On the
useof explicit bounds
onresidues of Dedekind zeta functions taking into account
the behavior of small primes
par STÉPHANE R. LOUBOUTIN
RÉSUMÉ. Nous donnons des majorants
explicites
des résidus aupoint s =1 des fonctions zêta
~K (s)
des corps de nombres tenantcompte du comportement des petits nombres premiers dans K.
Dans le cas où K est
abélien,
de telles majorations sont déduitesde majorations de
IL (1, x) tenant
compte du comportement de Xsur les petits nombres premiers, pour X un caractère de Dirichlet
primitif.
De nombreusesapplications
sont données pour illustrer l’utilité de tels majorants.ABSTRACT.
Lately, explicit
upper bounds onx)1 1 (for
prim-itive Dirichlet characters
x) taking
into account the behaviorsof x
on a given finite set of primes have been obtained. Thisyields explicit
upper bounds on residues of Dedekind zeta func- tions of abelian number fieldstaking
into account the behavior ofsmall primes, and it as been
explained
how such boundsyield
im- provements on lower bounds of relative class numbers of CM-fields whose maximaltotally
real subfields are abelian. We present heresome other
applications
of such boundstogether
with new boundsfor non-abelian number fields.
1. Introduction
Let K be a number field of
degree n
= ri +2r2
> 1. LetdK,
wK,hK
and be the absolute value of itsdiscriminant,
the numberof
complex
roots ofunity
inK,
theregulator,
classnumber,
and residue ats = 1 of the Dedekind zeta function
~K(s)
of K. Recall the class number formula:Manuscrit reçu le 5 mars 2004.
Mots clefs. L-functions, Dedekind zeta functions, number fields, class number.
In order to obtain bounds on
hK
we need bounds on The bestgeneral
upper bound is(see [Lou01,
Theorem1]):
If K is
totally
realcubic,
then we have the better upper bound(see [Lou01,
Theorem
2]):
Finally,
if K isabelian,
then we have the even bettergeneral
upper boundwhere K =
(5 - 2 log 6)/2
=0.70824 ~ ~ ~ , by [Ram01, Corollary 1].
However,
from the Eulerproduct of (K (s)
weexpect
to have better upper bounds forprovided
that the smallprimes
do notsplit
inK. For any
prime p > 1,
we setwhere ’P runs over all the
primes
ideals of K above p. A carefulanalysis
ofthe
proofs
of all theprevious
boundssuggests
that we shouldexpect
that there exists some r,’ > 0 such thatNotice that the factor
IIK (2)/IIQ (2)
isalways
less than orequal
to1,
but isequal
to1 / (2n -1 ),
hencesmall,
if theprime
p = 2 is inert in K. Combined with lower bounds forRess==l ((K (s) ) depending
on the behavior of the smallprimes
in K(see [Lou03,
Theorem1]),
we would as a consequence obtain better lower bounds for relative class number of CM-fields. The aim of this paper is to illustrate on variousexamples
the use of such better bounds onTo
begin with,
we recall:Theorem 1.
(See [Ram04] 1;
see also[Lou04a]
and[Lou04d]).
Let S be agiven finite
setof pairwise
distinct rationalprimes.
Set ~s :=#S ~ log
4 +2
Then, for
anyprimitive
Dirichlet character Xof
conductorqx > 1 such that p E S
implies
that p does not divide qx, we haveif x
isodd,
andif
X is even and qx > 6.2 .4#s.
We refer the reader to
[BHM], [Le], [MP], [Mos], [MR], [SSW]
and[Ste]
for various
applications
of suchexplicit
bounds on L-functions.They
arenot the best
possible theoretically. However,
if such better bounds aremade
explicit,
we end up with useless ones in a reasonable range for q.(see [Lou04a]
and[Boo]). Therefore, applications
of these better bounds topractical problems
are notyet possible.
2.
Upper
bounds for relative class numbersCorollary
2. Let q - 5(mod 8), q =1- 5,
be aprime,
let Xq denote any-one
of
the twoconjugate
oddquartic
charactersof
conductor q and lethq
denote the relative class nurraber
of
theimaginary cyclic quartic field Nq of
conductor q.
Then,
which
implies hq
qfor
qC.,
whereAx, Bx
andCx
are asfollows:
’Note however the misprint in [Ram04, Top of page 143] where the term
should be
Proof. Since q -
5(mod 8),
we haveYq(2)2
=(~) =
-1 andXq(2) =
+1.Set S =
{p E {2,3,5}; x(p) -E-l.
