the bound on r is slightly weaker. (See Remark 6.2.) Konyagin s theorem uses deep results in algebraic number theory such as Lehmer s conjecture on the heights of algebraic integers which are not roots of unity. There are several motivations to consider this type of problems. Konyagin s motivation was to prove the Heilbronn conjecture on the Waring problem and certain partial cases of the Stechkin conjecture on Gauss sums for composite moduli (see [KS, 6]). This is also related to the work of Robinson on the distribution of the solutions of x m a in residue classes (see [R]). The method we use here is totally different from Konyagin s. The main ingredients of the proof are Freiman s theorem and certain geometric techniques from Bilu s proof of Freiman s theorem (see [Bi]). 1 Notation. 1. For q N, Z q = Z/qZ. 2. Let x = (x 1,..., x d ), y = (y 1,..., y d ) R d. Then xy T = i x iy i, where y T is the transpose of the matrix y. If x Z d and y Z d q, then the matrix multiplication is done over Z q.

4 438 Mei-Chu Chang 3. Let ξ = (ξ 1,..., ξ d ) Z d q, P = d i=1 [A i, B i ] R d, and P = P Z d. A generalized arithmetic progression is P = {xξ T : x P}. When a progression P is given, P and P are used with the above meaning. Sometimes we refer to a progression by (ξ, P ), or (ξ, P). 4. A progression P given by ξ, P = d i=1 [1, J s ] is proper if P = P. We say P is proper with respect to L if is proper. 5. For A, B Z q, and k N, { xξ T : x d [1, LJ s ] Z d} i=1 A + B = {a + b : a A, b B}, ka = (k 1)A + A, 6. For q N, e q (θ) = e 2πiθ. AB ={ab : a A, b B}, A k = A k 1 A, a B = {a}b (mod q) for a Z, ab = {a}b for a Z q. 7. x = the distance from x to the nearest integer. Lemma 1.0. Let y = (y 1,..., y d ) Z d with gcd(y 1,..., y d ) = 1. Then there exists S SL d (Z) with y as an assigned row or column. Proof. We do induction on d. Let a = gcd(y 2,..., y d ). The assumption implies that gcd(a, y 1 ) = 1. Hence there exist b, c Z with b a and c y 1 such that y 1 b ac = 1. Let y i = ay i for i = 2,..., d, and let S = (s i,j ) SL d 1 (Z) be given by induction with (y 2,..., y d ) as the first row. Then y 1 y 2 y d c y 2 S = b y d b 0 s 2,1 s 2,d 1 0 SL d(z). 0 s d 1,1 s d 1,d 1 Remark It is clear from our proof that S(i, j), the (i, j)-cofactor of S, is bounded by y 1 ŷ j y d. To prove the next lemma, we need Lemma 6.6 and part of the proof of Theorem 1.2 from Bilu s work [Bi] on Freiman s theorem. We include them here for the reader s convenience.

5 Sum-product representations in Z q 439 B1. For x R m, B R m, x B := inf{λ 1 : λx B}. B2. Let e 1,..., e m be a basis of R m, W = e 1,..., e m 1, and π : R m W be the projection. Let B be a symmetric, convex body. Then where vol m (B) is the volume of B R m. vol m 1 (π(b)) m 2 e m B vol m (B), B3. Let λ 1,..., λ m be consecutive minima related to B. Then there exists a basis f 1,..., f m Z (called a Mahler basis) such that f 1 B λ 1, f i B i 2 λ i for i = 2,..., m. B4. Let f 1,..., f m Z be the Mahler basis as given in B3 and ρ i = f i B. Then for x = i x if i, x ρ := max ρ i x i. i B5. For x R m, we have Lemma 1.1. m 1 x B x ρ m!2 2 m 1 x B. Let a progression P be given by ξ Z d q and P = d i=1 [ J i, J i ]. Assume there exists L > 0 such that the progression (ξ, d i=1 [1, LJ i ]) is not proper. Then there exist v N and a progression P given by ξ Z d 1 q, P = d 1 i=1 [ J i, J i ] satisfying (i) v < L min i J i and v q, d 1 (ii) J i < C L d [ (d 1)! 2 ] d 1 d J i, where C d = d v 2 d 2 (d 1), i=1 i=1 (iii) v {xξ T : x P} {x ξ T : x P }. Proof. Let v = gcd(y 1,..., y d ). We may clearly assume that v q. Let y = (y 1,..., y d ) = (y 1 /v,..., y d /v). Hence gcd(y 1,..., y d ) = 1. Let e 1,..., e d be the standard basis of R d, and let S SL d (Z) with e d S = y be given by Lemma 1.0. For x P, let x Z d 1 and ξ Z d 1 q be defined by xs 1 = ( x, ), (1.1) vsξ T = ( ξ, 0) T. (1.2) Hence Let vxξ T = (xs 1 )(vsξ T ) = x ξ T. (1.3) B = P S 1. (1.4)

