BURGESS INEQUALITY IN F p 2. Mei-Chu Chang. 0 Introduction

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1 Geom. Funct. Anal. Vol DOI 0.007/s Published online November 2, The Authors GAFA Geometric And Functional Analysis This article is published with open access at Springerlink.com BURGESS INEQUALITY IN F p 2 Mei-Chu Chang Abstract. The purpose of the paper is to present new estimates on incomplete character sums in finite fields that are of the strength of Burgess celebrated theorem for prime fields. More precisely, an inequality of this type is obtained in F p 2 and also for binary quadratic forms, improving on the work of Davenport Lewis and on several results due to Burgess. The arguments are based on new estimates for the multiplicative energy of certain sets that allow us to improve the amplification step in Burgess method. 0 Introduction The paper contributes to two problems on incomplete character sums that go back to the work of Burgess and Davenport Lewis in the sixties. Incomplete character sums are a challenge in analytic number theory. By incomplete, we mean that the summation is only over an interval I. Typical applications include the problem of the smallest quadratic non-residue mod p and the distribution of primitive elements in a finite field. Recall that Burgess bound [B] on multiplicative character sums x I x inaprimefieldf p provides a nontrivial estimate for an interval I [,p ] of size I >p /4+ε,withanygivenε>0. Burgess result, which supersedes the Polya Vinogradov inequality, was a major breakthrough and remains unsurpassed. It is conjectured that such result should hold as soon as I >p ε. The aim of this paper is to obtain the full generalization of Burgess theorem in F p 2.Thus Theorem 5. Given ε>0, thereisδ>0such that if ω F p 2\F p and I,J are intervals of size p /4+ε p sufficiently large, then x + ωy <p δ I J 0. x I y J for a nontrivial multiplicative character. The importance of the statement is its uniformity in ω. Both Burgess [B2] and Karacuba [K] obtained the above statement under the assumption that ω satisfies a given quadratic equation ω 2 + aω + b =0 modp 0.2 with a, b Q. Keywords and phrases: Character sums, finite fields, multiplicative energy, Burgess, Davenport Lewis 2000 Mathematics Subject Classification: Primary: L40, L26, Secondary: A07, B75

2 002 M.-C. CHANG GAFA In the generality of Theorem 5, the best known result in F p 2 was due to Davenport and Lewis [DL], under the assumption I, J >p /3+ε. More generally, they consider character sums in F p n of the form x ω + + x n ω n, 0.3 x I,...,x n I n where I,...,I n [,p ] are intervals. It is shown in [DL] that x ω + + x n ω n <p δε I I n 0.4 x I,...,x n I n provided for some ε>0, I i >p ρ+ε n with ρ = ρ n = 2n In [C2], newly developed sum-product techniques in finite fields were used to establish 0.4 under the hypothesis I i >p 2 5 +ε for some ε> Hence [C2] improves upon 0.5 provided n 5 and Theorem 5 in this paper provides the optimal result for n =2. We will briefly recall Burgess method in the next section. It involves several steps. As in [C2], the novelty in our strategy pertains primarily to new bounds on multiplicative energy in finite fields see section for definition. The other aspects of Burgess technique remain unchanged. We also did not try to optimize the inequality qualitatively, as our concern here was only to obtain a nontrivial estimate under the weakest assumption possible. The new estimates on multiplicative energy are given in Lemma 2 and Lemma 3 in section. Contrary to the arguments in [C2] that depend on abstract sum-product theory in finite fields, the input in this paper is more classical. Lemma 2 is based on uniform estimates for the divisor function of an extension of Q of bounded degree. In Lemma 3, we use multiplicative characters to bound the energy EA, I = { x,x 2,t,t 2 A 2 I 2 : x t x 2 t 2 mod p }, 0.7 where A F p n is an arbitrary set and I [,p ] an interval. The underlying principle is actually related to Plunnecke Ruzsa sum-set theory [TV] here in its multiplicative version, but in this particular case may be captured in a more classical way. Closely related to Theorem 5 is the problem of estimating character sums of binary quadratic forms over F p, x 2 + axy + by 2, 0.8 x I,y J where x 2 +axy+by 2 F p [x, y] is not a perfect square and a nontrivial multiplicative character of F p. Theorem. Given ε>0, thereisδ>0such that if x 2 +axy +by 2 is not a perfect square mod p, andifi,j [,p ] are intervals of size I, J >p 4 +ε, 0.9

