Short character sums for composite moduli

Transcription

1 Short character sums for composite moduli Mei-Chu Chang Department of Mathematics University of California, Riverside Abstract We establish new estimates on short character sums for arbitrary composite moduli with small prime factors. Our main result improves on the Graham- Ringrose bound for square-free moduli and also on the result due to Gallagher and Iwaniec when the core q = p q p of the modulus q satisfies log q log q. Some applications to zero free regions of Dirichlet L-functions and the Pólya and Vinogradov inequalities are indicated. Introduction. In this paper we will discuss short character sums for moduli with small prime factors. In particular, we will revisit the arguments of Graham-Ringrose [GR] and Postnikov [P]. Our main result is an estimate valid for general moduli, which improves on the known estimates in certain situations. It is well known that non-trivial estimates on short character sums are important to many number theoretical issues. In particular, they are relevant in establishing density theorems for the corresponding Dirichlet L-functions. In the literature, several bounds on short incomplete character sums to some modulus q may be found, depending on the nature of q. Burgess bound applies for moduli q that are cube free, provided the summation interval I has size N q 4 +ɛ. Assuming q has small prime factors, nontrivial estimates may 2000 Mathematics Subject Classification. L40, M06. Key words. character sums, zero free regions Research partially financed by the NSF Grant DMS

2 be obtained under weaker assumptions on N. There are two classical results in this aspect, based on quite different arguments. Citing from [IK], the Graham-Rignrose theorem see [IK], Corollary 2.5) makes the assumptions that q is square-free and 4 N q log log q + P 9 0.) with P the largest prime factor of q. On the other hand, Iwaniec s generalization of Postnikov s theorem see [IK], Theorem 2.6) comes with a condition of the form N > q ) 00 + e log q)3/4 log log q 0.2) with q = p q p the core of q. The main purpose of this work is to formulate a condition on N as weak as possible, for general modulus q with small prime factors, providing at least subpower savings. That this is possible assuming log N > φlog q) for some function φ satisfying φx) 0 as x ) was probably known to experts, x though no result of this kind seems to appear in the literature. More specifically, we prove the following. Theorem 5. Assume N satisfies and q > N > max p 03 p q log N > log q) c + C log 2 log q ) log q log q log log q, 0.3) where C, c > 0 are some constants, e.g. we may take c = 0 3 and C 0 3 ) and q = p q p. Let χ be a primitive multiplicative character modulo q and I an interval of size N. Then χx) Ne log N. 0.4) Remark. By adjustment of the constants c and C in the statement, the bound 0.4) can be improved to N e log N) ɛ for any fixed ɛ > 0. 2

3 Note that assumption 0.3) of Theorem 5 is implied by the stronger and friendlier assumption log N > C log P + log q ), 0.5) log log q where P = max p q p. Assumption 0.5) is weaker than Graham-Ringrose s condition log q ) log N > C log P +, log log q which moreover assumes q square-free. Many techniques used in the paper are just elaborations of known arguments. Two distinct methods are involved in order to treat small and big prime factors. Small prime powers are dealt with using the standard Postnikov argument combined with Vinogradov s estimate while for big primes we also rely on Weyl s bound as used in Graham-Ringrose s argument. In addition to the basic techniques introduced in the work of Graham-Ringrose and Postnikov, we introduce one further ingredient which is a new) mixed character sum estimate see Theorem ). It allows to merge more efficiently Postnikov s procedure of replacing multiplicative characters to a powerful modulus by additive characters with a polynomial argument and the Weyl differencing scheme which is the basis of the Graham-Ringrose analysis. Note that the replacement of log log q) 2 in 0.) by log log q is achieved by a more economical variant of the Graham-Ringrose argument based on a notion of admissible pair f, q) with f Z[x], q Z). But this is a technical point with no essentially new ideas. The condition 0.3) in Theorem 5 is the best we could do. But we did not try to optimize the power c in the first term nor the saving in 0.4). The main interest of 0.3) compared with 0.2) is that the assumption is weakened to log N > C log q log q log2 ) log q log N > C log log q log q leading to an improvement when q is relatively large. This is the place where the mixed character sum Theorem ) comes into play. 3

4 Next, we turn to some consequences of Theorem 5 that are elaborated in the last section of the paper. Following well-known arguments cf. [ I ]), Theorem 5 implies the following zero-free regions for the corresponding Dirichlet L-functions. Theorem 0. Let χ be a primitive multiplicative character with modulus q, P = max p q p, q = log q p q p, and K =. For T > 0, let log q θ = c min log P, log log q log q ) log 2K, ). log qt ) c Then the Dirichlet L-function Ls, χ) = n χn)n s, s = ρ + it has no zeros in the region ρ > θ, t < T, except for possible Siegel zeros. In the theorem above, one may take c = /0. In certain ranges of q, Theorem 0 improves upon Iwaniec s condition [ I ] { } θ = min c,. log qt ) 2 3 log log qt ) 3 log q Using the zero-free region above and the result from [HB2] on the effect of a possible Siegel zero, we obtain the following. Corollary. Assume q satisfies that log p = olog q) for any p q. If a, q) =, then there is a prime P amod q) such that P < q 2 5 +o). Using Theorem 5, we may also obtain a slight improvement of the following result by Goldmakher [G], corollary to Theorem ) on the Pólya- Vinogradov inequality. Goldmakher) Given χmod q) primitive, with q square-free. Then χn) q log q log log log q n<x log log q + log P log q ) ) What we obtain is the following. 4

5 Theorem 2. Let χ be a primitive multiplicative character with modulus q, and let P be the largest prime divisor of q, q = log q p q p and K =. Let log q M = log q) c log q + log 2K + log P. Then log log q χn) q log q M log log log q. n<x In particular, Theorem 2 gives the bound ) log P q log q log log log q + log log q log q for arbitrary q. 0.7) Clearly, 0.7) is a stronger bound than 0.6). The paper is organized as follows. In Section, we state the mixed character sum theorem with square-free modulus and indicate where the changes are in the proof for prime modulus. In Section 2, we give a version of Postnikov s Theorem, using it to derive a non-trivial character sum bound for the modulus q0 m q, q square-free. Section 3 contains the notion of admissible pair and an improved version of Graham-Ringrose Theorem. Section 4 is the Graham-Ringrose version of mixed character sum estimate. Section 5 contains our main theorem, discussion of our assumption and comparison of it with the assumptions in known results. The proof of the main theorem is in Section 6 and Section 7, its applications in Section 8. Our main interest is the general form of the bounds as a function of the modulus. Constants may often be improved and we did not put emphasis on those. Notations and Conventions.. eθ) = e 2πiθ, e p θ) = e θ p ). 2. ωq) = the number of prime divisors of q. 3. τq) = the number of divisors of q. 4. q = p, the core of q. p q 5. P = Pq) = max p q p. 5