6 440 Mei-Chu Chang Then vol(b) = 2 d d i=1 J i. (1.5) Denote by π the orthonormal projection on [e 1,..., e d 1 ]. Let f 1,..., f d 1 be a Mahler basis for π(b) [e 1,..., e d 1 ]. For x = d 1 i=1 x i f i π(b), the second inequality in B5 implies x i (d 1)!2 2 d 2 x π(b) ρi 1 c d ρi 1, i, (1.6) where c d = (d 1)! 2 /2 d 2. Let Define x = (x 1,..., x d 1 ), F = J i dρi 1, (1.7) P d 1 = [ J i i ]. (1.8) i=1 f 1. f d 1 Then x ξ T = (x F ) ξ T = x ξ T. Hence (1.3) implies This is property (iii) in our conclusion. From the choice of S, we have Hence (cf. B1) Combining (1.11), B2, and (1.5) we have Hence GL d 1 (Z), ξ T = F ξ T. vxξ T = x ξ T. (1.9) e d = y S 1 = y v S 1 L v P S 1 = L B. (1.10) v e d B L/v. (1.11) vol(π(b)) < dl dl vol(b) = 2d 2v 2v d J i. (1.12) i=1 On the other hand, the first inequality in B5 on π(b) gives {x : x i ρ 1 i } {x : x π(b) d 1}. 2 d 1 d 1 i=1 ρ 1 i (d 1) d 1 vol(π(b)). (1.13)

7 Sum-product representations in Z q 441 Putting (1.7), (1.13) and (1.12) together, we have d 1 i=1 This is (ii) in the lemma. J i = cd 1 d ρ 1 i cd d 1 2 (d 1) (d 1) d 1 vol(π(b)) cd d 1 2 (d 1) (d 1) d 1 2 d dl 2v d J i. (1.14) Remark In Lemma 1.1, we take P = d i=1 [ J i, J i ] for the convenient notation, because we need a symmetric body to use Bilu s result. Clearly, we can apply the lemma to the progression P = (ξ, P ) with P = d i=1 [1, J i ]. Then P is given by (ξ, d 1 i=1 [1, J i ]). i=1 2 Lemma 2.1. Let P = (ξ, P ) be the progression with ξ = (ξ 1,..., ξ d ) Z d q and P = d s=1 [1, J s ] R d, where the integers J s satisfy J 1 J d > 0. Assume there exist ε > 0 and a Z q satisfying P ap > ε P. (2.1) Then for any index i = 1,..., d, one of the following alternatives hold. (i) J i < 2/ε, (ii) P is not proper with respect to 9/ε, (iii) there exist k i Z and k = (k 1,..., k d ) Zd such that Proof. Define 0 < k i < 1/ε, k s < 8/ε2 for all i s d, ak i ξ i = k ξ T. = {x P : axξ T P ap}. (2.2) Assume (ii) fails. In particular, the arithmetic progression P in Z q is proper. It follows from (2.1) that > ε P = εj 1 J d. (2.3) Hence there exist x 1,..., x i,..., x d Z such that {x i : x = (x 1,..., x d ) } > εj i. Assume (i) fails as well. Then εj i 2 and there is k i Z with 0 < k i < 1/ε (2.4)

8 442 Mei-Chu Chang and ak i ξ i P P. Hence d ak i ξ i = k ξ T with k = (k 1,..., k d ) [ J s, J s ] Z d. (2.5) To show assumption (iii) holds, we need to show k s < 8/ε2 for i s d. We assume Let s=1 k t 8/ε2 for some t {i,..., d}. (2.6) R = [4/ε], (2.7) l = 2 min s J s / k s = 2J t/k t, (2.8) S = {xξ T + rlk i ξ i : x, r = 1,..., R}, S = {x + rlk i e i : x, r = 1,..., R}. For r N, 1 r R, by (2.7), (2.8), (2.4) and (2.6), we have (2.9) rlk i < 4 ε 2 J t 1 k t ε J t J i. (2.10) Hence S P + [0, J i ]e i. This inclusion and the assumption that P is proper with respect to 9/ε imply as = S = S < 2J 1 J d. (2.11) On the other hand, for any x, by (2.2) we have axξ T = xξ T P (2.12) for some x = x(x) P. Let P be the set of all such x. Then there is a one-toone correspondence between and. Putting (2.5) and (2.12) together, we have (as any element in a S) (cf. (2.9)) Since for s {1,..., d}, we have x + rlk axξ T + arlk i ξ i = ( x + rlk )ξ T. (2.13) rlk s < 4 ε 2 J s k s k s < 8 ε J s, (2.14) d [ ( ) ( J s, ] )J s. (2.15) ε ε The failure of assumption (ii) and (2.15) imply that a S is proper. s=1

9 Sum-product representations in Z q 443 Let σ be such that J σ / k σ = min s J s / k s. Then lk σ = 2J σ and the sets P + lk, P + 2lk,..., P + Rlk are disjoint. Hence the sets + lk, + 2lk,..., + Rlk are all disjoint. Therefore, R as = ( + rlk )ξ T = R > εj 1 J d R > 3J 1 J d, r=1 which contradicts (2.11). Proof of Theorem B. We will use the notation c(k ) for various (maybe different) constants depending on K. Assume A Z p is such that K < A < p/k and A + θa < K A, where θ Z p satisfies the assumption of Theorem B. By Ruzsa s inequality for A = B we have A A A + B 2, B A A < (K ) 2 A. Identifying Z q {0, 1,..., q 1}, we apply Freiman s theorem to A, first considered as a subset of Z with doubling constant 2K 2 and A = A. It follows from Freiman s theorem that A P, where P is a generalized d-dimensional progression with d < c(k ) and P / A < c(k ). Since A + θa < K A, there is c F p such that and thus (c A) θa A 2 A + θa > A K, (A A) θ(a A) > A K. Let ˆP = P P. Then ˆP = c(k ) P. We get ˆP θ ˆP > c(k ) ˆP. (2.16) Our aim is to apply Lemma 2.1 with ɛ = c(k ) and a = θ. Some simplifications occur because q is prime. We want to rule out alternatives (i) and (ii). If (i) holds for some i = 1,..., d, we may clearly replace ˆP by a progression P 1 ˆP of dimension d 1 with (c(k )/2) ˆP P 1 and still satisfying P 1 θp 1 > c 1 (K ) P c 1 (K ) P 1. (2.17) If (ii) holds, apply Lemma 1.1 to obtain a reduction from d to d 1. Observe that since the integer v in Lemma 1.1 satisfies v p and v < c(k ) min J i < c(k ) A < c(k ) p K < p,