Then 2 E S andaccording
to Theo-rem 1 we may choose
and
Remarks 3.
Using Corollary 2
to alleviate the amounto f required
rela-tive class numbers
computation,
several authors have beentrying
to solve in[JWW]
the openproblem
hinted at in[Lou98]:
determine the least(or
atleast
one) prime
q - 5(mod 8) for
whichhq
> q.Indeed, according
toCorollary 2, for finding
such a q in the range q 5 .1010,
we may assume thatxq(3) _ +1,
which amounts toeliminating
threequarters of
theprimes
q in this range. In the same wag, in the range q 3 -
lOl3
we may assumethat
xq(3) _
+1 orXq(5) = +1,
which amounts toeliminating 9/16 of
theprimes q
in this range.3. Real
cyclotomic
fields oflarge
class numbersIn
[CW],
G. Cornell and L. C.Washington explained
how to usesimplest
cubic and
quartic
fields toproduce
realcyclotomic
fields(~+(~p)
ofprime
conductor p and class
number hp
> p.They
could findonly
one such realcyclotomic
field. Weexplain
how to use our bounds on L-functions to findmore
examples
of such realcyclotomic
fields.However,
it is much moreefficient to use
simplest quintic
and sextic fields toproduce
realcyclotomic
fields of
prime
conductors and class numbersgreater
than their conductors(see [Lou02a]
and[Lou04c]).
3.1.
Using simplest
cubic fields. Thesimplest
cubic fields are the realcyclic
cubic number fields associated with theQ-irreducible
cubicpoly-
nomials
m n . ,
of discriminants
dm
whereAm := m~+3m+9.
SinceP-m-3 (x),
we may assume that m > -1. We letdenote the
only positive
root ofMoreover,
we will assume thatthe conductor of
Km
isequal
toAm,
which amounts toasking
that(i)
m 0
0(mod 3)
and~m
issquarefree,
or(ii)
m -0,
6(mod 9)
and~m/9
issquarefree (see [Wa, Prop.
1 andCorollary]).
In thatsituation, {20131, pm, - 1 / (pm + 1) 1 generate
the full group ofalgebraic
units ofKm
andthe
regulator
ofKm
iswhich in
using (1) yields
Lemma 4. The
polynomials Pm(x)
has no root mod2,
has at least one rootmod 3
if
andonly if m -
0(mod 3),
and has at least one root mod 5if
andonly i f m
-1(mod 5) . Hence, if ~m
issquare-free,
then 2 and 3 are inertin
Km,
andif m t
1(mod 5)
then 5 is also inert inKm .
As in
[Lou02a,
Section5.1],
we letxKm
be theprimitive,
even, cubic Dirichlet characters moduloA,
associated withKm satisfying
Since the
regulators
of theseKm’s
aresmall, they
should havelarge
classnumbers. In
fact,
weproved (see [Lou02c, (12)]):
Corollary
5. Assume that m > -1 is such that~m
=m2 +
3m + 9 issquarefree. Then,
Proof.
If aprime
1 > 2 is inert inK7"
thenXKm(l)
Eexp(4i7r/3)1. According
to Lemma 4 and to Theorem 1(with
S ={2,3}
and S =
{2,3, 5}),
we haveNow, according (4)
and(3),
the desired results follow for m ~ 95000. The numericalcomputation
of the class numbers of theremaining K~, provides
us with the desired bounds
(see [Lou02a]).
DFrom now on, we assume
(i)
that p =Am
=m 2
+ 3m + 9 isprime (hence m ~
0(mod 3))
and(ii)
that p - 1(mod 12),
which amounts toasking
that m -0,
1(mod 4).
In that case, bothKm
and :=are subfields of the real
cyclotomic
fieldQ+((p)
and theproduct h2h3
ofthe class numbers
h2
:= andh3
:= of and divides the classnumber hp
ofQ+(~p). Now, h3 An/60
andh2h3 > An imply h2 > 60,
henceh2 >
61(for h2
isodd),
and Cohen-Lenstra heuristicspredict
that realquadratic
number fields ofprime
conductors with class numbersgreater
than orequal
to 61 are few and far between.Hence,
suchsimplest
cubic fieldsKm
ofprime
conductorsAn
=m2 + 3m + 9 == 1 (mod 4)
withh2h3
>àn
are few and far between. As we have at hand avery efficient method for
computing
class numbers of realquadratic
fields(see [Lou02b]
and[WB]),
we used thisexplicit
necessary conditionh2 >
61 to
compute (using [Lou02a])
the class numbers ofonly
584 out of the46825
simplest
cubic fieldsI~m
ofprime
conductorsAm m
1(mod 12)
with -1 m 1066285 to obtain the
following
Table.(Using
the factthat
h2 > 61,
the class number formula forkm
and Theorem 1 for S= 0 imply Reg2 Am) /244,
whereReg2
denotes theregulator
of thereal
quadratic
fieldkm
= andtaking
into account the fact thatReg2
is much faster tocompute
thanh2,
we could stillimprove
thespeed
ofthe
required computations).