6 6. When there is no ambiguity, p ε = [p ε ] Z. 7. Modulus q is always sufficiently large. 8. ɛ, c, C = various constants, and ɛ is particularly small. 9. All characters are non-principal. 0. For polynomials fx) and gx) with no common factors, the degree is deg fx) + deg gx). of fx) gx). A B and A = OB) are each equivalent to that A cb for some constant c. If the constant c depends on a parameter ρ, we use ρ. Otherwise, c is absolute. Mixed character sums. Theorem. Let P x) R[x] be an arbitrary polynomial of degree d, p a sufficiently large prime, I [, p] an interval of size I > p 4 +κ.) for some κ > 0) and χ a multiplicative character mod p). Then χn)e ip n) < cκ) d 2 I p κ 2 0d 2 +2d+3)..2) n I In the proof of Theorem, the assumption that p is a prime is only used in order to apply Weil s bound on complete exponential sums. For the Weil s estimate below, see Theorem.23 in [IK]) Weil s Theorem. Let p be a prime, f Z[x] a polynomial of degree d, and χ a multiplicative character mod p) of order r >. Suppose fmod p) is not an r-th power. Then we have p χfx)) d p. x= 6

7 The assumption of fmod p) in Weil s Theorem holds if fmod p) has a simple root or a simple pole. For q square-free, one can derive the following theorem. Weil s Theorem. Let q = p p k be square-free, f Z[x] a polynomial of degree d, and χ a multiplicative character mod q). Let q q be such that for any prime p q, fmod p) has a simple root or a simple pole. Then q χfx)) d ωq q ). q x= Proof. Let χ = i χ i, where χ i is a multiplicative character mod p i ). Then q χfx)) x= k p i χ i fx)) d p i p i = d ωq) q. q x= p i q p i q i= Therefore, we have the following. Theorem. Let P x) R[x] be an arbitrary polynomial of degree d, q Z square-free and sufficiently large, I [, q] an interval of size I > q 4 +κ.3) for some κ > 0) and χ a multiplicative character mod q). Then χn)e ip n) < cκ) d 2 I q cκ2 d 2 τq) 4log d)d 2..4) n I Here c is an absolute constant. Remark.. In the proof of Theorem in [C], the assumption that p is a prime is only used to derive display 4) from display 3) by applying Weil s Theorem. For q square-free, the same argument works if Weil s Theorem is replaced by Weil s Theorem. Remark.2. In Theorem if d < log q) /3, then the factor d 2 can be dropped. 7

8 2 Postnikov s Theorem. An immediate application is obtained by combining Theorem with Postnikov s method See [P], [Ga], [ I ], and [IK] 2.6). Postnikov s Theorem. Let χ be a primitive multiplicative character mod q), q = q m 0. Then χ + q 0 u) = e q F q0 u)). Here F x) Q[x] is a polynomial of the form with ) F x) = BD x x2 2 + ± xm m D = k m k,q 0 )= k, m = 2m and B Z, B, q 0 ) =. Note that F q 0 x) Z[x].) 2.) Remark. In [IK] the above theorem was proved for χ + q u) = e q F q u)), where q = p q p is the core of q. That argument works verbatim for our case. Theorem 2. Let q = q m 0 q with q 0, q ) = and q square-free. Assume I [, q] an interval of size I > q 0 q 4 +κ. 2.2) Let χ be a multiplicative character mod q) of the form χ = χ 0 χ with χ 0 mod q0 m ) arbitrary and χ mod q ) primitive. Then χn) I q cκ2 m 2 τq ) clog m)m ) n I 8

9 Proof. For a [, q 0 ], a, q 0 ) = fixed, using Postnikov s Theorem, we write χ 0 a + q 0 x) = χ 0 a)χ 0 + q 0 āx) = χ 0 a)e q m 0 F q0 āx) ), 2.4) where Hence n I χn) a,q 0 )= aā = mod q m 0 ). a+q 0 e q m 0 F q0 āx) ) χ a + q 0 x). 2.5) Writing χ a + q 0 x) = χ q 0 )χ a q 0 + x), q 0 q 0 mod q ), the inner sum in 2.5) is a sum over an interval J = J a of size I q 0, and Theorem applies. 3 Graham-Ringrose Theorem. As a warm up, in this section we will reproduce Graham-Ringrose s argument. With some careful counting of the bad set, we are able to improve their condition on the size of the interval from q / log log q to q C/ log log q. Theorem 3. Let q Z be square-free, χ a primitive multiplicative character mod q), and N < q. Assume. For all p q, p < N log N > C log q log log q. Then N χx) Ne log N/5)3/4. x= We will prove the following stronger and more technically stated theorem. Theorem 3 Assume q = q... q r with q i, q j ) = for i j, and q r squarefree. Factor χ = χ... χ r, 9

10 where χ i mod q i ) is arbitrary for i < r, and primitive for i = r. We further assume i). For all p q r, p > log q r. ii). For all i, q i < N 3. iii). r < c log log q for some c < /4 ɛ. Then N χx) Ne log qr) c / log log q r. x= Remark 3.. Instead of proving Theorem 3, we will prove Theorem 3. To see that Theorem 3 implies Theorem 3, we write where Hence q = p p l p p 2, p,..., p l < log q, and p > p 2 > log q. l p i < e 2 log q < q 0. 3.) i= Let q = k i= p i, where we let k be maximum as to ensure that q < N 3. Therefore, p k+ q > N 3. By ), q > N 3 0 > N 5. We repeat this process on q q to get q 2 such that N 3 > q 2 > N 5. Then, we repeat it on etc. After re-indexing, we have q q q 2 q r > q r > > q 2 > N 5. Hence q > ) N 5 r, which together with 2) gives iii). Theorem 3 now can be applied. Remark 3.2. It follows from ii) and the argument in Remark 3. that in the proof of Theorem3, one may choose q r satisfying N /5 < q r < N /3. Hence log N log q r. 0

11 The following definition will be used frequently throughout the rest of the paper. Definition. Let p be a prime and f Z[x]. We say p is good or f is p-good, if f mod p has a simple root or a simple pole. Otherwise it is called bad or p-bad. For q q r satisfying q > q r, the pair f, q) is called q r -admissible or admissible when there is no ambiguity) if p > log q r for all p q, and p q p is good p > q q τ r, where τ = 0 log log q r. Remark 3.3. Let f, q) be admissible, and let χ be primitive mod q, where q is square-free and a multiple of q. Assume log d < τ = log log q r, where d = deg f. 3.2) 0 Then we have a bound on the complete sum q χfx)) < q ˆq) 3 0 < q q) q x= where ˆq is the product of the good primes p q. Proof of Remark 3.3. For p q q r, our assumptions imply 0 τ r, p /5 > log q r ) /0 > d. 3.3) To prove the remark, we factor χ = χ χ 2, where χ respectively, χ 2 ) is q a character mod q resp. mod ). Weil s estimate gives a bound on the q complete sum of χ in the following estimate. q x= q x= χfx)) χ fx)) q q x= χ 2 fx)) < ˆq q d ωˆq) q q = q ˆq d ωˆq). 3.4)

