CONDITIONAL BOUNDS FOR THE LEAST QUADRATIC NON-RESIDUE AND RELATED PROBLEMS

Transcription

1 CONDITIONAL BOUNDS FOR THE LEAST QUADRATIC NON-RESIDUE AND RELATED PROBLEMS YOUNESS LAMZOURI, IANNAN LI, AND KANNAN SOUNDARARAJAN Abstract This paper studies explicit and theoretical bounds for several interesting quantities in number theory, conditionally on the Generalized Riemann Hypothesis Specifically, we improve the existing explicit bounds for the least quadratic non-residue and the least prime in an arithmetic progression We also refine the classical conditional bounds of Littlewood for L-functions at s = In particular, we derive explicit upper and lower bounds for L, χ and ζ + it, and deduce explicit bounds for the class number of imaginary quadratic fields Finally, we improve the best known theoretical bounds for the least quadratic non-residue, and more generally, the least k-th power non-residue Introduction Let q > be a natural number and let G = Z/qZ denote the group of reduced residues mod q Given a proper subgroup H of G, two natural and interesting questions in number theory are: Determine or estimate the least prime p not dividing q and lying outside the subgroup H, and Given a coset of H in G, determine or estimate the least prime p lying in that coset These problems have a long history, and have attracted the attention of many mathematicians There are two particular cases that are especially well known If q is prime and H is the group of quadratic residues the first problem amounts to the famous question of Vinogradov on the least quadratic non-residue For any q, if H = {} is the trivial subgroup, then the second problem is that of estimating the least prime in an arithmetic progression In this paper we assume the truth of the Generalized Riemann Hypothesis and obtain explicit as well as asymptotic bounds on these problems Our work improves upon previous results in this area, notably the works of Bach [], and Bach and Sorenson [4] The least prime outside a subgroup Assuming GRH, Ankeny [] established that the least prime lying outside a proper subgroup H of Z/qZ is Olog q Ankeny s work was refined by Bach [], who gave explicit as well as asymptotic bounds for the least such prime Thus Bach showed, on GRH, that the least prime p that does not lie in H is less than log q, and if the additional natural requirement that p q is imposed then p is less than 3log q He further Date: September 0, Mathematics Subject Classification Primary N60; Secondary R4 The first author is supported in part by an NSERC Discovery grant The third author is supported in part by NSF grant DMS-00068, and a Simons Investigator grant from the Simons Foundation

2 established that the least prime p q with p not lying in H satisfies the asymptotic bound p + olog q In the special case when H consists of the square residue classes mod q, Wedeniwski [6] has recently given the improved explicit bound 3log q 44 log q + 3 We begin with an explicit result on this question 5 We shall assume that q 3000, since questions for smaller values of q may easily be settled by direct calculation As usual we denote by ωq the number of distinct prime factors of q Theorem Assume GRH Let q 3000 be an integer, and let H be a proper subgroup of G = Z/qZ Let Aq = max0, log log q 8/5 p q log p/p, and put Bq = max0, log log q ωqlog log q / log q Aq The least prime l with l q and l not lying in the subgroup H satisfies the bound l log q + Bq Suppose q is not divisible by any prime below log q Then there is a prime l log q with l not lying in the subgroup H The quantity Bq in Theorem is zero when q is large and does not have too many small prime factors Asymptotically Bq = olog log q, and in any given situation it may easily be directly estimated Consider now the special case when q is prime and H is the subgroup of quadratic residues mod q Combining Theorem with a computer calculation for the primes below 3000 we obtain the following corollary Corollary Assume GRH If q 5 is prime, the least quadratic non-residue mod q lies below log q Theorem establishes an explicit bound of the same strength as the asymptotic bound given in Bach [] Interestingly it turns out that the asymptotic bound in this question may also be improved Theorem Assume GRH Let q be a large integer and let H be a subgroup of G = Z/qZ with index h = [G : H] > Then the least prime p not in H satisfies p < αh + olog q, where α = 08, α3 = 07 and in general αh = 066 for all h > 3 In particular the least quadratic non-residue is bounded by 04 + olog q Theorem gives good bounds when h is small, but when h is very large the following stronger result holds, which appears to be the limit of our method Theorem 3 Assume GRH Let q be a large integer and let H be a subgroup of G = Z/qZ with index h = [G : H] 4 Then the least prime p not in H satisfies p < 4 + o logh log q h logh 4

