J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
K
EN
Y
AMAMURA
Maximal unramified extensions of imaginary quadratic
number fields of small conductors
Journal de Théorie des Nombres de Bordeaux, tome
9, n
o2 (1997),
p. 405-448
<http://www.numdam.org/item?id=JTNB_1997__9_2_405_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Maximal unramified extensions of
imaginary
quadratic
number fields of small conductorspar KEN YAMAMURA
RÉSUMÉ. Nous déterminons la structure du groupe de Galois
de l’extension maximale non ramifiée Kur de chaque corps quadratique
imaginaire de
conducteur ~
420(~
719 sousGRH).
Pour tous ces corps K,l’extension Kur coïncide avec K, ou avec le corps de classes de Hilbert de K,
ou avec le second corps de classes de Hilbert de K ou avec le troisième corps
de classes de Hilbert de K. Les bornes d’Odlyzko sur les discriminants et
les informations sur la structure des groupes de classes obtenues par l’action
du groupe de Galois sur les groupes de classes sont ici essentielles. Nous utilisons aussi des relations sur le nombre de classes et un ordinateur pour
le calcul du nombre de classes de corps de bas degré pour obtenir le nombre
de classes de corps de degré plus élevé. Nous utilisons aussi des résultats
sur les tours de corps de classes, ainsi que notre connaissance des 2-groupes
d’ordre ~
26 et des groupes linéaires sur des corps finis.ABSTRACT. We determine the structures of the Galois groups
Gal(Kur/K)
of the maximal unramified extensions Kur of imaginary quadratic number fields K of
conductors ~
420(~
719 under the Generalized RiemannHy-pothesis).
For all such K, Kur is K, the Hilbert class field of K, the second Hilbert class field of K, or the third Hilbert class field of K. The use of Odlyzko’s discriminant bounds and information on the structure of class groups obtained by using the action of Galois groups on class groups ises-sential. We also use class number relations and a computer for calculation
of class numbers of fields of low degrees in order to get class numbers of fields of higher degrees. Results on class field towers and the knowledge of the 2-groups of orders ~ 26 and linear groups over finite fields are also used.
1. INTRODUCTION
Let K be an
algebraic
number field(of
finitedegree)
and itsmax-imal unramified extension. Then the Galois group
Gal(Kur/ K)
can be1991 Mathematics Subject Classification. Primary 11R32,11R11.
Key words and phrases. Maximal unramified extension, imaginary quadratic number
field, discriminant bounds, class field tower.
both finite and infinite and in
general
it isquite
difficult to determine thestructure of this group. If .K has
sufficiently
small rootdiscriminant,
thenKur =
K,
thatis,
K has no nontrivial unramified extension. This is thecase, for
example,
for theimaginary
quadratic
number fields with class number one, thecyclotomic
number fields with class number one, the real abelian number fields ofprime
powerconductors ~
67(see
[57,
Appen-dix]).
For some fields K with small rootdiscriminant,
we can determineGal(Kur/ K).
The purpose of this paper is to determine the structure of ofimaginary quadratic
number fields K of small conductors. Forimaginary quadratic
number fields K of conductors ~ 420(~
719 under the Generalized RiemannHypothesis
(GRH))
we determineand tabulate them for K with
Kl,
whereKl
denotes the Hilbert class field of K.(If
Kur =
Kl,
thenGal(Kur/ K) =
Cl(K),
the class group of Kby
class fieldtheory.)
For all suchK,
is one ofK, Kl, .K2,
orK3,
whereK2
(resp. K3 )
is the second(resp. third)
Hilbertclass field of .K. In other
words,
coincides with thetop
of the class field tower of K and thelength
of the tower is at most three. Ifpossible,
we
give
alsosimple expressions
ofK1
andK2.
Also for K = with723 ~
d~
1000,
we tabulateexcept
for some d.From now on let K =
Q(U2)
be animaginary quadratic
number fieldwith discriminant d 0. J. Martinet stated in
[34]
that if~d~
250,
thenKur = Kl
except
for 7fields,
for which he gave the structure of(We
note thatH24
for K =Q( 248)
in[34]
is
falser)
He also stated that this fact isproved by using
the methods which J.Masley
[35] (and
later F. J. van der Linden[50])
used for calculation ofclass numbers of real abelian number fields of small conductors.
They
usedOdlyzko’s
discriminant bounds and information on the structure of class groups obtainedby using
the action of Galois groups on class groups.In addition to their
methods,
we use acomputer
for calculation of classnumbers of fields of low
degrees
(we
useKANT)
and then use class numberrelations to
get
class numbers of fields ofhigher degrees
(see §3).
Resultson class field towers
[5,
25, 29, 31,
49]
and theknowledge
of the2-groups
oforders
26
[17]
and linear groups over finite fields(see §4)
are also used.We know that if 499
(jd)
__
2003 underGRH),
then thedegree
K~
is finite(see
§2).