10 444 Mei-Chu Chang necessarily v = 1 (assuming K is large enough). Thus by Lemma 1.1, P P 1, where P 1 < c(k ) P and P 1 is of dimension d 1. In both cases (either (i) or (ii)), we obtain P 1 of dimension d 1 such that c(k ) P < P 1 < c(k ) P and (2.17) holds. Continuing the process, we get a progression P satisfying (2.17) and alternative (iii) of Lemma 2.1 for all i = 1,..., d 1, where d 1 is the dimension of P and ɛ = ɛ(k ). Thus P = { d 1 i=1 } x i ξ i : 0 x J i, x i Z and for all i = 1,..., d 1 there are k i Z and k i,s Z (1 s d 1) satisfying d 1 θk i ξ i = k i,s ξ s, (2.18) s=1 0 < k i < c(k ), (2.19) k i,s < c(k ) J s J i for all s = 1,..., d 1. (2.20) For (2.20), we use (2.10) (cf. (2.8)) which is valid for all s = 1,..., d 1 (rather than (2.6) which is a consequence). Returning to (2.18), it follows that the polynomial [ p(x) = det (xk i k i,i )e i,i ] k i,j e i,j Z p [x] (2.21) j =i has θ as a root, where e i,j is the matrix with (i, j)-entry 1 and 0 elsewhere. Clearly p(x) is of degree d 1 d c(k ) with non-vanishing x d 1-coefficient by (2.19). By (2.19), (2.20) all coefficients of (2.21) are bounded by d 1 π Sym(d 1 ) i=1 ( k i δ i,π(i) + k i,π(i) ) < d 1 c(k ) d 1 This contradicts the assumption on θ for K sufficiently large. π i=1 J π(i) J i < c (K ). Remark. Quantitatively speaking, the previous argument will require K to be at most sublogarithmic in p, since we do rely on Freiman s theorem (cf. [C]). Thus we may ask how large the quantity min A + θa / A (2.22) p ε < A <p1 ε can be made for some θ F p. Considering sets A of the form A = { d i=1 x i θ i : 0 x i M}, it is easily seen that (2.22) is less than exp( log p).

11 Sum-product representations in Z q Lemma 3.1. Let P = (ξ, P ) be a progression with ξ = (ξ 1,..., ξ d ) Z d q and P = ds=1 [1, J s ] R d, where the integers J s satisfy J 1 J d > 0. Assume δ 0 Ji < P < q 1 3γ (3.1) with γ > 0 a constant. Let ε, M > 0 (ε small, M large) satisfy Assume Let B Z q be such that δ 1 0 (1/ε)3d+10 < M < q γ /2. (3.2) π q (P) > M for all q q, q > q γ. (3.3) π q (B) > M for all q q, q > q γ /10d. (3.4) Here π q : Z q Z q denotes the quotient map mod q. Then there is a B such that ap P < ε P. (3.5) Lemma 3.1 will be proved by assuming ap P > ε P for all a B, applying Lemma 2.1 (on a progression which may have fewer generators) and ruling out alternatives (i) (iii) to get a contradiction. We will first make a possible reduction of the number d of generators of P to ensure properness with respect to some constant, using Lemma 1.1. The reduction. We take Assume ε 0 = ε. (3.6) δ 0 P P, (3.7) and P is not proper with respect to 9/ε 0. Lemma 1.1 then allows a reduction of the dimension of the progression P in the following sense: there are v 0 N and ξ (1) Z d 1 q, P 1 = d 1 i=1 [1, J 1,i] Z d 1 satisfying v 0 < 9 ε 0 min J i, v 0 q, P 1 < C P, ε 0 v 0 (3.8) v 0 P P 1. (3.9) By (3.7) (3.9), with P 1 P 1 P v 0 > δ 0 v 0 P > δ 1 P 1 (3.10) δ 1 = cε 0 δ 0. (3.11)

12 446 Mei-Chu Chang Take ε 1 = εδ 1. (3.12) and repeat the preceding. If P 1 is not proper with respect to 9/ε 1, apply one more time Lemma 1.1 to obtain v 1 N and ξ (2) Z d 2 q, P 2 = d 2 i=1 [1, J 2,i] Z d 2 satisfying By (3.14), (3.10) and (3.13), with Notice that Take v 1 < 9 ε 1 min J 1,i, v 1 q, P 2 < C P 1, ε 1 v 1 (3.13) v 1 P 1 P 2. (3.14) P 2 P 2 P 1 v 1 > δ 1 v 1 P 1 > δ 2 P 2, (3.15) δ 2 = cε 1 δ 1. (3.16) P 2 v 0 v 1 P. (3.17) ε r 1 = εδ r 1. (3.18) After applying Lemma 1.1 r times, we have v r 1 N and ξ (r) Z d r q, P r = d r i=1 [1, J r,i] Z d r satisfying Same reasoning as before yields v r 1 < 9 min J r 1,i, ε r 1 v r 1 q, (3.19) C P r < P r 1, ε r 1 v r 1 (3.20) v r 1 P r 1 P r. (3.21) P r P r > δ r P r (3.22) with Also, We have the following: (1) cε 0 ε 1 ε r 1 δ 0 = δ r. (2) δ r = cεδr 1 2. δ r = cε r 1 δ r 1. (3.23) v 0 v 1 v r 1 P P r. (3.24)