3 The least prime in a given coset Now we turn to the second of our questions: the problem of determining the least prime lying in a given coset of H Theorem 4 Assume GRH Let q 0000 and let H be a subgroup of G = Z/qZ with index h = [G : H] > Let p be the smallest prime lying in a given coset ah Then either p 0 9, or p h log q + 3h log log q From the proof one may obtain more precise estimates, but the form above seems easiest to state We next single out the case when H is the trivial subgroup comprising of just the identity Here the problem amounts to getting bounds on the least prime P a, q in the arithmetic progression a mod q where we assume that a, q = A celebrated unconditional theorem of Linnik gives that P a, q q L for some absolute constant L The work of Heath-Brown [0] shows that L = 55 is an admissible value for Linnik s constant, and this has been recently improved to L = 5 by ylouris [7] Under the assumption of the Generalized Riemann Hypothesis, Bach and Sorenson [4] showed that P a, q + oφq log q, and they also derived the explicit bound P a, q q log q Our work leading up to Theorem 4 and some computation for q 0000 permits the following refinement of these bounds Corollary Assume GRH Let q > 3, and let a mod q be a reduced residue class The least prime in the arithmetic progression a mod q satisfies P a, q φq log q By modifying the argument for Theorems and 3, and using the Brun- Titchmarsh theorem, one may derive an asymptotic bound of the form P a, q δ + oφq log q for some small δ > 0 3 Bounds for values of L-functions at s = The Generalized Riemann Hypothesis implies the Generalized Lindelöf Hypothesis, thus furnishing good upper bounds for the values of L-functions Such results go back to Littlewood [3], [4], and the best known asymptotic bounds may be found in [7] for L-values on the critical line and [5] for L-values inside the critical strip Explicit bounds on GRH for L-values on the critical line may be found in [6] The methods of this paper allow us to give, assuming GRH, new explicit upper and lower bounds for L-values at the edge of the critical strip Theorem 5 Assume GRH Let q be a positive integer and χ be a primitive character modulo q For q 0 0 we have L, χ e γ log log q log + +, log log q 3

4 and L, χ eγ π log log q log + + log log q 4 log log q + log q By Dirichlet s class number formula, we can deduce explicit bounds for the class number of an imaginary quadratic field under the assumption of GRH More specifically, let q be a fundamental discriminant with q > 4 and χ q n = q n be the Kronecker symbol, which is a primitive Dirichlet character mod q Moreover, denote by hq the class number of the imaginary quadratic field Q q Then, the Dirichlet class number formula reads q 3 hq = π L, χ q For discussions on the computation of these class numbers, we refer to [3], [9], [], [5] From Theorem 5 we obtain the following corollary Corollary 3 Assume GRH Let q be a fundamental discriminant, and let hq denote the class number of Q q If q 0 0 then hq π q log log q log + e γ + 4 log log q, + log log q log q and hq eγ q log log q log + π + log log q The work of [] computes class numbers and class groups for all imaginary quadratic fields with absolute discriminant below 0 From our corollary it follows that on GRH hq 9053 when q 0 Our methods would also lead to explicit bounds for more general L-functions We do not pursue this here, and content ourselves by stating that for real numbers t 0 0 we have assuming RH ζ + it e γ log log t log + +, log log t and ζ + it eγ π log log t log + + log log t Preliminary lemmas 4 log log t + log t This section collects together several preliminary results that will be useful in our subsequent work We first introduce a convenient notation that will be in place for the rest of the paper We let θ stand for a complex number of magnitude at most one In each occurrence θ might stand for a different value, so that we may write θ θ = θ, θ θ = θ and so on Recall from Chapter of [8] the following properties of the Riemann zetafunction and Dirichlet L-functions Set ξs = ss π s s Γ ζs, 4

5 which is an entire function of order, satisfies the functional equation ξs = ξs, and for which the Hadamard factorization formula gives ξs = e Bs s e s/ρ ρ ρ Above ρ runs over the non-trivial zeros of ζs, and B is a real number given by B = ρ Re ρ = log4π γ = Let χ mod q denote a primitive Dirichlet character, and let Ls, χ denote the associated Dirichlet L-function Put a = χ/ = 0 or depending on whether the character χ is even or odd Let ξs, χ denote the completed L-function s + a q π s Γ ξs, χ = Ls, χ The completed L-function satisfies the functional equation ξs, χ = ɛ χ ξ s, χ where ɛ χ is a complex number of size The zeros of ξs, χ are precisely the nontrivial zeros of Ls, χ, and letting ρ χ = + iγ χ denote such a zero throughout the paper we assume the truth of the GRH, we have Hadamard s factorization formula: 3 ξs, χ = expaχ + sbχ s e s/ρχ ρ ρ χ χ Above, Aχ and Bχ are constants, with ReBχ being of particular interest for us Note that 4 ReBχ = ReBχ = Re ξ ξ 0, χ = ρ χ Re ρ χ Recall the digamma function ψ 0 z = Γ z, and its derivative the trigamma Γ function ψ z = ψ 0z The following special values will be useful: ψ 0 = γ, ψ = ζ = π 6, ψ 0/ = log γ, and ψ / = π Lemma Assume RH For x > there exists some θ such that x k Λn logx/n = x log π log x + + θ B x + 4k Lemma Assume GRH Let q 3 and let χ mod q be a primitive Dirichlet character Define k= Sx, χ := Λnχn logx/n For x >, there exists some θ such that Sx, χ = ReBχ θ x + θ ξ ξ 0, χ log x + log q log x + π Ẽax, 5