For thesed,
we want to determineThe
key
fact is that any unramified(finite)
extension L of K has the sameroot discriminant as K:
rdL - ~
rdK = M.
Thus,
if wehave
rdK
B(2N),
whereB(2N)
denotes the lower bound for the root1 The referee pointed out that this was corrected by Martinet in a supplement to his
discriminants of the
totally imaginary
number fields of(finite) degrees >
2N,
then weget
K]
N. We do not know the real values ofB (2N)
(except
forN -
4),
however,
some lower bounds forB(2N)
are known.The best known unconditional lower bounds for
B(2N)
can be found inthe tables due to F. Diaz y Diaz
[12].
If we assume the truth ofGRH,
much better lower bounds can be obtained. The best known conditional
(GRH)
lower bounds are found in theunpublished
tables due to A. M.Odlyzko
[38],
which arecopied
in Martinet’sexpository
paper[34].
LetKi
be thetop
of the class field tower of K: K =Ko 9 K, 9 K2 C
...is the Hilbert class field of
Ki),
thatis,
I is the smallest number with=
Kl.
If we cannotget
Ki]
60,
whichimplies
=Ki,
from available lower bounds for
B(2N),
we need tojudge
whetherKi
hasan unramified nonsolvable Galois extension and this is
quite
difficult. Forthe fields
Q( -423)
andG~( -723),
we haveh(Ki)
=1,
thatis,
I = 1 and we cannotget
~Kur : Kil
60 from available lower bounds forB(2N)
(even
under GRH forQ(~/~723)).
420(idl
~ 719 underGRH),
we
get
[Kur : Kl]
60 and our mainproblem
is to determine thedegree
[Kl : Q].
Ingeneral,
it is difficult to determine[K2 : Q],
because it is veryhard to calculate the class number
h(Ki)
ofKl.
(Of
course, for K withsmall
Cl(K),
we can calculateh(Kl )
with thehelp
of acomputer.)
Nowlet
Kg
be the genus field ofK,
thatis,
the maximal unramified abelian extension of K which is abelian overQ.
If d is the discriminant of K andd =
did2 ... dt
is the factorization of d into theproduct
of fundamentalprime discriminants,
thend2,
... ,dt ),
and we havewhich
implies
As
K9
is amulti-quadratic
numberfield,
h(Kg)
can be calculatedby
themethod in
[51],
and we mayexpect
that[K2 :
(Kg)l]
is small for fields weconsider on the
ground
of thefollowing
proposition
(§4,
Proposition
2).
PROPOSITION. Let L be the Hilbert class
field of
the genusfield
Kg
of
anirnaginary
abelian numberfield
K. Thenfor
anyprime
number p withp{
[L : Q],
thep-class
groupCl(P)
(L)
of
L is trivial ornoncyclic.
For all K with
idi
1000 such thath(Kg)
>K],
which isequivalent
toKl,
we haveK2
= For >(at
least)
two distinctprime
factors, however,
for mostK,
thisinequality
holds. Infact,
if aquadratic
subfield #
K ofKg
has class number divisibleby
an oddprime
p, then we haveph(K)/[Kg :
K~.
Thus,
thefollowing
problem
arisesnaturally:
PROBLEM. Characterize the
imaginary quadratic
numberfields
K withK2=(K9)1
The author
expects
that thisproblem
can be settledgroup-theoretically
and a similar
problem
can beposed
for realquadratic
number fields. Wewill discuss this
problem
in theAppendix
2.For 12 fields K with
~d~
1000 for which we have verifiedKur ~
there exists an
S’4-extension
M ofQ
such thatKgM/Kg
is an unramifiedA4-extension,
whereS4
(resp. A4)
denotes thesymmetric
(resp.
alternat-ing)
group ofdegree
four,
and thereforeKur D
Except
for
Q(V-856)
andQ(B/2013996),
we can charaterize such Ksimply
asdE
d
for some
quartic
number field E: If the discriminant of d of K isdivisi-ble
by
the discriminantdE
of aquartic
number fieldE,
thenKg
has anunramified
A4-extension.
Then the normal closure of E is anS’4-extension
of
Q
unramified at all finiteprimes
over itsquadratic
subfieldSince
dE
I
d,
is unramifiedby
genustheory
and thereforeis an unramified
A4-extension.
Also for the fieldsQ( -856)
and
Q(B/2013996),
we can findS4-extensions
ofQ
thatgive
their unramifiedS4-extensions
by composition.
(For
details,
see§7.)
Therefore,
data forquartic
number fields are useful for ourstudy.
Thus, K
=Q(J2)
withIdl
1000,
can be classifiedsimply
as follows:Here, by
"dE
d" (resp.
d")
we mean "d is divisibleby
dE
for somequartic
number field"(resp.
