ON SUM-PRODUCT REPRESENTATIONS IN Z q. Mei-Chu Chang

Transcription

1 ON SUM-PRODUCT REPRESENTATIONS IN Z Mei-Chu Chang Abstract The purpose of this paper is to investigate efficient representations of the residue classes modulo, by performing sum and product set operations starting from a given subset A of Z. We consider the case of very small sets A and composite for which not much seemed known (nontrivial results were recently obtained when is prime or when log A log ). 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 and log k log, provided the residue sets π (A) are large for all large divisors of. Even in the special case of prime modulus, 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 log A with similar restriction on k, something not covered by earlier work of Konyagin and Shparlinski(see KS). 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 max x H ax 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 ) to composite modulus. 0. Introduction In this paper, we consider the following problem. Consider a subset H Z ( N arbitrary) such that π p (H) > 1 for all prime divisors p. Let kh be the k-fold sum set, and H r the r-fold product set of H. Then kh = Z for some k N. One may for instance take k = 3 (see proof of Theorem 2). Assume now we allow both addition and multiplication and seek for a representation Z = 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, M with the following property. Let N be odd and H Z such that π p (H) > 1 for all prime divisors p of (0.1) π (H) > M for all divisors, > κ (0.2) Then Z = kh r with k < κ and r < κ log. (0.3) 1 Typeset by AMS-TEX

2 (This will be proved in 6). 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 ɛ for some fixed (and arbitrary) ɛ > 0, then ka k = F p provided k > k(ɛ) and also max (a,p)=1 x 1,...,x k A e p (ax 1... x k ) < p δ(ɛ) A k for some δ(ɛ) > 0. This and related estimates had very significant application 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 ɛ = ɛ(p) be small. The boundary of the assumption 1 here is ε log log p, which is likely much stronger than necessary for such results to hold. More precisely, letting H < F p, one could expect an euidistribution result of the form max (a,p)=1 x H e p (ax) < o( H ) (*) to hold whenever log H log log p, which at this stage we can only establish if log H > (see [BGK]). log p (log log p) ε 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 uestions we are interested in, such as understanding when (*) holds, but bring new techniues into play through related and more modest aims. Our bound in (0.3) is essentially optimal. Consider a composite = p 1 p 2, where p 1 and p 2 are prime. Let p κ. Define H = {1, θ} + p 1 {0, 1,..., p 2 1} where θ is of multiplicative order 2 (mod p 1 ). Hence H Z. 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) reuires k p κ, hence κ κ. The argument used to prove Theorem A has the following interesting conseuence for subsets A Z p, p prime. 2

3 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 Laba, 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 ( arbitrary) and assume H M > 1. Then max e (ax) < H c δ(m) (0.4) (a,)=1 x H where δ(m) 0 for M (independently of ). 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 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 is a prime, Theorem A and Theorem C may be gotten 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 Warring problem and certain partial cases on 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 solution 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 techniues from Bilu s proof of Freiman s Theorem (see [Bi]). 1. Notation. 1. For N, Z = Z/Z. 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, then the matrix multiplication is done over Z. 3

4 3. Let ξ = (ξ 1,..., ξ d ) Z d, 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 give, 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. {xξ T : x 5. For A, B Z, and k N, d [1, LJ s ] Z d } i=1 A + B ={a + b : a A, b B}, ka = (k 1)A + A, AB ={ab : a A, b B}, A k = A k 1 A, a B ={a}b (mod ), for a Z, ab ={a}b, for a Z. 6. For N, e (θ) = e 2πiθ. 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. exists S SL d (Z) with y as an assigned row or column. Then there Proof. We do induction on d. Let a = gcd(y 2,, y d ). The assumption implies gcd(a, y 1 ) = 1. Hence there exist b, c Z, b a, 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 2b y d b S = 0 s 2,1 s 2,d 1 SL d(z). 0 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. 4

5 To prove the next lemma, we need the following from Bilu s work on Freiman s Theorem. These are Lemma 6.6 and part of the proof of Theorem 1.2 in [Bi]. We include them here for the reader s convenience. 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 vol m 1 (π(b)) m 2 e m B vol m (B), where vol m (B) is the volume of B R m. B3. Let λ 1,, λ m be consecutive minima related to B. Then there is a basis f 1,, f m Z (called 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 m 1 x B x ρ m!2 2 m 1 x B. Lemma 1.1. Let a progression P be given by ξ Z d, 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 exists v N, and a progression P given by ξ Z d 1, P = d 1 i=1 [ J i, J i ], satisfying (i). (ii). v < L min i J i, and v d 1 i=1 J i < C d L d v i=1 J i, where C d = d [ (d 1)! 2 (d 1) ] d 1 2. d 2 (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. Let y = (y 1,..., y d ) = ( 1 v y 1,..., 1 v y d). 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

