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 and Postnikov. 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 free regions for the corresponding Dirichlet L-functions. More specifically, we prove the following. Let χ be a primitive multiplicative character to the modulus q, and let p be the largest prime divisor of q, q = log q p q p and K =. Let I be an log q interval of size I = N. We will denote various constants by C Mathematics Subject Classification. L40, M06. Key words. character sums, zero free regions Research partially financed by the National Science Foundation.

2 Theorem 5. Assume q > N > p C and Then log N > log q) C log q log 2K ) log log q χx) < Ne log N. 2) Note that assumption ) of Theorem 5 is implied by the stronger and friendlier assumption log N > C log p + log q ). 3) log log q Assumption 3) is weaker than Graham-Ringrose s condition log N > C log p + log q ). log log q The Graham-Ringrose estimate was established only for square free moduli.) Also, condition ) is weaker than Postnikov-Gallagher-Iwaniec s assumption log N log q + log q) 2 3 +ɛ in certain cases, namely when log q becomes comparable with log q. Many techniques used in the paper are just elaborations of known arguments. What differs from methods previously used on similar problems is the use of a new) mixed character sum estimate in conjunction with Postnikov s argument for powerful moduli, whose usual treatment exploits Vinogradov s exponential sum estimate at the end. This allows us to combine efficiently the methods introduced by Graham-Ringrose and Postnikov and obtain nontrivial bounds under less restrictive conditions for a large class of moduli. Next, we turn to some routine) consequences of Theorem 5. Theorem 8. Let M = log q) 9 log q 0 + log 2K + log p. Then log log q χn) q log q M. n<x 2

3 If q is square free, then the bound in Theorem 8 is log q q + log q ) log p. log log q It gives an improvement on Goldmakher s result. See [G], Theorem.) Theorem 5 implies the following zero-free regions for the corresponding Dirichlet L-functions. Theorem. Let θ = c min log p, log log q log q ) log 2K, log qt ) 9/0 ). Then the Dirichlet L-function Ls, χ) = n χn)n s, s = ρ + it has no zeros in ρ > θ, t < T except for possible Siegel zeros. In certain ranges, this 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. cf. the discussion in [HB]) Corollary 2. Assume q satisfies that log p = olog q) for p q. If a, q) =, then there is a prime P amod q) such that P < q 2 5 +O). Notation and Convention.. 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. When there is no ambiguity, p ε = [p ε ] Z. 6. Modulus p or q) is always sufficiently large. 7. For polynomials fx) and gx), the degree of fx) gx) 8. c, C = various constants. is deg fx)+deg gx). 3

4 Mixed character sums. The following theorem is from [C] for prime modulus. 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 nontrivial multiplicative character mod p). Then χn)e ip n) < I p c κ 2 d 2..2) n I In the proof of Theorem, the assumption p being prime is only used in order to apply Weil s bound on complete exponential sums. For Weil s bound, see Theorem.23 in [IK]) Let f Z[x] be a polynomial of degree d and let χ be a multiplicative character mod q) and of order r >. Weil s Theorem. power. Then we have Let q = p be prime. Suppose fmod p) is not a r-th p χfx)) d p. x= For q = p p k square free, since q x= χfx)) k i= p i x= χ ifx)), where χ i is a multiplicative character mod p i ), we have the following version of Weil s estimate. Weil s Theorem. Let q be square free and 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= Therefore, we have the following. 4

5 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 nontrivial multiplicative character mod q). Then χn)e ip n) < I q cκ 2 d 2 τq) 4log d)d 2..4) n I 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) Z[X] is a polynomial of the form with and B Z, B, q 0 ) =. ) F x) = BD x x2 2 + ± xm m D = k m k,q 0 )= k, m = 2m 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) 5

6 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 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. 6

7 Then N χx) < Ne log N. 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 square free. 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. ii). For all i, q i < N 3. iii). r < c log log q. Then N χx) < Ne log q r. x= Remark 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 log q < q 0. 3.) i= Let q r = k i= p i such that k is the maximum subject to the condition that q r < N 3 Therefore, p q r > N 3. By ), q r > N 3 0 > N 5. We repeat this process on q q r 7