3 GAFA BURGESS INEQUALITY IN F p then for p sufficiently large, we have x 2 + axy + by 2 <p δ I J, 0.0 x I,y J where δ = δε > 0 does not depend on the binary form. This is an improvement upon Burgess result [B3], requiring the assumption I, J >p /3+ε. We will not discuss in this paper the various classical application of Theorem to primitive roots, quadratic residues, etc as the arguments involved are not different from the ones in the literature. Preliminaries and Notation In what follows we will consider multiplications in R = F p d and R = F p F p.denote by R the group of invertible elements of R. LetA, B be subsets of R. Denote AB := {ab : a A and b B}. 2 ab := {a}b. Intervals are intervals of integers. 3 [a, b] :={n Z : a n b}. 4 The multiplicative energy of A,...,A n R is defined as EA,...,A n := {a,...,a n,a,...,a n :a a n = a a n } with the understanding that all factors a i,a i are in A i R. Using multiplicative characters of R, one has 5 EA,...,A n = n R i= ξ i A i ξ i 2. Energy is always multiplicative energy in this paper. 6 Burgess method. In this paper we will apply Burgess method several times. We outline the recipe here, considering intervals in F p 2. For details, see section 2 of [C2]. Suppose we want to bound x + ωy,. x I,y J where I,J are intervals. We translate x, y bytu, tv TM, wherem = I J is aboxinf p 2,andT =[,T] such that T I <p ε I and T J <p ε J for some small ε>0. Therefore, it suffices to estimate the following sum T M t T x I u,v M y J Let wµ = { x, y, u, v I J M : µ = x+ωy u+ωv}. x + tu +y + tvω..2

4 004 M.-C. CHANG GAFA Then the double sum in.2 is bounded by wµ t + µ wµ 2k 2k 2k µ F p 2 t T µ F p 2. }{{} α µ F p 2. µ + t 2k 2k t T } {{ } β.3 where k is a large integer to be chosen. By Hölder s inequality and the definition of wµ, α [ wµ ] k [ wµ 2 ] 2k = I J I J k E I + ωj, I + ωj 2k. A key idea in Burgess approach is then to estimate.3 using Weil s theorem for multiplicative characters in F p n here n = 2, leading to the bound, β kt /2 p n/2k +2Tp n/4k. So the remaining problem to bound the character sum. is reduced to the bounding of multiplicative energy EI + ωj, I + ωj. We will describe a new strategy., 2 Multiplicative Energy of Two Intervals in F p 2 The first step in estimating the multiplicative energy is the following: Lemma. Let ω F p 2\F p and Then Q = { x + ωy : x, y [, 0 p/4]}. max {z,z 2 Q Q : ξ = z.z 2 } < exp c ξ F p 2 An essential point here is that the bound is uniform in ω. Also, the specific size of Q is important. Note that for our purpose, any estimate of the type p o would do as well. Proof. For given ξ F p 2, assume that ξ can be factored as products of two elements in Q in at least two ways. We consider the set S of polynomials in Z[X] y y 2 y y 2X 2 +x y 2 + x 2 y x y 2 x 2y X +x x 2 x x 2, 2. where x i + ωy i,x i + ωy i Q for i =, 2, and x + ωy x 2 + ωy 2 =ξ =x + ωy x 2 + ωy in F p 2. Let gx =X 2 + ax + b F p [X] be the minimal polynomial of ω. Then it is clear that every fx ins, when considered as a polynomial in F p [X], is a scalar multiple of gx. Next, observe that, by definition of Q, the coefficients of 2. are integers bounded by 25 p/2. Therefore, since the coefficients of two non-zero polynomials 2. are proportional in F p, they are also proportional in Q. Thus the polynomials.

