AN EXPLICIT UPPER BOUND FOR THE LEAST PRIME IDEAL IN THE CHEBOTAREV DENSITY

ANNALES DE

L’INSTITUT FOURIER

LesAnnales de l’institut Fouriersont membres du Centre Mersenne pour l’édition scientifique ouverte

Jeoung-Hwan Ahn & Soun-Hi Kwon

An explicit upper bound for the least prime ideal in the Chebotarev density theorem

Tome 69, n^o3 (2019), p. 1411-1458.

<http://aif.centre-mersenne.org/item/AIF_2019__69_3_1411_0>

© Association des Annales de l’institut Fourier, 2019, Certains droits réservés.

Cet article est mis à disposition selon les termes de la licence Creative Commons attribution – pas de modification 3.0 France.

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AN EXPLICIT UPPER BOUND FOR THE LEAST PRIME IDEAL IN THE CHEBOTAREV DENSITY

THEOREM

by Jeoung-Hwan AHN & Soun-Hi KWON (*)

Abstract. — Lagarias, Montgomery, and Odlyzko proved that there exists an effectively computable absolute constantA1such that for every finite extensionK ofQ, every finite Galois extensionLofKwith Galois groupGand every conjugacy classCofG, there exists a prime idealpofKwhich is unramified inL, for which

L/K p

=C, for whichN_K/Qpis a rational prime, and which satisfiesN_K/Qp6 2dLA₁. In this paper we show without any restriction thatN_K/Qp6dL12577 if L6=Q, using the approach developed by Lagarias, Montgomery, and Odlyzko.

Résumé. — Lagarias, Montgomery, et Odlyzko ont démontré qu’il existe une constante absolue effectivement calculableA1telle que pour chaque extension finie K de Q, chaque extension galoisienne finie L de K à groupe de Galois G, et chaque classe de conjugaison C deG, il existe un idéal premier pdeK qui est nonramifié dans L, pour lequelL/K

p

=C, pour lequelN_K/Qpest un nombre premier rationel, et qui satisfaitN_K/Qp62dLA₁. Dans cet article nous démontrons sans aucune restriction que NK/Qp6dL12577 siL6=Q, en suivant la méthode developpée par Lagarias, Montgomery, et Odlyzko.

1. Introduction

LetKbe a finite algebraic extension ofQ, andLa finite Galois extension of K with Galois group G. Let dL and dK denote the absolute values of discriminants ofLandK, respectively, and letn_L= [L:Q],n_K = [K:Q].

To each prime ideal p of K unramified in L there corresponds a certain

Keywords: The Chebotarev density theorem, Dedekind zeta functions, the Deuring–

Heilbronn phenomenon.

2010Mathematics Subject Classification:11R44, 11R42, 11M41, 11R45.

(*) The first author was supported by Basic Science Research Program through the National Research Foundation of Korea funded by the Ministry of Education(NRF- 2013R1A1A2061231) and a Korea University Grant. The second author was supported by NRF-2013R1A1A2007418.

conjugacy classC of Gconsisting of the set of Frobenius automorphisms attached to the prime idealsPofLwhich lie overp. Denote this conjugacy class by the Artin symbolh_L/K

p

i

. For a conjugacy classC ofGlet π_C(x) =|{p |pa prime ideal ofK, unramified inL,

L/K p

=C, andN_K/Qp6x}|.

The Chebotarev density theorem states that π_C(x)∼|C|

|G|Li(x)

as x → ∞. (See [15], [53], [28], [39], and [50]. See also [47] for some ex- tensions of Chebotarev’s theorem and applications.) The error term of this theorem was estimated in [24], [41], and [59]. Lagarias, Montgomery, and Odlyzko estimated upper bound for the least prime idealpwithh_L/K

p

i

=C under the Generalized Riemann Hypothesis (GRH), and unconditionally, in [24] and [23], respectively.

Theorem A(Lagarias and Odlyzko [24]). — There exists an effectively computable positive absolute constantA0 such that if the GRH holds for Dedekind zeta function of L6= Q, then for every conjugacy class C of G there exists an unramified prime idealpin Ksuch thath_L/K

p

i

=C and N_K/Qp6A₀(logdL)².

Oesterlé ([41]) has stated that if GRH holds, then one may haveA0= 70.

Bach and Sorenson ([4]) has improved this result in two ways: If GRH holds, then for any classC ofGthere is a primepinKof degree 1 overQ withh_L/K

p

i

=C andN_K/Qp 6(4 logd_L+ 2.5n_L+ 5)². (See also [3], [38], and [22].) Let

P(C) =

p

pa prime ideal ofK, unramified inL, of degree one overQand

L/K p

=C

. Theorem B (Lagarias, Montgomery, and Odlyzko [23]). — There is an absolute, effectively computable constantA1 such that for every finite extensionKofQ, every finite Galois extensionLofK, and every conjugacy classCofG, there exists a prime pinP(C)which satisfies

NK/Qp62dLA₁

.