6 For x P, let x Z d 1 and ξ Z d 1 be defined by xs 1 = ( x, ), (1.1) and Hence Let Then vsξ T = ( ξ, 0) T. (1.2) vxξ T = (xs 1 )(vsξ T ) = x ξ T. (1.3) B = P S 1. (1.4) d vol(b) = 2 d J i. (1.5) i=1 Denote 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 ineuality in B5 implies x i (d 1)!2 2 d 2 x π(b) ρ 1 i c d ρ 1 i, i (1.6) where c d = (d 1)!2 2 d 2. Let and Denote and Then Hence (1.3) implies This is property (iii) in our conclusion. J i = c d ρ 1 i, (1.7) d 1 P = [ J i, J i]. (1.8) i=1 x = (x 1,, x d 1), f 1 F = GL d 1(Z), f d 1 ξ T = F ξ T. x ξ T = (x F ) ξ T = x ξ T. vxξ T = x ξ T. (1.9) 6

7 From the choice of S, we have e d = y S 1 = y v S 1 L v P S 1 = L B. (1.10) v Hence (cf B1) e d B L v. (1.11) Combining (1.11), B2, and (1.5) we have vol(π(b)) < dl dl vol(b) = 2d 2v 2v On the other hand, the first ineuality in B5 on π(b) gives Hence d J i. (1.12) i=1 {x : x i ρ 1 i } {x : x π(b) d 1} d 1 2 d 1 i=1 Putting (1.7), (1.13) and (1.12) together, we have d 1 i=1 J i = c d 1 d This is (ii) in the Lemma. ρ 1 i ρ 1 i (d 1) d 1 vol(π(b)). (1.13) c d 1 d 2 (d 1) (d 1) d 1 vol(π(b)) c d 1 d 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 ]) 2 Lemma 2.1. Let P = (ξ, P ) be the progression with ξ = (ξ 1,, ξ d ) Z d, and P = d s=1 [1, J s] R d, where the integers J 1 J d > 0. Assume there exists ε > 0 and a Z satisfying i=1 P ap > ε P. (2.1) Then for any index i = 1,..., d, one of the following alternatives hold. 7

8 (i). J i < 2 ε (ii). P is not proper with respect to 9 ε (iii). there exists k i Z, and k = (k 1,, k d ) Zd, such that 0 < k i < 1 ε, k s < 8 ε 2 for all i s d, and ak i ξ i = k ξ T. Proof. Denote Ω = {x P : axξ T P ap}. (2.2) Assume (ii) fails. In particular, the arithmetic progression P in Z is proper. It follows from (2.1) that Ω > 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 ) Ω} > εj i. Assume (i) fails as well. Then εj i 2 and there is k i Z, 0 < k i < 1 ε (2.4) with ak i ξ i P P. Hence ak i ξ i = k ξ T with k = (k 1,, k d) d [ ] Js, J s Z d. (2.5) To show assumption (iii) holds, we need to show k s < 8 ε, for i s d. We 2 assume k t 8 for some t {i,, d}. (2.6) ε2 s=1 Let R = [ ] 4, (2.7) ε l = 2 min s J s k s, (2.8) S = {xξ T + rlk i ξ i : x Ω, r = 1,, R}, (2.9) and S = {x + rlk i e i : x Ω, r = 1,, R}. 8

9 For r N, 1 r R, by (2.7), (2.8), (2.4) and (2.6), we have Hence rlk i < 4 ε 2 J t k t S P + [0, J i ]e i. 1 ε J t J i. (2.10) The above 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 Ω, we have by (2.2) axξ T = xξ T P (2.12) for some x = x(x) P. Let Ω P be the set of all such x. Then there is a one-to-one correspondence between Ω and Ω. 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 [ s=1 ( ) ( J s, ] )J s. (2.15) ε ε The failure of assumption (ii) and (2.15) imply that a S is proper. 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 Ω + lk, Ω + 2lk,, Ω + Rlk are all disjoint. Therefore, as = R ( Ω + rlk )ξ T = Ω R > εj1 J d R > 3J 1 J d, r=1 which contradicts (2.11). Proof of Theorem B. We will use the notion c(k ) for various (maybe different) constants depending on K. Assume A Z p such that K < A < p K and A + θa < K A, where θ Z p satisfies the assumption of Theorem B. By Ruzsa s ineuality A + B 2 A A B 9