5 GAFA BURGESS INEQUALITY IN F p are multiples of each other in Q[X] and therefore have a common root ω C. Since x + ωy x 2 + ωy 2 =x + ωy x 2 + ωy in Q ω whenever 2.2 holds, it suffices to show that if we fix some ξ Q ω, then {z,z 2 Q Q : ξ = z z 2 } < exp c, 2.4 where { [ Q = x + ωy : x, y, 0 p/4]}. This is easily derived from a divisor estimate. Let ux 2 + vx + w be a nonzero polynomial in S, then u ω 2 + v ω + w =0. Note that η = u ω is an algebraic integer, since it satisfies η 2 + vη + uw =0. Thus u 2 ξ =ux + ηy ux 2 + ηy 2 is a factorization of u 2 ξ in the integers of Qη. Since the height of these integers is obviously bounded by p, 2.4 is implied by the usual divisor bound in a quadratic number field which is uniform for extensions of given degree. This proves Lemma. As an immediate consequence of Lemma, we have the following: Lemma 2. Let Q be as in Lemma. Then the multiplicative energy EQ, Q satisfies EQ, Q < exp c Q and Lemma 2. Let Q be as in Lemma and z,z 2 F p 2.Then Ez + Q, z 2 + Q < exp c Q Proof of Lemma 2. We have Ez + Q, z 2 + Q Q 2 + EQ + Q, z 2 + Q and by Cauchy Schwarz see [TV, Cor. 2.0] EQ + Q, z + Q EQ + Q, Q + Q /2 Ez + Q, z + Q /2. Hence 2.6 follows from Further Amplification The second ingredient is provided by Lemma 3. Let Q be as in Lemma, and let I =[,p /k ],wherek Z +. Let z,z 2 F p 2.Then EI,z + Q, z 2 + Q < exp c p + 3 2k. 3.

6 006 M.-C. CHANG GAFA Proof. Denote the multiplicative characters of F p 2.Thus EI,z + Q, z 2 + Q = t 2 ξ + z 2 ξ + z 2 p 2 t I } {{ } A 2 ξ Q } {{ } B 2 ξ Q Here the sum over ξ Q is such that ξ + z i 0,fori =, 2. Hence by Hölder s inequality, EI,z + Q, z 2 + Q { [ A 2 ] } BC 2 { k k k = { p 2 p 2 } {{ } 3.3 p 2 A 2k B 2 C 2} { k p 2 B 2 C 2} k. 2 } {{ } C 2 [ ] BC 2 2 k } k k k. 3.2 Since the second factor is equal to Ez + Q, z 2 + Q k, 2.6 applies and we obtain the bound 3.3 exp c Q 2 k. 3.4 Estimate 3.3 as 3.3 Q 2/k{ t 2k ξ + z 2} /k < exp c =exp p 2 t I ξ Q p 2 {. Q 2 k t 2 ξ + z 2 } /k t F p ξ Q Q 2 k EFp,Q+ z k. 3.5 c The second inequality is by definition of I and the divisor bound. Next, let z = a+ωb, with a, b F p and let Q = J + ωj, withj =[,p /4 ]. Then EF p,q+ z = {t,t 2,ξ,ξ 2 F 2 p Q 2 : t ξ + z =t 2 ξ 2 + z 0} = {t,t 2,x,x 2,y,y 2 F 2 p J 4 : t x + a+ωy + b = t 2 x 2 + a+ωy 2 + b 0}. 3.6 Equating coefficients in 3.6, we have { t x + a =t 2 x 2 + a, t y + b =t 2 y 2 + b, Therefore, x + a y + b = x 2 + a y 2 + b.