See also [57]. WhenK=QandL=Q(e^2πi/q), the conjucacy classes ofG correspond to the residues classes moduloqand Theorem B gives an upper bound for the least prime in an arithmetic progression ([24] and [23]). In this case Theorem B is weaker than Linnik’s theorem ([29], [30], [5]). For the least prime in an arithmetic progression, see for example [7], [8], [13], [14], [17], [18], [42], [43], [55], [56], and [61]. IfK=Q, L=Q(√

D), andρ is the non identity inGal(L/Q), Theorem B gives an upper bound for the least quadratic nonresidue moduleD. For this case no upper bound better than Theorem B is known ([54], [6], [24], [23], [2], [25], [26]). In this paper we compute the constantA₁.

Theorem 1.1. — For every finite extensionK ofQ, every finite Galois extensionL(6=Q)ofK with Galois groupG, and every conjugacy classC ofG, there exists a prime idealp inP(C)which satisfies

N_K/Qp6dLA₁

withA1= 12577.

To compute the constantA₁we follow the method developed by [23]. In particular, we express zero-free regions for Dedekind zeta functions, density of zeros of Dedekind zeta functions, and Deuring–Heilbronn phenomenon with explicit constants in Sections 5-7 below. Zaman showed in [63] that N_K/Qp dL40 for sufficiently large dL. See also [51]. Winckler proved A₁= 27175010 without any restriction in [60].

2. Outline of Lagarias–Montgomery–Odlyzko’s method Let<zand=zdenote the real part and imaginary one ofz∈C, respec- tively. We review the procedure for the proof of Theorem B in [23]. Let g∈C and

FC(s) =−|C|

|G|

X

ψ

ψ(g)L⁰

L(s, ψ, L/K),

whereψruns over the irreducible characters of GandL(s, ψ, L/K) is the Artin L-function attached toψ. The main parts of [23] consist of estimates of inverse Mellin transforms

1 2πi

Z 2+i∞

2−i∞

FC(s)k(s) ds

wherek(s) is a kernel function. The main steps of the proof of Theorem B in [23] are as follows:

(i) From the orthogonality relations for the charactersψit follows that for<s >1

FC(s) =X

p

∞

X

m=1

θ(p^m)(logNK/Qp)(NK/Qp)^−ms

where for prime idealspofK unramified inL

θ(p^m) =







1 ifh_L/K

p

im

=C, 0 otherwise,

and|θ(p^m)|61 ifp ramifies inL. So we can separate the p^m with h_L/K

p

im

=Cfrom the others. (See [24, Section 3].)

(ii) Using a method due to Deuring ([10] and [35])FC(s) can be writ- ten as a linear combination of logarithmic derivatives of Hecke L- functions instead of Artin L-functions. LetH =< g >be the cyclic subgroup generated byg,E the fixed field ofH. Then

(2.1) FC(s) =−|C|

|G|

X

χ

χ(g)L⁰

L(s, χ, E),

whereχruns over the irreducible characters ofH, andL(s, χ, E) is a Hecke L-function attached to fieldE withχ(p) =χh_L/E

p

i for all prime idealspofE unramified in L. (See [24, Section 4].) So, all the singularities ofF_C(s) appear at the zeros and the pole ofζ_L(s).

(iii) The kernel functions which weight prime ideals of small norm very heavily are used. Set

k0(s;x, y) =

y^s−1−x^s−1 s−1

²

fory > x >1, k₁(s) =k₀(s;x, x²) forx>2,

and

k₂(s) =k₂(s;x) =x^s2+s forx>2.

In the case that ζL(s) has a real zero very close to 1 we use the kernel k₂(s). Otherwise we use the kernel k₁(s). The use of the kernel functions is the main innovation of [23].

(iv) For u > 0 we denote by bk(u) the inverse Mellin transform of the kernel functionk(s). Then, for<s >1,

I= 1 2πi

Z 2+i∞

2−i∞

FC(s)k(s) ds

=X

p

∞

X

m=1

θ(p^m)(logN_K/Qp)bk(N_K/Qp^m),

where the outer sum is over all prime ideals ofK. An upper bound E(logdL) for

(2.2)

I− X

p∈P(C)

(logN_K/Qp)bk(N_K/Qp)

6E(logdL) was estimated in [23, (3.15) and (3.16)].

(v) The integralIis evaluated by contour integration:

I= |C|

|G|k(1)−|C|

|G|

X

χ

χ(g)X

ρ_χ

k(ρχ) +O

|C|

|G|n_Lk(0) +|C|

|G|k

−1 2

logd_L

, where ρ_χ runs over the zeros of L(s, χ, E) in the critical strip.