10 for A = B, we have A A < (K ) 2 A. Identifying Z {0, 1,..., 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 ), 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 being 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, 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 of 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, necessarily v = 1 (assuming K large enough). Thus by Lemma 1.1, P P 1, where P 1 < c(k ) P and P 1 of dimension d 1. In both cases (either (i) or (ii)), we obtain P 1 of dimension d 1 such that and (2.17) holds. c(k ) P < P 1 < c(k ) P 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 { d1 } P = x i ξ i : 0 x J i, x i Z i=1 and for all i = 1,, d 1 there are k i Z, k i,s Z(1 s d 1) satisfying d 1 θk i ξ i = k i,sξ s (2.18) s=1 10

11 and 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) which is valid for all s = 1,, d 1 ( rather than (2.6) which is a conseuence ). Returning to (2.18), it follows that the polynomial [ p(x) = det (xk i k i,i)e i,i ] k i,je i,j Z p [x] (2.21) i j 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 π Sym(d 1 ) d 1 i=1 ( k i δ i,π(i) + k i,π(i) ) < d 1 c(k ) d 1 π i=1 J π(i) J i < c (K ). This contradicts to the assumption on θ for K sufficiently large. Remark. Quantitatively speaking, the previous argument will reuire K to be at most sublogarithmic in p, since we do rely on Freiman s theorem (cf [C]). Thus we may ask the uestion how large the uantity A + θa min p ε < A <p 1 ε A (2.22) can be made to some θ F p. Considering sets A of the form { d A = i=1 } x i θ i : 0 x i M, it is easily seen that (2.22) is less than exp( log p). 3. Lemma 3.1. Let P = (ξ, P ) be a progression with ξ = (ξ 1,, ξ d ) Z d, and P = d s=1 [1, J s] R d, where the integers J 1 J d > 0. Assume with γ > 0 a constant. δ 0 Ji < P < 1 3γ (3.1) 11

12 Let ε > 0, M > 0 (ε small, M large) satisfy δ 1 0 ( ) 3 d+10 1 ε < M < γ/2. (3.2) Assume Let B Z such that π (P) > M for all, > γ. (3.3) π (B) > M for all, > γ/10d (3.4) denoting the uotient map mod. Then there is a B such that π : Z Z 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 and ε 0 = ε. (3.6) δ 0 P P, (3.7) P is not proper with respect to 9 ε 0. Lemma 1.1 allows then a reduction of the dimension of the progression P in the following sense: There is v 0 N, and ξ (1) Z d 1, P 1 = d 1 i=1 [1, J 1,i] Z d 1 satisfying v 0 < 9 ε 0 min J i, v 0, P 1 < C ε 0 v 0 P, (3.8) v 0 P P 1. (3.9) 12

13 with By (3.9), (3.7) and (3.8), Take and repeat the preceding. 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) ε 1 = εδ 1. (3.12) 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, P 2 = d 2 i=1 [1, J 2,i] Z d 2 satisfying v 1 < 9 ε 1 min J 1,i, v 1, By (3.14), (3.10) and (3.13), P 2 < C ε 1 v 1 P 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) with Notice that Take 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, P r = d r i=1 [1, J r,i] Z d r satisfying Same reasoning as before, with v r 1 < 9 ε r 1 min J r 1,i, v r 1, (3.19) P r < C ε r 1 v r 1 P r 1, (3.20) v r 1 P r 1 P r. (3.21) P r P r > δ r P r, (3.22) δ r = cε r 1 δ r 1. (3.23) 13

14 Also, v 0 v 1 v r 1 P P r. (3.24) We have the following 1. cε 0 ε 1 ε r 1 δ 0 = δ r 2. δ r = cεδ 2 r 1 3. ε r 1 = c(εδ 0 ) 2r 1 3. Assume δ 0 > ε. Then ε r 1 > c ε 2r > (cε) 2d 3. ε 0 ε 1 ε r 1 > c εε r > c ε 2r+1 c ε 2d+1 4. P > c ε 0 ε 1 ε r 1 v 0 v 1... v r 1 P r 4. C ε 0 ε 1 ε r 1 5. ( P > v 0 v 1 v r 1 ) 2 d+1 C ε P > v 0 v 1... v r 1 To see the above (in)eualities 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 and P ap > ε P. By (3.24), (3.7), (4), (1), and (3.18), 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) where ε r = εδ r. (3.26) We need the following little fact from algebra to prove Lemma 3.2. Fact A. Let A Z, k Z, and = gcd(k,). Then π (A) = k A. 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 ordered decreasingly J 1 J 2 J d. 14