"d is never divisibleby
the discriminantdE
forany
quartic
numberfield" ),
and in the factorization of d =d’dE, d’
denotesa fundamental
quadratic
discriminant. Weexpect
thatKur = Kl
holdsfor all I~ with
Idl
1507,
because weexpect
thatQ ( -1507)
is the firstK
having
an unramified nonsolvable Galois extension(see
below).
Thisfor all fields .K with
dE
f
d 0
-856, -996,
= holds and that allinequalities
for 1 areequalties.
If ourexpectation
istrue,
the classificationabove can be
replaced by
thefollowing:
Though,
there arepossible
exceptions,
the author think that thisclassifi-cation is
meaningful
because this iscomplete
forIdi
~
719 and even if anexception
would occur forIdl
>719,
modification is easy.For most K we
considered,
Kur = Kl
is verified.Thus,
thefollowing
natural
question
arises: What is the firstimaginary quadratic
number fieldhaving
an unramified nonsolvable Galois extension ?(What
is the first Kwith
Kur
#
Ki
?)
Recent data forquintic
number fields[3, 44]
enable usto
give
apartial
answer(§8,
Proposition
8):
PROPOSITION. The
field
Q("; -1507)
is thefirst
imaginary
quadratic
num-ber
field having
anunramified
A.5 -extension
which is normal overQ
in the sense that noneof
of
discriminants d with 0 > d > -1507 has suchan extension.
We
expect
that the fieldQ( -1507)
gives
the answer to thequestion
above.For the determination of the structure of
Gal(Kur/K),
the results on2-class field towers due to H.
Kisilevsky
[25],
F.Lemmermeyer
[29,
31],
and E.Benjamin,
F.Lemmermeyer,
and C.Snyder
[5]
are veryhelpful. They
give
us information on the structure of the Galois groupGal(K 2 (2)
ofthe second Hilbert 2-class field
K22~
of K over K in many cases.Now we
explain
the notations in our table. In thesimple
expressions
of
Kl
andai,,3i and Ti
denote anyalgebraic
numbersgenerating
theith cubic number field of
signature
(1, ),
the ithquartic
number field ofsignature
(2, 1)
with Galois groupisomorphic
toS4,
and the ithquintic
number field of
signature
(1, 2)
with Galois groupisomorphic
toD5,
re-spectively,
where we consider that the number fields of eachsignature
andeach
type
(of
Galois group of normalclosure)
are numbered up toconju-gacy
by
absolute values of discriminants.(Here
we do not need to considernonisomorphic
fields with samediscriminants.)
G denotes the Galois group
Gal(Kur/ K).
Asusual,
Cn
is thecyclic
group of order n,
V4
is the four group, thatis,
V4 -
C2
=C2
xC2,
Dn
(n > 3)
is the dihedral group of order2n,
Q4n
(n > 2)
is thegeneralized
quaternion
group of order4n,
and,S’D8n
(n >_ 2)
is the semi-dihedral group of order 8n:(m ~
2, n >
3)
denotes the group of order 2mngiven by
M2n
(n >_ 4)
denotes the modular group of order 2ngiven by
A4
is the double cover ofA4 : A4 ^--’
SL(2, 3).
For some
2-groups
we usedesignations given
in the tableby
M. Hall andJ. K. Senior
[17].
We note that T. W.Sag
and J. W.Wamsley give
minimalpresentations
forall 2-groups
of orders26[42].
Forsimplicity,
for some ofthem we use the
following
designations
used in[5].
rl t
denotes the groupof order
2m+t+l
given by
denotes the group of order
given by
We note that
r~,2
=641,3p,
r~,3
=64rsC2,
r!,2
=64r3n2,
r~,3
64r 8 e.
The
organization
of this paper is as follows: In Section2,
we review howto obtain an upper bound for
K]
by using
discriminant bounds. InSection
3,
we review some results on class number relations. In Section4,
we describe the information on the structure of class groups that canbe obtained
by considering
the action of Galois groups on class groups.In Section
5,
we review some results on class field towers. In Section6,
we describe how to determine for selected values of d. In Section
7,
we describe fieldshaving
an unramified extension not containedin
(Kg) 1 -
In Section8,
we describe unramified nonsolvable Galoisexten-sions of
imaginary quadratic
number fields. In theAppendix
1,
weexplain
how to calculate class numbers of
S4-extensions
ofQ.
In theAppendix
2,
we discuss the
problem
ofcharacterizing quadratic
number fields I~ withKJ
Acknowledgements.
The author thanks Prof. R. Schoof for useful ad-vices[43].
He also thanks Dr. F.Lemmermeyer
for information on hisresults
[32].
Table of imaginary quadratic number fields K = 719 with Kl
Continued
Supplements.
Weexpect
that allinequalities
for l in our table areequal-ities. For this
expectation
wegive
thefollowing supplemental
data., , .