(See [23, Section 3].) So we get (2.3) |G|

|C|I>k(1)−X

ρ

|k(ρ)| −c6

nLk(0) +k

−1 2

logdL

, whereρ runs over the zeros of ζ_L(s) in the critical strip and c₆ is some constant. Note thatζL(s) =Q

χL(s, χ, E), whereχruns over the irreducible characters ofH =Gal(L/E). From (2.2) and (2.3) it follows that

(2.4) X

p∈P(C)

(logNK/Qp)bk(NK/Qp)> |C|

|G|k(1)−|C|

|G|

X

ρ

|k(ρ)|

−c₆|C|

|G|

n_Lk(0) +k

−1 2

logd_L

− E(logd_L).

(vi) The sum

k(1)−X

ρ

|k(ρ)|

is estimated from below. To do this we need to know the location and the density of the zeros of ζ_L(s). If the possible exceptional zero exists, say β0, then k(β0) is large. The term k(1)− |k(β0)|

must be controlled compared toP

ρ6=β₀|k(ρ)|. We need an enlarged zero-free region which makes possibleP

ρ6=β0|k(ρ)|to be small. The Deuring–Heilbronn phenomenon guarantees that the other zeros of ζL(s) can not be very close to 1.

(vii) We choosexof the kernelk(s) in terms ofdL so that the right side of (2.4) is positive.

Then Theorem B follows. In the remaining sections of this paper we will make explicit numerically the constants intervening in the zero free regions, the density of zeros, and Deuring–Heilbronn phenomenon of ζL(s), and ultimatelyA₁.

3. Prime ideals in P(C) In this section we will estimate from above

I− X

p∈P(C)

(logN_K/Qp)bk(N_K/Qp) .

We will treat carefully their bounds in [23, Section 3]. We begin by recalling the inverse Mellin transform of the kernel functions. They can be easily computed. Forx>2 andu >0 we have

kb1(u) = 1 2πi

Z a+i∞

a−i∞

x^2(s−1)−x^s−1 s−1

² u^−sds

=











u⁻¹log^x_u⁴ if x³6u6x⁴, u⁻¹log_x^u₂ if x²6u6x³,

0 otherwise,

and

kb2(u) = 1 2πi

Z a+i∞

a−i∞

x^s2+su^−sds= (4πlogx)⁻¹²exp (

− log^u_x2

4 logx )

, wherea >−¹₂.

Lemma 3.1. — Let PR

denote summation over the prime idealsp of Kthat ramify inL. Forx>2 we have then

(1) X^R

∞

X

m=1

θ(p^m)(logNK/Qp)kb1(NK/Qp^m)6 2 logx x² logdL;

(2) X^R X

m>1 N_K/Qp^m6x⁵

θ(p^m)(logNK/Qp)kb2(NK/Qp^m)

6 5

2√

πlog 3(logx)¹²logdL. Proof.

(1). — Let p be a prime ideal of K that is ramified in L. Note that N_K/Qp>2 andPR

logN_K/Qp6logd_L. We have X^R

∞

X

m=1

θ(p^m)(logNK/Qp)kb1(NK/Qp^m) 6logxX^R

logN_K/Qp X

m>1 N_K/Qp^m>x²

(N_K/Qp^m)⁻¹

6logxX^RlogN_K/Qp NK/Qp^mp

1 1−NK/Qp⁻¹

62 logx x² logdL, wheremp=l _log(x2)

logN_K/Qp

m .

(2). — Let N_R be the number prime ideals of K that are ramified in L/K. Note thatd_L>3^NR. (See [46, Chapters III and IV]).) We have

X^R X

m>1 N_K/Qp^m6x⁵

θ(p^m)(logN_K/Qp)kb2(N_K/Qp^m)

6(4πlogx)⁻¹²X^R

logN_K/Qp X

m>1 N_K/Qp^m6x⁵

1

6(4πlogx)⁻¹²X^R 5 logx

6 5

2√

πlog 3(logx)¹²logd_L. Lemma 3.2.

(1) (Rosser and Schoenfeld[44]) Forx >1, π(x)< α₀ x

logx

with α₀ = 1.25506, where π(x) is the number of primes p with p6x.

(2) Forx >1,

S(x)6 2α₀ log 2

√x,

where S(x) is the number of prime powers p^h with h > 2 and p^h6x.

(3) Forx>101

X

p prime p^h>x²,h>2

p^−h64.02α0

xlogx. Proof.

(1). — See [44, Corollary 1].

(2). — We have

S(x)6π √

xlogx log 2 6 2α0

log 2

√x by (1).

