A Bombieri-Vinogradov theorem for all number fields

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1 A Bombieri-Vinogradov theorem for all number fields M. Ram Murty and Kathleen L. Petersen January 30, 0 Abstract The classical theorem of Bombieri and Vinogradov is generalized to a non-abelian, non-galois setting. This leads to a prime number theorem of mixed-type for arithmetic progressions twisted by splitting conditions in number fields. One can view this as an extension of earlier work of M. R. Murty and V. K. Murty on a variant of the Bombieri-Vinogradov theorem. We develop this theory with a view to applications in the study of the Euclidean algorithm in number fields and arithmetic orbifolds. Dirichlet s density theorem gives an asymptotic estimate for the density of primes in arithmetic progressions. Let π(x denote the number of primes p x, and for positive integers a q such that (a, q =, denote by π(x, q, a the number of primes p x which satisfy the congruence p a (mod q. Dirichlet s theorem indicates that as x π(x, q, a π(x φ(q where φ is Euler s totient function, and f g means that f/g. The Riemann hypothesis for all Dirichlet L-functions implies that the error term satisfies the estimate π(x π(x, q, a φ(q x log qx, where f g (equivalently f = O(g means that f/g is bounded, and will be referred to by saying that f is of order g. The celebrated theorem of Bombieri [3] and Vinogradov [4] shows that this estimate holds on the average. Theorem 0. (Bombieri, Vinogradov. Let A > 0 be given. Then there is a B = B(A > 0 so that for Q = x (log x B π(y, q, a π(y x (a,q= y x φ(q (log x A. In [7], a variant of Theorem 0. is proven for K/Q a Galois extension of number fields. The goal of this paper is to prove an analogous theorem without the Galois assumption. Specifically, let K be a number field and M a subfield of K (possibly M = K such that K/M is Galois. Let G = Gal(K/M and let C be a conjugacy class in G. Let p be a prime ideal of M unramified in K and let σ p denote the conjugacy class of Frobenius automorphisms corresponding to prime ideals q of K lying over p. Define π(x, C to be the number of prime ideals p of M unramified in K with σ p = C and Np x where N = N M/Q. With a, q as above let π(x, C, q, a denote the number of primes ideals p of M unramified in K with σ p = C, Np x and Np a (mod q. By the Chebotarev density theorem [3] π(x, C, q, a d(c, q, aπ(x for a density d(c, q, a 0. If K Q(ζ q = Q where ζ q is a primitive q-th root of unity, then d(c, q, a = C G φ(q. We prove the following theorem, which is equivalent to Theorem 4. proven in 4.

2 Theorem 0.. Let K/M be a Galois extension of number fields, with group G and let C be a conjugacy class in G. Let H be the largest abelian subgroup of G such that H C and let E be the fixed field of H. Let η = {[E : Q], } and for any ɛ > 0 let Q = x η ɛ. Then for any A > 0, (a,q= y x π(y, C, q, a d(c, q, aπ(y x (log x A where the decoration indicates that the summation is over only those q such that K Q(ζ q = Q. In comparison with Theorem 0., a value of η = would be ideal, but is not within the grasp of current methods. The coarsest interpretation of our results yields an η value of {[K : Q], }. There are several motivations for deriving this theorem. One of them is to the study of the Euclidean algorithm in number fields as developed in []. Another is to the study of arithmetic orbifolds dealing with the structure of quotients of P SL (O K as in [8]. In many applications we have in mind, this theorem will be used in conjunction with the lower bound sieve, where it is often required to produce infinitely many primes p with a certain splitting type in a fixed (not necessarily Galois extension with some restriction on the prime divisors of p. We will not pursue any of these applications in this paper. Our purpose here is to derive the result with all of the technical details clearly presented. As is well-known, the celebrated theorem of Bombieri-Vinogradov has had an immense number of applications in analytic number theory and we foresee a similar role for our theorem. The desirable goal of course would be to have our theorem with the value η = but this seems to be far into the future. 0. Outline of Proof Let K be a number field with n K = [K : Q], ring of integers O K, and let d K be the absolute value of the discriminant of K. Chebyshev s ψ function for the number field K is, in its simplest incarnation ψ(x = log Np Np m x where the sum is over all powers of unramified prime ideals p of K which satisfy the bound Np m x where N = N K/Q. It is well known that an estimate of ψ(x can be translated to an estimate of π(x, the number of prime ideals in K of norm at most x. We will indicate the inclusion of ramified primes in a summation by the decoration over the function. We write ψ(x = Λ(c Nc x where Λ(c is the von Mangoldt function. That is, { log Np if c is a power of the prime ideal p Λ(c = 0 otherwise. We write ψ(x, q, a for the sum ψ(x restricted to those p m which satisfy Np m a (mod q. For ease of exposition, we will omit possible conjugacy class restrictions in this overview. Precise definitions will be given in subsequent sections. In particular, 0.6 contains definitions of the various Chebyshev functions. The aim of this paper is to prove a statement of the form (a,q= y x ψ(y, q, a y which is equivalent to the statement in the introduction. φ(q x (log x A To refine the summation, we rewrite the average as a sum over twisted Chebyshev functions, ψ(x, χ. In the simplest manifestation, χ is a Hecke character. We reserve the notation χ 0 for the trivial character. The twisted Chebyshev function is ψ(x, χ = Λ(cχ(c Nc x (

3 and is the partial coefficient sum of the logarithmic derivative, L L (s, χ, of the L-series L(s, χ = χ(c Nc s. c O K This connection with L-functions will be used extensively. We first convert the average on the left hand side of ( to one of the form Q ψ k(y, χ y x χ where the decoration indicates that the summation is over primitive characters only. That is, to prove ( we show that it is sufficient to prove that this new summation is O(x(log x A. We then reinterpret this summation so that we can reduce to the case where the associated L-functions are all abelian, and estimate this new summation. We break the estimate into two cases, depending on the size of Q. We estimate the initial range Q (log x γ for any positive constant γ using classical techniques in. In 3 we estimate the terminal range, for Q in the range (log x γ Q min{x ɛ, x d ɛ } (with d = n E for E K as described in Proposition.8 and γ defined in Proposition 3. using the large sieve. To do so we will smooth the Chebyshev functions with an inverse Mellin transform. Specifically, for f : R R define f 0 = f and for k N recursively define the inverse Mellin transforms f k of f as We will use the notation exp(x = e x. f k (x = x f k (t dt t. ( Class Functions If f : G C is a C-valued function on a group G and σ G we define f σ : G C by f σ (g = f(σ gσ. We say that f is a class function if f σ = f for all σ G. We use class functions to impose conditions on the primes in our summation using the Frobenius, which we now define. Let L/M be Galois with group G and N = N M/Q. For each prime ideal p of M and prime ideal q of L lying over p the decomposition group, D q, is the stabilizer of q in G, {g G : g(q = q}. The decomposition group is isomorphic to Gal(L q /M p where L q (resp. M p is the completion of L (resp. M at q (resp. p. Let m p denote the residue field O M /p and l q the residue field O L /q. By Hensel s lemma there is a surjective map from D q to Gal(l q /m p, giving the exact sequence I q D q Gal(l q /m p where I q is the kernel of this map. The group Gal(l q /m p is cyclic with generator x x Np and Np is the cardinality of m p. The element of D q whose image in Gal(l q /m p is a generator is well defined modulo I q. Denote this element of D q /I q by σ q. If p is unramified in L, then I q = and we can consider σ q as an element of G, the Frobenius element. This element is characterized by σ q (x = x Np (mod q for all x L. For another choice q above p, I q and D q are conjugates of I q and D q respectively in G. We let σ p denote this conjugacy class in G. If p is ramified, and f is a class function then we define f(σ m p by extending f to powers of the ramified primes by setting the value to be the average f(σp m = f(g I q 3