13 Sum-product representations in Z q 447 (3) ε r 1 = c(εδ 0 ) 2r 1. (3 ) Assume δ 0 > ε. Then ε r 1 > cε 2r > (cε) 2d. (3 ) ε 0 ε 1 ε r 1 > cεε r > cε 2r+1 cε 2d+1. (4) P > c ε 0 ε 1 ε r 1 v 0 v 1 v r 1 P r. (4 C ) P > v 0 v 1 v r 1. ε 0 ε 1 ε r 1 ( ) C 2 d+1 (5) P > v 0 v 1 v r 1. ε To see the above (in)equalities hold, we note that our notations (3.11), (3.16),..., (3.23) imply (1); (3.18) and (3.23) imply (2); (3.18) and (2) imply (3); (3.8), (3.13),..., (3.20) imply (4); (4 ) and (3 ) imply (5). Assume a Z q and P ap > ε P. By (3.24), (3.7), (4), (1), and (3.18), where P r ap r (v 0 v 1 v r 1 P) a(v 0 v r 1 P) (v 0 v 1 v r 1 ) 1 P ap > (v 0 v r 1 ) 1 εδ 0 P > cεε 0 ε r 1 δ 0 P r = cεδ r P r = cε r P r, (3.25) ε r = εδ r. (3.26) We need the following little fact from algebra to prove Lemma 3.1. Fact A. Let A Z q, k Z, and q = q/gcd(k, q). Then π q (A) = ka. Proof of Lemma 3.1. We assume after r reductions P r is proper with respect to 9/ε r. (If P is already proper with respect to 9/ε, then r = 0 and P 0 = P.) We apply Lemma 2.1 to P = P r, replacing ε by ε r. By our construction, alternative (ii) in Lemma 2.1 is ruled out. Assume J i := J r,i are ordered decreasingly, J 1 J d. Let ξ i = ξ (r) i {1,..., q 1} Z q \{0} and define Claim. q d q 1 γ. q 1 = gcd(ξ 1, q), q 2 = gcd(ξ 1, ξ 2, q),. q d = gcd(ξ 1,..., ξ d, q). Proof of Claim. Assume q d > q 1 γ. Let w = q/q d < q γ. Then (5), (3.1) and (3.2) imply v 0 v 1 v r 1 w (C/ε) 2d+1 P q γ < (C/ε) 2d+1 1 δ 0 q 1 2γ < q 1 γ. (3.27)

14 448 Mei-Chu Chang Also, hence, from (3.24), By Fact A, with (by (3.27)) This contradicts (3.3), proving the Claim. wξ 1 = = wξ d = 0 (mod q), v 0 v 1 v r 1 wp = 0 (mod q). π q (P) = 0 (3.28) q = q/gcd(q, v 0 v r 1 w) > q γ. Therefore, there is i {1,..., d } such that q i 1 > q γ /d, q i (3.29) q, q 1,..., q i 2 q γ /d. q 1 q 2 q i 1 (3.30) Apply Lemma 2.1 considering this particular index i. Alternative (ii) is ruled out by construction. Claim. Alternative (i) fails. Proof. If (i) holds, we get (C/ε) 2d+1 > 2/ε r > J i J i+1 J d. (3.31) Let By (3.30), (5), (3.1) and (3.2), v = v 0 v r 1 q/q i 1. v v 0 v r 1 q γ < (C/ε) 2d+1 P q γ < cq 1 3γ +γ /2 q γ < q 1 γ Hence, from the definition of q i 1, q vp = v 0 v 1 v r 1 P q i 1 (3.31), (3.32) imply q q i 1 P r = q { x s ξ s q : x s J s i 1 s i vp J i J i+1 J d < (C/ε)d2d+1 < M. }. (3.32) Hence Fact A implies π q (P) = vp < M with q contradicting (3.3). = q/gcd(q, v) > q γ, again

15 Sum-product representations in Z q 449 So alternative (iii) holds and there are k i, (k s ) 1 s d Z such that 0 < k i < 1/ε r, (3.33) k s < 8/ε2 r for i s d, (3.34) ak i ξ i d = k s ξ s (mod q). (3.35) s=1 Since q i 1 = gcd(ξ 1,..., ξ i 1, q), (3.35) implies ak i ξ i = k s ξ s (mod q i 1). (3.36) s i By (3.33), (3.34), (3 ) and (3.2), the coefficients (k i, k i,..., k d ) in (3.36) range in a set of at most (1/ε r )(8/ε 2 r )d < (C/ε) (2d+1)2d+1 < M 1/2 elements. Recalling (3.29) and (3.4) we have π qi 1 /q i (B) > M (3.37) and we may consider elements B B, B > M, such that π qi 1 /q i B is one-to-one. Assuming P ap > ε P for all a B, we have for all a B (cf. (3.25)), P ap > cε r P (3.38) and the preceding applies, providing in particular a representation (3.36). In view of the bound on the number of coefficients in (3.36), there is B B with B > M 1/2 such that for all a B, (3.36) holds with the same coefficients k i, k s (s i). Taking any a 1, a 2 in B we obtain (a 1 a 2 )k i ξ i = 0 (mod q i 1), (a 1 a 2 )k i = 0 (mod q i 1 /q i ), (3.39) implying 1 = π qi 1 /q i (k i B ) 1 k i π q i 1 /q i (B ) = B k i > ε rm 1/2, a contradiction. This proves Lemma 3.1. Following the same arguments as in Lemma 3.1, we also obtain: Lemma 3.1. Under the assumptions of Lemma 3.1, there exist elements a 1,..., a R in B with R M 1/10 such that a s P a s P < ε P for s = s. (3.40)