Notice that the authors of[CW]
and[SWW]
only
came up with one suchKm,
the one for m = 106253.Least values of m > -1 for which àn = m2 + 3m + 9 is prime and h2h3 > ~m
3.2.
Using simplest quartic
fields. The so calledsimplest quartic
fields
(dealt
with in[Lazl], [Laz2]
and[Lou04b])
are the realcyclic quartic
number fields associated with the
quartic polynomials
of discriminants
dm
=4A3
where :=(m2
+16 3.
SincePm(-x) = P-m (x),
we may and we will assume that m > 0. The reader willeasily
check
(i)
that has no rational root,(ii)
thatPm(x)
isQ-irreducible,
except for m = 0 and m =3,
and(iii)
thatPm(x)
has aonly
one root/3m
=(m+B/A~)/2.
Inparticular, km
=Q( @)
is thequadratic
subfieldof the
cyclic quartic
fieldKm.
It is known thathkm
dividesh Km’
and weset
hKm
= Since pm > 1 andpm - 0,
we obtainand
Proposition
6. Assurrze that m > 1 is odd andthat Llm
=m 2+16
isprime.
Then,
the discriminantof
the realquadratic sub fceld km
=of Km
is
equal
toOm,
the discriminantof Km
isequal
toà3
@ its conductor isequal
toLlm,
the class numbersof Kn
andkm
areodd, RegKm/Regkm
=Reg*
and(see [Lou04b,
Theorem9])
where
xKm is
angoneof
the twoconjugate primitive,
even,quartic
Dirichletcharacters modulo
Om
associated withKm. Moreover, XKm (2) = -l,
andm > 5
implies
"~
Proof. According (6),
to Theorem 1(with
S ={2})
whichyields
and to
(5),
we have~m/(36+0(1))
andhKm A,/26
for 3000.The numerical
computation
of the class numbers of theremaining Km provides
us with the desired bound(see [Lou02a]).
Dis
odd),
and Cohen-Lenstra heuristicspredict
that realquadratic
numberfields of
prime
discriminants with class numbersgreater
than orequal
to 27are few and far between. Hence such
simplest quartic
fieldsKm
ofprime
conductors
Am
=m 2 +
16 with >Am
are few and far between. Aswe have at hand a very efficient method for
computing rigorously
classnumbers of real
quadratic
fields(see [Lou02b]
and[WB]),
we used thisexplicit
necessary condition 27 tocompute only
1687 of the class numbers of the 86964simplest quartic
fieldsKm
ofprime
conductorsOm
=m2
+ 16 =1(mod 4)
with 1 m 1680401 to obtain thefollowing
Table.Notice that G. Cornell and L. C.
Washington
did not find any suchKm (see [CW,
bottom of page268] ).
Least values of m > 1 for which Am = m2 + 16 is prime and
hKm
> ~m4. The
imaginary cyclic quartic
fields with ideal class groups ofexponent
2We
explain
how one could alleviate the determination in[Lou95]
of allthe
non-quadratic imaginary cyclic
fields of2-power degrees
2n = 2r > 4 with ideal class groups ofexponents
2(the
timeconsuming part bieng
thecomputation
of the relative class numbers of the fields sievedby Proposition
8 or Remark 9
below).
Tosimplify,
we will nowonly
deal withimaginary cyclic quartic
fieldsof
odd conductors...’ ...
Theorem 7. Let K be an
imaginary cyclic quartic field of
odd conductorfK;
Let and xK denote the realquadratic subfield of K,
the conductorof k,
and anyoneof the
twoconjugate primit2ve quartic
Dirichlet characters modulof K
associated with K.Then,
where
and
Proof.
Use[Lou03, (34)], [Lou03, (31)]
with[Ram01, Corollary 1]
and[Ram01, Corollary 2]’s
values for Kk = KX’ where X is theprimitive
evenDirichlet character of conductor
dk
associated withk,
and4/12 - xK(2) ~2.