8 to get q r such that N 3 > q r > N 5. Then, we repeat it on After re-indexing, we have q q rq r etc. q r, q r,..., q 2 > N 5. Hence q > N 5 ) r, which together with 2) give iii). Under the assumption of Theorem 3, we give the following Definition. For q q r satisfying q > q r, and f Z[x] of degree d, the pair f, q) is called admissible if and *) means p q f mod p satisfies *) p > q q τ r, where τ = 0 log log q r, *) The polynomial) has a simple root or a simple pole. Remark 3.2. If f, q) is admissible, χ primitive mod q and then log d < τ = log log q r 0 q x= χfx)) < q) q 0 τ r. 3.2) Proof of Remark 3.2. We say p is good, if f mod p satisfies *). For p q q r, assumption i) implies that p > log q. In particular, 3.2) implies p /5 > log q) /0 > d. 3.3) Weil s estimate gives q x= χfx)) < q q d ωq ) 3.4) 8

9 where q is the product of the good primes p. Using 3.3), we bound the character sum above by q p q d p < q p q p 3/0 = qq 3/0. Since f, q) is admissible, we have q χfx)) < 3 qq 0 < q) 7 x= 3 0 q 0 τ r. Proof of Theorem 3. Take M = [ N ]. Shifting the interval [, N] by yq i for any y M, we get n χx) x= n χx + yq ) 2yq Mq. x= Averaging over the shifts gives N Let N χx) NM x= N M ) Mq χx + yq ) + O. 3.5) N x= y= χ = χ 2 χ r. Using the q -periodicity of χ and Cauchy-Schwarz 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 admissible. Note that if f y,y, q r ) is not admissible, then the product of bad prime 9

10 factors of q r is at least qr τ. We will estimate the size of the set of bad y, y ) and use trivial bound for the inner sum in 3.6). { 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 log log M r = M 2 q 7 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 0 τ r. 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)) N = max f y,y χ f y,y x)). 3.9) x= Thus, there exists q q r, q > qr τ *). f y,y,q r) admissible x= and for any p q, f mod p has property We will use induction to bound the second term in 3.8). After s steps, we reduce the problem to bounding the character sum N χ N sf s x)), 3.0) x= where χ s = χ s+ χ r, f s Z[x] with deg f s = 2 s. Since f s, q s ) is admissible, there is q s q r such that q s > qr sτ and p q s is good. As before, the q s+ -periodicity of χ s+ and Cauchy-Schwarz give a bound on 3.0) by [ M N ) ] χ fs x + q s+ y) /2 NM 2 s+. 3.) f s x + q s+ y ) y, y = x= 0

11 Set f s+ x) = f sx + q s+ y) f s x + q s+ y ), where y, y ) is chosen 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) ± j x b j ) c j, where a b j. mod p Hence ) ± x + qs+ y a ) cj x + qs+ y b j f s+ x) =. x + q s+ y a x + q s+ y b j j For y y mod p, if a q s+ y is not a simple root or pole, then a q s+ y = b j q s+ y, mod p for some j. Therefore, { y, y ) [, M] 2 : f y,y, q s ) is not admissible } { y, y ) [, M] 2 : p Q, f y,y mod p satisfies *) } 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) By assumptions i) and iii), 2 s ) ωq) Q and 3.2) is bounded by p Q 2 r p < p Q p = Q < q τ 2 r, 3.3) M 2 2ωqr) q τ/2 r < M 2 q τ/5 r.

12 At the last step, we are bounding N χ r f r x)), 3.4) x= where f r Z[x] with deg f r = 2 r and there is q r q r such that q r > qr r )τ > q r and p q r is good. In particular, f r, q r ) is admissible and Remark 3.2 applies. Recall that q r < N /3.) Hence, we have N χ r f r x)) < N q 4 r < Nq 8 x= and we reach the final bound N N χx) x= q τ/5 r q 2 ) /2 r + log log qr 2c log log qr r = e log qr log log qrlog qr) c < e log q r q /8 r r, ) /2 r The proof of Theorem 3 also gives an argument for the following theorem. Theorem 3 Assume q = q... q r with q i, q j ) = for i j, and q r square free. 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. ii). For all i, q i < N /3. iii). r < c log log q. Let fx) = x b j ) c j, j c = ±, d = deg f = c j. Suppose that f, q r ) is admissible as defined after the statement of Theorem 3 ). Furthermore, assume 2