15 Let ξ i = ξ(r) i {1, 2,..., 1} Z \{0} and define Claim. d 1 γ. Proof of Claim. Assume Let w = d 1 = gcd (ξ 1, ) 2 = gcd (ξ 1, ξ 2, ). d = gcd (ξ 1,, ξ d, ). d > 1 γ. < γ. Then (5), (3.1) and (3.2) imply v 0 v 1 v r 1 w ( ) 2 d+1 C P γ < ε ( ) 2 d+1 C 1 1 2γ < 1 γ. (3.27) ε δ 0 Also, hence, from (3.24) By Fact A, wξ 1 = = wξ d = 0 (mod ), v 0 v 1 v r 1 wp = 0 (mod ). π (P) = 0 (3.28) with ( by (3.27) ) = This contradicts (3.3). gcd (, v 0... v r 1 w) > γ. Therefore, there is i {1,..., d } such that i 1 i > γ d. (3.29) 1, 1 2,..., i 2 i 1 γ d (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 ( ) 2 d+1 C ε > 2 > J i J i+1 J d. (3.31) ε r Let v = v 0... v r 1. i 1 15

16 By (3.30), (5), (3.1) and (3.2), v v 0 v r 1 γ < Hence, from the definition of i 1 (3.31), (3.32) imply vp = v 0 v 1 v r 1 P i 1 ( ) 2 d+1 C P γ < c 1 3γ+ γ 2 γ < 1 γ ε P r i 1 { x s ξ s : x s J s = i 1 s i }. (3.32) vp J ij i+1... J d < ( C ε ) d2 d+1 < M. Hence Fact A implies with = π (P) = vp < M gcd(,v) > γ again contradicting (3.3). 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 ε 2 r for i s d (3.34) ak i ξ i = k sξ s (mod ). (3.35) d s=1 Since i 1 = gcd(ξ 1,..., ξ i 1, ), (3.35) implies ak i ξ i = s i k sξ s (mod i 1 ). (3.36) By (3.33), (3.34), (3 ) and (3.2), the coefficients (k i, k i,..., k d ) in (3.36) ranges in ( a set of at most 1 8 ) d ( ε r ε < C ) (2d+1)2 d+1 2 r ε < M 1/2 elements. Recalling (3.29) and (3.4) π i 1 (B) > M (3.37) i and we may consider elements B B, B > M, such that π i 1 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). 16

17 In view of the bound on the number of coefficients in (3.36), there is B B, 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 (a 1 a 2 )k i ξ i = 0 (mod i 1 ) (a 1 a 2 )k i = 0 ( i 1 ) mod i (3.39) implying a contradiction. 1 = π i 1 (k i B ) 1 i k i This proves Lemma 3.1. π i 1 (B ) B = i k i > ε rm 1/2 Following the same arguments as in Lemma 3.1, we also obtain: Lemma 3.1. Under the assumptions of Lemma 3.1, there exit 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) Proof. Let B B be the set constructed in the proof of Lemma 3.1. Assume a 1,..., a r B obtained satisfying (3.40) and suppose max a sp ap > ε P for all a B. 1 s r Hence, there is some s = 1,..., r and B 1 B with B 1 > 1 r B > M R > M 9/10 and such that P a a s P > ε P for all a B 1. (3.41) It follows that all elements a a s, a B 1, have a representation (3.36). Passing again to a subset B 1, B 1 > M 9/10 1/2 = M 2/5, we may ensure the same coefficients k i, (k s) s i and get a contradiction as before. 4. and Let A Z such that A + A < K A (4.1) 1 A < 1 4γ. (4.2) Identifying Z {0, 1,..., 1}, we apply Freiman s theorem to A, first considered as a subset of Z (with doubling constant 2K). 17

18 From [C], we obtain d 2K and a progression P given by ξ = (ξ 1,..., ξ d ) Z d ; P = d i [0, J i], with J 1 J 2 J d in N, such that A P = { xξ T : x P } (4.3) and P < C K3 A. (4.4) Applying π : Z Z, P becomes a progression in Z containing A Z. Assuming by (4.5), (4.4), (4.2) and (4.4), we have C K3 < γ/2, (4.5) 1 3γ > C K3 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 ) 10 K+5 < M < γ/2 (4.7) and moreover ε < C K3. Assume B Z such that π (B) π (A), (e.g. B,contained in a translate of A) for all and > γ. Furthermore, assume B satisfies Since by assumption also π (B) > M, if and > γ 20K. (4.8) π (P) π (A) π (B) > M, if and > 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 R r s R A r s a r P a s P > R A R 2 εc K3 A. (4.9) Taking R = 1 2 ε C K3, (4.9) implies AB a r A > R 2 A > 1 ε C K3 A. (4.10) r R 18