6 where and Ẽ 0 x = π 4 γ log x log x k= x k k, Ẽ x = π 8 log + γ/ log x x k k + In particular, ReSx, χ = ReBχ θ x + θ + log x + log q log x + π Ẽax We shall confine ourselves to proving Lemma ; the proof of Lemma is essentially the same Proof of Lemma We begin with πi +i i k=0 ξ s, χxs ξ s ds, and first evaluate the integral by moving the line of integration to the left There are poles at the non-trivial zeros ρ χ of Ls, χ, and a double pole at s = 0 Evaluating the residues here we find that our integral equals 0, χ ρ χ x ρχ ρ χ ξ ξ ξ 0, χ log x ξ Invoking GRH, ρ χ x ρχ ρ χ = θ x ρ χ ρ χ = θ x ReBχ Further ξ 0, χ = = θ ReBχ ξ ρ ρ χ χ Putting these observations together we obtain +i 5 ξ s, χxs πi ξ s ds = ReBχ θ x + θ ξ 0, χ log x ξ i On the other hand, the LHS of 5 equals +i 6 πi log q π s + a ψ 0 L x s L s, χ s ds i The first and third terms above contribute log q log x + Λnχn logx/n π If a =, the middle term in 6 equals here there are simple poles at the negative odd integers, and a double pole at 0 x k k + 0 ψ log x 4 ψ = Ẽx k=0 6

7 If a = 0 the middle term in 6 equals here there are simple poles at the negative even integers, and a triple pole at 0 x k k + log x ψ 0 log x 4 ψ = Ẽ0x k= This completes our proof of the first assertion The second assertion follows by taking real part, and noting that Re ξ 0, χ = ReBχ from 4 ξ Lemma 3 Assume GRH Let q 3 and let χ mod q be a primitive Dirichlet character For any x > we have, for some θ, where and ξ ξ 0, χ x ξ θ 0, χ + ReBχ ξ x = E 0 x = log γ E x = log q x π Λn n χn n + E a x, x k=0 In particular, ReBχ equals + θ + x x x + log x + x x k= x k kk +, x k k + k + γ + log x x log q π Re Λn n χn n + E a x x Proof The proof is similar to that of Lemma, and so we will be brief We consider 7 πi +i i ξ ξ s x s, χ ss ds, and begin by evaluating the integral by moving the line of integration to the left There are poles at s =, s = ρ χ and s = 0 and therefore our integral above equals ξ ξ, χ ξ x ξ 0, χ + ρ χ x ρχ ρ χ ρ χ Note that ξ ξ, χ = 0, χ Further, since we are assuming GRH, we have for ξ ξ some θ, ρ χ xρχ ρ χ ρ χ = θ x ρ ρ χ = θ ReBχ x χ Thus the quantity in 7 equals the LHS of the first identity claimed in our lemma On the other hand, we may rewrite the integral 7 as +i πi i log q π + s + a ψ 0 + L x s 8 L s, χ ss ds 7

8 Now the first and the last terms above give x log q π Λn n χn n x If a = then the middle term in 8 equals there are simple poles at s =, 0, and the negative odd integers x k k + k + + ψ 0 0 x ψ = E x k=0 If a = 0 then the middle term equals here there are simple poles at s = and at the negative even integers, and a double pole at s = 0 0 ψ log x + + γ/ x k + x kk + = E 0x This proves the first part of our lemma Now, by the functional equation and 4, we have Re ξ ξ k= ξ, χ = Re 0, χ = ξ ReBχ Thus taking real parts in the identity just established, we deduce the stated identity for ReBχ Lemma 4 Assume RH For x > we have, for some θ, Λn n = log x + γ + logπ x n B + θ n x x nn + x Proof We argue as in Lemma 3, starting with πi +i ζ i x s ζ s ss ds = n= Λn n n x Moving the line of integration to the left, and computing residues, we find that the above equals The lemma follows log x + γ + logπ x ρ x ρ ρρ x n nn + Lemma 5 Let q 3 and let χ mod q be a primitive Dirichlet character Suppose that GRH holds for Ls, χ For any x, there exists a real number θ such that χnλn log x n n log n log x + log x log q π + + a ψ 0 log L, χ = Re Proof For any σ, we consider πi +i i log x + θ xlog x L σ + s, χxs L s ds = 8 n= ReBχ + χnλn n σ logx/n θ x log x