(3). — We have

X

p prime p^h>x²,h>2

p^−h= X

p prime

p^−hp 1−p⁻¹, whereh_p= maxl_log(x2)

logp

m,2

for each primep. We observe that X

p6x

p^−hp 1−p⁻¹ 6 2

x²π(x)6 2α₀ xlogx. Forx>101

X

p>x

p^−hp

1−p⁻¹ 6X

p>x

p⁻²

1−p⁻¹ 6 x x−1

X

p>x

p⁻²61.01X

p>x

p⁻². By using the integration by parts and (1) we estimate P

p>xp⁻² from above. Namely,

X

p>x

p⁻²6 Z ∞

x

1

t²dπ(t)6 Z ∞

x

2π(t) t³ dt 6

Z ∞ x

2α0

t²logtdt6 2α0

logx Z ∞

x

dt

t² = 2α0

xlogx. Hence,

X

pprime

p^−hp

1−p⁻¹ 64.02α₀ xlogx,

which yields (3).

Lemma 3.3. — Fory6∞, letPP

y denote summation over those(p, m) for whichN_K/Qp^mis not a rational prime andN_K/Qp^m6y. Then

(1) forx>101 X^P

∞θ(p^m)(logN_K/Qp)kb1(N_K/Qp^m)616.08α0nK

logx x ; (2) forx>10¹⁰

X^P

x⁵θ(p^m)(logNK/Qp)kb2(NK/Qp^m)6α1nKx³⁴(logx)³² with

α1= α0

3√ πlog 2

15

10⁴⁷² log 10+ 7 + 37 10⁵²

= 2.4234· · ·. Proof.

(1). — Since for a positive integerqthere are at mostnK distinct prime power idealsp^m withN_K/Qp^m=q, it follows that

X^P

∞θ(p^m)(logN_K/Qp)kb₁(N_K/Qp^m)6logxX^P

∞(logN_K/Qp)(N_K/Qp^m)⁻¹ 64(logx)²nK

X

pprime x²6p^h6x⁴,h>2

p^−h.

Hence, by Lemma 3.2 (3) we obtain (1).

(2). — We have X^P

x⁵θ(p^m)(logN_K/Qp)kb₂(N_K/Qp^m)6n_K X

pprime p²6p^h6x⁵

(logp^h)kb₂(p^h)

6nK

Z x⁵ 4

(logu)kb2(u)dS(u), whereS(u) is as Lemma 3.2 (2). According to Lemma 3.2 (2), we have

S(u)6 2α₀ log 2

√u.

Hence, Z x⁵

4

(logu)kb2(u) dS(u) 6(logx⁵)kb2(x⁵)S(x⁵) +

Z x⁵ 4

kb2(u)

logulog^u_x 2 logx −1

S(u)du u 6 5α₀

√πlog 2x⁻³²(logx)¹² + Z 4 logx

log_x⁴

kb₂(xe^t)

(t+ logx)t 2 logx

S(xe^t) dt

6 α₀

3√ πlog 2

15

x⁹⁴logx+ 7 + 37 x¹⁴

x³⁴(logx)³². Lemma 3.4. — Forx>2, we have

X

p

X

m>1 N_K/Qp^m>x⁵

θ(p^m)(logN_K/Qp)kb2(N_K/Qp^m)6α2nKx(logx)¹²

withα2= ^√5_π. Proof. — We have X

p

X

m>1 N_K/Qp^m>x⁵

θ(p^m)(logNK/Qp)kb2(NK/Qp^m)6nK

X

pprime p^h>x⁵

(logp^h)kb2(p^h)

6n_K Z ∞

x⁵

(logu)kb₂(u) dT(u), whereT(u) is the number of prime powersp^hwithh>1 andp^h6u. Since T(u)6uforu >0, we have

Z ∞ x⁵

(logu)kb₂(u) dT(u)6 Z ∞

x⁵

kb₂(u)

logulog^u_x 2 logx −1

T(u)du u 6

Z ∞ 4 logx

kb₂(xe^t)

(t+ logx)t 2 logx −1

T(xe^t) dt

6α₂x(logx)¹².

From Lemmas 3.1, 3.3, and 3.4 we deduce an upper bound for

I_j− X

p∈P(C)

(logN_K/Qp)kb_j(N_K/Qp) forj = 1,2 as follows.

Proposition 3.5. — Letk_j(s)be as above. Let Ij= 1

2πi

Z 2+i∞

2−i∞

FC(s)kj(s) ds.

Assume thatL6=Q. Then (1) forx>101 (3.1)

I₁− X

p∈P(C)

(logN_K/Qp)kb₁(N_K/Qp) 6 2 logx

x² logdL+ 16.08α0nK

logx x 6α3

logx x logdL

with

α₃= 2

101 +32.16α₀

log 3 = 36.759· · · ; (2) forx>10¹⁰

(3.2)

I2− X

p∈P(C) N_K/Qp6x⁵

(logN_K/Qp)kb2(N_K/Qp)

6 5

2√

πlog 3(logx)¹² logdL+α₁nKx³⁴(logx)³² +α₂nKx(logx)¹² 6α₄x(logx)¹²logd_L

with

α₄= 1 log 3

10⁻⁹ 4√

π +α₁log 10 5√

10 + 2α₂

= 5.4567· · ·.