19 Assume M > C 50K+10 (4.11) ( which implies the first ineuality of (4.7), hence it also implies (4.5) ) and take 1 ε = M 10 K 6. From (4.10) and (4.11) AB > M 10 K 7 A. (4.12) Replacing 4γ by γ and summarizing, we proved the following: Lemma 4.1. Let A Z satisfy A < 1 γ (4.13) A + A < K A. (4.14) Let M satisfy C 50K+10 < M < γ/8. (4.15) Let B Z such that π (B) π (A), if and > γ, also Then π (B) > M, if and > γ/80k. (4.16) AB > M 10 K 7 A. (4.17) 5. Proposition 1. Let κ > 0 be a small and M a large constant. Let H Z ( large) satisfy π (H) > M whenever, > κ. (5.1) Then there is k, r N such that k < κ (5.2) r < log 2 κ (5.3) kh r > 1 κ (5.4) where κ = κ (κ, M) 0 for κ 0, M (independent of ). 19

20 Proof. We describe the construction. Given any and denote 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 < 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 < 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 ineuality 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. with Therefore A α+1 > K A α, (5.7) k α+1 2k α (5.8) r α+1 r α+1. To reach size 1 κ 1, the number of steps is at most log log K, because after s steps, by (5.7) A α+1 > K s H K s. By (5.8), we have in (5.2) Hence by (5.5) and (5.6). k 2 log log K = κ 2. κ 2 = log 2 log K 1 min{log log log M, log 1 κ }, We conclude the proof of Proposition 1 by taking κ = max{κ 1, κ 2 }. 20

21 6. We need the following to prove Theorem A. Let ν, ν : Z R be functions. We recall F1. ˆν(ξ) = x Z ν(x)e ( xξ). If ν is a probability measure, i.e. 0 ν(x) 1, then ˆν(ξ) 1. F2. ν ν (x) = y Z ν(x y)ν (y). F3. supp (ν ν ) Supp ν + Supp ν. F4. ν(x) = 1 ξ Z ˆν(ξ)e (xξ). F5. ν ν (ξ) = ˆν(ξ)ˆν (ξ). Let 0 x 5π 6. Then T1. sin x > x 2π. Therefore, e (1) 1 > 1. T2. cos x < 1 x2 4π. Therefore, e (1) + 1 < 2 π 2 T3. e (x) e (y) = 2 sin 2π 2 (x y). Proof of Theorem A. Let κ > 0, M be constants as in Proposition 1. Let N be odd. Let H Z satisfy the following conditions: 1 2. π p (H) 2 for all primes p (6.1) π (H) > M for all, > κ (6.2) We want to show that for some k 1, r N satisfying k 1 H r = Z r < log κ (6.3) k 1 < 5κ. (6.4) By (6.2), Proposition 1 applies. Let k, r satisfy (5.2)-(5.4). Denote D = { N : 1 and }, hence For D, we have 1 D < log log. (6.5) π (kh r ) khr / > 1 κ / = κ (6.6) while by (6.1), also π (kh r ) π (H) 2. (6.7) 21

22 Take a subset Ω kh r such that π Ω is one-to-one and Define the probability measures Ω max {2, κ }. (6.8) µ = 1 Ω x Ω δ x being the indicator function, and their convolution δ x, (6.9) µ(x) = * µ (x) D = µ D (x y 1 y D 1 ) µ 2 (y 2 )µ 1 (y 1 ). y 1,,y D 1 (6.10) Then by F3, supp µ supp µ Ω D kh r. (6.11) D D We estimate the Fourier coefficients ( ) a ˆµ = e ( ax)µ(x) x Z for 0 < a <. Let a = a where and (a, ) = 1. From (6.10) and F5 ( ) ˆµ(a ) a µ = 1 Ω x Ω e (a x). (6.12) Claim 1. ˆµ ( a ) < κ. Proof of Claim 1. Assume ˆµ ( a ) > 1 τ. We want to find a lower bound on τ. Suaring both sides of (6.12), we obtain x,y Ω Choose an element x 0 Ω By T3, we write y Ω e (a x 0 ) e (a y) 2 = cos 2πa (x y) > (1 τ) 2 Ω 2. (6.13) such that cos 2πa (x 0 y) > (1 τ) 2 Ω. (6.14) ] 2 [2 sin 2πa (x 0 y) = 2 2 cos 2πa 2 22 (x 0 y).