DProposition
8. Assume that theexponent of
the ideal class groupof
animaginary cyclic quartic field
Kof
odd conductorfK
is less than orequal
to 2.
Then, fk
1889 andfK 107 (where
k is the realquadratic subfield
of K). Moreover,
whereas there are 1 377 361imaginary cyclic fields
Kof
odd conductors
fK 107
and such thatfk 1889, only
400 outof
themmay have their ideal class groups
of eiponents 2,
thelargest possible
conductor
being fK
= 5619(for fk
= 1873 andfKIk
:=fK/.fk
=3).
Proof.
It is known that if theexponent
of the ideal class group of K of odd conductorfK
is2,
thenfk =-
1(mod 4)
isprime
andwhere denotes the number of
prime
ideals of k which are ramified inKlk (see [Lou95,
Theorems 1 and2]). Conversely,
for agiven
realquadratic
field k ofprime
conductorfk ==
1(mod 4),
the conductorsfK
ofthe
imaginary cyclic quartic
fields K of odd conductors andcontaining k
are of the form
,fk
=fkfK/k
for somepositive square-free integer fKIk
> 1relatively prime
withfk
and such that(in
order to havexK (-1 ) - -1,
i.e. in order toguarantee
that K isimaginary). Moreover,
for such agiven
and such agiven
there existsonly
oneimaginary cyclic quartic
field Kcontaining
and of conductorf K
= and for this K we havewhere
( fk )
denote theLegendre’s symbol. Finally,
if we letÇk
denote any-one of the two
conjugate quartic
characters modulo aprime fk
-1(mod 4),
then - , K/k where
(-.-2013)
K/k denote the Jacobi’ssymbol,
andHence,
we mayeasily compute
Kk, cK and fromfk
and Inparticular,
weeasily
obtain that there are 1 377 361imaginary cyclic
fields.K of odd conductors
f K 107
and such thatfk 1889,
and that cK = 32 for 149 187 out ofthem,
CK =32/5
for 938 253 out ofthem,
and cK =32/9
for 289 921 out of them.
Now,
letPn
denote theproduct
of the first n oddprimes
3 = pi 5 = p2 ... Pn ...(hence, Po
=1, Pi
=3, P2 = 15, ... ).
There are two cases to consider:(1)
IfXk(2) =
+1. 1(mod 8)
isprime,
Kk =0, 32/9,
f K
= where is aproduct
of n > 0 distinct oddprimes.
Hence, Pn,
1 +2n, hK
=2~/~’~ ~
4n andusing (7)
we obtain
Assume that 36. Then
3/~~ ~ 5~ and for n >
1 we have
P2 =
5, Pn 2: Pi
= 3 andSince we
clearly
haveF~(1) Fk(0),
we obtainminn>o Fk(n)
=Fk(1)
and
-
which
implies fk 1899,
hencefk
1889(for fk -
1(mod 8)
mustbe
prime). Hence, using (7),
we obtainLet now n denote the number of distinct
prime
divisors off K.
Thenf K > Pn, 2(n -1) + 1
andhK = 4n-l. Hence, using (7),
we obtainwhich
imPlies n 7, 4,
and
yields fK 107.
(2)
IfXk (2) = - 1.
5(mod 8)
isprime, 2.78, cK >
32/5
and we follow theprevious
case. We obtainfk 1329,
hence1301
(for fk m
5(mod 8)
must beprime), n 7, 46
and7. 106.
Hence,
the first assertionProposition
8 isproved. Now,
for agiven
oddprime fk
1889equal
to 1 modulo4,
and for agiven
oddsquare-free
integer 107/ f~. relatively prime
withfk,
wecompute (using
(10)),
cK(using (11))
and use(7)
and(8)
to deduce that if the exponent of the ideal class group of K is less than orequal
to2,
thenNow,
an easy calculationyields
thatonly
400 out of 1 377 361imaginary cyclic
fields K of odd conductors and such thatfk
1889 andf K 107 satisfy (12),
and the second assertion of theProposition
isproved.
DRemarks 9. Our
present
lower bound(7)
should becompared
with thebound
obtained in
[Lou97]. If
we used this worse lower boundfor hh
then wewould end up with the worse
following
result:If
theexponent of
the idealclass group
of
animaginary cyclic quartic field
Kof
odd conductorfK
isless than or
equal
to2,
thenfk
4053 andfK 2. 107. Moreover,
whereasthere are 2 946 395
imaginary cyclic fields of
odd conductorsfK
2 .107
and such that
fk 4053, only
1175 outof
therra may have their ideal class groupsof exponents 2,
thelargest possible
conductorbeing fK
= 11667(for fk
= 3889 andfKIk
=3).