4 where the sum is over all g D q whose image in D q /I q is σ m q. For a subgroup H of G, and a class function f : H C we define the induced class function Ind G Hf : G C as follows. First, we extend f to a function on G by setting f to be trivial outside of H. Let g,..., g R be coset representatives for H in G. We define ( Ind G H f (g = R r= f gr (g = H f s (g. For a finite group G and f, f : G C two C-valued functions on G, we define their inner product 0.3 Artin L-series f, f = G s G f (gf (g. g G Our references for Artin L-series are [5] and [6]. Let L/M be Galois with group G. (We suppress the field extension in the notation unless there is need for clarification. Consider a representation, ρ : G GL n (C with character χ. For Re(s > we define the unramified L-series associated to ρ (or χ, L u (s, χ, to be a product of local Euler factors, where the product is taken over the primes ideals of M unramified in L. Specifically, L u (s, χ = L p (s, χ where the decoration indicates that the product is over those p in M which are unramified in L. The Euler factors are L p (s, χ = det(i ρ(σ q Np s where N = N M/Q. We extend this L-series to ramified primes as follows. Let p be a prime ideal of M ramified in L and q a prime ideal in L above p. Let V be the underlying complex vector space on which ρ acts. Restricting this action to the decomposition group D q, the quotient D q /I q acts on the subspace V Iq of V on which I q acts trivially. All σ q have identical characteristic polynomials on this subspace, and we define the Euler product at the ramified prime ideals p to be L p (s, χ = det(i ρ(σ q V Iq Np s. For all primes ideals p, each factor L p depends only on p and has degree χ(, the dimension of ρ. For each factor L p (s, χ, log L p (s, χ = χ(σp m m Np ms m= where χ(σ m p is the trace of ρ(σ m p if p is unramified in L, and we extend χ to the ramified prime ideals as discussed in 0.. The (full Artin L-series is p L(s, χ = p L p (s, χ where the product is over all primes ideals p in M. We can rewrite this product as L(s, χ = p L p (s, χ 4

5 where this product is over rational primes p, and L p (s, χ = p p L p(s, χ. Each L p is a polynomial in p s of degree, say m p. Note that m p n M for all p and m p = n M for p unramified in M. The Euler factors can be written as m p ( L p (s, χ = j= π p,j p s (0.3. where π p,j =. From the logarithms of the Euler factors it is apparent that the logarithmic derivative of L(s, χ is L L (s, χ = (log Npχ(σp m Np ms. (0.3. p m= With the induced function defined in 0., if H is a subgroup of G and L H is the subfield of L fixed by H then for a character χ of H L(s, Ind G Hχ, L/M = L(s, χ, L/L H. ( Functional Equations Our reference for this material is [5]. For L/M Galois we now define an enlarged L-function, Λ, from which a functional equation will follow. First, we define the gamma factors. Let γ(s = π s Γ ( s. For each infinite complex place ν of M, define the local factor γ ν (s, χ = [γ(sγ(s + ] χ(. Now assume that ν is real. To a place of L above ν there corresponds a decomposition group G( = {g G : g( = } which is of order one or two. The generator σ is well defined up to conjugation by ν. Let V be the underlying representation space for ρ. The vector space V decomposes as a direct sum, V = V ν + Vν corresponding to the + and eigenvalues of ρ(σ. We define the local factor γ ν (s, χ = γ(s dim V + ν γ(s + dim V ν and define γ(s, χ = ν γ ν(s, χ. We can write γ(s, χ = γ(s a(χ γ(s + b(χ for integers a(χ + b(χ = n M χ(. Before we define Λ we need to introduce the notion of a conductor. The Artin conductor f(χ associated to χ is an ideal of O M defined as follows. Let p be a prime ideal of M and q a prime ideal of L lying over p. Let G 0 denote the inertia group I q at q. There is a descending filtration of higher ramification groups G 0 G.... The quantity n(χ, p = i=0 G i G 0 codim(v Gi is an integer which is independent of the choice of q above p. It is equal to zero for all prime ideals p unramified in L, and therefore vanishes for all but finitely many p. The Artin conductor is the ideal f(χ = p pn(χ,p and the conductor is the constant A(χ = d χ( M N M/Qf(χ. If χ is an Artin character induced by a Dirichlet character of modulus q, with A = A(χ 0 it follows that A(χ Aq n M. In this case, χ( = and the Artin conductor is q, so A(χ d M N M/Q q d M q n M q n M. If L = K(ζ q, χ is a character induced by Gal(K/Q, and χ is induced by a Dirichlet character modulo q, then in this case A(χ χ = A(χ A(χ. Therefore, A(χ χ A(χ q n M. A simple case of the conductor discriminant formula is the relation where d L/M is the relative discriminant, d L = d [L:M] M N M/Q(d L/M (0.4. d L/M = χ f(χ χ( and the product is over all irreducible characters χ of Gal(L/M. (See [9] Proposition