12 where ˆq is the product of the good primes p q. Using 3.3), we bound the character sum above by q p ˆq d p < q p ˆq p 3/0 = qˆq 3/0. Since f, q) is admissible, we have q χfx)) < qˆq 3 0 < q q) 3 x= Proof of Theorem 3. We will use Weyl differencing. 3 0 q 0 τ r. Take M = [ N ]. Shifting the interval [, N] by yq for any y M, we get N N χx) χx + yq ) 2yq Mq. x= x= Averaging over the shifts gives N N χx) NM x= N M ) Mq χx + yq ) + O. 3.5) N x= y= Let χ = χ 2 χ r. Using the q -periodicity of χ and Cauchy-Schwarz inequality on the double sum in 3.5), we have NM [ N M χx + yq ) NM 2 x= y= For given y, y ), we consider M y, y = N x= f y,y x) = x + q y x + q y χ ) ] x + q y /2. 3.6) x + q y and distinguish among the pairs f y,y, q r ) by whether or not they are q r - admissible. Note that if f y,y, q r ) is not admissible, then the product of bad 2

13 prime factors of q r is at least qr τ and this product must divide y y. We will estimate the size of the set of bad y, y ) and use trivial bound for the inner sum in 3.6) corresponding to such bad y, y ). { y, y ) [, M] 2 : f y,y, q r ) is not admissible } { y, y ) [, M] 2 : Q y y } Q q r Q q τ r Q q r Q q τ r M 2 Q < 2ωqr) M 2 q τ r < M 2 q 7 0 τ r. For the second inequality, we note that M > Q.) Hence 3.6) is bounded by q 7 20 τ r + N N χ f x)) x= 2 3.7), 3.8) where f is the f y,y with the maximal character sum among all admissible pairs. i.e. N χ f x)) = x= Thus, there exists q q r, q > q τ r max f y,y f y,y,q r) admissible N χ f y,y x)). 3.9) x= and for any p q, f is p-good. To bound the second term in 3.8), we will do induction on the number of characters in the factorization of χ and first prove the following. Claim. For s =,, r, denote χ s = χ s+ χ r. Let f s x) be of the form f s x) = j x b j) c j, where b j, c j Z and deg f s 2 s. Denote q 0 = q r. Assume there is q s q r such that q s qr s )τ and f s, q s ) is admissible. Then N N χ sf s x)) q τ 5 2 x= r + N N χ s+f s+ x)) x= 2, 3.0) 3

14 where f s+ is of the same form as f s with deg f s+ 2 s+, and there is q s q r such that q s > qr sτ and f s+, q s ) is admissible. Proof of Claim. As before, the q s+ -periodicity of χ s+ and Cauchy-Schwarz inequality give a bound on the character sum in the left-hand-side of 3.0) by Set [ NM 2 M y, y = N x= χ s+ ) ] fs x + q s+ y) /2. 3.) f s x + q s+ y ) f s+ x) = f sx + q s+ y) f s x + q s+ y ), where y, y ) is chosen among all good pairs as in 3.9), such that the inner character sum in 3.) is the maximum. We want to bound the set of bad y, y ). For p q s, f s x) = x a) ɛ j x b j ) c j for some ɛ {, }, where a b j mod p. Hence ) ɛ x + qs+ y a ) cj x + qs+ y b j f s+ x) =. x + q s+ y a x + q s+ y b j For y y mod p, if a q s+ y is not a simple root or pole, then j a q s+ y = b j q s+ y mod p for some j. Therefore, by the same reasoning as for 3.7), { y, y ) [, M] 2 : f s+, q s ) is not admissible } { y, y ) [, M] 2 : p Q, f s+ is p-bad } Q q s Q>q τ r Q q s Q>q τ r M 2 Q 2 s ) ωq) = M 2 Q q s Q>q τ r 2 s ) ωq) Q. 3.2) 4

15 In the above bound, the factor 2 s comes from the choices of b j.) By assumptions i) and iii), ) 2 s ωq) Q p Q and 3.2) is bounded by and the claim is proved. 2 r p < p Q M 2 2ωqr) qr τ/2 = < q τ 2 r, 3.3) p Q < M 2 q τ/5 r, At the last step of our induction, we are bounding N χ r f r x)), 3.4) x= where f r is of the form as in the claim with deg f r 2 r and there is q r 2 q r such that q r 2 > qr r 2)τ > q r and p q r 2 is good. In particular, f r, q r 2 ) is admissible and Remark 3.3 applies. Note that q q τ) r 2 < q 4 r 2. ) Hence, we have N χ r f r x)) < N q 4 r 2 < Nq 8 x= and we reach the final bound N χx) N x= qr 7τ/20 + qr τ/0 + + q τ/5 r ) /2 r + log log qr 2c log log qr r q 2 = e log qr log log qr log qr) c < e log qr) c / log log q r. q /8 r q τ/5 r ) /2 r ) /2 r + q /8 r r, ) /2 r 3 0 r 2 q 3 0 τ r < ) Mq + O N 3.5) 5

16 The proof of Theorem 3 also gives an argument for the following theorem. log N/5)3/4 Remark 3.4. In Theorem 3 the saving of e is certainly better than the e log N we use in the theorems thereafter optimizing the exponent of log N is not our focus). Theorem 3 Assume q = q... q r with q i, q j ) = for i j, and q r squarefree. Factor χ = χ... χ r, where χ i mod q i ) is arbitrary for i < r, and primitive for i = r. We further assume i). For all p q r, p > log q r. ii). For all i, q i < N /3. iii). r < c log log q for some c < /4 ɛ. Let fx) = x b j ) c j, c {, }, d = deg f = c j. j Suppose that f, q r ) is admissible as defined after the statement of Theorem 3 ). Furthermore, assume Then iv). d = deg f < log q r ) 8. N χfx)) Ne log qr) c / log log q r x= Remark 3.5. To prove Theorem 3, one only needs to modify the proof of Theorem 3 slightly by multiplying M 2 by Q dωq) in 3.7) and replacing 2 s respectively, 2 r ) by 2 s d resp. 2 r d) in 3.2) resp. 3.3)). 4 Graham-Ringrose for mixed character sums. The technique used to prove Theorem may be combined with the method of Graham-Ringrose for Theorem 3 to bound short mixed character sums with highly composite modulus see also [IK] p ). Let q = q... q r with q i, q j ) = for i j, and q r square-free, such that i) and iii) of Theorem 3 hold. 6