23 Together with (6.14) give e (a x 0 ) e (a y) 2 2 Ω 2(1 τ) 2 Ω < 2τ Ω. (6.15) y Ω From (6.15), { y : e (a x 0 ) e (a y) > 2 τ } 1 2 Ω. So there is a subset Ω Ω, Ω > 1 2 Ω such that for all y Ω hence T1 implies Therefore Since π Ω is one-to-one, also e (a x 0 ) e (a y) < 2 τ, a x 0 a y π (Ω ) = π (a Ω ) 2 τ κ by (6.8). This gives a lower bound < 2 τ. (6.16) < Ω 2 τ + 1 (6.17) τ > κ. Take l = [ 3κ ] (6.18) Claim 2. Let µ (l) be the l-fold convolution of µ. Then supp µ (l) = Z. Proof of Claim 2. For x Z, write µ (l) (x) = ( a µ (l) a=1 By Claim 2 and (6.18), the second term in (6.19) is at most Hence µ (l) (x) > 0. ) e (ax). (6.19) ( ) ( ) max a 1 a< µ (l) = max a l ˆµ 1 a< ( < 1 1 ) 3κ < e κ < κ. Hence, putting together Claim 2, (6.11), (6.18), (6.5) and (5.2), we have Z = l supp µ = l D kh r = k 1 H r 23

24 with k 1 3κ 1/ log log κ < 5κ. Remark 6.1. It is much simpler to prove the following weaker bound Proof. ( a µ ) < 1 π 2 1 ( ) 3. Since Ω 2, there are elements x 1, x 2 Ω with π (x 1 ) π (x 2 ). Since (a, ) = 1, also π (a x 1 ) π (a x 2 ). Therefore, by T2, e (a x 1 ) + e (a x 2 ) e (1) + 1 < 2 π 2 1 ( ) 2 Write x Ω e (a x) ( Ω 2) + e (a x 1 ) + e (a x 2 ) < Ω π 2 1 ( ) 2. Therefore µ ( a ) < 1 π 1 2 ( ) 3. Remark 6.2. For prime, Theorem A has a simpler proof, which gives a slightly weaker bound on r. In this case, Z is a cyclic group. The condition on H Z 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. Case 1. t log (log log ) 1/2. Take r 0 log (log log ) 4. Using the ineuality max (a,)=1 r 0 x=1 {1, θ,, θ r } H r. e (aθ x ) < r 0 ( 1 ) c (log ) 2 due to Konyagin (see [KS], p. 26), simple application of the circle method implies kh r 0 = Z with k < C(log ) 3. Case 2. t < log (log log ) 1/2. 24

25 Denote ϕ(t) = Z t. Use then Theorem 4.2 from [KS] max (a,)=1 t e (aθ x ) < t c(ρ) 2/ρ for 2 ρ ϕ(t). (6.20) x 1 Hence kh t = Z with k < 1 c(ρ) (log )2 2/ρ. Since ϕ(t) for M, we may achieve k < c(κ ) κ M. with κ (M) 0 as 7. Corollary Let H Z satisfy the assumption (5.1) of Proposition 1 and κ be as in Proposition 1. Let, > κ and (a, ) = 1. Let r N, r > κ log. Then max x,y H r a (x y) >. (7.1) 2κ 2. Let H < Z be a multiplicative subgroup satisfying assumption (5.1) of Proposition 1. Let, > κ and (a, ) = 1. Then max x,y H a (x y) >. (7.2) 2κ Proof. By (5.4) π (kah r ) = π (kh r ) > 1 κ / Hence there are elements z, w kh r s.t. = κ > 1. (7.3) a (z w) κ. (7.4) Writing z = x x k, w = y y k with x i, y i H r, it follows that max x,y H r a (x y) 1 k κ > 2κ. by (5.2). 25

26 Corollary Let H Z satisfy conditions (6.1), (6.2) and κ be as in Theorem A. Let 1 a <. Then for r > κ log max x,y H r a (x y) 5κ. (7.5) 2. If moreover H < Z is a group, we get max x,y H a (x y) 5κ (7.6) Proof. Write a = a, (a, ) = 1. Since π (k 1 H r ) = Z, max z,w k 1 H r a (z w) 1 2 (7.7) hence max x,y H r a (x y) 1 > 5κ. (7.8) k 1 8. The case of subgroups The main result of this section is the following (for prime this issue was considered in [P]) Theorem 5. Let H < Z, H > M > 1. Then min max a Z x,y H where δ = δ(m) 0 for M (independently of ). a (x y) > δ (8.1) 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 be of order t (large). Then min max a (a,)=1 j,k (θj θ k ) > 1 c(r) r 1 (8.2) for 1 < r < ϕ(t). 26