6 We now define Λ(s, χ = A(χ s γ(s, χl(s, χ for Re(s >, which has a meromorphic continuation to the whole complex plane and functional equation Λ(s, χ = W (χλ( s, χ where W (χ, the Artin number, is a complex number of absolute value. The character χ is the complex conjugate of χ. That is, if χ is the character of the representation ρ : G GL(V, then χ is the character of the contragradient representation ρ : G GL(V where V is the dual of V and ρ is defined by for all g G, v V and v V. It will be convenient to set ρ g (f, v = f, ρ g (v With this notation we can rewrite the functional equation as since A(χ = A(χ. Θ(s, χ = W (χa(χ s γ( s, χγ(s, χ. (0.4. L(s, χ = Θ(s, χl( s, χ (0.4.3 We will use Stirling s inequality in various forms. For σ fixed, as t, (see [6] ex For s = σ + it, Γ(σ + it e π t t σ π. (0.4.4 Θ(s, χ [A(χ( t + n M ] σ. (0.4.5 Logarithmically differentiating (0.4. and applying Stirling s formula, we also have 0.5 Congruences and Characteristic Functions Θ Θ (s, χ n M (log( t + + s. (0.4.6 Let K/M be Galois with group G. In the average, we wish to estimate the number of prime ideals p in M which are unramified in K such that σ p = C for a fixed conjugacy class C of G. We extend class functions to the ramified primes as in 0.. Let δ C : G {0, } denote the characteristic function of C. By the orthogonality relations of characters, for a fixed element g of C, δ C = C G η(gη (0.5. where the sum is over irreducible characters η of G. An alternate decomposition is as follows. Let H be a subgroup of G such that H C. Let h H C and let C H be the conjugacy class of h in H. With λ = C H G C H, and the induced character as defined in 0., η δ C = λind G Hδ CH. We wish to count only those prime powers p m so that N M/Q p m a (mod q, and σ m p = C. If K Q(ζ q = Q, with G q = Gal(K(ζ q /M, G q = G Gal(Q(ζq /Q = G (Z/qZ. Fixing this isomorphism, let the conjugacy class C be the class associated to (C, C where C is the preferred class in G and C is a class in (Z/qZ corresponding to the congruence condition. Let δ C and δ a,q = δ C be characteristic functions on G q. (For an element (g, t G q, define δ C (g, t = if g C and 0 otherwise. The function δ a,q is defined in a similar manner. 6

7 Definition 0.. Let ξ = ξ(c, a, q = δ C = δ C δ a,q. For a subgroup H of G with H C set H q = H Gal(Q(ζ q /Q. If χ is a Dirichlet character of modulus q, then χ induces an Artin character by defining χ(σ m p = χ(np m since as M Q(ζ q = Q in the extension M(ζ q /M, the conjugacy class of Np determines the Frobenius element. By Mackey s induction theorem, δ C χ = λind Gq H q (δ CH χ where we have used χ for the character on both G q and H q induced by the character χ modulo q. As in (0.3.3, where L H is the fixed field of H. L(s, δ C χ, L/M = L(s, Ind Gq H q (δ CH χ, L/M λ = L(s, δ CH χ, L/L H λ ( Chebyshev ψ functions For L/M Galois and a class function f of Gal(L/M we define ψ(x, f, L/M = (log Npf(σp m Np m x where the sum is over powers of those prime ideals p of M unramified in L, and N = N M/Q. We suppress the field extension unless there is need for clarification. When f is trivial this is ψ(x. After smoothing with the inverse Mellin transform (0.. ψ k (x, f = k! Np m x (log Np(log x Np m k f(σ m p. (0.6. To make the connection with L-functions precise, we include ramified primes in our summation. We do so by extending the class function f to the ramified primes as in 0.. We denote the inclusion of ramified primes by the decoration over the function. For an Artin character χ, if Re(s >, by examining the logarithmic derivative of L(s, χ (see (0.3. ψ(x, χ is the partial sum of the coefficients, and ψ k (x, χ = πi ( where the integration is on the line Re(s =. L L xs (s, χ ds (0.6. sk+ We will begin estimating an average involving the error term ψ(x, C, q, a d(c, q, ax, and show that it is equivalent to one which includes ramified primes, corresponding to the function ψ(x, C, q, a = ψ(x, ξ, K(ζ q /M, where ξ is as in Definition 0.. In turn, we will rewrite this average as one with functions of the form ψ(x, δ C χ, K(ζ q /M = (log Npδ C (σp m χ(np m Np m x where χ is induced by a Dirichlet character of modulus q and the sum is over powers of prime ideals p in M. These functions can be smoothed by the inverse Mellin transform. Taking coefficient sums of (0.5. with the notation from 0.5 ψ k (x, δ C χ, L/M = ψ k (x, Ind Gq H q (δ CH χ, L/M λ = ψ k (x, δ CH χ, L/L H λ. (0.6.3 Writing the characteristic function ξ = C G η(gη (see 0.5, for Re(s >, using (0.3. η L C (s, ξ = L G η η(g L (s, η. L 7

8 By (0.6. ψ k (x, ξ, L/M = πi ( L L where the integration is on the line Re(s =. xs (s, ξ ds = sk+ πi C G η η(g ( L L xs (s, η ds (0.6.4 sk+ 0.7 Abelian L-functions In our previous definitions, the Artin characters are, in general, non-abelian characters of Gal(L/M. We now rewrite the L-functions in terms of abelian L-functions. We follow [3], using the methods of Deuring [8] and MacCluer [4]. In (0.6.4 we rewrote ξ as a sum over irreducible characters using a preferred element g of G = Gal(K/M. Let H = g, the cyclic group generated by g and let E be the fixed field of H. Since H is cyclic, the irreducible characters of H are one-dimensional. From (0.6.3 ψ(x, δ C χ, L/M = λ ψ(x, δ CH χ, L/E where the characters on the right are one-dimensional by construction. We can further write the character δ CH as the sum over irreducible characters from (0.5. with g H C. We conclude that δ CH = C H H η(gη ψ k (x, δ C χ, L/M = λ ψ k (x, δ CH χ, L/E = C G η η(g ψ k (x, η χ, L/E. η By (0.6.4 ψ k (x, δ C χ, L/M = πi C G η η(g ( L L xs (s, η χ, L/E ds. (0.7. sk+ The motivation behind rewriting our functions in terms of abelian characters is the following, which are known to hold only in the abelian case. In the following, L(s, η is an abelian L-function and η is assumed not to be the trivial character. (See [3] The Hadamard factorization, for constants B (η and B(η is Λ(s, η = e B(η+B(ηs ( s ρ e s ρ ρ where ρ runs through all of the zeros of Λ(s, η. These are the non-trivial zeros of L(s, η, those for which 0 < Re(ρ <. By logarithmic differentiation, using the above, L L (s, η = B(η + ρ ( s ρ + ρ γ log A(η (s, η. (0.7. γ From [3] Lemma 5., Re ( B(η = ρ Re( ρ. Therefore, L L (s, η + L L (s, η = ρ ( s ρ + s ρ γ log A(η (s, η (0.7.3 γ where the sum runs through all non-trivial zeros ρ of L(s, η. 8