9 Shifting the contour to the left, we find that the integral also equals L L σ, χ log x σ, χ xρχσ L L ρ χ σ x naσ n + a + σ Therefore L L σ, χ = ρ χ χnλn logx/n + n σ log x log x + θx σ log x n=0 L σ, χ L ρ ρ χ + θxσ log x χ n=0 x n n + Integrating both sides over σ from to and taking real parts we conclude that log L, χ = Re χnλn logx/n n log n log x log x ReL, χ L θ + xlog x ρ ρ χ + θ xlog x χ The lemma follows upon noting that Re L, χ = Reξ L ξ, χ + log q π + + a ψ 0 = ReBχ + log q π + ψ 0 Lemma 6 Assume RH For all x e we have Λn logx/n n log n log x = log log x + γ + γ log x + + a Bθ xlog x + θ 3x 3 log x Proof Analogously to the proof of Lemma 5 we begin with, for any σ >, πi +i i ζ s + σxs ζ s ds = Λn logx/n, n σ and moving the line of integration to the left, this equals x σ σ ζ ζ σ ζ σ log x ζ ρ x ρσ ρ σ n= x nσ n + σ Let σ 0 denote a parameter that will tend to from above, and integrate the two expressions above for σ from σ 0 to Thus Λn x σ logx/n = n σ 0 log n σ dσ + log x log ζσ 0 + ζ ζ σ 0 σ 0 + Bθ θ + x log x 3x 3 log x 9

10 Now, with α = σ 0 log x, σ 0 x σ dσ = log x σ and upon integrating by parts we have α e y eα dy = y α e y y dy + α α e y y dy, e y dy + log α y Now let σ 0 +, so that α 0 +, and note that log ζσ 0 = logσ 0 +Oσ 0, ζ /ζσ 0 = /σ 0 γ +Oσ 0, and that 0 ey dy/y e y dy/y = γ The lemma then follows 3 Proof of Theorem Unlike the other results of this paper, the following simple lemma is unconditional Lemma 3 Let m 3 be an integer and x be a real number Then Λn logx/n ωmlog x, and n,m> n,m> n,m> Λn n Λn logx/n = p x p m n x n n,m> α log x/ log p Λn n = p m log p p Proof If n, m > and n x is a prime power, then n = p α where p m and α log x/ log p Hence x log p log p α = p x p m p x p m log p log x log p log x The second part of the lemma is evident [ ] log x log p p x p m log p α log x/ log p [ ] log x log p ωmlog x Let q 3000 and let H be a proper subgroup of G = Z/qZ Let be such that all primes l q with l lie in the subgroup H Let H denote the group of Dirichlet characters χ mod q such that χn = for all n H Thus H is a subgroup of the group of Dirichlet characters mod q, and note that H equals [G : H] The assumption that all primes l with l q lie in H is equivalent to χl = for all χ H Since H is a proper subgroup, there exists a non-principal character χ mod q in H Our strategy is to compare upper and lower bounds for ReS, χ 0 α

11 where, as defined earlier, S, χ = Λnχn log/n Note that the character χ need not be primitive, and we let χ mod q denote the primitive character that induces χ 3 Proof of Part Here we assume that = log q + Bq log q 64, and seek a contradiction The assumption that χl = for all primes l below with l q gives S, χ = Λn log/n n,q> Λn log/n Using Lemma and Lemma 3 recall that χn = for all n we obtain the lower bound 3 ReS, χ log log q B + ωq logπ log B, log q where we used that log q and so log / 4log log q / log q Next we work on the upper bound for ReS, χ Note that 3 S, χ S, χ = θ n,q/ q> Λn log/n = θ ωq/ q log Using Lemma for χ, and since Ẽa /4 for 64, we find 33 ReS, χ + + log ReB χ + log q log π 4 We shall bound ReB χ above using Lemma 3 First note that by Lemmas 4 and 3 Λn χn n = Λn n Λn n n n n n, q> max 0, log 8 5 log p p p q max 0, log log q 8 5 log p = Aq p p q For 64, we have that + 5 and E a < /4 Using this in Lemma 3, we get ReB χ log q π Aq + E a log q + 5 Aq We use this bound in 33 and combine that with 3 to obtain an upper bound for ReS, χ Since ωq/ q logq/ q/ log, the resulting upper bound is largest