17 Let χ = χ... χ r, where χ i mod q i ) is arbitrary for i < r, and primitive for i = r. Let I [, q] be an interval of size N < q, and let fx) = α d x d + +α 0 R[x] be an arbitrary polynomial of degree d. Assuming ii) of Theorem 3 and an appropriate assumption on d, we establish a bound on χx)e ifx). 4.) The case f = 0 corresponds to Theorem 3. The main idea to bound 4.) is as follows. First, we repeat part of the proof of Theorem 3 in order to remove the factor e ifx) at the cost of obtaining a character sum with polynomial argument. Next, we invoke Theorem 3 to estimate these sums. Write q = q Q with Q = q 2... q r, and denote Y = χ 2... χ r. Choose M Z such that M max q i < N, and M < N. 4.2) Using shifted product method as in 3.5), we have M χx)e ifx) = M 0 y<m Next, we write 0 y<m χx + q y)e ifx+q y) = M χx + q y)e ifx+q y) + Oq M), 4.3) 0 y<m M χ x)y x + q y)e ifx+q y) 0 y<m fx + q y) = f 0 x) + f x)y + + f d x)y d. Y x + q y)e ifx+q y). 4.4) Subdivide the unit cube T d+ into cells U α = B ξ α, M d+ ) T d+, ξ α T d+. 7

18 Denote Ω α = {x I : f 0 x),..., f d x) ) U α mod }. Hence, for x Ω α ) fx + q y) = ξ α,0 + ξ α, y + + ξ α,d y d + O M e ifx+qy) = e iξ α,0+ +ξ α,d y d) ) + O. 4.5) M The number of cells is Substituting 4.5) in 4.4) gives M = M where C α y) =. 0 y<m α x Ω α M d+) d+. 4.6) Y x + q y)e ifx+q y) 0 y M N ) C α y)y x + q y) + O, M 4.7) Next, applying Hölder s inequality to the triple sum in 4.7) with k N, we have C α y)y x + q y) α x Ω α N 2k 0 y M α 0 y M C α y)y x + q y) 2k ) 2k. Therefore, up to an error of O N ), 4.7) is bounded by M N [ ) M d+ d+ NM 2k = N d+ M d+) 2k [ NM 2k 0 y,...y 2k <M 0 y,...,y 2k <M 8 x + q y ) x + q y k ) )] 2k Y x + q y k+ ) x + q y 2k ) Y Ry,...,y 2k x) ) ] 2k, 4.8)

19 where R y,...,y 2k x) = x + q y ) x + q y k ) x + q y k+ ) x + q y 2k ). To bound the double sum in 4.8), we apply Theorem 3 with fx) = R y,...,y 2k x) for those tuples y,..., y 2k ) [0, M ] 2k for which R y,...,y 2k, q r ) is admissible. For the other tuples, we use the trivial bound. If R y,...,y 2k, q r ) is not admissible, then there is a divisor Q q r, Q > q τ r, such that for each p Q, the set {π p y ),, π p y 2k )} has at most k elements. Here π p is the natural projection from Z to Z/pZ. We will estimate the contributions of y,..., y 2k ) for which R y,...,y 2k, q r ) is not admissible by distinguishing the tuples y,..., y 2k ) according to the relative size of Q and its prime factors. a). Suppose that there is p Q with p > M. Then the number of p-bad tuples y,..., y 2k ) is bounded by ) 2k M k k k + M ) k < 4k) k M 3 2 k. k p Indeed, one chooses a set I of k indices and specify the corresponding y j ; for the other indices, π p y j ) is taken within {π p y j ) : j I}, and summing over the prime divisors of q r gives ωq r )4k) k M 3 2 k < M 7 4 k, 4.9) provided and 50 k < M 5, 4.0) log q r < M. 4.) b). Suppose Q > M and p M for each p Q. Take Q Q such that M < Q M. The number of tuples y,..., y 2k ) that are p-bad for each p Q is at most M ) 2k ) 2k p k k k Q k p Q M ) 2k < 4kp) k < ck) kωq) M 2k < M 2k < M Q Q k p Q Q k k,

20 provided k < min p q r p ) Summing over all Q as above gives the contribution c). Suppose q τ r < Q < M. M 6 k+ < M 5 8 k. 4.3) The number of tuples y,..., y 2k ) that are p-bad for all p Q is at most M 2k /Q k 3, and summation over these Q gives the contribution 2k ωqr) M 2 Q k 3 < M 2k q r 4 kτ. 4.4) Hence, in summary, the number of y,..., y 2k ) for which R y,...,y 2k, q r ) is not admissible is at most M 2k M k 8 + q 4 kτ r ). From 4.3)-4.4) and 4.7)-4.8) we obtain the estimate χx)e ifx) < N d+ [ M d+) 2k M k 8 ] + q 4 kτ r + e log q 2k r using Theorem 3 for the contribution of good tuples y,..., y 2k ). Here we need to assume 2k < ) log q 8 r, 4.5) which also implies 4.0) and 4.2), under assumption i) and if 4.) holds. Take k = 50d 2, assuming which implies 4.5),) then d < 0 χx)e ifx) < NM d+)2 2k log q r ) 6, 4.6) M 6 + e log qr ) < N M M d+) 2 ) 00d e. log q r 20 2k )

21 Choose M = So 4.) is also satisfied.) We have Thus we proved [ log q ) ] r exp. 2d + ) 2 χx)e ifx) < Ne log qr 200d 2. Theorem 4. Assume q = q... q r with q i, q j ) = for i j, and q r squarefree. Factor χ = χ... χ r, where χ i mod q i ) is arbitrary for i < r, and primitive for i = r. Then We further assume i). For all p q r, p > log q r. ii). For all i, q i < N /3. iii). r < c log log q for some c < /4 ɛ. Let fx) R[x] be an arbitrary polynomial of degree d. Assume n I d < 0 where I is an interval of size N. log q r ) 6. e ifn) χn) < CNe log qr 200d 2, 4.7) Combined with Postnikov as in the proof of Theorem 2), Theorem 4 then implies Theorem 4 Suppose q = q 0... q r with q i, q j ) = for i j, and q r squarefree. Assume q 0 q 0 and q 0 q 0 ) m for some m N, and m < 20 log q r ) 6. Factor χ = χ 0... χ r, where χ i mod q i ) is arbitrary for i < r, and primitive for i = r. 2