9 Rewriting the Average We continue to consider K/M Galois with group G, and let L = K(ζ q. If a < q and (a, q = and K Q(ζ q = Q, then L/M is also Galois with group isomorphic to G q = Gal(K/M (Z/qZ. We let ξ = ξ(c, q, a = δ C δ a,q be the class function from Definition 0.. Therefore, ψ(x, C, q, a = ψ(x, ξ, L/M. We wish to estimate the average (for a suitable choice of Q ψ(y, C, q, a d(c, q, ay (a,q= y x where the decoration on the summation indicates that the sum is over those q so that K Q(ζ q = Q. For these q, d(c, q, a = C G φ(q. To simplify expressions, we will let r k(x, C, q, a = ψ k (x, C, q, a d(c, q, ax and decorate r to mirror the corresponding ψ function. The goal of this section is to prove Proposition.8, which states that in order to show that the above average is O(x(log x A it suffices to show that for all values Q Q there is a k 0 so that the average where K/E is abelian. Q χ ψ k (y, χ, K(ζ q /E y x x (log x A First, we show that the contribution of the ramified primes is negligible. Lemma.. Assume that a < q and (a, q = with K Q(ζ q = Q. Then with the following hold.. ψ k (x, C, q, a ψ k (x, ξ E.. r k(y, C, q, a y x (a,q= E = E (x, k = k! (log xk+ (log d K + n M x (a,q= y x r k(y, ξ Q E. Proof. First we prove. The number of prime ideals p of M ramified in K is at most p d K/M log Np log log N(d K/M, log with N = N M/Q and d K/M the relative discriminant of K/M. By the conductor discriminant formula (0.4., d K = d [K:M] M N(d K/M. We conclude that the number of ramified primes is at most log d K. Using definition (0.6. the difference in part is the following sum over ramified prime powers k! ( (log Np log x kξ(σ m Np p k! (log xk+ ξ ( + Np m x Np x Np m x m> k! (log xk+( log d K + n M x, where ξ = p,m ξ(σ m p since ξ is a characteristic function. This proves part. For part, from, the triangle inequality implies that ψ k (y, C, q, a d(c, q, ay ψ k (y, ξ d(c, q, ay = r k (y, C, q, a r k (y, ξ E. Hence, it suffices to estimate E (y, k Q E (x, k. y x (a,q= 9

10 We now show that the difference between iterations of the Mellin transform, even incorporating ramified primes is negligible. Lemma.. Assume that a < q and (a, q = with K Q(ζ q = Q. Let A > and a positive integer n be given. Then with E = E (x, k, n, A, Q = the following hold.. r k (x, C, q, a. (a,q= (k + n! (log x n(n+a+k+n+ Q (log Q + n M x x C (log Q (log log Q + G (log x A d(c, q, ax (log x A + (log x n(n+a y xe n(log x A r k+ (y, C, q, a. y x r k(y, C, q, a (log x n(n+a (a,q= r k+n (y, ξ + E. y xe n(log x A Proof. Let ψ k (x = ψ k(x, C, q, a. For the first part of the proof we will let d = d(c, q, a. Since ψ k is increasing, for any 0 < α, using the mean value thoerem one can show the following identity We can rewrite this as First, we will show x ψ α k(t dt e α x t ψ k(x α e α x x ψ k(t dt t. α (ψ k+(x ψ k+(e α x ψ k(x α (ψ k+(e α x ψ k+(x. r k(x αd x + α Substituting the remainder terms into ( we have rk+(y. y=x,xe ±α ( ( αψ k(x ψ k+(e α x ψ k+(x = d e α x + r k+(e α x d x r k+(x = (r k+(e α x r k+(x + d x(e α = (r k+(e α x r k+(x + αd x + O(α d x. The last equality can be verified using the Taylor series expansion of e α. We conclude that rk (x α (r k+ (eα x rk+ (x + O(αd x. Similarly, rk (x α (r k+ (x r k+ (eα x + O(αd x. Together these imply (3... Iteratively applying (3.. with α, α,..., α n we have r k(x α d x + α y xe α 3 α = α + αe α + αe α+α + + αe α+ +αn α(ne n α α = α n(n+ α 3 = α + α + + α n nα. r k+(y where This implies that rk(x αd x + α n(n+ y xe r k+n(y. Choosing α = (log x A completes the proof nα of. Summing implies that (a,q= y x r k(y (log x n(n+a (a,q= rk+n(y + d(c, q, ax y xe n(log x A (log x A. 0

11 Focusing on the error term, since d(c, q, a = C G (see [] 8.4 we conclude that φ(q and φ(q (log Q (log log Q Therefore (a,q= y x r k(y (log x n(n+a d(c, q, ax (log x A C x(log Q (log log Q G (log x A. (a,q= rk+n(y + C x(log Q (log log Q y xe n(log x A G (log x A. With Lemma. part this directly proves part as e (log x A is bounded when A >. Using the following identity for Dirichlet characters χ modulo q, { φ(q if n a (mod q χ(aχ(n = 0 otherwise, χ it follows that ψ k (x, ξ = φ(q χ(a ψ k (x, δ C χ. χ(q The trivial character contributes the term ψ k (x, δ C and r k (x, ξ = ψ k (x, ξ d(c, q, ax = φ(q χ(a ψ k (x, δ C χ + φ(q χ χ 0 ( ψk (x, δ C C G x. (.0.4 We bound the second term on the right by the effective Chebotarev density theorem in the form of Lemma.5 part. The first term is estimated in Lemma.5 part. The Chebotarev density theorem we require is a combination of a slight modification of Theorem.3 of [3] and Theorem of [] (as mentioned in [3]. We state these results below. Theorem.3 ([3] Theorem.3. Assume K Q. There is at most one zero of ζ K (s in the region defined by s = σ + it with (4 log d K σ, t (4 log d K. If such a zero exists, it is real and simple. Let β denote this zero if it exists. If x exp(0n K (log d K then r(x, δ C E 3 (x = C G xβ + x exp( cn K (log x, where the β term in E 3 is present only if such an exceptional zero exists, and the implied constant is effective and absolute. We will combine this with a result of Stark s to bound such a β. These two theorems give an effective bound on the error term. Theorem.4 ([] Theorem. There is an effective constant c 3 such that β < { (4n K!(log d K, (c 3 d n K K }. Lemma.5. With E 4 = E 4 (x, k = (log Q (log log Q (E (x, k + E 3 (x, (with E and E 3 as in Lemma. and Theorem.3 respectively the following hold.