27 Proof. For j = 1,..., t, let b j Z such that and extend periodically with period t for j Z. b j = aθ j (mod ) (8.3) Claim. Let c Z and 2 r < ϕ(t). Then max j b j c > c(r) r 2 r 1. proof of Claim. Let B = max 1 j t b j c. (8.4) Denote b = (b 1,, b r ), and 1 = (1,, 1). We consider the lattice L = { l = (l 1,..., l r ) Z r : b l T = 0, 1 l T = 0 } = { l = (l 1,..., l r ) Z r : (b c1) l T = 0, 1 l T = 0 } (8.5) We considering 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 For this vector l = (l 1,..., l r ), Hence, multiplying with θ j, for all j. Since also l l r = 0 The left side is bounded by by (8.4), (8.6). Assume c(r)b r 1 r 2 max l j < c(r)b 1 r 2. (8.6) 1 j r b 1 l b r l r = 0. b j+1 l b j+r l r = 0 (mod ) (b j+1 c)l (b j+r c)l r = 0 (mod ). (8.7) <, hence It follows then from (8.7) that rbc(r)b 1 r 2 < c(r)b r 1 r 2 B < c(r) 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. 27

28 Hence (b j ) is a periodic linearly recurrent seuence of order at most r and smallest period t. Let ψ(x) be the minimal polynomial of (b j ). (See [KS].) Then form (8.9) implying deg ψ r 1. Obviously ψ(x) (x t 1). Assume ψ(x) (l 1 + l 2 x + + l r x r 1 ) ψ(x) 1 r<t (1 x τ ). Since ψ(θ) = 0 (mod ), it would follow that θ τ 1 ( mod ) for some τ < t, contradicting ord (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 a contradiction. Hence (8.8) fails. φ(t) deg ψ < r Suppose (8.2) fails. Letting c = aθ k Z = {0, 1,..., 1}, Hence max j max j aθ j c < 1 c(r) r 1. dist(aθ j c, Z) < c(r) 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 ), and b j c < c(r) r 2 r 1. This contradicts the Claim and proves the lemma. Proof of Theorem 5 Let H < Z, H > M. Fix κ > 0 (small) and 1 M 1 < M κ 2. By Lemma 8.2, we may assume ord (x) < M 1 for all x H. (8.11) If π 1 (H) > M 1 for all 1, with 1 > κ, then (8.1) holds for δ = δ(κ, M 1 ) 0 for κ 0, M 1 ( by Corollary 3(2) ). 28

29 Assume thus there exists 1, with 1 > κ and π 1 (H) M 1. Hence H 1 = H π 1 1 (1) < Z satisfies H 1 > M M 1. Consider the set H 1 = {x Z 1 : x H 1 } = H Assume there is 2 1, with 2 > κ and π 2 (H 1 ) < M 1. Hence π 1 2 (H 1 ) < M 1 and defining H 2 = H 1 π (1), we have Considering the set we repeat the process. and H 2 > H 1 > M M 1 M1 2. H 2 = { x Z 1 2 : x H 2 } = H , At some stage s 1 κ, the process has to stop. Thus Define H s = {x Z 1 s H s = H π 1 1 s (1), (8.12) : s x H s } = H s 1 1 s, (8.13) H s = H s > M M 1/κ 1 > M 1/2, (8.14) π (H s ) > M 1 for all, with > κ. (8.15) 1 s Q 1 = 1... s and Q 2 = Q 1. Case 1. Q 2 < 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 2 > 1 > κ. Q 2 29

30 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 = a( x ȳ), we obtain a( x ȳ) and hence (8.1) holds. Case 2. Q 2 κ. > κ First, note that if there is no ambiguity, we use the notation (A, B) = gcd(a, B). Claim 1. (Q 1, Q 2 ) 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 Q1 (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 π Q1 (Q 1,Q 2 )(H s ) M 1. Therefore By construction of H s, (8.16) implies π (Q1,Q 2 )(H s ) M 1. (8.16) (Q 1, Q 2 ) κ. (8.17) Let Q 1 = Q 1 (Q 1,Q 2 ), Q 2 = Q 2 (Q 1,Q 2 ). Hence (Q 1, Q 2) = 1 and Q 2 > κ κ by case assumption and (8.17). We want to apply Corollary 3(2) to π Q 2 (H s ) < Z Q 2 Let Q 2 with > (Q 2) 4 κ > 2κ, and let = > κ. with 4 κ and M 1. (,Q 1,Q 2 ). Thus by (8.17), Claim 2. π (H s ) > M 1. Proof of Claim 2. It follows from (8.15) that π (H s ) > M 1. 30