13 Then iv). d = deg f < log q r ) 8. N χfx)) < Ne log q r. x= Remark 3.3. To modify the proof of Theorem 3, one only needs to multiply M 2 by Q dωq) in 3.7) and replaced 2 s respectively, 2 r ) by 2 s d resp. 2 r d) in 3.2) resp. 3.3)). Assumptions i) and iii) imply 2 r < p /4, while assumptions i) and iv) imply d < p /4. 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. 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 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. 3

14 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 χx)e ifx) = χx + q y)e ifx+qy) + Oq M), 4.3) M M 0 y<m Next, we write 0 y<m χx + q y)e ifx+q y) = M 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). Fix θ = θq) > 0 and subdivide T d+ in cells U α = B ξ α, θ M d ) T d+. Denote Hence, for x Ω α The number of cells is Ω α = {x I : f 0 x),..., f d x) ) U α mod }. fx + q y) = ξ α,0 + ξ α, y + + ξ α,d y d + Oθ) 4.4) e ifx+q y) = e iξ α,0+ +ξ α,d y d) + Oθ) 4.5) M d ) d+. 4.6) θ Substituting 4.5) in 4.4) gives M = M 0 y<m α x Ω α Y x + q y)e ifx+q y) 0 y M C α y)y x + q y) + OθN), 4 4.7)

15 where C α y) =. Next, applying Hölder to the triple sum in 4.7) with k Z +, we have α x Ω α 0 y M C α y)y x+q y) N 2k α 0 y M Therefore, up to an error of OθN), 4.7) is bounded by N [ ) M d d+ NM 2k θ 0 y,...y 2k <M C α y)y x+q y) 2k ) 2k. x + q y ) x + q y k ) )] 2k Y x + q y k+ ) x + q y 2k ) = N M d θ ) d+ 2k [ NM 2k 0 y,...,y 2k <M Y Ry,...,y 2k x) ) ] 2k, 4.8) 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 tuplets y,..., y 2k ) {0,..., M } 2k for which R y,...,y 2k, q r ) is admissible. For the other tuplets, we use the trivial bound. If R y,...,y 2k, q r ) is not admissible, then there is a divisor Q q r, Q > qr τ, 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 distinguish the tuplets y,..., y 2k ) in the following contributions. a). Suppose that there is p Q with p > M. Then the number of p-bad tuplets y,..., y 2k ) is bounded by ) 2k M k k k + M ) k < 4k) k M 3 2 k, k p 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 k < M 5, 4.0) 5

16 and 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 tuplets 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 3 6 k, 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 tuplets 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 ) d+ M χx)e ifx) d 2k ] < N [M k 8 + q 4 kτ r + e log q 2k r θ 6

17 using Theorem 3 for the contribution of good tuplets 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 θ = M, k = 50d2, assuming which implies 4.5),) then Choose d < 0 χx)e ifx) < NM d+)2 2k M = So 4.) is also satisfied.) We have Thus we proved log q r ) 6, 4.6) M 6 + e < N M 20 M d+) 2 + e log q r [ log q ) ] r exp. 2d + ) 2 χx)e ifx) < Ne log qr 200d 2. log qr 2k ) ) 00d 2 ). Theorem 4. Assume q = q... q r with q i, q j ) = for i j, and q r square free. 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. ii). For all i, q i < N /3. iii). r < c log log q. Let fx) R[x] be an arbitrary polynomial of degree d. Assume d < 0 log q r ) 6. 7

18 Then e ifn) χn) < CNe log qr 200d 2, 4.7) n I where I is an interval of size N. 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 square free. Assume q 0 q 0 and q 0 q 0 ) m for some m Z +, 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. We further assume i). For all p q r, p > log q. ii). For all i, q i < N/ q 0 ) /3. iii). r < c log log q. 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. Proposition 7.) Our goal is to develop this approach further in order to prove the following stronger result. 8

19 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, 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) We will prove Theorem 5 in the next section. In this section, we will set up the proof and discuss the implication of assumption 5.2). Claim. We may assume the following.) 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 and with q r = q m 0, q 0 square free 5.6) e log N) 3 4 < q 0 < N 0, 5.7) m log N) 3κ, 5.8) where κ > 0 a sufficiently small constant e.g. 0 3 ). Also, the core of Q s satisfies Q s < N 5 for s =,..., r. 5.9) 3.) There exist q,, q r, q i, q j ) = q i, q r ) =, max i q i < N /2, 5.0) 9