12 . r k (x, δ C E (x, k + E 3 (x.. (a,q= y x r k(y, ξ characters χ modulo q. φ(q χ χ 0 y x ψ k (y, δ C χ + E 4, where the final summation is over Proof. First we prove part. From Theorem.3, r(x, δ C, K/M E 3. This implies that x r(t, δ C dt t = x ( ψ(t, δ C C G t dt t x ( C G tβ + t exp( cn K (log t dt t. Since the left hand side is r (x, δ C, upon evaluating the integral, and using L Hospital s rule on the quotient of exp( cn K (log t dt by x exp( cn K (log x for the last estimate, r (x, δ C C G xβ + x exp( cn K (log x = E3 (x. Repeating the inverse Mellin transform implies that for any k, r k (x, δ C E 3 (x. By Lemma. part, r k (x, δ C E + E 3. Now we prove part. Combining the estimate in part and (.0.4, Let S = (a,q= r k (x, ξ φ(q y x r k(y, ξ. Therefore S φ(q χ χ 0 χ(a ψ k (x, δ C χ + φ(q (E + E 3. ψ k (y, δ C χ + y x φ(q (E + E 3. χ χ 0 Since E and E 3 are independent of q, with the estimate φ(q (log Q (log log Q (see [] 8.4, we conclude that S φ(q χ χ 0 y x ψ k (y, δ C χ + (log Q (log log Q (E + E 3. Now it suffices to estimate φ(q ψ k (y, δ C χ. The next step in our refinement is to y x χ χ 0 replace the character sum by the sum over primitive, non-trivial characters. We indicate this restriction on a summation by the decoration. Lemma.6. With E 5 = E 5 (x, k, Q = k! Q log Q (log x k+, φ(q ψ k (y, δ C χ (log Q y x φ(q χ χ 0 χ ψ k (y, δ C χ + E 5. y x Proof. By definition (0.6., since χ(σ m p = χ(np m when considered as a Dirichlet character, ψ k (x, δ C χ = k! Np m x (log Np(log x Np m k δ C (σ m p χ(np m. If χ is a character modulo q induced by the character χ modulo q with q q and q q then ψ k (x, δ C χ ψ k (x, δ C χ = k! Np m x (log Np(log x Np m k δ C (σ m p (χ(np m χ (Np m.

13 The characters χ(np m = χ (Np m unless Np m divides q/q. Here χ(q = 0 (by the definition of nonprimitive character. Therefore ψ k (x, δ C χ ψ k (x, δ C χ = k! For a fixed modulus q, φ(q χ χ 0 y x ψ k (y, δ C χ = x (log Np(log Np m k δ C (σp m (χ(np m χ (Np m Np m q q k! (log xk( log q q k! (log xk+. χ χ 0 φ(q ψ ( k (y, δ C χ + O y x k! (log xk+ φ(q where χ is the primitive character of modulus q which induces χ and the decoration on the second summation indicates that the summation is over only those characters not primitive to the modulus q. Let S = φ(q ψ k (y, δ C χ. Then, it follows that y x χ χ 0 S = φ(q ψ ( k (y, δ C χ + O y x k! (log xk+ χ χ 0 χ(q χ. φ(q Rewriting this as a sum over the conductors, (instead over the moduli of the non-primitive characters, S = q Q χ (q y x ψ k (y, δ C χ q q φ(q + O ( k! (log xk+ q Q χ (q q q φ(q and since φ(q log Q φ(q q q S log Q q Q χ (q (see [7] Chapter 8 page 63, φ(q y x ψ k (y, δ C χ + k! (log xk+ log Q q Q χ(q φ(q. Since there are φ(q characters modulo q, the right hand term is of the order k! (log xk+ log Q q Q k! (log xk+ Q log Q E 5. We sum over dyadic intervals and use a standard estimate to replace the φ(q in the summation in Lemma.6 with a Q, before at last combining the estimates of this section. Lemma.7. φ(q χ ψ k (y, δ C χ (log Q (log log Q y x Q Q Q χ ψ k (y, δ C χ. y x Proof. Let S be the double summation on the left side of the asymptotic in the statement of the lemma. Break q Q into intervals of the form ( Q, Q], so S = n q ( Q,Q] φ(q y x χ ψ k (y, δ C χ 3

14 where n ranges from 0 to [log Q ] and Q = n+. Since there are log Q of these intervals The bound log log q φ(q q S log Q S log Q Q Q q ( Q,Q] φ(q y x (see [] 8.4 gives the new estimate Q Q q ( Q,Q] log log q q y x χ χ ψ k (y, δ C χ. ψ k (y, δ C χ. Since Q < q Q and Q Q, S (log Q (log log Q Q Q Q q ( Q,Q] y x which is clearly less than the summation over the expanded range q Q. χ ψ k (y, δ C χ We combine Lemmas.,.5 part,.6, and.7 to transform the average into a more usable form. Proposition.8. Let K/M be a Galois extension of number fields, with group G and let C be a conjugacy class in G. Let H be an abelian subgroup of G such that H C and let E be the fixed field of H. For any ɛ > 0 and Q x ɛ in order to prove that for any D > 0 (a,q= it suffices to prove either of the following y x r 0(y, C, q, a. For all Q Q and any F > 0, there is a k 0 such that Q χ x (log x D ψ k (y, δ C χ, K(ζ q /M y x x (log x F where the second summation is over all primitive non-trivial Dirichlet characters of modulus q.. For all Q Q and any F > 0, there is a k 0 such that Q ψ k (y,, K(ζ q /E y x x (log x F where the second summation is over irreducible non-trivial characters of Gal(K(ζ q /E such that A( Aq n E where A is the conductor of L(s, K/M. Moreover, one may assume that there are O(φ(q characters in the second summation. Proof. First, we prove part. Let E = E (x, k, 0, A,(as in Lemma. f = exp(k(log x A, A = k(k + A, and S = r 0(y, C, q, a. We will show that for every B > 0 y x (a,q= Q S (log x A +3 Q Q Q χ ψ k (y, δ C χ + y xf x (log x B. This proves the proposition as we can replace xf with x in the imum since f (e k! when x. It then suffices to choose F > D + A

15 We now estimate S. S (log x A (a,q= (log x A φ(q (log x A log Q φ(q χ y xf r k(y, ξ + E Lemma. part ψ k (y, δ C χ + E 4 (xf + E Lemma.5 part y xf (log x A (log Q (log log Q Q Q Q χ ψ k (y, δ C χ + E 5 (xf + E 4 (xf + E Lemma.6 y xf χ ψ k (y, δ C χ y xf + E 5 (xf + E 4 (xf + E. Lemma.7 Let A = ( ka + (k + A. Since Q x ɛ, f is bounded, and from Theorem.4, β < { (4n K!(log d K, (c 3 d n K K }, by direct analysis the error E 5 (xf + E 4 (xf + E x ɛ (log x A + x(log x A + x β (log x + x(log x exp( cn K (log xf x(log x D choosing A D. (Note that exp( cn K (log xf (log x B for any B > 0. This proves part. Now we prove part. By the equation before (0.7. ψ k (x, δ C χ, K(ζ q /M = C G η(g ψ k (x, η χ, K(ζ q /E where the summation is over irreducible characters η of G. Since η is irreducible, for χ a primitive character of modulus q, η χ is also irreducible. By 0.4, if = η χ then A( = A(η χ = A(ηA(χ A(ηAq n E, which is O(Aq n E with the implied constant depending only on K. Therefore, ψ k (x, δ C χ, K(ζ q /M C G η ψ k (x,, K(ζ q /E where the sum is over irreducible of Gal(K(ζ q /E such that A( Aq n E. Since this is a finite sum over the by part it suffices to bound the imum. Definition.. For K/M Galois and C a conjugacy class of G = Gal(K/M, let H be an abelian subgroup of G so that H C and let E be the fixed field of H. Letting H be the largest such subgroup, define d = n E. We will prove the above estimate for small values of Q, those at most (log x γ, in Proposition. and for large values of Q, those between (log x γ and min{x ɛ, x d ɛ }, in Proposition 3.. The Initial Range By Proposition.8 part we can reduce to the case of K/E abelian, and consider the functions ψ(x,, K(ζ q /E where is a non-trivial irreducible Hecke character of Gal(K(ζ q /E and ψ = ψ 0. We will assume that K/E is abelian for the remainder of the section and that A( Aq n E where A is the conductor of L(s, K/M, which we can do without loss of generality by Proposition.8 part. We let L = K(ζ q. We will suppress the notation of K(ζ q /E in writing ψ functions and associated L-functions unless needed for clarity. This section will be devoted to the proof of the following proposition. 5