22 We further assume i). For all p q r, p > log q r. ii). For all i, q i < N/ q 0 ) /3. iii). r < c log log q for some c < /4 ɛ. Then χn) < CNe log qr 800m 2, 4.8) n I where I is an interval of size N. Note that for Theorem 4 to provide a nontrivial estimate, we should assume at least r log log q r and log m log log q r. 5 The main theorem. Theorem 4 as a consequence of Theorem 4 was stated mainly for expository reason. cf. 7. Proposition 7.) Our goal is to develop this approach further in order to prove the following stronger result. Theorem 5. Assume N satisfies and q > N > max p 03 5.) p q log N > log q) c + C log 2 log q ) log q log q log log q, 5.2) where C, c > 0 are some constants, e.g. we may take c = 0 3 and C 0 3 ) and q = p q p. Let χ be primitive mod q) and I an interval of size N. Then χx) Ne log N. 5.3) 22

23 We will prove Theorem 5 in the next two sections. In this section, we make some further technical specifications which will be important in the proof. Also, we discuss assumption 5.2) and compare it with the assumptions in known results. Some of the remarks at the end of this section Remarks 5.2 and 5.3) will be used in the proof as well. Claim. We may make the following assumptions..) q > N ) 2.) q = Q q r = Q Q r q r, where Q i, Q j ) = Q i, q r ) =, [ ] log Q r = + 0, 5.5) log N where Q is the core of Q,) and q r = q m 0, q 0 square-free 5.6) with e log N) 3 4 < q 0 < N 0, 5.7) m log N) 3c. 5.8) Also, the core of Q s satisfies Moreover, we may assume Q s < N 5 for s =,..., r. 5.9) 3.) There exist q,, q r, q i, q j ) = q i, q r ) =, p q r, p > log q r. 5.0) max i q i < N /2, 5.) such that with q = Q Q r q r q m q m r r q r, 5.2) [ ] log Qs m s = ) log N 23

24 Proof of Claim. To verify Assumption ), we will obtain the bound 5.3) for the case q N 200 by using Theorem 2.6 in [IK]. The latter provides the following bound χx) < C slog s)2 M c s 2 log s 5.4) M<x M with s = log q, assuming that log M q 00 < M < M 2M. This gives a nontrivial bound M c s 2 3 log s provided log M log q) 4 +ɛ. We will use this result by dividing our interval [, N] dyadically. Let M = Ne log N, M i = 2 i M and s i = log N. We divide [, N] into subintervals log q log M i, for i = 2,, m = [, N] = [, M ] m i= M i, M i+ ], and note that M i > N /2 > q 00, and log M i log N. Hence s i log q log N := r for i. We bound the character sum N i= χx) by bounding subsum over each subinterval, using the trivial bound for the interval [, M ] and Theorem 2.6 in [IK] for the intervals M i, M i+ ]. It is straightforward to check that the sum of all bounds is bounded by Ne log N + m M r 2 log r m Ne log N, if log N log q) 4 5 +ɛ, which follows from 5.2). To see Assumption 2), we first note that log N) 3c p < q < e log N) c < q 200, 5.5) ν p>log N) 3c where ν p is the exponent of p in the prime factorization of q. In the display above, the second inequality follows from 5.2) which implies that log q < log N) +2c ), and the last inequality follows from Assumption ) in the claim and provided log q is sufficiently large. 24 c

25 From log N) 3c m= there exists m log N) 3c such that ν p=m ν p=m p ) > q 99 00, 5.6) p > q ) 2 log N) 3c > e 400 log N) 3c > e log N) 3 4. The second inequality in Assumption 5.) that max p q p < N 03 ensures that we may take q 0 p such that q 0 < N 0. ν p=m Therefore, there exists a q r which satisfies 5.6)-5.8). To see 5.0), we note that in the inner product in 5.6), we may impose the condition that p > log q r, because p< log q r p < e 2 The second inequality follows from 5.6)-5.8).) Write q = Q q r, and log qr < e log N) 2 3 < q ) Q = p ν i i, where ν i := ν pi and ν ν 2. Let Q = ν i p i be the core of Q, and factor Q = Q Q r such that Q s < N 5 and r = + 0 [ log Q For each s, define Q s = p Q s p νp and q s = p Q s p νp as follows log N ]. ν p = {[ νp m s ] +, if νp > m s, otherwise. 5.8) Denote m s = 0 log Q s log N. 25

26 It follows that Q s q s ) ms and q s < Q sq 2 ms s < N 2, which are 5.)-5.2). Remark 5.. Assumption 5.2) can be reformulated as 3 log q ) 0 log Nq log N log q N < log N) c. 5.9) Remark 5.2. Using 5.4), 5.9) and the inequality of arithmetic and geometric means, one can show that r m i < ) 75 log q ) i= Remark 5.3. It is easy to check that 5.2) and 5.7) imply Remark 5.4. If log q log N, 5.9) becomes r < 0 3 log log q ) log N > log q) c, which is similar to Theorem 2.6 in [IK]. Remark 5.5. If q = q i.e. q is square-free), condition 5.9) becomes log q log log q < c log N. This is slightly better than Corollary 2.5 in [IK] and essentially optimal in view of the Graham-Ringrose argument. 6 The proof of Theorem 5. The following lemma is the technical part of the inductive step. It is based on the techniques proving Theorem 2 and Theorem 4, which are shifting product and averaging Graham-Ringrose), replacing χ i by an additive character of a polynomialpostnikov), and the mixed character technique to drop the additive character. 26

27 Lemma 6. Assume a). q = q m ˆq, where ˆq = ˆqq r, with q, ˆq, q r mutually coprime. b). χ = χ ˆχ, with χ mod q m ) and ˆχmod ˆq). c). fx) = β x a α ) dα, d = d α, with a α Z distinct and d α Z\{0}. α= d). there exists q q r such that for each p q, p > log q r and f is p-good. e). I an interval of length N, q 2 < N < q, 440 m 2 d < log N) 5. f). M Z, log N + log q < M < N 0, M < q τ. Then χ fx) ) < M / M N N where f x) is of the form ˆχ f x) ) 60m 2, 6.) f x) = k ν= fx + q β t ν ) 2k ν=k+ fx + q t ν ) = x b α ) d α 6.2) α = with b α, d α Z, 2k = 60m 2 and Furthermore, f, q) is admissible. d ) := d α 60 d m ) Proof. Take t [, M]. Clearly, χ fx + tq ) ) ) = χ fx) χ + fx + tq ) ) fx) ˆχ fx + tq ) ). 6.4) fx) Hence, as in the proof of Theorem 2, χ fx + tq ) ) = χ fx) )ˆχ fx + tq ) ) m e q m j= Q j x) q j t j ), 6.5) where Q j x) = d j { fx + t) fx) F j! dt j fx) 27 )} t=0 6.6)