20 such that with q = Q Q r q r q m q m r r q r, 5.) [ ] log Qs m s = ) log N Proof of Claim. The validity of Assumption ) follows from Theorem 2.6 in [IK], which gives a bound χx) < C r log log r N c r 2 log r 5.3) with r = log q logn q 00 < N. This gives a nontrivial result provided log N log q) 3 4 +ɛ. To see Assumption 2), we first note that log N) 3κ p < q < e log N) κ < q 00, 5.4) ν p>log N) 3κ where ν p is the exponent of p in the prime factorization of q. From log N) 3κ m= there exists m log N) 3κ such that ν p=m ν p=m p ) > q 99 00, p > q ) 2 log N) 3κ > e 200 log N) 3κ > e log N) 3 4. Therefore, there exists q r satisfies 5.6)-5.8). Write q = Q q r, and Q = p ν i i, where υ i := ν pi and ν ν 2. Let Q = ν i p i and factor Q = Q Q r such that Q s < N 5 and r = + 0 [ log Q 20 log N ].

21 For each s, define Q s = p Q s p νp and q s = p Q s p νp as follows ν p = {[ νp m s ] +, if νp > m s, otherwise 5.5) Denote m s = 0 log Q s log N. It follows that Q s q s ) ms and q s < Q sq 2 ms s < N 2, which are 5.0)-5.). Remark 5.. Assumption 5.2) can be reformulated as 3 log q ) 0 log Nq log N log q N < log N) c. 5.6) Remark 5.2. Using 5.4), 5.6) 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, 5.6) becomes r < 0 3 log log q ) log N > log q) c, which is similar to Theorem 2 in [IK]. Remark 5.5. If q = q i.e. q is square free), condition 5.6) 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. 2

22 6 The proof of Theorem 5. The proof will use the technique from the previous sections. The following lemma is the technical part of the inductive step. Lemma 6. Assume a). q = q m q, where q = q q r, with q, q, q r mutually coprime. b). χ = χ χ, with χ mod q m ) and χ mod q ). c). fx) = β x a α ) dα, d = d α, with a α Z distinct and d α Z\{0}. α= d). q q r such that for each p q, p > log q and f mod p satisfies ) i.e. has a simple zero or a simple pole). e). I an interval of length N, q 2 < N < q, 440 m 2 d < log N) 0. f). M Z, log N + log q < M < N 0, M < q τ. Then χ fx) ) < M / M χ f x) ) 60m 2, 6.) N N where f x) is of the form 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) χ fx + tq ) ). 6.4) fx) 22

23 Hence, as in the proof of Theorem 2, χ fx + tq ) ) ) = χ fx) χ fx + tq ) ) m e q m j= Q j x) q j t j ), 6.5) where with Q j x) = d j { fx + t) fx) F j! dt j fx) F x) = 2m s= )} t=0 6.6) ) s s xs up to a factor). 6.7) By the same technique as used in the proof of Theorem 4 See 4.3)-4.8)), after averaging and summing over t M and x I, we remove the last factor in 6.5). Thus, our goal is to show that after t = t,, t 2k ) [, M] 2k is chosen for 6.2) such that f maximizes the character sum in 6.) among all admissible f, q), the first term in 6.) accounts for those t for which f is not q-admissible. The zeros or poles of f x) are of the form b α = a α t ν q with t ν [, M]. 6.8) Here, while applying Hölder, we take k Z + satisfying 48kd < log N ) 0 and k > ) To count the set of bad t, we fix p q. By assumption on f, we may also assume d = and a a α mod p) for any α >. Recalling 6.8), assume that 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). 6.0) 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.) 23

24 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 K \ S, equations 6.0) will determine the remaining values, after specification of αν) and σν). An easy count shows that {πp t) : f mod p) does not satisfy *) } ) ) ) k ) k 2k k 3 k k 2 2 k 2 β 2 ) 3k M p 2 ) < 48kd) k 3k 6.2) 2 M p 2. 2 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 d)-f) to 6.2), we obtain {πp t) : f mod p) does not satisfy *) } < M p ) 8 5 k. 6.3) If f, q) is not admissible, there is some Q q, Q > q τ r > q τ such that for each p Q, f 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 : f is p-bad } < M 8 5 k and summing over p gives the contribution M 8 5 k log q. a ). M < max p Q p < M. Then {t [, M] 2k : f is p-bad } M p + ) 2k {πp t) : f 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. b). max p Q p M and Q > M. 24