12 when q = q Thus ReS, χ log q + + log + 5 Aq + log log q π 4 log q + 4 log Aq + log q + log q log q + log log q Aq 39 0 Comparing the upper bound above with the lower bound 3 gives the desired contradiction 3 Proof of Part Here we suppose that log q and seek a contradiction The proof follows the same lines as Part, with simplifications due to the assumption that q has no prime factors below, and a little more care with constants Thus using Lemma we have the lower bound 34 ReS, χ = Λn log/n B + logπ log Now we turn to the upper bound Since q has no prime factors below we have S, χ = S, χ From Lemma 3 and Lemma 4 we have that ReB χ log q π log E a For 64 we have that / + 5/, and we may check that for a = 0 or, log π log E a log Therefore, as log q, and q q, ReB χ + 5 log q log log q log log q log 4 7, where the last bound follows upon using 64 together with a little calculus Using this in Lemma, and noting that Ẽax < 0 for a = 0 or and x 3, we get recall log q ReS, χ + + log log q 7 log 4 + log q log π log q + log logπ log q 7 log 8 7 Comparing the bounds 34 and 35 gives a contradiction, which proves the claimed result

13 4 The least prime in a given coset As before, let H denote the group of characters χ mod q with χn = for all n H Recall that H = [G : H] = h, and given a coset ah we have the orthogonality relation { if n ah 4 χaχn = h 0 if n / ah χ H Note that q 0000 Let be such that no prime below lies in the coset ah, and we assume below that max0 9, h log q Set S, χ = Λnχn log/n, so that 4 Λn log/n = χas, χ h n ah Our strategy is again to obtain upper and lower bounds for the quantity above, and then to derive a contradiction 4 Preliminary bounds First, consider the principal character χ 0 mod q which certainly belongs to the group H Since S, χ 0 = χ H Λn log/n n,q> Λn log/n, using Lemmas and 3, we obtain that for maxh log q, 0 9, 43 S, χ 0 B + + log π log + + ωq log 0 + ωq log For a non-principal character χ H, let χ mod q denote the primitive character that induces χ By Lemma we find that 44 S, χ + ReB χ + ξ ξ 0, χ log + log q log + π Ẽa Now for 0 9 we know that Ẽa < 0, and examining the definition of Ẽa we find that 45 log q log + π Ẽa q log log max π, Recall from 3 that log 46 S, χ S, χ + ωq/ q We now invoke Lemma 3, taking there x = 00 Since ξ /ξ0, χ = ξ /ξ0, χ ReB χ we obtain ξ ξ 0, χ log q π + Λn n n 00 3 n 00

14 It follows with a little computation that 47 ξ ξ 0, χ 3 log q π + 4 Further, using Lemma 3 taking there x =, and since E a /7 for 0 9, we find that ReB χ Further, Re log q π Re Λn n χn n q 7 Λn n χn n Re Λn n χn n Re Λn n χn n n,q/ q> Λn n n p q/ q log p p Therefore, we find that for 0 9, the quantity + ReB χ is bounded by 9 + log q 6 π + log p p Re Λn n χn n 4 7 p q/ q Consider the quantity above together with the term ωq/ qlog / appearing in 46 Since log q log q p q/ q log p, and 09, we may check that the sum of these two quantities is largest when q/ q is 6 Putting in this worst case bound, we find that log + ReB χ + ωq/ q log log q π + log 4 7 Re Λn n χn n Combining this estimate with 44, 45, 46, and 47 we conclude that 9 S, χ + log q log q π + 4 log log + log max 50 Re Λn n χn n + log q π, We sum the above over the h non-principal characters of H Note that, using the orthogonality relations and Lemma 4, Re χ H χ χ 0 Λn n χn n 4 Λn n log 3 n