28 with F x) = 2m s= ) s s xs up to a factor). 6.7) We estimate χfx)) by the same technique as used in the proof of Theorem 4 with d = m see 4.3)-4.8)). After averaging and summing over t M, and applying Hölder s inequality, we remove the last factor in 6.5) and obtain [ M m 2 χfx)) N ˆχ R NM 2k t,...,t 2k x) ) ] 2k. 6.8) 0 t,...,t 2k <M Here R t x) := R t,...,t 2k x) = fx + q t ) fx + q t k ) fx + q t k+ ) fx + q t 2k ). Choose t = t,, t 2k ) [, M] 2k such that f x) = R t x) maximizes the inner character sum in the right-hand-side of 6.8) among all admissible R t, q). Let B be the set of t such that R t, q) is not admissible. Hence 6.8) gives 60 χfx)) M 2k B 60 + M N N We want to give an upper bound on B. Claim. B M 2k k 6.. ˆχ f x) ) 60m ) Proof of Claim. First, we observe that the zeros or poles of R t x) are of the form b α := a α t ν q with t ν [, M]. 6.0) Second, we note that while applying Hölder s inequality to obtain 6.8), we take k Z + satisfying 48kd < log N ) 0 and k > ) To bound B, we fix p q. In R t x) = β α = x b α )d α, we may assume d = and a a α mod p) for any α >. Recalling 6.0), assume that 28

29 none of the a t ν q, ν 2k, is simple mod p). This means that for each ν there is a pair αν), σν)) in {,..., β} {,..., 2k} such that αν), σν) ν and a t ν q a αν) t σν) q mod p). 6.2) The important point is that σν) ν for all ν, by assumption on a. One may therefore obtain a subset S {,..., 2k} with S = k such that there exists S S with S = k 2 and S = {ν S : σν) / S }. 6.3) The existence of S and S satisfying this property is justified in Fact 6. following the proof of this lemma.) Specifying the values of t ν for those ν {,, 2k}\S, equations 6.2) will determine the remaining values, after specification of αν) and σν). An easy count shows that {πp t) : R t ) ) 2k k 3 k k 2 2 k is p-bad } ) k ) k 2 β 2 ) 3k M 2 ) < 48kd) k 3k 2 M 2, 2 6.4) where M = minm, p). The first factor counts the number of sets S, the second the number of sets S, and the third and the forth the numbers of maps σ S and α S. Applying assumptions e)-f) to 6.4), we obtain {π p t) : R t is p-bad } < M ) 8 5 k. 6.5) If R t, q) is not admissible, there is some Q q, Q > q τ r > q τ such that for each p Q, R t is p-bad. As in the proof of Theorem 4, we distinguish several cases. a). There is p Q with p > M. Hence, {t [, M] 2k : R t is p-bad } < M 8 5 k and summing over p gives the contribution M 8 5 k log q. b). M < max p Q p < M. 29

30 Then {t [, M] 2k : R t is p-bad } M p + ) 2k {πp t) : R t is p-bad } M ) 2kp p + 8 p ) 2 5 k < M 2k 5 k < 4M) 9 5 k. 32 Summing over p gives the contribution 4M) 9 5 k log q. c). max p Q p M and Q > M. Take Q Q such that M < Q < M. Then {t [, M] 2k : R t is p-bad for each p Q } M Q + ) 2k {πq t) : R t is p-bad for each p Q } M ) 2k + p 8 Q p Q 5 k < M 2k Q 32 Summing over Q gives the contribution 4M) 9 5 k+. ) 2 5 k < 4M) 9 5 k. Summing up cases a)-c) and recalling assumption f), we conclude that B = {t [, M] 2k : R t, q) is not admissible } < M 2k M k τk ) 6.6) 6 + q 5 M 2k k 6. Putting 6.9) and 6.6) together, we obtain 6.) and the lemma is proved. Fact 6.. Let K = {,, 2k} and σ : K K be a function such that σν) ν for all ν K. Then there exist subsets S S K with S = k, S = k and σν) S 2 for any ν S. Proof. Since the subset of elements of K with more than one pre-image of σ has size k, there exists a subset S K with S = k such that every ν S has at most one pre-image. To construct S S, we choose ν i S inductively, such that ν i {ν,..., ν i, σν ),..., σν i )} σ {ν,..., ν i }) and σν i ) S. 30

31 Proof of Theorem 5. Following the claim in 5, we will prove the theorem for χ = χ χ r, where χ i mod q m i i ) is arbitrary for i < r, and primitive for i = r, and q = q m q m r r q r satisfies 3) of the claim. We will apply lemma 6 repeatedly. First, we choose M < M 2 < < M r a sequence of values of M, which will satisfy condition 6.9). Then we will iterate 6.), losing a factor χ i in χ and adding an error term M /2 i in the bound each time. In order to satisfy Assumption e), we assume r r m 2 i ) < log N ) 5, 6.7) i= which follows from 5.20) and 5.2). Let and for s =,, r, take M s, such that M = e log qr)9/0, 6.8) M 60 M 60 2 m 2 2 M 60 s m 2 m2 s 2 s M 5 =M 5 60 M s m 2 m2 s s =M ) One checks recursively that Indeed, from 6.9), M s M 5s 5 s m 2 m2 s. 6.20) M 5 60 s 2 m 2 m2 s 2 s = M 60 s m 2 m s 2 s M s m 2 m2 s s. Therefore, by induction M s = M 75m2 s s M 5s 2 5 s m 2 m2 s 2 75m2 s ) = M 5s 5 s m 2 m2 s. In order to satisfy the last condition in Assumption f) of Lemma 6, we assume r log m i < 40 log log q r, 6.2) i= 3

32 Clearly, this follows from 5.20).) and note that M 5 5)r m 2 m2 r 2 < q τ log log qr r = qr. 6.22) Since by 5.2), r < 0 3 log log q r.) By 6.) and iteration, N x= χx) is bounded by N 0 M 5 + M 60 M 5 60m M M 2 m 2 2 M m 2 m M M 2 m M r 2 m 2 m r 3 r 2 M r 2 m 2 m2 r 2 r +M M 2 m 2 60 r m 2 m2 r 2 60 r S r m 2 m2 r, 2 M 6.23) where S is of the form S = N χ r fx)), 6.24) N x= with χ r primitive modulo q r, and fx) = β x a α ) dα, a α, d α Z, α= and f, q) admissible for some q q r, q > q r. d = d α < 60 r m 2... m 2 r, 6.25) Condition 6.9) ensures that each of the r fist terms in 6.23) is bounded by M. Applying 6.20) to 6.23) gives N χx) r < N M + M 5 x= 4 )r S 60 r m 2 m2 r. 6.26) We will continue the proof of the theorem in the next section by distinguishing two cases. 32