16 Proposition. (Initial Range. Let K/E be an abelian Galois extension, and let D and γ be any positive constants. For all Q (log x γ we have Q ψ(y, x y x (log x D where the inner summation is over all irreducible, non-trivial characters of K(ζ q /E such that A( Aq d. Proposition. follows from the following proposition. Proposition.. For K/E abelian, there is a positive constant c = c(k, E so that for any non-trivial irreducible character of Gal(K(ζ q /E ψ(x, (log A(x exp( c(log x. Before proving Proposition., we show that it implies Proposition.. Proof of Proposition.. Let D > 0 be fixed. By Proposition., there is a c > 0 so that ψ(x, (log Aq n E x exp( c(log x. Assume that Q (log x γ, then for some positive N, A( Aq n E (log x N. We have, Q Therefore it suffices to prove Proposition.. χ ψ(y, φ(q(log A(x exp( c(log x y x Q Q(log x N x exp( c(log x (log x γ+n x exp( c(log x x(log x D. The remainder of the section will be devoted to the proof of Proposition.. In. we prove preliminary results about zero free regions and Siegel zeros. In the following section,., we provide the framework to prove Proposition. by connecting ψ(x, with the zeros of L(s, using the method of contour integration. Finally, in.3, we prove Proposition... Zero Free Regions We wish to bound ψ(x,, K(ζ q /E, With (0.6. we will do so by showing the existence of a region with at most one zero for L(s, and provide bounds for this possible zero. This closely follows the bounds in [7] and []. In what follows, we use the notation s = σ + it. Definition.. For K/E abelian and parameter t, let L(t = log A( + n E log( t +. Since A( Aq d, L(t log(aqd + n E log( t +. We now compile some facts, mainly from [3] (pages and 0.7. We will assume that is an irreducible non-trivial Artin character. First, we prove the following lemma. Lemma.3. Let s = σ + it. For a constant C with < σ < C and non-trivial irreducible there is a positive absolute constant C such that Re L L (s, < C L(t ρ ( Re s ρ 6

17 Proof. By (0.7.3 Re L L (s, = ρ ( s ρ + + Re( s ρ γ log A( (s,. γ Stirling s formula, for σ > (see [3] Lemma 5.3 implies that Re γ γ (s, n E log( t + which proves the result. The following proposition gives a region with at most one zero, which must be real and simple. Proposition.8 gives more precise bounds for such zero. Propositions.4 and.8 will be used in the proof of Proposition.. Proposition.4. There is an absolute positive constant C such that L(s, has at most one zero σ + it in the region C L(t σ. If such a zero exists, then it is real and simple and is a character of order or. (That is, is a quadratic Hecke character, since is assumed not to be the trivial character. We will call this zero β if it exists, and refer to β as a Siegel zero of L(s,. By the functional equation (0.4.3 this implies that there is at most one zero of L(s, in the region 0 σ C/L(t as well. This zero, β, must be simple as well. If the constant C is large enough that C/L(0, then Proposition.4 implies that there can be no zeros in the range C/L(t σ. If D < C the region D/L(t is contained in the region C/L(t. Therefore, we can assume that C < L(0 = ( log A( + n E log. Proof. We will use the notation ρ = β + iγ for zeros and s = σ + it for a complex parameter. First, we will collect some useful facts. Recall that (if we extend multiplicatively from (0.3. L L (s, = p (log Np(σp m Np ms = a m= Λ(N a(a Na s = a Λ(Na(a(Na σ it(log Na e where the first sum is over prime ideals, p of E, N = N E/Q and the last sum is over ideals a of E. We can rewrite the trigonometric inequality (which holds for θ R cos θ + cos θ = ( + cos θ 0 with cos θ = Re (ae it log Na. With 0 denoting the principal character of conductor f( we have (See [7] 4 page 88, [] page 95. [ 3Re L L (σ, 0 4Re L L ] (σ + it, ReL L (σ + it, 0. (.. By Lemma 3 of [], for a number field F with r real and r complex places and s = σ + it with σ > Re ζ F (s < Re ζ F s + Re s + ( log d F r π n F Also, from Lemma of [] (with f(s = s(s ζ F (s ρ + r ReΓ Γ ( s + r Re Γ (s. (.. Γ s ρ = s + ( log d F + s n F log π + r Γ ( s ( Γ + r Γ Γ (s log + ζ F (s. ζ F If s = σ with < σ < then all terms after log d F are negative, and the left hand side is positive. Therefore, ζ F (σ ζ F σ + log d F. (..3 7

18 The proof will be based on estimates for the terms in (.., heavily relying on (.. and (..3. We will now estimate the first two terms of (... Writing out the summation for σ > by direct comparison of terms, L L (σ, 0 ζ E (σ. ζ E By this and (..3, L L (σ, 0 < σ + C L(0 (..4 for some positive constant C. For any s = σ + it and zero ρ = β + iγ in the range 0 < β < < σ we have ( Re = σ β s ρ s ρ 0. Due to this positivity, Lemma.3 implies that for such a ρ, there is a positive absolute constant C so that ( Re L L (s, < C L(t Re. (..5 s ρ Choose t = γ, so that Re /(s ρ = /(σ β. Combining this and (..4 into inequality (.. there is a positive C 3 so that 4 σ β < C 3L(t + 3 σ ReL L (σ + it,. (..6 It remains to estimate this remaining L /L term. Many of the auxiliary results we will use require to be irreducible. If is induced by, the difference between the corresponding L /L factors is small. This difference is Re L L (s, L L (s, = Re p f( m= log Nf( log A( (log Np(σp m p f( m= p f( Np ms Np mσ ( + Np σ log A( p d K d K/E log A( L(t. Therefore, replacing by in (..6 only changes the positive constant in the equality. The argument now breaks into two cases, depending on whether or not is a principal character. First, assume that is not a principal character. We may assume that is irreducible. By (..5 Re L L (s, < C 3 L(t. Inequality (..6 now becomes, for < σ <, and a positive C 4, Writing σ = + D/L(t, this is β < + D L(t 4 σ β < C 4L(t + 3 σ. 4D (3 + DC 4 L(t = L(t ( D(C4 D. C 4 D + 3 Therefore, if D < C 4 we conclude that there is an absolute constant C > 0 such that β < C L(t. 8