15 Thus, we conclude that S, χ χ H χ χ h log q h + + log + h log + h log 3 log q π q log max π, Using that q 0000 and max0 9, h log q, we may simplify the above bound, and obtain that S, χ h log q h log + h log 49 χ H χ χ 0 + h log q log max π, Combining this with 43 we obtain a lower bound for the LHS of 4 4 Upper bound for 4 Now we turn to the upper bound By assumption there are no primes p with p ah Therefore Λn log/n log p log/p k + log p log/p k+ n ah p k p k ah p k+ The second term above is, using Lemma, and a little computation using 0 9, k + log p log /k+ /p k log / log / p /k+ k + /k+ + /4k+ 0 7, so that 40 k log / log / n ah Λn log/n p k p k ah log p log/p k The least prime in an arithmetic progression Using a computer we checked Corollary when q 0000 Suppose now that q > 0000 and let a mod q be an arithmetic progression with a, q = such that no prime below = φq log q is a mod q If q has at least six distinct prime factors then φq φ = 5760, while if q has at most five prime factors then φq qφ / > 455 In either case, φq 456 By similar elementary arguments we may check that for q > 0000 we have ωq q 3/7, and that φq q 5/6 Thus = φq log q 0 9, and we may use our work above; note that here H is the trivial subgroup consisting of just the identity, and h = φq First we work 5

16 out the lower bound for the quantity in 4 Using 43 and 49, together with the bounds ωq q 3/7 and q log q, we find that φq Λn log/n φq log q 4 5 φq+ 3 +3log n a mod q To obtain a corresponding upper bound, it remains to estimate the first term in 40 Note that the number of square roots of a mod q is bounded by ωq+ q 3/7 Since log plog /p k log /8 we find that p k p k ah Therefore log p log/p k n a mod q log 8 Λn log/n n n a mod q 7 3/7 log q 4 log + q3/7 4 q q + + Using our bounds φq q 5/6, q > 0000 and q log q 0 9, a little computation shows that the upper and lower bounds derived above give a contradiction 44 The general case In general we bound the first term in 40 crudely by Λn log n /4 + +, 0 so that n n ah Λn log/n 5 On the other hand, using 43 and 49 together with the bounds ωq q 3/7 and max0 9, h log q, we obtain h Λn log/n h log q 4 5 h log log + 3h n ah Comparing this with our upper bound, we must have 5 h log q + 4 h log log + 3h Since 0 9, we have 7 log log + /00, so that from the above estimate 3 we may first derive that h log q Now inserting this bound into our estimate, we obtain the refined bound 5 4logh log q h log q + h logh log q + 4 3h The bound stated in Theorem 4 follows from this with a little calculation 6

17 5 Explicit bounds for L, χ : Proof of Theorem 5 5 Upper bounds for L, χ Let q 0 0 be a positive integer and x 00 be a real number to be chosen later Lemma 5 gives log L, χ Re χnλn logx/n + log q + a n log n log x log x π + ψ 0 log x ReBχ + xlog x xlog x Invoking Lemma 3 with the same value of x above, we have ReBχ + Λnχn x x n Now for x 00, E a x + x log x xlog x log q π Re + log x ψ 0 + a + n + E a x x xlog x 0, and therefore log L, χ Re χnλn logx/n logq/π + n log n log x + x + log x log x + log x + Re Λnχn n xlog x x n x The right hand side above is largest when χp = for all p x, and so log L, χ Λn logx/n n log n log x + Λn n log q + log x n x + x + log x log x Appealing now to Lemmas 4 and 6 we conclude that log L, χ log log x + γ log x + log q + x log x log x Choosing x = log q /4, so that x 30 for q 0 0, we deduce that L, χ e γ log x e γ log log q log + log x + log log q This proves the stated upper bound for L, χ 5 Lower bounds for L, χ The argument proceeds similarly to the one for upper bounds Let q 0 0 be a positive integer As before choose x = log q /4 so that x 3 Lemma 5 gives log L, χ Re Λnχn logx/n n log n log x log x + xlog x + log q log x π + ψ 0 ReBχ 7 xlog x + a

18 From Lemma 3 with the same value of x above we find that ReBχ Λnχn x x n Now for x 00 E a x x log x + xlog x and therefore 5 log L, χ Re + log q π Re Λnχn logx/n n log n log x + log x ψ 0 x log x + xlog x + a From Lemma 4 it follows that for x 3, Re Λnχn n log x +, n x and therefore, with a little calculation, x log x + xlog x log x n x + E a x xlog x 0, logq/π + + / x x log x log x Re Λnχn n n x Re Λnχn n n x x Using this bound in 5 we obtain log L, χ Re Λnχn n log n x log x logq/π + + / x 5 x log x log x x The next lemma shows that the sum over n above is smallest when χp = for all p x Lemma 5 For x 00 we have Re Λnχn n log n Λp k k x log x p k log p k x log x p k x Proof Consider the contribution of the powers of a single prime p x to both sides of the inequality above; we claim that the contribution to the left hand side is at least as large as the contribution to the right hand side If χp = 0 then the contribution to the left hand side is zero, and the contribution to the right hand side is negative If χp 0 then writing χp = eθ, we see that the difference in the contributions to the left hand side and right hand side equals log p k log x/ log p k coskθ 8 p k log p k x log x