25 Take Q Q such that M < Q < M. Then {t [, M] 2k : f is p-bad for each p Q } M Q + ) 2k {π Q t) : f 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+. ) 5 k < 4M) 9 5 k. Summing up cases a)-b) and recalling assumption f), we conclude that {t [, M] 2k : f, q) is not admissible } < M 2k M k τk 6 + q 5 ) 6.4) Taking d = m in 4.8), by assumption f), we show that χ fx) ) N q M ) < Oθ) + O + [M k6 M m ) m + χ f x) ) ] 6.5) 2k. N θ N We obtain 6.) by letting θ = M in 6.5). 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 2, S = k and σν) S for any ν S. Proof. Since the subset of elements of K with more than one pre-image of σ has size k, there exist S K with S = k, and every ν S has at most one pre-image. Therefore, σ makes sense on S. To construct S S, we choose ν i for S inductively, such that ν i {ν,..., ν i, σν ),..., σν i ), σ ν ),..., σ ν i )} and σν i ) S. Proof of Theorem 5. Choose a sequence M < M 2 < < M r of M values and iterate 6.) in Lemma 6. 25

26 Since p< log q p < e log q, we use trivial bound for small p and we may assume Assumption d). In order to satisfy Assumption e), we assume r r m 2 i ) < log N ) 0, 6.6) which follows from 5.7) and 5.8). By 6.) and iteration, N x= χx) is bounded by M 2 + M 60 N i= M 2 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 M r m 2 m2 r 2 60 r S r m 2 m2 r, where S is of the form S = N with χ r primitive modulo q r, and gx) = 6.7) N χ r gx)), 6.8) x= β x a α ) dα, a α, d α Z, α= and g, q) admissible for some q q r, q > q r. Take M s, for s =,, r such that d = d α < 60 r m 2... m 2 r, 6.9) M 60 M 60 2 m 2 2 M 60 s m 2 m2 s 2 s M 2 =M 2 60 M s m 2 m2 s s =M ) to ensure that each of the r fist terms in 6.7) is bounded by M. 26

27 One checks recursively that and hence 6.7) implies M s < M 6s 2 s m 2 m2 s 6.2) N χx) r < N M + M 6 x= 5 )r S 60 r m 2 m2 r. 6.22) In order to satisfy the last condition in Assumption f) of Lemma 6, we impose The above holds, if we take and assume M 6 2)r m 2 m2 r < q τ log log qr r = qr. 6.23) 0 M = e log qr)9/0, 6.24) r log m i < 40 log log q r. 6.25) i= Clearly, this follows from 5.7).) We will prove the theorem by distinguishing two cases in the next section. 7 The two cases. To finish the proof of Theorem 5, we need to bound S in 6.22). Case. m =. Since g, q) is admissible and q is square free, S may be bounded by Remark 3.2. S < q q 0 τ r < q 7 r, 7.) and by 6.22) N χx) r < N M + M 6 x= 5 )r 7 60 q r m 2 m2 r 27 r < r M. 7.2)

28 The last inequality is by Remark 5.2 and 6.23). Now we obtain 5.3) by combining 6.24), 7.2) and 5.8). 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. ii). For all i, q 2 i < N < q. iii). r < 0 3 log log q r. 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 6.7), We will use Postnikov and Vinogradov rather than Weil. 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. α= 28

29 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 ) We assume that fx) satisfies the following property. Q j x)q j 0t j ). 7.5) For each prime divisor p of q 0, there is α such that d α = and β α a α a β ) relative prime to p. This will provide some information on the coefficients Q j in 6.5). Assume α = in fx) and a α = 0 which we may). Thus 0 is a simple zero or pole of f; replacing f by which we may by replacement of χ by χ), we can f assume fx) = xgx) = x x a α ) dα mod p 7.6) a α 0 with g0) defined and non-vanishing mod p). From 6.6), 6.7), and 7.6), we have j!q j x) = s ) s sxgx)) s d j dt j [ x + t)gx + t) xgx) ) s ] t=0. 7.7) Clearly only the terms s j contribute and Q j has a pole at 0 of order j, C Q j x) = x j + A jx) B j x) 7.8) with A j x), B j x) Z[X] and B j x) = x k ˆBj x), k < j, ˆB j 0) 0 and hence ˆB j 0) 0mod p) 7.9) 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 7.0) j x) 29