16 450 Mei-Chu Chang Proof. Let B B be the set constructed in the proof of Lemma 3.1. Assume a 1,..., a r B are already obtained satisfying (3.40) and suppose max a sp ap > ε P 1 s r for all a B. Hence, there is some s = 1,..., r and B 1 B with and such that B 1 > 1 r B > M R > M9/10 P a P a > ε P for all a B 1. (3.41) s It follows that all elements a/a s, a B 1, have a representation (3.36). Passing again to a subset B 1 with B 1 > M9/10 1/2 = M 2/5, we may ensure the same coefficients k i, (k s ) s i and get a contradiction as before. 4 Let A Z q be such that A + A < K A, (4.1) 1 A < q 1 4γ. (4.2) Identifying Z q {0, 1,..., q 1}, we apply Freiman s theorem to A, first considered as a subset of Z (with doubling constant 2K). From [C], we obtain d 2K and a progression P given by ξ = (ξ 1,..., ξ d ) Z d and P = d i=1 [0, J i ], with J 1 J d in N, such that A P = {xξ T : x P}, (4.3) P < C K3 A. (4.4) Applying π q : Z Z q, P becomes a progression in Z q containing A Z q. Assuming by (4.2) (4.5), we have C K3 < q γ /2, (4.5) q 1 3γ > C K3 q 1 4γ > P P A > C K3 P. (4.6) Thus assumption (3.1) in Lemma 3.1 holds with δ 0 = C K3. Let ε, M satisfy (3.2), i.e. C K3 (1/ε) 10K+5 < M < q γ /2 (4.7)

17 Sum-product representations in Z q 451 and moreover ε < C K3. Assume B Z q is such that π q (B) π q (A) (e.g. B contained in a translate of A) for all q q and q > q γ. Furthermore, assume B satisfies Since by assumption also π q (B) > M if q q and q > q γ /20K. (4.8) π q (P) π q (A) π q (B) > M if q q and q > q γ, conditions (3.3), (3.4) of Lemma 3.1 are satisfied. Apply Lemma 3.1. Let a 1,..., a R B satisfy (3.5). (We take R < M 1/10.) Write a r A R A a r A a s A R A a r P a s P r R r =s r =s > R A R 2 εc K3 A. (4.9) Taking R = 2ε 1 C K3, (4.9) implies AB a r A > R 2 A > 1 ε C K3 A. (4.10) r R Assume M > C 50K+10 (4.11) (which implies the first inequality of (4.7), hence it also implies (4.5)) and take From (4.10) and (4.11), 1/ε = M 10 K 6. AB > M 10 K 7 A. (4.12) Replacing 4γ by γ and summarizing, we have proved the following: Lemma 4.1. Let A Z q satisfy Let M satisfy Let B Z q be such that A < q 1 γ, (4.13) A + A < K A. (4.14) C 50K+10 < M < q γ /8. (4.15) Then π q (B) π q (A) if q q and q > q γ, π q (B) > M if q q and q > q γ /80K. (4.16) AB > M 10 K 7 A. (4.17)

18 452 Mei-Chu Chang 5 Proposition 1. satisfy Let κ > 0 be a small and M a large constant. Let H Z q (q large) π q (H ) > M whenever q q, q > q κ. (5.1) Then there is k, r N such that where κ = κ (κ, M) 0 Proof. We describe the construction. Given any set k < q κ, (5.2) r < log 2 q κ, (5.3) kh r > q 1 κ, (5.4) as κ 0, M (independently of q). κ 1 > κ 1/2, (5.5) { } K = min (log log M) 1/2 κ 1,. (5.6) 100κ Let A 0 = H and A α = k α H r α be the set obtained at stage α. Assume A α < q 1 κ 1. We distinguish the following cases. (i) A α + A α > K A α. Take then k α+1 = 2k α and r α+1 = r α. (ii) A α + A α K A α. Apply Lemma 4.1 with A = A α, B = H, γ = κ 1. In (5.1) we can assume M < q γ /10. Conditions (4.15) and (4.16) clearly hold, because of (5.6). Hence A α H > M 10 K 7 A α > K A α. The second inequality is again by (5.6). Hence k α H r α+1 > K A α. In this case we take k α+1 = k α and r α+1 = r α + 1. Therefore A α+1 > K A α, (5.7) with k α+1 2k α, r α+1 r α+1. (5.8) To reach size q 1 κ 1, the number of steps is at most log q/log K, because after s steps, by (5.7), q A α+1 > K s H K s. By (5.8), in (5.2) we have k 2 log q/log K = q κ 2. Hence κ 2 = log 2 log K 1 min{log log log M, log 1/κ}, by (5.5) and (5.6). We conclude the proof of Proposition 1 by taking κ = max{κ 1, κ 2 }.