Note thatdL>3^nL/2fornL>2. It follows from the Hermite–Minkowski’s inequality dL > ^π₃ ^3π₄ ⁿL−1

for nL > 1. For nL = 2, dL > 3, and for nL > 3, ^π₃ ^3π₄ nL−1

= ⁴₉ ^3π₄nL

> 3^nL/2. (See also [48, p. 140] and [23, p. 291].)

4. The Contour integral

In this section we will evaluate the integralsI₁andI₂by contour integra- tion. We will useL(s, χ) to denoteL(s, χ, E). LetF(χ) be the conductor

ofχandA(χ) =d_EN_E/QF(χ). Let δ(χ) =

(1 ifχis the principal character, 0 otherwise.

We recall that for eachχthere exist non-negative integersa(χ),b(χ) such that

a(χ) +b(χ) = [E:Q] =n_E, and such that if we define

γχ(s) =n

π^−s2Γs 2

o^a(χ) π^−s+12 Γ

s+ 1 2

b(χ)

and

ξ(s, χ) ={s(s−1)}^δ(χ)A(χ)^s/2γ_χ(s)L(s, χ), thenξ(s, χ) satisfies the functional equation

ξ(1−s, χ) =W(χ)ξ(s, χ),

whereW(χ) is a certain constant of absolute value 1. Furthermore,ξ(s, χ) is an entire function of order 1 and does not vanish ats= 0. By Hadamard product theorem we have for everys∈C

−L⁰

L(s, χ) =1

2logA(χ) +δ(χ) 1

s+ 1 s−1

+γ_χ⁰

γχ

(s)

− B(χ)− X

ρχ∈Z(χ)

1

s−ρ_χ + 1 ρ_χ

, whereB(χ) is some constant andZ(χ) denotes the set of nontrivial zeros ofL(s, χ). (See [48] and [24].) According to [40, (2.8)]

<B(χ) =− X

ρχ∈Z(χ)

< 1 ρ_χ. Hence, for everys∈C

(4.1) <

−L⁰ L(s, χ)

= 1

2logA(χ) +δ(χ)<

1 s + 1

s−1

+<γ⁰_χ γχ

(s)

− X

ρχ∈Z(χ)

< 1 s−ρχ

. Forj = 1, 2 we have

Ij =|C|

|G|

X

χ

χ(g)Jj(χ) by (2.1),

where

J_j(χ) = 1 2πi

Z 2+i∞

2−i∞

−L⁰

L(s, χ)k_j(s) ds.

Assume that T > 2 does not equal the ordinate of any of the zeros of L(s, χ). Consider

J_j(χ, T) = 1 2πi

Z

B(T)

−L⁰

L(s, χ)k_j(s) ds

forj = 1, 2, whereB(T) is the positively oriented rectangle with vertices 2−iT, 2 +iT,−¹₂+iT, and−¹₂−iT. By Cauchy’s theorem

(4.2) J_j(χ, T) =δ(χ)k_j(1)− {a(χ)−δ(χ)}k_j(0)− X

ρ_χ∈Z(χ)

|=ρχ|<T

k_j(ρ_χ)

forj = 1, 2.

Lemma 4.1. — Let Vj(χ) = 1

2πi

Z −¹₂−i∞

−¹₂+i∞

−L⁰

L(s, χ)kj(s) ds forj = 1,2. Then

(1) forx>101

|V1(χ)|6k₁

−1 2

{µ1logA(χ) +n_Eν₁}, whereµ₁= 0.75296· · · andν₁= 19.405· · ·; (2) forx>10¹⁰

|V₂(χ)|6k₂

−1 2

{µ₂logA(χ) +n_Eν₂}, whereµ2= 0.058787· · · andν2= 1.4793· · ·. Proof. — Lets=−¹₂+it. By [59, Lemme 5.1]

−L⁰ L

−1 2 +it, χ

6logA(χ) +nEv(t), where

v(t) = log r1

4+t²+ 2

!

+19683 812 .

Moreover, forx>101

k₁

−1 2+it

6x⁻³(1 +x⁻³²)²

9 4+t²

=k1

−1 2

1 +x⁻³² 1−x⁻³²

!2

9 9 + 4t²

6k1

−1 2

v1(t) withv1(t) =

1+101⁻³2

1−101⁻³2

²

9 9+4t²

and forx>10¹⁰

k₂

−1 2 +it

=x^−14−t2=k₂

−1 2

x^−t26k₂

−1 2

v₂(t) withv2(t) = 10^−10t2. Hence,

1 2πi

Z −¹₂−iT

−¹₂+iT

−L⁰

L(s, χ)kj(s) ds 6 1

πkj

−1 2

Z T 0

{logA(χ) +nEv(t)}vj(t) dt.

Set

µj= 1 π

Z ∞ 0

vj(t) dt and νj = 1 π

Z ∞ 0

v(t)vj(t) dt.