19 This concludes the proof in the case where is not principal, as this is not in the prescribed range. We now consider the case where is a principal character. As we can reduce to considering irreducible characters, we may assume that is the trivial character. Inequality (..4 with the fact that < σ < implies that there is a positive constant D so that ( Re L L (σ + it, < Re Substituting this into (..6, there is a positive D so that σ + it + D L(0. 4 ( σ β < Re + D L(t + 3 σ + it σ. With σ = + D/L(t, if t = γ > D/L(t we conclude that and therefore 4 σ β < 6 5 β < L(t D + D L(t, D ( 4 5D D. L(t 6 + 5D D For a sufficiently small D depending on D, the value D(4 5D D/(6 + 5D D is bounded below by a constant. Choosing t = γ > D/ log A( satisfies the bound γ > D/L(t. We have shown that there are positive absolute constants C, and D such that if is a non-trivial irreducible character, then any zero β + iγ of L(s, with γ D / log A( satisfies β < C/L(t. It now remains to assume that is non-principal and D t = γ < log A( for a small constant D. We wish to show that there is a positive constant F such that there is at most one zero β + iγ with β > F/L(t. It suffices to show that there is at most one zero with β > F/ log A( since log A( < L(t. Such a zero must be real, as otherwise there would be a pair of conjugate zeros. By Lemma.3, for σ > there is a positive constant F such that L L (σ, < F log A( ρ σ ρ where the last summation is real as the zeros occur in complex conjugate pairs. We have, L L (σ, = a (σ a Λ(NaNa σ a Λ(NaNa σ ζ E ζ E (σ. Combining this with (..3, L (σ, L σ log d E. Therefore, there is a positive constant F so that σ F log A( ρ σ ρ. (..7 Now we consider a few cases, depending on the type of possible zeros. Specifically, we will first assume that there is a complex zero (and hence its conjugate is also a zero and then assume that there are at least two (possibly identical real zeros. In each case, we will conclude that there is a positive constant F such that the real parts of the zeros are less than F/L(t. Therefore, there is a positive constant C such that there can be at most one zero with real part between C/L(t and. 9

20 Assume that there are zeros β ± iγ with γ 0. Inequality (..7 implies that With σ = + D/ log A( we have Therefore there is a constant F 3 > 0 so that σ F (σ β log A( (σ β + γ. D γ < log A( = (σ < (σ β. σ < F 8 3 log A( 5(σ β. This is equivalent to β < σ 8 σ 5 + F 3 log A((σ = D ( 3 0DF3 log A( + DF 3 we conclude that there is a positive constant F so that β < F/L(t if 3 0DF 3 is positive. Choosing D < 3/0F 3 suffices. It remains to consider the case where there are at least two real zeros, β β. Inequality (..7 implies that σ F log A(. σ β With σ = + D/ log A( as before β < D ( F D log A( + F D and if D < /F we conclude that there is a positive constant F so that β < F/L(t. The following lemma is a refinement of the Brauer-Siegel theorem, which gives a bound for β. We require a bound for β that has minimal dependence on the field L = K(ζ q. The following facilitates such a bound, with only a logarithm term depending on q. This is a refinement of Lemma 0 in []. Lemma.5. Let Q = F 0 F F F t = F be a sequence of number fields such that for i t, F i /F i is Galois. If there is a real β in the range 4(n F! log d F β < such that ζ F (β = 0, then there is a quadratic field S F with ζ S (β = 0. This proof will make use of Theorem 3, Lemma 3, and Lemma 8 of [], which we summarize in the following theorem. Theorem.6 (Stark. a If K 4 /K is a Galois extension of number fields and α is a simple zero of ζ K4 then there is a field K cyclic over K and contained in K 4 such that for any field K 3 between K and K 4, ζ K3 (α = 0 if and only if K K 3. If α is real then K = K or is quadratic over K. b If K is a number field then ζ K has at most one zero σ + it in the region σ (4 log d K, t (4 log d K. If such a zero exists, it is real and simple. 0

21 c Let K be a number field. If there is a real α in the range (4n K! log d K α < such that ζ K (α = 0, then there is a quadratic field S contained in K so that ζ S (α = 0. Proof of Lemma.5. We will first prove the following claim. Claim.7. Let A B C with C/B and B/A Galois. Assume that there is a real β such that (4 log d C β < and ζ C (β = 0. Either ζ A (β = 0 or there is a quadratic extension A of A with A C such that ζ A (β = 0. First, we show that Claim.7 implies Lemma.5. Since (4(n! log d F β and 4 log d F 4(n! log d F, (4 log d F β. By Claim.7 ζ Ft (β = 0 or there is a quadratic extension, Q t of F t such that ζ Qt (β = 0. Since both 4 log d Qt and 4 log d Ft are less than 4 log d F we can proceed inductively and conclude that ζ F (β = 0 or there is a quadratic extension Q of F such that ζ Q (β = 0. Let S be this field. Therefore, n S n. Discriminant relations imply that d S d F so that 4n S! log d S 4(n! log d S 4(n! log d F. Therefore (4n S! log d S (4(n! log d F β and by Theorem.6 c there is a quadratic field S F such that ζ s (β = 0. Proof of Claim.7. By Theorem.6 b, β is simple. By Theorem.6 a either ζ B (β = 0 or there is a quadratic extension B of B contained in C such that ζ B (β = 0. If ζ B (β = 0 we can apply Theorem.6 b and a again and conclude that either ζ A (β = 0 or that there is a quadratic extension A of A contained in B C such that ζ A (β = 0. Otherwise, assume that ζ B (β = 0 and ζ B (β 0. If B is Galois over A we can reapply Theorem.6 b and a to again conclude that either ζ A (β = 0 or there is a quadratic extension A of A contained in B C such that ζ A (β = 0. Therefore, it suffices to show that B is Galois over A. We will assume that B is not Galois over A. By Theorem.6 a, for any field E between B and C, ζ E (β = 0. Let B be the conjugate of B over A, so [B : B] = as B/A is Galois. Both B and B are Galois over B, so the composite B = B B is Galois and degree 4 over B. Since B is the A conjugate of B, ζ B (β = ζ B (β = 0. By the Aramata Brauer theorem [, 5] ζ B (β = 0. In fact (see [] Theorem 3, or Lemma the multiplicity of the zero is at least two. Since B C, d B d C and (4 log d B (4 log d C β, which contradicts Theorem.6 b. We have L(β,, K(ζ q /E = 0. By Proposition.4, is a real quadratic character. Since is a principal character, must evaluate to ± on all elements g G = Gal(K(ζ q /E. Let H < G consisting of all g G such that (g =, so G/H = Z/Z. Let N K(ζ q be the fixed field of H. By the Galois correspondence, [N : E] =. By construction, L(β,, N/E = 0. Since [N : E] =, if σ p is the identity, (σ p = and otherwise, (σ p =. The splitting type of the non-ramified prime ideal p in E is easily determined by the Frobenius elements σ p. Let q be a prime ideal of N lying over p. The unramified prime ideal p splits exactly when N N/Q q = N E/Q p, and is inert when N N/Q q = N E/Q p. Therefore, since σ p is defined as σ p (x = x N E/Qp (mod q, we see that σ p is the identity exactly when p is split in N and σ p is not the identity when p is inert. The unramified Euler product for L(s,, N/E (indicated with a subscript u is L u (s,, N/E = ( (σp N E/Q p s = ( N E/Q p s ( + N E/Q p s. p The unramified Euler product for ζ N (s is (ζ N u (s = q ( N N/Q q s = q p split p split p split ( N N/Q q s q p inert p inert p inert ( N N/Q q s = ( N E/Q p s ( N E/Q p s = L u (s,, N/E p inert ( N E/Q p s.