31 Let x 1,..., x n H s, n > M 1 such that Since (, Q 1) = (, Q 1) = 1 and we also have Hence Since 1 + Q 1 x i H s, it follows that x i x j 0 (mod ). ( ) (Q 1, Q 2 ), ( = 1,, Q 1, Q 2 ) Q 1 (, Q 1, Q 2 ) (x i x j ) 0 (mod ). Q 1 (x i x j ) 0 (mod ). π (H s ) > M 1. Apply Corollary 3 (2) to the group π Q 2 (H s ) < Z Q 2. Claim 2 implies that for all Q 2 with x, ȳ H s such that π ( πq 2 (H s) ) = π (H s ) > M 1 > (Q 2) 4 κ. Hence for any a, (a, Q 2) = 1, there are thus a ( x ȳ) > (Q 2) κ > κ (8.18) Q 2 where κ = κ (4 κ, M 1 ) 0 for κ 0, M 1. Write x = 1 + Q 1 x, ȳ = 1 + Q 1 y with x, y H s. From (8.18) a Q 1 Q 2 Recalling that (Q 1, Q 2) = 1, we may choose a satisfying Therefore (8.19) gives Hence, by Claim 1, (x y) >. (8.19) κ a Q 1 a (mod Q 2). a(q 1, Q 2 ) 2 (x y) Q 2 > κ. a (x y) Q 2 > κ (Q 1, Q 2 ) 2 > κ 2κ, and a ( x ȳ) = a (Q 1x Q 1 y) > 2κ κ. 31

32 Therefore max x,y H a (x y) > 2κ κ, where κ, κ may be made arbitrary 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 conseuence of Theorem 5. Proof of Theorem C. For a Z, let {x 1,, x H } = ah, and let x 1 = ax, x 2 = ay be given in Theorem 5 such that x 1 x 2 > κ where κ = κ(m). Let Then H S = e (x i ). i=1 S 2 = H + 2 ( ) 2π(xi x j ) cos i j [( ) ] ( ) H 2π(x1 x 2 ) H cos 2 ( H π x 1 x 2 2) < H 2 2π 2κ. References [BIW]. B. Barak, R. Impagliazzo, A. Wigderson, Extracting randomness using few independent sources, Proc of the 45th FOCS (2004), [BKSSW]. B. Barak, G. Kindler, R. Shaltiel, B. Sudakov, A. Wigderson, Simulating Independence: New Constructions of Condensers, Ramsey Graphs, Dispersers, and Extractors, STOC (to appear). [B]. J. Bourgain, Mordell s exponential sum estimate revisited, (preprint 2004, submitted to JAMS). [BC]. J. Bourgain, M. Chang, Exponential sum estimates over subgroups and almost subgroups of Z, where is composite with few prime factors, submitted to GAFA. [BGK]. J. Bourgain, A. Glibichuk, S. Konyagin, Estimate for the number of sums and products and for exponential sums in fields of prime order, submitted to J. London MS. [BKT]. J. Bourgain, N. Katz, T. Tao, A sum-product estimate in finite fields and their applications, GAFA 14 (2004), n1, [C]. M. Chang, A polynomial bound in Freiman s theorem, Duke Math. J. 113 (2002), no 3,

33 [Bi]. Y. Bilu, Structure of sets with small sumset, in Structure theory of et additions, Asterisue 258, SMF (1999), [KL]. S. Konyagin, I. Laba, Distance sets of well-distributed planar sets for polygonal norms, Israel J. Math (to appear). [KS]. S. Konyagin, I. Shparlinski, Character sums with exponential functions and their applications, Cambridge Tracts in Math., 136 (1999). [N]. M.B. Nathanson, Additive Number Theory: Inverse Problems and the Geometry of Sumsets, Springer (1996).. [P]. C. Powell, Bounds for multiplicative cosets over fields of prime order, Math. Comp. 66 (1997), no 218, [R]. R. Robinson, Number having m small mth roots mod p, Mathematics of Computation Vol 61, no 203, (1993), Mathematics Department, University of California, Riverside CA address: mcc@math.ucr.edu 33