The result follows.

On the two segments from 2±iT to−¹₂±iT we proceed with the same way as [24, Section 6]. (See [23, Section 3], [59, Section 5], and [27].) Let Hj(T) = 1

2πi Z −¹₄

−¹₂

L⁰

L(σ+iT, χ)kj(σ+iT)−L⁰

L(σ−iT, χ)kj(σ−iT)

dσ and

H^∗_j(T) = 1 2πi

Z 2

−¹₄

L⁰

L(σ+iT, χ)k_j(σ+iT)−L⁰

L(σ−iT, χ)k_j(σ−iT)

dσ.

Then

Hj(T) +H^∗_j(T)

= 1 2πi

(Z −¹₂+iT 2+iT

−L⁰

L(s, χ)kj(s) ds+ Z 2−iT

−¹₂−iT

−L⁰

L(s, χ)kj(s) ds )

. Lemma 4.2. — Forj= 1,2we have

Hj(T) |kj(iT)|(logA(χ) +nElogT).

Proof. — Lets=σ±iT with−¹₂ 6σ6−¹₄. Then L⁰

L(s, χ)logA(χ) +n_ElogT

by [24, Lemma 6.2] andkj(s) |kj(iT)|. The result follows.

Lemma 4.3. — Let−¹₄6σ62. Then, we have L⁰

L(σ±iT, χ)− X

ρχ∈Z(χ)

|=ρχ∓T|61

1

σ±iT −ρ_χ logA(χ) +n_ElogT.

Proof. — See [24, Lemma 5.6]. (See also [59, Lemma 4.8].) Therefore, forj= 1, 2

H^∗_j(T)− 1 2πi

Z 2

−¹₄











kj(σ+iT) X

ρχ∈Z(χ)

|=ρχ−T|61

1 σ+iT−ρ_χ

−kj(σ−iT) X

ρ_χ∈Z(χ)

|=ρχ+T|61

1 σ−iT−ρχ









 dσ

|kj(iT)|(logA(χ) +nElogT) sincek_j(σ±iT) |k_j(iT)|for−¹₄ 6σ62.

Lemma 4.4. — Letρχ∈Z(χ)witht6==ρχ. If|t|>2, then Z 2

−¹₄

kj(σ+it) σ+it−ρχ

dσ |kj(it)|

forj = 1,2.

Proof. — Suppose first that =ρχ > t. Let Bt be the positive oriented rectangle with vertices 2 +i(t−1), 2 +it,−¹₄+it, and −¹₄+i(t−1). By Cauchy’s theorem,

Z

B_t

k_j(s) s−ρχ

ds= 0

for j = 1,2. However, on the three sides of the rectangle other than the segment from−¹₄+itto 2 +it, the integrand is majorized by

α5|kj(it)|

for some positive constantα5 depending onx, which proves the result for

=ρ_χ> t. A similar proof for=ρ_χ< tuses the rectangle with vertices 2 +it, 2 +i(t+ 1),−¹₄+i(t+ 1), and −¹₄+it.

Forj= 1, 2 we have

1 2πi

Z 2

−¹₄











k_j(σ+iT) X

ρ_χ∈Z(χ)

|=ρχ−T|61

1 σ+iT−ρχ

−kj(σ−iT) X

ρχ∈Z(χ)

|=ρχ+T|61

1 σ−iT−ρ_χ









 dσ

|k_j(iT)|{n_χ(T) +n_χ(−T)}

|kj(iT)|(logA(χ) +nElogT) by [24, Lemma 5.4], wherenχ(T) denotes the number of zeros ρχ ∈Z(χ) with|=ρχ−T|61.

We may then conclude as follows.

Lemma 4.5. — Forj= 1,2we have

H^∗_j(T) |kj(iT)|(logA(χ) +nElogT).

Lemma 4.6. — Forj= 1,2we have lim

T→∞

1 2πi

(Z −¹₂+iT

2+iT

−L⁰

L(s, χ)k_j(s) ds+ Z 2−iT

−¹₂−iT

−L⁰

L(s, χ)k_j(s) ds )

= 0.

Proof. — By Lemmas 4.2 and 4.5

Hj(T) +H^∗_j(T) |kj(iT)|{logA(χ) +nElogT}.

Since

|kj(iT)|6 ( ₉

4x²(1+T²) ifj= 1, x^−T2 ifj= 2,

the result follows.

LettingT → ∞in (4.2) and combining this and Lemmas 4.6 yield Jj(χ) +Vj(χ) =δ(χ)kj(1)− {a(χ)−δ(χ)}kj(0)− X

ρ_χ∈Z(χ)

kj(ρχ)

forj = 1, 2. Hence, we have

|G|

|C|Ij=X

χ

χ(g)Jj(χ)

=kj(1)−kj(0)X

χ

χ(g){a(χ)−δ(χ)} −X

χ

χ(g)



 X

ρ_χ∈Z(χ)

kj(ρχ)





−X

χ

χ(g)Vj(χ) forj= 1, 2. Note that by the conductor-discriminant formula ([46, Chap- ter VI, Section 3])

X

χ

logA(χ) = logd_L. We therefore conclude as follows.