19 If p 3 then using coskθ k cos θ we see that the quantity above is log p cos θ p log p x log x j 0 p j log p j For p = we use the bound 0 coskθ k cos θ for k 6 and compute explicitly the trigonometric polynomial arising from the first five terms With a little computer calculation the lemma follows in this case Note that Λp k k p k log p = k x log x p k x 53 log x j= Λn logx/n n log n log x Λn n + Λm n x m x The third term above is log ζ Λm m log m Λm x log x m >x m x m log m x log x log ζ 3 x, where we bound the sums over prime powers trivially by replacing them with sums over odd numbers and powers of The first two terms in 53 are handled using Lemmas 4 and 6 Putting these together, and using x 3, we conclude that Λp k k p k log p log log x γ + log ζ + k x log x log x 8 5 x p k x Inserting this estimate and Lemma 5 into 5, and noting that logq/π x, we obtain with a little calculation log L, χ log log x γ + log ζ log x log x 4 x Exponentiating this, and using x 3, we obtain the stated lower bound 6 Asymptotic bounds Let δ be a fixed positive real number, and let Ks denote a function holomorphic in a region containing / δ < Res + δ, save for possibly a simple pole at s = / with residue r Further suppose that for all s in this region bounded away from / we have Ks / + s For ξ > 0 define the Mellin transform Kξ = πi c+i ci Ksξ s ds, where the integral is over any vertical line with < c + δ If ξ then by taking the c = / in the integral above, and using Ks / + s, we find that Kξ ξ If ξ then moving the line of integration to c = / δ/ encountering potentially a pole at / we find that Kξ / ξ Thus 6 Kξ minξ, /ξ Finally we assume that K is such that Kξ > 0 for all ξ > 0 9

20 As before, let G = Z/qZ and H be a proper subgroup of G Let be such that all primes l q with l lie in the subgroup H We may assume that log q Further, let H denote the group of characters χ mod q with χn = for all n H Recall that H = [G : H] = h, and that the orthogonality relation 4 holds Lemma 6 Assume GRH Let χ mod q be in H If χ is the principal character then Λnχn Kx/n = K/ x + O + log q log x n x If χ is non-principal then Re n n Λnχn n Kx/n = θ + o log q π Kit dt + O + log q log x x Proof Consider first the case when χ is non-principal Let χ mod q denote the primitive character that induces χ and let ξs, χ denote the corresponding completed L-function We start with 6 I = πi /+δ ξ ξ s +, χksxs ds, which we evaluate, as usual, in two ways Note that πi log q Ksx s ds log q, π x /+δ upon shifting the integral to Res = / δ and possibly picking up a pole at s = / Further, by moving the line of integration to Res = 0, we find that Γ / + s + a Ksx s ds log + t Kit dt πi /+δ Γ Now by expanding L /L into its Dirichlet series we have L πi /+δ L + s, χksxs ds = Λn χn n n Note that n Λn χn n Kx/n and using 6 this is x log p p min k/ p p q k= n k/, pk/ Λnχn n Kx/n x p q From these observations, we conclude that 63 I = n log p n,q> Kn/x x + log x x log p Λnχn n Kx/n + O + log q log x x 0 Λn n Kx/n, log q log x x

21 Now evaluate the integral in 6 by shifting the line of integration to Res = δ Thus, with γ running over the ordinates of zeros of ξs, χ, I = Kiγx iγ r ξ ξ 0, χx/ + πi γ ξ δ ξ + s, χksxs ds Using now the functional equation for ξ, we find that the integral on the right hand side is bounded by x /δ Recall that Re ξ 0, χ log q, so that ξ ReI = θ γ Kiγ + O log q log q = θ + o x π log q Kit dt + O, x where the final estimate follows from an application of the explict formula see Theorem 5 of [] This establishes our lemma for non-principal characters For principal characters, we have that n Λn n Kn/x = πi ζ /+δ = K/ x πi = K/ x + O On the other hand, similar to before Λnχ 0 n Kn/x n n ζ / + sksxs ds /δ n ζ ζ / + sksxs ds γ Λn n Kn/x log q log x x From these observations, we conclude that 64 I = Λnχn Kn/x + O + log q log x n n x Kiγx iγ Proposition 6 Assume GRH Keep the notations above, and recall that is such that all primes l q with l lie in the subgroup H Then, for any λ > 0 we have h λ 0 Ku du K/ + o λh log q Kit dt u π Proof We may assume that log q We shall use Lemma 6 for each character χ H, and taking x = /λ Thus we obtain that Re Λnχn Kx/n K/ x++oh log q Kit dt χ H n n π From 4, and since Kx/n 0 for all n, we see that the left hand side above is h Λn Kx/n + O h Λn Kx/n = h Kx/t dt + oh log q n n t n,q>