5. The non-abelian case
We showed in
[Lou03]
howtaking
into account the behavior of theprime
2 in CM-fields can
greatly improve
upon the upper bounds on the root num-bers of the normal CM-fields with abelian maximal
totally
real subfields ofa
given (relative)
class number. We nowexplain
how we canimprove
uponpreviously
known upper bounds for residues of Dedekind zeta functions ofnon-necessarily
abelian number fieldsby taking
into accound the behavior of theprime
2:Theorem 10. Let K be a number
field of degree
3 and root d,iscrim-Proof.
Weonly
prove(13),
theproof
of(14) being
similar. We set(which
is > 1 for s >0). According
to[Lou01,
Section6.1]
butusing
thebound
instead of the bound
~K(s) 1er’~~(s) ,
we havewhere sK = 1 + E
~1, 6~
andand
(13)
follows. DCorollary
11.(Compare
with[Lou01,
Theorems 12 and14]
and[Lou03,
Theorems 9 and
22]).
Set c =2(J$ - 1)2
= 1.07 ... and Vm :=(m/(m - 1 ))"L-1
E[2, e).
Let N be a normalCM-field of degree
2m >2,
relative class number
h-
and root discriminant pnr =dN2"’ >
650. Assumethat N contains no
irraaginary quadratic subfield (or
that the Dedekind zetafunctions of
theimaginary quadratic subfields of
N have no real zero in therange 1 -
(c/ log dN)
s1). Then,
Hence, hN
> 1for
m > 5 and14610,
andfor
m > 10 and PN 2 9150.Moreover,
oo as[N : Q]
= ocfor
such normalCM-fields
Nof
rootdiscriminants PN >
3928.Proof.
To prove(15),
follow theproof of [Lou01,
Theorems 12 and14]
and[Lou03,
Theorems 9 and22],
but now make use of Theorem 10 instead of[Lou01,
Theorem1]
andfinally
notice that(x
is thequadratic
character associated with thequadratic
extensionN/K,
and ’P ranges over all the
primes
ideals of Klying
above the rationalprime
2).
0We also refer the reader to
[LK]
for a recent paperdealing
with upper bounds on thedegrees
and absolute values of the discriminants of the CM- fields of class number one, under theassumption
of thegeneralized
Riemannhypothesis.
Theproof
relies on ageneralization
ofOdlyzko ([Odl]),
Stark([Sta])
and Bessassi’s([Bes])
upper bounds for residues of Dedekind zeta functions oftotally
real number fields oflarge degrees,
thisgeneralization taking
into account the behavior of smallprimes.
All these bounds arebetter than ours, but
only
for numbers fields oflarge degrees
and smallroot
discriminants,
whereas ours aredevelopped
to deal with CM-fields of smalldegrees.
6. An open
problem
Let k be a non-normal
totally
real cubic field ofpositive
discriminantâ~.
It is known that
(see [Lou01 ,
Theorem2]):
This bound has been used in
[BL02]
totry
to solve the class number oneproblem
for the non-normal sextic CM-fields Kcontaining
noquadratic
subfields.
However,
to date thisproblem
is in fact notcompletely
solvedfor we had a much too
large
bound 2 -1029
on the absolute valuesdK
of their discriminants(see [BL02,
Theorem12]).
In order togreatly improve
upon this upperbound,
we would like to prove that there existssome
explicit
constant r, such thatholds true for any non-normal
totally
real cubic field I~.However, adapting
the
proof
of[Lou01,
Theorem2]
is not that easy and we have not come upyet
with such aresult,
the hardest cases to handlebeing
the cases(2) = ’P
or
PiP2
inl~,
where we wouldexpect
bounds of thetype
At the
moment,
we canonly
prove thefollowing
result whichalready yields
a 1000-fold
improvement
on ourprevious
bounddK
2 ~1029:
Theorem 12. Let k be a
totally
reol cubic numberfield. Then,
where
As a consequence,
if
K is a non-normal sexticCM-field containing
no qua- draticsubfield
andif
the class numberof
K isequal
to one, thendK 2.1026.
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Stéphane R. LOUBOUTIN
Institut de Mathématiques de Luminy, UNIR 6206 163, avenue de Luminy, Case 907
13288 Marseille Cedex 9, FRANCE E-mail : louboutiOiml. univ-mrs. fr