19 Sum-product representations in Z q We need the following to prove Theorem A. Let ν, ν : Z q R be functions. We recall F1. ˆν(ξ) = x Z q ν(x)e q ( xξ). If ν is a probability measure, i.e. 0 ν(x) 1, then ˆν(ξ) 1. F2. ν ν (x) = y Z q ν(x y)ν (y). F3. supp(ν ν ) supp ν + supp ν. F4. ν(x) = q 1 ξ Z q ˆν(ξ)e q (xξ). F5. ν ν (ξ) = ˆν(ξ)ˆν (ξ). Let 0 x 5π/6. Then T1. sin x > 2π x. Therefore, e q(1) 1 > q 1. T2. cos x < 1 4π x2. Therefore, e q(1) + 1 < 2 π 2 1. q 2 T3. e q (x) e q (y) = 2 sin 2π 2q (x y). Proof of Theorem A. Let κ > 0 and M be constants as in Proposition 1. Let q N be odd. Let H Z q satisfy the following conditions: π p (H ) 2 for all primes p q, (6.1) π q (H ) > M for all q q, q > q κ. (6.2) We want to show that k 1 H r = Z q for some k 1, r N satisfying By (6.2), Proposition 1 applies. Let k, r satisfy (5.2) (5.4). Define D = {q N : q = 1 and q q}, hence For q D, we have while by (6.1), also π q (kh r ) kh r q/q r < log q κ, (6.3) k 1 < q 5κ. (6.4) D < q 1/log log q. (6.5) > q1 κ q/q = q q κ, (6.6) π q (kh r ) π q (H ) 2. (6.7) Take a subset q kh r such that π q q is one-to-one and q max {2, q q κ }. (6.8)

20 454 Mei-Chu Chang Define the probability measures µ q = 1 q x q δ x, (6.9) δ x being the indicator function, and their convolution Then by F3, µ(x) = q * µ q (x) D = µ q D (x y 1 y D 1 ) µ q2 (y 2 )µ q1 (y 1 ). (6.10) y 1,...,y D 1 We estimate the Fourier coefficients supp µ supp µ q q D kh r. (6.11) q D q D ˆµ(a/q) = x Z q e q ( ax)µ(x) for 0 < a < q. Let a/q = a /q where q q and (a, q ) = 1. From (6.10) and F5, ( ) ( a ˆµ a ) q µ q q = 1 e q (a x). (6.12) q Claim 1. ˆµ(a/q) < q 2κ. x q Proof of Claim 1. Assume ˆµ(a/q) > 1 τ. We want to find a lower bound on τ. Squaring both sides of (6.12), we obtain x,y q Choose an element x 0 q such that By T3, we write y q e q (a x 0 ) e q (a y) 2 = cos 2πa q (x y) > (1 τ) 2 q 2. (6.13) cos 2πa q (x 0 y) > (1 τ) 2 q. (6.14) ] [2 sin 2πa (x 0 y) 2 q = 2 2 cos 2πa 2 q (x 0 y). Together with (6.14) this gives e q (a x 0 ) e q (a y) 2 2 q 2(1 τ) 2 q < 2τ q. (6.15) y q

21 Sum-product representations in Z q 455 From (6.15), {y : e q (a x 0 ) e q (a y) > 2 τ} 1 2 q. So there is a subset q with > 1 2 q such that for all y, hence T1 implies Therefore Since π q is one-to-one, also e q (a x 0 ) e q (a y) < 2 τ, a x 0 /q a y/q < 2 τ. (6.16) π q ( ) = π q (a ) 2 τq q q κ < 2 τq + 1 (6.17) by (6.8). This gives a lower bound τ > 1 16 q 2κ, proving Claim 1. Take l = [q 3κ ]. (6.18) Claim 2. Let µ (l) be the l-fold convolution of µ. Then supp µ (l) = Z q. Proof of Claim 2. For x Z q, write µ (l) (x) = 1 q + 1 q q ( ) a µ (l) e q (ax). (6.19) q a=1 By Claim 2 and (6.18), the second term in (6.19) is at most ( ) ( ) max µ a (l) = max a l q 1 a<q ˆµ < q 1 a<q Hence µ (l) (x) > 0, proving Claim 2. ( 1 1 ) q 3κ < e qκ < 1 16q 2κ q. Putting together Claim 2, (6.11), (6.18), (6.5) and (5.2), we have Z q = l supp µ = l D kh r = k 1 H r with k 1 q 3κ q 1/log log q q κ < q 5κ, which completes the proof of Theorem A. Remark 6.1. It is much simpler to prove the weaker bound ( a ) µ q q < 1 π 1 2 (q ) 3. Indeed, since q 2, there are x 1, x 2 q with π q (x 1 ) = π q (x 2 ). As (a, q ) = 1, also π q (a x 1 ) = π q (a x 2 ). Therefore, by T2, e q (a x 1 ) + e q (a x 2 ) e q (1) + 1 < 2 π 2 1 (q ) 2

22 456 Mei-Chu Chang Write x q This yields the stated bound. e q (a x) ( q 2) + e q (a x 1 ) + e q (a x 2 ) < q π 2 1 (q ) 2. Remark 6.2. For q prime, Theorem A has a simpler proof, which gives a slightly weaker bound on r. In this case, Z q is a cyclic group. The condition on H Z q is simply H > M with M large. It follows that H contains an element θ H of multiplicative order t > M. Assuming (as we may) that 1 H, it follows that We distinguish 2 cases. {1, θ,..., θ r } H r. Case 1: t log q/(log log q) 1/2. Take r 0 log q(log log q) 4. Using the inequality max (a,q)=1 r 0 x=1 ) e q (aθ x c ) < r 0 (1 (log q) 2 due to Konyagin (see [KS, p. 26]), simple application of the circle method implies kh r 0 = Z q with k < C(log q) 3. Case 2: t < log q/(log log q) 1/2. Define ϕ(t) = Z t. Use Theorem 4.2 from [KS] to get max (a,q)=1 t e q (aθ x ) < t c(ρ)q 2/ρ for 2 ρ ϕ(t). (6.20) x 1 Hence kh t = Z q with k < 1 c(ρ) (log q)2 q 2/ρ. Since ϕ(t) for M, we may achieve k < c(κ )q κ M. with κ (M) 0 as 7 Corollary 3. (1) Let H Z q satisfy assumption (5.1) of Proposition 1 and κ be as in that proposition. Let q q, q > q κ and (a, q ) = 1. Let r N with r > κ log q. Then max a x,y H r (x y) q > q 2κ. (7.1)