22 Since Ku = K/u, after a change of variables u = t/x, the above becomes h x λ 0 Ku du u + oh log q The proposition follows with a little rearrangement 6 Limitations of the method By Mellin inversion we have that K/ = Kudu/ u, and note also that Ku Kit dt Therefore 0 π λ h Ku du λ K/ h / Ku du h λ Kit dt 0 u 0 u π Hence Proposition 6 cannot lead to a bound for that is better than 4 + olog q 6 Bounds for large h Take Ks = eαs e αs and λ = Note that the s Mellin transform for K is Ku = max0, α log u Now π Kit dt = π sinαt dt = α, t and Ku du α + log u = du = 4α 4 + 4e α 0 u e u α Thus Proposition 6 implies that h4α 4 + 4e α 4e α/ e α/ + oαh log q Selecting α = logh, it follows that for h o logh log q h logh 4 For large h, the above bound is about /4 + olog q which is of the same quality as the limit of the method, but the convergence to this limit is quite slow 63 Bounds for small h For smaller values of h, better estimates may be obtained using the kernel Ks = Γs + / Note that Ks satisfies the conditions stated at the beginning of this section and by a change of variables Ku = ue u 0 for all u Noting that K/ = Γ =, we apply Proposition 6 to see that λ h e u log q du λh Γ/ + it dt, π 0 so that λh log q Γ/ + it dt πh he λ When h =, we choose λ = 45, which is more or less optimal, and find that < 0794+olog q For h = 3, we choose λ = 05, and find that < 07+ olog q For h = 4, we choose λ = 85 and get that < olog q We may apply this for other smaller values of h and get progressively better bounds as h increases, with 0545log q being the limit for this test function

23 References [] NC Ankeny, The least quadratic non-residue, Ann of Math 55 95, pp 65-7 [] E Bach, Explicit bounds for primality testing and related problems, Math Comp , no 9, [3] A Booker, Quadratic class numbers and character sums, Math Comp , no 55, [4] E Bach and J Sorenson, Explicit bounds for primes in residue classes, Math Comp , no 6, [5] E Carneiro and V Chandee, Bounding ζs in the critical strip, J Number Theory 3 0, no 3, [6] V Chandee, Explicit upper bounds for L-functions on the critical line, Proc Amer Math Soc , no, [7] V Chandee and K Soundararajan, Bounding ζ/ + it on the Riemann hypothesis, Bull Lond Math Soc 43 0, no, [8] H Davenport, Multiplicative Number Theory, vol74, Springer-Verlag GTM, 000 [9] J L Hafner and K S McCurley, A rigorous subexponential algorithm for computation of class groups, J Amer Math Soc 989, no 4, [0] D R Heath-Brown, Zero-free regions for Dirichlet L-functions, and the least prime in an arithmetic progression, Proc London Math Soc , no, [] H Iwaniec and E Kowalski, Analytic Number Theory, Colloquium Publications, vol 53, American Mathematical Society, Providence, 004 [] MJ Jacobson Jr, S Ramachandran, and HC Williams, Numerical results on class groups of imaginary quadratic fields Algorithmic Number Theory Lecture notes in Computer Science Vol [3] J E Littlewood, On the zeros of the Riemann zeta-function, Math Proc Cambridge Philos Soc [4] J E Littlewood, On the function /ζ + it, Proc London Math Soc [5] M Watkins, Class numbers of imaginary quadratic fields Math Comp , [6] S Wedeniwski, Primality Tests on Commutator Curves Dissertation: Tübingen, Germany, 00 [7] T ylouris, On Linnik s constant, Acta Arith 50 0, 65 9 Department of Mathematics and Statistics, York University, 4700 Keele Street, Toronto, ON, M3JP3 address: lamzouri@mathstatyorkuca Department of Mathematics, University of Illinois at Urbana-Champaign, 409 W Green Street, Urbana, IL, 680 address: xiannan@illinoisedu Department of Mathematics, Stanford University, Stanford, CA address: ksound@mathstanfordedu 3