30 where P j x) Z[X] is of degree at most dj, P j 0) 0 mod p). It follows that { x p : Qj x) 0 mod p)} dj. 7.) and { x q0 : Q j x) 0 mod q 0 )} dj) ω q 0 ) q 0 q ) whenever q 0 q 0. Taking j = m and fixing x, we will apply Vinogradov s lemma [Ga], Lemma 4) to bound N/q 0 t= m e q m 0 j= Q j x)q j 0t j ). 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 Then 2 < P b P k 7.3) efn)) < C klog k)2 P n I for any interval I of size P c, C are constants). Write Q m x) q 0 ā Zmod q m 0 0 ), q 0 q 0 and ā, q 0 ) =. c k 2 log k 7.4) If q 0 q 0, then the coefficient of t m in 7.5) is a m = Q m x) q m q0 m 0 = ā q 0, with q 0 = q 0 q 0 > q 0. Applying 7.4) with P = q 0 gives N/q 0 t= m e q m 0 j= Q j x)q j 0t j ) < N C mlog m) 2 q c m 2 log m) ) It remains to estimate the contribution of those x q 0 such that Q j x) 0 in Z/ q 0 Z for some q 0 > q 0. This number is by 7.2) at most dm) q 0 q 0 q 0 > q 0 ω q 0 ) q 0 < 2 ωq0) dm) ωq 0) q 0 < 2dm) 2 log q0 log log N q 0 7.6) q 0 30

31 since all prime divisors of q 0 are at least log N) 2. Note that the degree d of fx) is bounded by 6.9). Applying 5.8) and 5.7), we have dm < 60 r log q r ) 2 75 log N ) 3κ < log N ) 5κ. 7.7) In particular, 7.7) will ensure that 7.6) is bounded by q 3/4 0 and hence N χfx)) < C mlog m)2 q c m 2 log m) 0 N. 7.8) x= This gives a bound on S in 6.22). 8 Applications. Following Goldmakher s argument [G] based on work of Granville and Soundararajan [GS]), and applying Theorem 5 instead of the character sum estimates developed by Graham-Ringrose [GR] and Iwaniec [ I ], we obtain the following improvement of the Pólya and Vinogradov bound. Theorem 8. Let χ be a multiplicative character with modulus q, and let p be the largest prime divisor of q, q = log q p q p and K =. Let M = log q log q) 9 log q 0 + log 2K) + log p. Then log log q χn) q log q M. 8.) n<x Remark 8.. If q is square free, then the bound in 8.) becomes q { log q log p + log q log log q }. This slightly improves on the corollary to Theorem in Goldmakher s paper [G]. Repeating the argument in deducing Theorem 4 from Theorem 3, we obtain the following mixed character sum estimate from the proof of Theorem 5. 3

32 Theorem 9. Under the assumptions of Theorem 5, χx)e ifx) < Ne log N 8.2) for fx) R[X] of degree at most log N) c. Corollary 0. 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. 8.3) Then for χ non-principal, we have χn)n it < Ne log N n I 8.4) Following [Ga] and [ I ], See in particular, Lemma in [ I ]), this implies Theorem. Let χ be a non-principal character mod q), p the largest prime divisor of q and q = p q p. Let θ = c min log p, log log q log q )log 2 log q ). ). 8.5) log qt ) c log q If Ls, χ), s = ρ + it, has a zero for ρ > θ, t < T, it has to be unique, simple and real. Moreover χ is real. It follows in particular that θ log QT, if log p log q 0. This allows us to state Corollary 2. Acknowledgement. The author would like to thank I. Shparlinski for helpful comments and the mathematics department of University of California at Berkeley for hospitality. 32

33 References [C] M.-C. Chang, An Estimate of Incomplete Mixed Character Sums, Duke Math. J. 45, No ), [Ga] [Go] [GR] [GS] [HB] [HB2] [IK] 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 Plya-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 ), D. R. Heath-Brown, Siegel zeros and the least prime in an arithmetic progression, Quart. J. Math. Oxford Ser. 2), 4 990), 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.) ),