33 7 The two cases. To finish the proof of Theorem 5, we need to bound S in 6.26). We will use Claim in 5. Recall 5.6) that We will do induction on m. Case. m =. q r = q m 0, q 0 square-free. Since f, q) is admissible and q is square-free, we may apply Remark 3.3 to bound S. Therefore, and by 6.26) N χx) r < N M + M 5 x= S < q 3 0 q 3 0 τ r < q 7 r, 7.) 4 )r 7 60 q r m 2 m2 r The last inequality is by 5.20) in Remark 5.2 and 6.22). r < r M. 7.2) Now we use 6.8) and 5.2) to bound 7.2) and 5.6)-5.8) to obtain 5.3). We state the above case as a proposition for its own interest. Proposition 7. Assume q = q m... q m r r q r with q i, q j ) = for i j, q r square-free and r ) 75 m i < log q r. 7.3) i= Factor χ = χ... χ r, where χ i mod q m i i ) is arbitrary for i < r, and primitive for i = r. We further assume i). For all p q r, p > log q r. ii). For all i, q 2 i < N < q. iii). r < 0 3 log log q r. 33

34 Then χx) < Ne log qr)4/5, 7.4) where I is an interval of size N. Case 2. m >. In this situation, we follow the analysis in the proof of Lemma 6. Particularly, see 6.4)-6.7).) To bound S in 6.23), we will use Postnikov s theorem and Vinogradov s lemma [Ga], Lemma 4) rather than Weil s estimate Remark 3.3) as we did in Case. Recall S = N with χ r primitive modulo q r, and fx) = N χ r fn)), n= β x a α ) dα with d 60 r m 2... m 2 r, 7.5) α= where fx) satisfies the property that for all p q 0, fx) is p-good. Write n [, N] as n = x + tq 0, with x q 0 and t N q 0. Then as in 6.4) and 6.6), N S = q 0 x= q 0 x= N/q 0 t= N/q 0 t= ) m χ r fx) eq m 0 m e q m 0 j= j= Q j x)q j 0t j ) Q j x)q j 0t j ). 7.6) We want to find some information on the coefficients Q j in 6.5). We may assume in 7.5) that a = 0 is a simple zero or pole of f; replacing f by f which we may by replacement of χ by χ), hence fx) = xgx) = x x a α ) dα mod p 7.7) a α 0 with g0) defined and non-vanishing mod p). 34

35 From 6.6), 6.7), and 7.7), we have j!q j x) = s ) s sxgx)) s d j dt j [ x + t)gx + t) xgx) ) s ] t=0. 7.8) Let Gt) = x + t)gx + t) xgx). Since Gt) divides dj Gt) s for s > j, and dt j G0) = 0, in 7.8) only the terms s j contribute and Q j has a pole at 0 of order j. Write C Q j x) = x + A jx) 7.9) j B j x) with A j x), B j x) Z[x] and B j x) = x k ˆBj x), k < j. Here ˆB j 0) 0mod p), 7.0) since B j x) is a product of monomials of the form x a α and a α 0mod p) for α. Thus C Q j x) = P jx) x j ˆB 7.) j x) where P j x) Z[x] is of degree at most dj, P j 0) 0 mod p). It follows that { x p : Q j x) 0 mod p)} dj, 7.2) and { x q0 : Q j x) 0 mod q 0 )} dj) ω q 0 ) q 0 q 0, 7.3) whenever q 0 q 0. Taking j = m and fixing x, we will apply Vinogradov s lemma [Ga], Lemma 4) to bound the following inner double sum in 7.6). N/q 0 t= m e q m 0 j= Q j x)q j 0t j ). 7.4) Lemma Vinogradov). Let ft) = a t + + a k t k R[t], k 2 and P Z + large. Assume a k rational, a k = a, a, b) = such that b 2 < P b P k 7.5) 35

36 Then efn)) < C klog k)2 P n I for any interval I of size P c, C are constants). Since the dominating saving in 7.6) is P of the leading coefficient a m = Q m x) q0 m c k 2 log k 7.6) c k 2 log k, we want the denominator q m 0 in 7.4) big. Let q 0 = Q m x), q 0 ). Write Qm x) q 0 ā Zmod q 0 ). Therefore a m = Q m x) q 0 = ā q 0, with q 0 = q 0 q 0 and ā, q 0 ) =. To sum x [, q 0 ] in 7.6), we distinguish the cases according to Q m x), q 0 ). Case i). q 0 = q 0 x) = Q m x), q 0 ) q 0. Hence q 0 > q 0. Divide the interval [, N/q 0 ] into subintervals of length q 0 each. Applying 7.6) with P = q 0 to the subsum over each subinterval and summing up the subsums give N/q 0 t= m e q m 0 j= ) Q j x)q0t j j N < mlog m)2 C q 0 The last inequality follows from 5.6)-5.8). q c 2m 2 log m) ) Case ii). q 0 = q 0 x) = Q m x), q 0 ) > q 0. We will use the trivial bound N/q 0 on 7.4). It remains to estimate the number of those x q 0 such that Q m x) 0 in Z/ q 0 Z for some q 0 > q 0. This number is by 7.3) at most ω q dm) 0 ) q 0 < 2 ωq0) dm) ωq 0) q 0 < 2dm) 2 log q0 log log N q 0, 7.8) q 0 q 0 q 0 q 0 > q 0 since all prime divisors of q 0 are at least log N) 2. Note that the degree d of fx) is bounded by 6.25). Applying 5.8) and 5.20), we have dm < 60 r log q r ) 2 75 log N ) 3c < log N ) 5c. 7.9) In particular, 7.9) will ensure that 7.8) is bounded by q 3/

37 Applying Case i) and Case ii) to 7.6), we have q 0 x= = N/q 0 t= q 0 x) q 0 m e q m 0 N/q 0 t= N mlog m)2 q 0 C q 0 j= Q j x)q j 0t j ) m e q m 0 j= q c 2m 2 log m) < NC m log m)2 q c 2m 2 log m) 0. Therefore Q j x)q j 0t j ) q N q 0 q 0 x)> q 0 c N/q 0 t= m e q m 0 j= Q j x)q j 0t j ) 7.20) S < C m log m)2 q 2m 2 log m) ) Now we want to apply??) to 6.26). In 6.26), using 5.2), we have a bound log q r ) ɛ on the exponent of M in the second term in 6.26). On the other hand, after applying??) in the second factor of the second term in 6.26), and using 5.20), 5.8) and 5.7), we bound the exponent of q 0 by log q r ) ɛ as well. Hence we have N N χx) < r M x= qr)9/0+ɛ + elog log qr) ɛ q0 7.22) log qr) ɛ By 5.6)-5.8), the factor q0 in 7.2) is bounded by exp log q r ) ɛ ). Hence 7.2) is bounded by e log qr)9/0 after applying 5.2) and 6.8). This proves the theorem. 8 Applications. Repeating the argument in deducing Theorem 4 from Theorem 3 and Theorem 3, we obtain the following mixed character sum estimate from the proof of Theorem 5. There is the following mixed character sum version of Theorem 5. 37