22 Therefore, L(s,, N/E divides ζ N (s and β is a zero of ζ N as well. Similarly, by Aramata Brauer [, 5] one can see that ζ N (s divides ζ KN (s, so β is a zero of ζ KN as well. We may also view as a (real quadratic Hecke character of E. The next proposition is an analog of Siegel s theorem giving a bound for the possible real Siegel zero in Proposition.4, and is an adaptation of Stark s [] method. The purpose of this proposition is to get a bound for β which only depends on K and E. Proposition.8. Let ɛ > 0 be given, and let be a non-trivial irreducible character of K(ζ q /E such that A( Aq n E. There is a positive constant C = C(ɛ so that if β is a Siegel zero of L(s,, K(ζ q /E then { } C β (d, E Aɛ 4(n E!n K log(d E A. Proof. We will assume that (4(n E!n K log(d E A β < and prove that β C(d E A ɛ. By the conductor discriminant formula (0.4. d KN d n K N. Moreover, since [N : E] =, d N/E = A, the conductor of ζ E, (since the only irreducible character is the principal character and d N = d EA. We conclude that d KN (d E An K, removing the dependence on N. Therefore, if β (4(n E!n K log(d E A then β (4(n E! log d KN. Applying Lemma.5 to the fields Q E N KN, we conclude that there is a quadratic subfield S of KN with ζ S (β = 0. By the classical Siegel theorem (cf. [7] there is a constant C = C(ɛ > 0 so that β Cd ɛ S. (The conductor discriminant formula implies that d S = d S/Q = f, the Artin conductor, since S is quadratic. Since d n K/ S d KN we conclude that β Cd ɛ/n K KN C(d E A ɛ using the above discriminant relations. We will need to estimate the contribution from the zeros close to the real line, and so require the following corollary to Proposition.4 and Proposition.8. Corollary.9. Let be a non-trivial irreducible character of K(ζ q /E such that A( Aq n E. If q (log x γ (log log x ρ ρ < where the sum is over all non-trivial zeros of L(s,. Proof. By [3] Lemma 5.4 the number of zeros in the region ρ < is O(log A( + n E log. Since A( Aq n E and q (log x γ, the number of zeros is O(log log x, where the implied constant depends only on K, E and γ. By Proposition.4 and the functional equation for L(s,, there is at most one zero, β in the region defined by Re(s C/L(t and by Proposition.8, / β is absolutely bounded. Therefore the summation is bounded by the number of zeros in the region ρ < multiplied by the imum of ρ over the zeros in the region. It suffices to show that ρ log log x for ρ and Re(ρ > C/L(Im(ρ. For such a ρ, ρ > s where s = σ + it is on the curve C/L(t when 0 t. The modulus s {t, C/L(t} where 0 t. As L(t = log A( + n E log( t +, we conclude that when 0 t, there is a positive constant D depending on K, E and γ so that L(t D log log x. Therefore, ρ > s C/L(t C/(D log log x and so ρ log log x.. Contour Integration We closely follow the proof of Theorem 7. in [3]. We sketch the ideas here as we need a slight modification of their work. Note that we are assuming that is an irreducible non-trivial character of Gal(K(ζ q /E and will leave off any terms which are only present when is the trivial character.

23 We will first obtain estimates by contour integration. Recall that from (0.3. ψ(x, = L (s, xs πi L s ds. This differs from a truncated inverse Mellin transform I(T = πi ( σ0+it σ 0 it L (s, xs L s ds for σ 0 > by a sufficiently small error term, E for T x. We will let σ 0 = + (log x. This truncated integral can be evaluated by contour integration and Cauchy s theorem. Specifically, we evaluate the contour integral I(B = πi B L (s, xs L s ds where B is the positively oriented box with vertices at σ 0 ± it, and U ± it where U = j + for some non-negative integer j. (We eventually let U. As in [3], we estimate I(T I(B = E. By Cauchy s theorem I(B equals the sum of the residues of the integrand at poles inside B. Therefore, the main terms of ψ(x, correspond to the residues of L /L(s, inside B. Since is not the trivial character, L(s, is analytic in B. As such, the main terms come from the zeros of L in B. Let S denote the contribution of the non-trivial zeros and S the contribution of the trivial zeros. We have ψ(x, = I(T + E = I(B + E + E = S + S + E + E. The estimate of E is very similar to the results in 3 of [3]. Therefore we only sketch the proof of this estimate. We use the following, which is Lemma 3. in [3]. Lemma.0. Let y > 0, σ > 0, and T > 0 and let I = πi σ+it σ it y s ds. Then s. I y σ min{, T log y } if y >.. I σt if y =. 3. I y σ min{, T log y } if y <. The estimates of E, E, S, and S are summarized in the following, which is the main result of this section. Proposition.. Let be a non-trivial irreducible character of K(ζ q /E such that A( Aq n E. If q (log x γ then for T x ψ(x, ρ x ρ ρ ρ ρ n E x(log x T ρ < where the summations are over the non-trivial zeros, ρ, of L(s,, and with G = Gal(K(ζ q /E, = g G (g. Proof. Using the definition of ψ from (0.6. and writing out the summation for L /L as in (0.3. we can write the error E = I(T ψ(x, as σ0+it πi σ 0 it ( (log Np(σp m (Np p,m ms xs s ds Np m x (log Np(σ m p. 3