7 GAFA BURGESS INEQUALITY IN F p and the number of x,x 2,y,y 2 satisfying 3.6 is bounded by Ea + J, b + J, which is bounded by p /2, by[fi].hence, EF p,q+ z p 3/2. By 3.5 and 3.4, 3.3 exp c Q 2/k p 3/2k, and EI,z + Q, z 2 + Q exp c Q 2 p 3/2k. This proves Lemma 3. Lemma 4. Let I j =[a j,b j ], where b j a j p /4 for j =,...,4. Denote R = I + ωi 2 and S = I 3 + ωi 4. Let I =[,p /k ] with k Z +. Then EI,R,S < exp c p 3 2k R 2 S Proof. Subdivide R and S in translates of Q and apply Lemma 3. Thus the left side of 3. needs to be multiplied with R / Q 2 S / Q 2 which gives Proof of Theorem 5 We now establish the analogue of Burgess for progressions in F p 2. Theorem 5. Given ρ>/4, thereisδ>0such that if ω F p 2\F p and I,J are intervals of size p ρ,then x + ωy <p δ I J 4. x I y J for a nontrivial multiplicative character. This estimate is uniform in ω. Proof. Denote I 0 =[,p /4 ]andk =[,p κ ], where κ is the reciprocal of a positive integer and ρ> 4 +2κ. 4.2 We translate I + ωj by KKI 0 + ωi 0 and estimate following the procedure sketched in section K 2 I 0 2 x + ωy + stx + ωy = x,y I 0 s K x I,y J K 2 I 0 2 t K x I,y J x,y I 0 s K t K t + x + ωy. 4.3 sx + ωy

8 008 M.-C. CHANG GAFA With the notation from section, we have α I 0 2 K I J k EK, I 0 + ωi 0,I + ωj 2k exp c I 0 2 K I J /2k K 3/2 I k 0 4 I 2 J 2 p =exp c I 0 2 I J K 4k p 2k, by Lemma 4, and β K /2 p /k + K p /2k. Hence, taking κ =/k, 4.3 is bounded by I J p /4k2 and the theorem is proved with any δ</4k 2 taking p large enough. Remark 5.. In [DL], the result 4. was obtained under the assumption that ρ>/3. In general, it was shown in [DL] that if ω,...,ω d is a basis in F p d then x ω + + x d ω d <p δ I I d, 4.6 x i I i provided I,...,I d are intervals in F p of size at least p ρ with d ρ> 2d For d 5, there is a better uniform result in [C2], namely ρ> ε. 4.8 As a consequence of Theorem 5, we have Corollary 6. Assume k F p is not a quadratic residue. Then x 2 + ky 2 <p δ I J 4.9 x I y J for nontrivial and I,J intervals of size at least p 4 +ε.hereδ = δε > 0 is uniform in k. Proof. Let ω = k. Since x 2 + ky 2 is irreducible modulo p, x 2 + ky 2 isa character mod p of x + ωy in the quadratic extension Qω. 5 Extension to F p d There is the following generalization of Lemma. Lemma 7. Let ω F p d be a generator over F p.given0 <σ</2 and let Q = { x 0 + x ω + + x d ω d : x i [,p σ ] } Then Q = { y 0 + y ω : y i [,p σ]} 2. max {z,z Q Q : ξ = zz } < exp ξ F p d c d. 5.

9 GAFA BURGESS INEQUALITY IN F p Proof. The proof is similar to that of Lemma. It uses the fact that if x 0 + x ω + + x d ω d y 0 + y ω=ξ =x x d ωd y 0 + y ω then the polynomial x 0 + x X + + x d X d y 0 + y X x 0 + x X + + x d Xd y 0 + y X is irreducible in F p [X], or vanishes. Hence the analogues of Lemmas 2, 2 hold. Thus Lemma 8. Let Q, Q be as in Lemma 7 and let z F p d.then Ez + Q, Q < exp c d Q Q + Q We need the analogue of Lemma 3, but in a slightly more general setting. Lemma 9. Let Q, Q be as in Lemma 7 with Q Q and let I s =[,p /ks ] for s =,...,r,withk s Z + and k + + k r <. Then EI,...,I r,z+ Q, Q < exp c d p +d 2σ+2 σ r s= /ks =exp c d Q Q I s 2 σ. 5.3 s Proof. The left of 5.3 equals r p t 2 z + ξ 2 ξ 2 d s= t I s ξ Q ξ Q which we estimate by Hölder s inequality as r { p d 2ks 2 2 } { ks p d 2 2 } ks. s= t I s ξ Q ξ Q } {{ } A /ks s ξ Q ξ Q } {{ } B /ks Here we denote t I s = t I s t, ξ Q = ξ Q z + ξ, etc. By Lemma 8 B = Ez + Q, Q < exp 5.4 Q Q. 5.5 c It is clear from the definition of multiplicative energy that A s Q 2 EI s,...,i }{{} s,z+ Q k s Q 2 exp c ks EF p,z+ Q. To bound EF p,z+ Q, we write z = a 0 + a ω + + a d ω d. Hence d EF p,z+ Q = Θ i, 5.6 i=0