Proposition 4.7. — Forj= 1,2we have (4.3) |G|

|C|I_j >k_j(1)− X

ρ∈Z(ζL)

|k_j(ρ)| −µ_jk_j

−1 2

logd_L

−nL

kj(0) +νjkj

−1 2

whereZ(ζL)denotes the set of all nontrivial zeros ofζL(s),µj andνj are as in Lemma 4.1.

5. Density of zeros of Dedekind zeta functions

To begin with, we recall that for every s∈Cwe have (5.1) <

−ζ_L⁰ ζ_L(s)

= 1

2logdL+<

1 s+ 1

s−1

+<γ_L⁰

γ_L(s)− X

ρ∈Z(ζL)

< 1 s−ρ, where

γ_L(s) =n

π^−s2Γs 2

or1+r2 π^−s+12 Γ

s+ 1 2

^r2

,

r1and 2r2are the numbers of real and complex embeddings ofL. (See [24, Lemma 5.1] or [48].)

For any real numbertwe let

nL(t) =|{ρ=β+iγ|ζL(ρ) = 0 with 0< β <1 and|γ−t|61}|.

For any complex numbersand positive real numberr >0 we let n(r;s) =|{ρ∈Z(ζL)| |ρ−s|6r}|.

From (4.1) Lagarias and Odlyzko deduced that nχ(t)logA(χ) +nElog(|t|+ 2)

for allt. (See [24, Lemma 5.4].) In this section we will bound n_L(t) and n(r;s) from above using (4.1). To do this we need some lemmas.

Lemma 5.1. — Lets=σ+itwithσ >1. We have X

ρ∈Z(ζL)

< 1

s−ρ >f₀(σ)nL(t), where

f0(σ) = 1 2min

( σ−1

(σ−1)²+ 1, σ−¹₂ σ−¹₂²

+ 1 )

+1 2min

( σ−¹₂ σ−¹₂²

+ 1 , σ

σ²+ 1 )

. Proof. — We have

X

ρ∈Z(ζL)

< 1 s−ρ >1

2 X

β+iγ∈Z(ζL)

|t−γ|61

σ−β

(σ−β)²+ 1+ σ+β−1 (σ+β−1)²+ 1

>f0(σ)nL(t).

Lemma 5.2. — If<s=σ >1, then

<ζ_L⁰ ζL

(s)6nLf1(σ), where

f₁(σ) =−ζ_Q⁰ ζ_Q(σ).

Proof. — For<s >1,

−ζ_L⁰ ζL

(s) =X

P

logNP NP^s−1 =X

P

logNP

∞

X

m=1

NP^−ms, whereP runs over all prime ideals ofL. Comparing−^ζ_ζ^0L

L(σ) with −^ζ

0 Q

ζ_Q(σ) yields

<ζ_L⁰ ζ_L(s)6

−ζ_L⁰ ζ_L(s)

6−ζ_L⁰

ζ_L(σ)6nL

−ζ⁰

Q

ζ_Q(σ)

.

(See [24, Lemma 3.2].)

See also [9], [31, Lemma (a)], [59, Lemma 3.2], [11, p. 184], and [33, Proposition 2].

Lemma 5.3. — Assume that<s > ¹₂. We have (1) <Γ⁰

Γ(s)6log|s|+1

3 6α₆log(|s|+ 2) withα6= 1.08;

(2) <Γ⁰

Γ(s)>log|s| −4

3 >log(|s|+ 2)−α₇ withα₇= ⁴₃+ log 5 = 2.9427· · ·. Proof. — For<s >0,

Γ⁰

Γ(s) = logs− 1 2s−2

Z ∞ 0

υ

(s²+υ²)(e^2πυ−1)dυ.

(See [58, p. 251].) Since|s²+υ²|>(<s)², we have

Z ∞ 0

υ

(s²+υ²)(e^2πυ−1)dυ 6 1

(<s)² Z ∞

0

υ

e^2πυ−1dυ= 1 24(<s)². If<s > ¹₂, then

<Γ⁰

Γ(s)6log|s|+ 1 12

1

(<s)² 6log|s|+1 3 and

<Γ⁰

Γ(s)>log|s| − 1 2|s| − 1

12 1

(<s)² >log|s| −4 3. Setϕ₁(υ) =α₆log(υ+ 2)−logυ−¹₃ forυ > ¹₂. Then,

ϕ⁰₁(υ) =(α₆−1)υ−2

υ(υ+ 2) andϕ₁(υ)> ϕ₁ 2

α₆−1

>0.