38 Theorem 8. Under the assumptions of Theorem 5, χx)e ifx) < Ne log N 8.) assuming fx) R[x] of degree at most log N) c for some c > 0. A more precise statement is again possible, but the above one is all we need for what follows. To prove Theorem 8, simply go back to the opening argument in Theorem 4 which removes the factor e ifx) at the cost of replacing χx) by χrx)) with Rx) a certain rational function of x. Note that this step is already part of the proof of Lemma 6. At this point, proceed further with 6 and 7 as in proving Theorem 5. Corollary 9. Assume N satisfies and q satisfies q > N > max p 03 p q log N > log qt ) c + C log 2 log q ) log q log q log log q. 8.2) Then for χ primitive, we have χn)n it < Ne log N. 8.3) n I From Corollary 9, one derives bounds on the Dirichlet L-function Ls, χ) and zero-free regions the usual way. See for instance Lemmas 8- in [ I ]. This leads to the following theorem. Theorem 0. Let χ be a primitive multiplicative character with modulus q, P = max p q p, q = log q p q p, and K =. For T > 0, let log q θ = c min log P, log log q log q ) log 2K, ). log qt ) c Then the Dirichlet L-function Ls, χ) = n χn)n s, s = ρ + it has no zeros in the region ρ > θ, t < T, except for possible Siegel zeros. 38

39 It follows in particular that θ log qt if log P log q 0. From Theorem 0, we have the following. Corollary. Assume q satisfies that log p = olog q) for any p q. If a, q) =, then there is a prime P amod q) such that P < q 2 5 +o). To deduce Corollary, we follow the exposition of Linnik s theorem in [IK] see p.440). Define ψx, q, a) = φq) χmod q χa) n x Λn)χn), where Λ is the von Mangoldt function. Assuming x > q 2 5 +ɛ, we have ψx, q, a) = x { σ xβ φq) β ) ) )} x cɛ x cη log q + O + O + O ɛ ɛ q 8.4) with ) log log q η > c min,, P = max p. 8.5) log P log q p q Here, on the right hand side of 8.4), the second term accounts for a possible Siegel zero s = β in Theorem 0 in which case σ =, otherwise σ = 0), while the third term is from Huxley s density estimate [H], and the fourth term from the zero-free region given by 8.5). Certainly, x cη 0, if log P log q 0 with ɛ fixed). Also, with γ = β ) log q. xβ β β q β ) = γ log q e γ We distinguish two cases. If γ is sufficiently small, then Corollary 2 in [HB2] applies. In this case there is a prime P a mod q) with P < q 2+δ < q 2 5. Otherwise, the right hand side of 8.4) is greater than zero. Hence the conclusion in Corollary holds in either case. Next, following Goldmakher s work [G], we will derive from Theorem 5 the following theorem. 39

40 Theorem 2. Let χ be a primitive multiplicative character with modulus q, P = max p q p, q = log q p q p and K =. log q Let M = log q) c log q + log 2K + log P. Then log log q χn) q log q M log log log q. n<x We will sketch the argument and refer to [G] for more details. Denoting S χ x) = n<x χn) with χ primitive modulo q. Proposition 2.2 in [G] states that S χ x) q log q) 2 L + log q, χ ξ) q log q) 7 8.6) for some primitive character ξ mod m) of conductor less than log q) /3. This result uses crucially the work of Granville and Soundararajan [GS]. Let ψ mod Q) be the primitive character which induces χξ. Then, by [G], Lemma 5. and Lemma 5.2, and q m Ls, χ ξ) Ls, ψ) Q qm 8.7) + log log m. 8.8) Taking s = +, we get by partial summation that log q ) Ls, ψ) = s ψn) dt. 8.9) t s+ n t For t > Q, estimate n t ψn) trivially by Q, which contributes in 8.9) for O). Next, Q t s+ ψn) dt log T + n t 40 Q T t 2 ψn) dt 8.0) n t

41 and we apply Theorem 5 to bound n t ψn) in the second term of the right hand side of 8.0). Note that by 8.7) the expression 0.3) in Theorem 5 is essentially preserved, if Q is replaced by q. Thus T = N has to be chosen to satisfy 0.3) and Theorem 2 follows from 8.6), 8.8), and 8.0). Acknowledgement. The author would like to thank the referee for very careful reading of the paper, which greatly improved its presentation. The author would also like to thank I. Shparlinski for helpful comments and the mathematics department of University of California at Berkeley for hospitality. Part of the work was finished when the author was in residence at the Mathematical Science Research Institute and supported by the NSF Grant References [C] [Ga] [Go] [GR] [GS] [HB] M.-C. Chang, An Estimate of Incomplete Mixed Character Sums, in An Irregular Mind. Editors: Brny, Imre, Solymosi, Jozsef. Bolyai Society Mathematical Studies. Budapest. Vol. 2, 200), P.X. Gallagher, Primes in progressions to prime-power modulus, Invent. Math ), L. Goldmakher, Character sums to smooth moduli are small, Can J. Math ), 099. S.W. Graham, C.J. Ringrose, Lower bounds for least quadratic nonresidues, In: Analytic number theory Allerton Park, IL, 989), Progr. Math., 85, Birkhauser, Boston, MA, 990), A. Granville, K. Soundararajan, Large Character Sums: Pretentious Characters and the P ólya-vinogradov Theorem, J. Amer. Math. Soc ), D. R. Heath-Brown, Zero-free regions for Dirichlet L-functions, and the least prime in an arithmetic progression, Proc. London Math. Soc. 3) 64, no ),

42 [HB2] [H] [IK] D.R. Heath-Brown, Siegel zeros and the least prime in an arithmetic progression, Quart. J. Math. Oxford Ser. 2), 4 990), M.N. Huxley, Large values of Dirichlet polynomials, III, Acta Arith., ), H. Iwaniec, E. Kowalski, Analytic number theory, Amer. Math. Soc., Providence RI, 2004). [ I ] H. Iwaniec, On zeros of Dirichlets L series, Invent. Math ), [P] A.G. Postnikov, On Dirichlet L-series with the character modulus equal to the power of a prime number, J. Indian Math. Soc. N.S.) ),