10 00 M.-C. CHANG GAFA where { Θ 0 = t, t,x 0,...,x d,x 0,...,x d F2 p [,p σ ] 2d : + x + a ω + + x d + a d ω d 5.7 x 0 + a 0 x 0 + a 0 = t + x + a x 0 + a ω + + x d + a } d 0 x 0 + a ω d and the other Θ i s are denoted similarly. Equating the coefficients of 5.7 and 5.8, we have t = t, x i + a i = x i + a i x 0 + a 0 x 0 + a, for i =,...,d For i =, the number of solutions x 0,x 0,x,x in 5.9 is bounded by E[,p σ ]+a 0, [,p σ ]+a, which is bounded by p 2σ. The choices of t and x 2,...,x d is bounded by pp σd 2. Therefore, EF p,z+ Q dp +σd, and A s Q 2 exp c ks p +σd. 5.0 Note that Q = p dσ and Q = p 2σ. Putting 5.4, 5.5 and 5.0 together, we have EI,...,I r,z+ Q, Q exp c d Q 2 /k s p +σd /k s Q Q /k s =exp Q + /k s Q /k s p +σd /k s =exp c d c d p + /k s 2σ+ /k sdσ++σd /k s, which is 5.3. We now estimate a character sum over F p d. Theorem 0. Let ω F p d be a generator over F p,andletj 0,...,J d be intervals of size at least p ρd+ε,where d ρ d = 2 +2d 7+3 d Denote Q = {x 0 + x ω + + x d ω d : x i J i, for i =0,...,d }. Then q <p δ J 0 J d, 5.2 q Q where δ = δε > 0 is independent of ω.

11 GAFA BURGESS INEQUALITY IN F p 2 0 Proof. First we denote ρ d by ρ. Note that, by 5. 4 ρ Let Q 0 = { y 0 + y ω : y i [,c d p ρ]} 2. Let further k,...,k r Z + satisfy 2ρ 2 2ε < k + + k r < 2ρ 2 ε, 5.4 where ε>0 will be taken sufficiently small and r<rε. Let I = [,p ε/2] and I s = [,p /ks] for s =,...,r. We then translate Q by r I I s Q 0 s= and carry out Burgess argument as outlined in section. The estimate of the left-hand side of 5.2 is q p ε 2 + ks + 2ρ αβ, 5.5 where q Q α Q Q 0 p /k s k EQ, Q 0,I,...,I r /2k Q Q 0 p /k s k exp β k I d d 2 p 2k +2 I p 4k and k Z + to be chosen. q Q c d Q Q 0 p 2 ρ /k s /2k, 5.6 <p ε 4 + d 2k + p ε 2 + d 4k, 5.7 Claim. Q Q 0 p 2 ρ ks < Q 2 Q 0 2 p 2 ks d 2 τ, for some τ> Proof of Claim. We will show dρ + 2ρ+2 ρ k s < 2dρ +2 2ρ+2 k s d This is equivalent to dρ + 2ρ+2ρ k s d 2 > 0. From 5.4, the choice of k,...,k r, and taking ε small enough, it suffices to show that dρ + 2ρ+2ρ 2ρ 2 d 2 > 0, namely, 4ρ 2 +d 3ρ d 2 > 0, 2 which is our assumption 5.. Putting together, we have q p ε ρ ks Q Q 0 p ks k Q 2 Q 0 2 p 2 ks d 2 τ 2k p ε 4 + d 2k + p ε 2 + d 4k