Hence

<Γ⁰

Γ(s)6α6log(|s|+ 2).

Setϕ₂(υ) = logυ−⁴₃−log(υ+ 2) +α₇ forυ > ¹₂. Then ϕ⁰₂(υ)>0 andϕ2(υ)> ϕ2

1 2

= 0.

Hence

<Γ⁰

Γ(s)>log(|s|+ 2)−α₇.

Lemma 5.4. — Lets=σ+it. Ifσ >1, then

<γ_L⁰

γ_L(s)6n_L

f₂(σ) log(|t|+ 2)−1 2logπ

, where

f2(σ) = α6

2

log(σ+ 5) log 2 −1

. Proof. — By definition and Lemma 5.3 (1) we have

<γ_L⁰ γL

(s) = (r₁+r₂) 2 <Γ⁰

Γ s

2

+r₂ 2<Γ⁰

Γ

s+ 1 2

−n_L 2 logπ 6α6

(r1+r2)

2 log

|s|

2 + 2

+α6

r2

2 log

|s+ 1|

2 + 2

−nL

2 logπ 6 nL

2

α6log

|s+ 1|

2 + 2

−logπ

. It is sufficient to verify that

(5.2) log

|s+ 1|

2 + 2

6

log(σ+ 5) log 2 −1

log(|t|+ 2).

Note that|s+ 1|>2|t|if and only if|t|6(σ+ 1)/√

3. If|t|>(σ+ 1)/√ 3, then (5.2) holds. We suppose now that |t| < (σ+ 1)/√

3. Set ϕ3(υ) = ϕ5(υ)/ϕ4(υ) withϕ4(υ) =υ+ 2 andϕ5(υ) = 2 +p

(σ+ 1)²+υ²/2. Then ϕ⁰₃(υ)60 andϕ₅(υ)6_ϕ

5(0) ϕ₄(0)

ϕ₄(υ) for 06υ <(σ+ 1)/√

3. For 06υ <

(σ+ 1)/√

3 we have then (5.3) logϕ5(υ)

logϕ₄(υ)6 logϕ4(υ) + logϕ5(0)−logϕ4(0) logϕ₄(υ)

6 logϕ5(0)

logϕ4(0) =log(σ+ 5) log 2 −1,

which yields (5.2).

We are now ready to bound nL(t).

Proposition 5.5. — For allt we have

(5.4) n_L(t)61.1 logd_L+ 2.09 log{(|t|+ 2)^nL}+ 0.56n_L+ 4.05.

In particular, ifL6=Q, then

(5.5) nL(t)62.72 log{dL(|t|+ 2)^nL}.

Proof. — Combining (4.1), Lemmas 5.1, 5.2, 5.3, and 5.4 yields f₀(σ)n_L(t)6 1

2logd_L+ 1 σ+ 1

σ−1 +nL

f2(σ) log(|t|+ 2)−1

2logπ+f1(σ)

forσ >1. We write

(5.6) n_L(t)6a₁(σ) logd_L+a₂(σ) log{(|t|+ 2)^nL}+a₃(σ)n_L+a₄(σ) forσ >1, where

a₁(σ) = 1

2f0(σ), a₂(σ) =f₂(σ)

f0(σ), a₃(σ) = 1 f0(σ)

f₁(σ)−1 2logπ

, and

a4(σ) = 1 f0(σ)

1 σ + 1

σ−1

. We choose now appropriateσ. Ifσ= (3 +√

17)/4, then (5.6) yields (5.4).

For the proof of (5.5), we choose σ = 2.45. In this case, a₃(σ) < 0 and 2a3(σ) +a4(σ)>0. SincenL >2, it follows from (5.6) that

n_L(t)6a₁(σ) logd_L+a₂(σ) log{(|t|+ 2)^nL}+ 2a₃(σ) +a₄(σ) 6B1logdL+B2log{(|t|+ 2)^nL},

whereB1=a1(σ) +_{log 3}¹ {2a3(σ) +a4(σ)}= 2.6885· · · andB2=a2(σ) =

2.7106· · ·. So, we obtain (5.5).

See also [21], [52], and [59, Lemme 4.6].

Proposition 5.6. — Letrbe a positive real number.

(1) Assume that

nL(t)6α8log{dL(|t|+ 2)^nL} for someα8>0. Then we have

n(r;σ+it)6α8(1 +r) log{dL(|t|+r+ 2)^nL}. (2) Assume thatL6=Q. Ifσ>1 and0< r61, then

n(r;σ+it)610

1 + 2f2(2)

5 rlog{dL(|t|+ 2)^nL}

. Proof. — Set

Z(r;s) ={ρ∈Z(ζ_L)| |ρ−s|6r}

and Z(t) ={β+iγ∈Z(ζL)| |γ−t|61}.

Note thatn(r;s) =|Z(r;s)|andnL(t) =|Z(t)|.