23 Sum-product representations in Z q 457 (2) Let H < Z q be a multiplicative subgroup satisfying assumption (5.1) of Proposition 1. Let q q, q > q κ and (a, q ) = 1. Then Proof. By (5.4), max x,y H a (x y) q > q 2κ. (7.2) π q (kah r ) = π q (kh r ) > q1 κ q/q = q q κ > 1. (7.3) Hence there are z, w kh r such that a (z w) q q κ. (7.4) Writing z = x x k and w = y y k with x i, y i H r, it follows that by (5.2). Corollary 4. max x,y H r a (x y) q 1 k q κ > q 2κ. (1) Let H Z q satisfy conditions (6.1), (6.2) and κ be as in Theorem A. Let 1 a < q. Then for r > κ log q, max a x,y H r (x y) q q 5κ. (7.5) (2) If moreover H < Z q is a group, we get max x,y H a (x y) q q 5κ. (7.6) Proof. Write a/q = a /q, (a, q ) = 1. Since π q (k 1 H r ) = Z q, we have max z,w k 1 H r a (z w) q 1 2, hence max x,y H r a (x y) q 1 > q 5κ. k 1

24 458 Mei-Chu Chang 8. The case of subgroups The main result of this section is the following (for q prime this issue was considered in [P]): Theorem 5. Let H < Z q with H > M > 1. Then max a x,y H (x y) q > q δ (8.1) min a Z q where δ = δ(m) 0 as M (independently of q). We first treat the case when H contains an element of large multiplicative order. The next result has a simple proof obtained by a straightforward modification of an argument in [KS] (see 4) for prime modulus. Lemma 8.1. Let θ Z q be of order t (large). Then min max a (a,q)=1 j,k q (θ j θ k ) > c(r)q 1/(r 1) (8.2) for 1 < r < ϕ(t). Proof. For j = 1,..., t, let b j Z be such that and extend periodically with period t for j Z. b j = aθ j (mod q) (8.3) Claim. Let c Z and 2 r < ϕ(t). Then max j b j c > c(r)q (r 2)/(r 1). Proof of Claim. Let B = max 1 j t b j c. (8.4) Set b = (b 1,..., b r ), and 1 = (1,..., 1). We consider the lattice L = {l = (l 1,..., l r ) Z r : bl T = 0, 1l T = 0} = {l = (l 1,..., l r ) Z r : (b c1)l T = 0, 1l T = 0}. (8.5) We consider all expressions (b i c)l i with l i = 0 and b i c B. From the pigeonhole principle and (8.4), there is (l 1,..., l r ) L \ {0} such that max l j < c(r)b 1/(r 2). (8.6) 1 j r For this vector l = (l 1,..., l r ) we have b 1 l b r l r = 0. Hence, multiplying with θ j gives b j+1 l b j+r l r = 0 (mod q)

25 Sum-product representations in Z q 459 for all j. Since also l l r = 0, The left side is bounded by (b j+1 c)l (b j+r c)l r = 0 (mod q). (8.7) by (8.4) and (8.6). Assume c(r)b (r 1)/(r 2) < q; hence It then follows from (8.7) that rbc(r)b 1/(r 2) < c(r)b (r 1)/(r 2) B < c(r)q (r 2)/(r 1). (8.8) (b j+1 c)l (b j+r c)l r = 0, b j+1 l b j+r l r = 0 (8.9) for all j. Hence (b j ) is a periodic linearly recurrent sequence of order at most r and smallest period t. Let ψ(x) be the minimal polynomial of (b j ). (See [KS].) Then from (8.9), ψ(x) (l 1 + l 2 x + + l r x r 1 ) implying deg ψ r 1. Obviously ψ(x) (x t 1). Assume ψ(x) (1 x τ ). 1 τ<t Since ψ(θ) = 0 (mod q), it would follow that θ τ 1 (mod q) for some τ < t, contradicting ord q (θ) = t. Therefore one of the roots of ψ is a primitive t th root and ψ is divisible by the t- cyclotomic polynomial. Hence ϕ(t) deg ψ < r, a contradiction. Hence (8.8) fails, proving the Claim. Hence Suppose (8.2) fails. Letting c = aθ k Z q = {0, 1,..., q 1}, we get max aθ j c j q < c(r)q 1/(r 1). max j dist(aθ j c, qz) < c(r)q (r 2)/(r 1). (8.10) From (8.10), we may for each j = 1,..., t take b j Z such that b j = aθ j (mod q) and b j c < c(r)q (r 2)/(r 1). This contradicts the Claim and proves the lemma.