12 02 M.-C. CHANG GAFA = Q p ε 4 + 2k d 2 τ + p τ 2k. Theorem 0 is proved, if we chose k>d/ε. Remark 0.. Returning to Remark., see 4.7, we note that d ρ d < 2d + with ρ 2 = 4,ρ 3 = 8,ρ 4 = 7 8, and ρ 5 = Character Sums of Binary Quadratic Forms Following a similar approach, we show the following: Theorem. Given ε>0, thereisδ>0such that the following holds. Let p be a large prime and fx, y =x 2 + axy + by 2 which is not a perfect square mod p. Let I,J [,p ] be intervals of size I, J >p 4 +ε. 6. Then fx, y <p δ I J 6.2 x I,y J for a nontrivial multiplicative character mod p. This estimate is uniform in f. Result was shown by Burgess assuming I, J >p 3 +ε instead of 6.. Proof. There are two cases. Case. f is irreducible mod p. Then fx, y is a character mod p of x + ωy, with ω = 2 a + 2 a 2 4b, in the quadratic extension Qω and the result then follows from Corollary 6 above. Case 2. fx, y is reducible in F p [x, y]. fx, y =x λ yx λ 2 y λ λ 2 mod p. We will estimate x λ yx λ 2 y. x I,y J The basis strategy is as in the F p 2-case cf. Theorem 5, but replacing F p 2 by F p F p with coordinate-wise multiplication. Let I 0 = [, 0 p/4] and K =[,p κ ], where κ = ε 4. We translate x, y bystx,sty withx,y I 0 and s, t K and estimate K 2 I 0 2 t + x λ y t + x λ 2y. 6.3 sx λ y sx λ 2 y x I,y J x,y I 0 s K t K For z,z 2 F p F p,denote { ωz,z 2 = x, y, x,y,s I J I 0 I 0 K : z = x λ y sx λ y,z 2 = x λ } 2y. sx λ 2 y

13 GAFA BURGESS INEQUALITY IN F p 2 03 Hence 6.3 = K 2 I 0 2 z F p z 2 F p ωz,z 2 t + z t + z 2, 6.4 which we estimate the usual way using Holder s inequality and Weil s theorem. The required property is a bound ωz,z 2 2 < I 2 J 2 K 2 p τ 6.5 z,z 2 for some τ>0 cf We may assume I, J <p.let R = { x λ y, x λ 2 y:x I, y J } T = { } x λ y,x λ 2 y :x,y I 0 S = { s, s :s K }, 6.6 considered as subsets of F p F p. Hence 6.5 is equivalent to ER, T, S <p τ I 2 J 2 K To establish 6.7, we prove the analogues of Lemmas 4. We first estimate ER, T. Lemma 2. Let R and T be defined as in 6.6. Then ER, T < exp c R Writing R as a union of translates of T R = T + ξ α we have Thus it will suffice to show that α R / T t K ER, T R 2 T 2 max ξ F p F p ET + ξ,t. max ET + ζ,t + ξ < exp ζ,ξ F p F p c T Using the same argument as in the proof of Lemma 2,itsufficestoprove6.9for ζ = ξ =0. Lemma 3. Let T be defined as in 6.6. Then ET,T < exp c T There is a stronger statement which is the analogue of Lemma. Lemma 4. Let T be defined as in 6.6. Then max {z,z 2 T T : ρ = z z 2 } < exp c. 6. ρ F p F p