26 460 Mei-Chu Chang Proof of Theorem 5. Let H < Z q, H > M. Fix κ > 0 (small) and 1 M 1 < M κ/2. By Lemma 8.1, we may assume ord q (x) < M 1 for all x H. (8.11) If π q1 (H ) > M 1 for all q 1 q with q 1 > q κ, then (8.1) holds, since δ = δ(κ, M 1 ) 0 as κ 0, M 1 (by Corollary 3(2)). Assume there exists q 1 q with q 1 > q κ and π q1 (H ) M 1. Hence satisfies H 1 > M/M 1. Consider the set H 1 = H π 1 q 1 (1) < Z q H 1 = {x Z q/q1 : 1 + q 1 x H 1 } = H 1 1 q 1. Assume there is q 2 q q 1 with q 2 > q κ and π q2 (H 1 ) < M 1. Hence π q1 q 2 (H 1 ) < M 1 and defining H 2 = H 1 π 1 q 1 q 2 (1), we have Considering the set H 2 > H 1 > M M 1 M1 2. H 2 = {x Z q/q1 q 2 : 1 + q 1 q 2 x H 2 } = H 2 1 q 1 q 2, we repeat the process. At some stage s 1/κ, the process has to stop. Thus Define H s = H πq 1 1 q s (1), (8.12) H s = {x Z q/q1 q s : 1 + q 1 q s x H s } = H s 1, q 1 q s (8.13) H s = H s > M M 1/κ 1 π q (H s ) > M 1 for all q Q 1 = q 1... q s and Q 2 = q/q 1. > M 1/2, (8.14) q q 1 q s with q > q κ. (8.15) Case 1: Q 2 < q κ. Since H s 2, there are elements x = y in H s Z Q2. Hence ax = ay (mod Q 2 ) and a(x y) Q > 1 > q κ. 2 Q 2 Let x = 1 + Q 1 x, ȳ = 1 + Q 1 y H s < H. Writing a(x y) = a(q 1x Q 1 y) = Q 2 q a( x ȳ), q

27 Sum-product representations in Z q 461 we obtain a( x ȳ) q > q κ and hence (8.1) holds. Case 2: Q 2 q κ. First, note that if there is no ambiguity, we use the notation (A, B) = gcd(a, B). Claim 1. (Q 1, Q 2 ) q κ. Proof of Claim 1. Observe that (1 + Q 1 x)(1 + Q 1 y) = 1 + Q 1 (x + y) (mod Q 2 1 ). Hence (1 + Q 1 x)(1 + Q 1 y) = 1 + Q 1 (x + y) (mod Q 1 (Q 1, Q 2 )). Consider π Q1 (Q 1,Q 2 )(H s ) < Z Q 1 (Q 1,Q 2 ). It follows from the preceding that π Q1 (Q 1,Q 2 )(H s ) = 1 + Q 1 S where S is an additive subgroup of Z (Q1,Q 2 ). Hence S < Z (Q1,Q 2 ), + and π Q1 (Q 1,Q 2 )(H s ) < Z Q 1 (Q 1,Q 2 ) are cyclic. By assumption (8.11), all elements of H s < H are of order M 1, implying Therefore By construction of H s, (8.16) implies and Claim 1 is proved. π Q1 (Q 1,Q 2 )(H s ) M 1. π (Q1,Q 2 )(H s ) M 1. (8.16) (Q 1, Q 2 ) q κ, (8.17) Let Q 1 = Q 1/(Q 1, Q 2 ) and Q 2 = Q 2/(Q 1, Q 2 ). Hence (Q 1, Q 2 ) = 1 and Q 2 > q κ κ by case assumption and (8.17). We want to apply Corollary 3(2) to π Q (H 2 s ) < Z Q with 4 κ and M 1. Let q Q 2 2 with q > (Q 2 )4 κ > q 2κ, and let q = q /(q, Q 1, Q 2 ). Thus by (8.17), q > q κ. Claim 2. π q (H s ) > M 1. Proof of Claim 2. It follows from (8.15) that π q (H s ) > M 1. Let x 1,..., x n H s, n > M 1, be such that x i x j = 0 (mod q ). Since (q, Q 1 ) = (q, Q 1 ) = 1 and ( ) q (Q 1, Q 2 ), (q = 1,, Q 1, Q 2 ) we also have Q 1 (q, Q 1, Q 2 ) (x i x j ) = 0 (mod q ).

28 462 Mei-Chu Chang Hence Q 1 (x i x j ) = 0 (mod q ). Since 1 + Q 1 x i H s, it follows that π q (H s ) > M 1, proving Claim 2. Apply Corollary 3(2) to the group π Q 2 (H s ) < Z Q 2. Claim 2 implies that π q (π Q 2 (H s )) = π q (H s ) > M 1 for all q Q 2 with q > (Q 2 )4 κ. Hence for any a with (a, Q 2 ) = 1, there are x, ȳ H s such that a ( x ȳ) > (Q 2 ) κ > q κ (8.18) Q 2 where κ = κ (4 κ, M 1 ) 0 as κ 0 and M 1. Write x = 1 + Q 1 x and ȳ = 1 + Q 1 y with x, y H s. From (8.18), a Q 1 Q (x y) > q κ. (8.19) 2 Recalling that (Q 1, Q 2 ) = 1, we may choose a satisfying a Q 1 a (mod Q 2 ). Then (8.19) gives a(q 1, Q 2 ) 2 (x y) Q > q κ. 2 Hence, by Claim 1, and Therefore a (x y) Q > q κ 2 (Q 1, Q 2 ) 2 > q κ 2κ, a ( x ȳ) q = a q (Q 1x Q 1 y) > q κ 2κ. max x,y H a (x y) q > q κ 2κ, where κ, κ may be made arbitrarily small by taking M large enough. This proves Theorem 5. Theorem C is an extension of Theorem 4.2 in [KS] for composite modules and is an immediate consequence of Theorem 5. Proof of Theorem C. For a Z q, let {x 1,..., x H } = ah, and let ax = x 1 and ay = x 2 be given in Theorem 5 such that x 1 x 2 q > q κ

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