14 04 M.-C. CHANG GAFA Proof. Writing z = x λ y,x λ 2 y, z 2 = x 2 λ y 2,x 2 λ 2 y 2 with x,x 2,y,y 2 I 0, we want to estimate the number of solutions in x,x 2,y,y 2 I 0 of { x λ y x 2 λ y 2 =ρ mod p 6.2 x λ 2 y x 2 λ 2 y 2 =ρ 2 mod p Let F be the set of quadruples x,x 2,y,y 2 I0 4 such that 6.2 holds. If x,x 2,y,y 2, x,x 2,y,y 2 F,thenλ,λ 2 are the distinct roots mod p ofthe polynomial y y y y 2X 2 +x y 2 + y x 2 x y 2 y x 2 X +x x 2 x x 2= By the definition of I 0, the coefficients in 6.3 are integers bounded by 25 p/2. Since all non-vanishing polynomials 6.3 are proportional in F p [X], they are also proportional in Z[X]. Hence they have common roots λ, λ 2 and there are conjugate ρ, ρ 2 Q λ such that { x λ y x 2 λ y 2 = ρ x λ 2 y x 2 λ y 2 = ρ 2 for all x,x 2,y,y 2 F. As in Lemma, we use a divisor estimate in the integers of Q λ toshow that there are at most exp c solutions of 6.4 in x λ y,x 2 λ y 2, x λ 2 y,x 2 λ 2 y 2. Since λ λ 2, these four elements of Q λ determine x,y,x 2,y 2. Therefore, F < exp c. This proves Lemma 4. Returning to 6.7, we proceed as in Lemma 3. Let κ =/k in the definition of K. Thus ER, T, S = p 2 [ = 2 p 2 z 2 z S z R z 2 z 2 z 2 T z 2k 2 2] k z S }{{} R 6.5 k T [ 2 p 2 2 2] R T k }{{} ER,T k < 6.5 k exp c R 2 k 6.6 the last inequality is by Lemma 2, where 6.5 = p z 2k z 2 2 z 2 2 T 2 p 2 < exp z S z R z 2k z c k z S T 2 p 2 z R 2 2 z 2 T t 2 t 2 x λ y 2 x λ 2 y t F p x I y J 2

15 GAFA BURGESS INEQUALITY IN F p 2 05 =exp c T 2 ER,, 6.7 where = {t, t :t F p }. The multiplicative energy ER, in 6.7 equals the number of solutions in x, x,y,y,t,t I 2 J 2 F p 2 of { tx λ y t x λ y modp tx λ 2 y t x λ 2 y modp 6.8 with the restriction that all factors are nonvanishing. Rewriting 6.8 as tx t x λ ty t y λ 2 ty t y modp and since λ λ 2 mod p tx t x mod p ty = t y mod p. Hence xy x y mod p 6.9 and the number of solutions of 6.9 is bounded by EI,J I J 6.20 since I, J <p. Once x, x,y,y is specified, the number of solutions of 6.8 in t, t isatmost p. Hence 6.8 has at most p I J solutions and substitution in 6.7 gives the estimate 6.5 < exp c p R T Substituting of 6.2 in 6.6 gives ER, T S < exp c p k R 2 2 k S k Recalling the definition of S, wehave S = I 0 2 = p /2. Also κ =/k, and K = p /k. Hence 6.22 = exp c p 2 2 k I J k and 6.7 certainly holds. This proves Theorem. =exp c K 2 I J 2 κ 6.23 The author would like to thank the referees for helpful com- Acknowledgement. ments.

16 06 M.-C. CHANG GAFA References [B] D.A. Burgess, On character sums and primitive roots, Proc. London Math. Soc , [B2] D.A. Burgess, Character sums and primitive roots in finite fields, Proc. London Math. Soc , 35. [B3] D.A. Burgess, A note on character sums of binary quadratic forms, JLMS , [C] M.-C. Chang, Factorization in generalized arithmetic progressions and applications to the Erdős Szemerédi sum-product problems, Geom. Funct. Anal , [C2] M.-C. Chang, OnaquestionofDavenportandLewisandnewcharactersum bounds in finite fields, Duke Math. J. 45:3 2008, [DL] H. Davenport, D. Lewis, Character sums and primitive roots in finite fields, Rend. Circ. Matem. Palermo-Serie II-Tomo XII-Anno 963, [FI] J. Friedlander, H. Iwaniec, Estimates for character sums, Proc. Amer. Math. Soc. 9:2 993, [K] A.A. Karacuba, Estimates of character sums, Math. USSR-Izvestija 4: 970, [TV] T. Tao, V. Vu, Additive Combinatorics, Cambridge University Press, Mei-Chu Chang, Mathematics Department, University of California, Riverside, CA 9252, USA mcc@math.ucr.edu Received: September, 2008 Revision: October 5, 2008 Accepted: October 23, 2008 Open Access This article is distributed under the terms of the Creative Commons Attribution Noncommercial License which permits any noncommercial use, distribution, and reproduction in any medium, provided the original authors and source are credited.