V4-extension
L which is anS4
xC3-extension
of K. IfL,
then~K.~,T:L~=2,or4.
For K =
Q( -916).
D3
xC5-
Put L =where denotes the maximal extension of
Q( V229)
unrami-fied at all finite
primes,
which is anS4-extension
ofQ.
(Cf.
For K =(~( -687),
Kur
= See§7.
The index "w-ur" means"weakly-unramified" . )
L is an unramifiedV4-extension
ofK2
which is anS4
xC5-extension
of K. IfL,
then[fur : L]
=2,
or 4.For K =
Q(ý-996). Gal(K2/K) c---
D3
xC6. K2
has an unramifiedV4-extension L which is an
S4
xC6-extension
of K. We haveL] __
322. UPPER BOUND FOR
K]
We review here how to obtain an upper bound for
~K~
for agiven
number field K
by using
discriminant bounds.In this section K denotes an
algebraic
number field of finitedegree.
If we denoteby
nK thedegree
ofK,
thenKth
root of the absolute value of the discriminantdK
of K is called root discriminant of K and denotedby
rdK:
The
following
lemma is fundamental for ourstudy.
LEMMA 1. Let
L/K
be afinite
extensionof algebraic
numberfields.
ThenrdL
=rdK
if
andonly if
L/K
isunramified
at allfinite
primes.
This can be
easily proved by
the transitive law of discriminants.The
following
estimates for rd =rdL
are known(see
[39]):
Uncondition-ally
we haveas n - oo, where C = 0.5772... denotes Euler’s
constant,
and n = nLand ri
(resp. r2)
denotes the number of real(resp. imaginary)
primes
of L. Fromthis,
for allimaginary quadratic
number fields K = d 0 with499,
[K,,,, : K]
oo. Under GRH we haveas n - oo. From
this,
for all K = 0 withI dl
~2003,
[Kur : K]
oo.By
Lemma 1 weimmediately
obtain thefollowing
proposition,
whichtells us how to
get
an upper bound for~Kur : K].
PROPOSITION 1. Let
B(n, rl, r2) (N 9 n
= rl +2r2,
rl, r2nonnegative
integers)
be the lower boundfor
the root discriminantsof algebraic
numberfields
Fof finite degrees
(>_ n)
such thatri(F)/nL
=ri /n
for
i =1, 2,
where
rl(F)
(resp. r2(F)) is
the numberof
real(resp. imaginary)
primes
of
F.Suppose
that K has anunramified
normal extension Lof degree
m.Let H be a
positive integer.
(i)
If rdK
then[Kur : L]
H andtherefore
h(L)
H. Inparticular, if rdK
B (2mnK, 2mrl (K), 2mr2(K)),
thenKur
= L.(ii)
If
h(L) -
1 andrdx
thenKur
= L.We note that the number 60 in
(ii)
is the minimal order of the finite nonsolvable groups, that is the order ofA5,
thealternating
group ofdegree
five.
Thus,
theknowledge
ofgood
lower bounds forB (n, rl, r2 )
is veryim-portant
for ourstudy.
The best known unconditional lower bounds forB(n,
rl,r2)
can be found in the tables due to F. Diaz y Diaz[12].
If we assumeGRH,
much better lower bounds can be obtained. The best knownconditional
(GRH)
lower bounds are found in theunpublished
tables due toA. M.
Odlyzko
[38],
which arecopied
in Martinet’sexpository
paper[34].
All the
imaginary quadratic
number fields K =Q(~), -d
=3, 4, 7, 8,11,
19,43, 67,163
satisfy
the condition in(ii)
for L = K. In factby
the tablein
[12]
we haverdK
163
B(60 -
2, 0, 60
1).
Therefore none of thesefields has any nontrivial unramified extension.
Let K be an
imaginary
quadratic
number field of small conductor. Whatunramified normal extension can we take as L when we
apply Proposition
1 ? Since known lower bounds for
B (n,
rl,r2 )
growlarge
as n growslarge
ifr2 /n
isfixed,
we need to take an extension L oflarge
degree.
By
the reasondescribed in the Introduction it seems that the best choice is L = the
Hilbert class field of the genus field
Kg
of Kexcept
when we caneasily
get
an unramified extension oflarger
degree.
In most cases we cannot conclude = Lonly by
(i)
ofProposition
1,
thatis,
weget
only
L]
Hfor some H >_ 3. Therefore we need to show
h(L)
= 1by
class numbercalculation or
using
criteria for class numberdivisibility.
3. CLASS NUMBER RELATIONS
In this section we review some results on class number relations on
nor-mal extensions of
algebraic
number fields. We will use them to calculate class numbers of fields oflarge degrees
from class numbers of subfields.Let
L/K
be a(finite)
normal extension ofalgebraic
number fields and Gits Galois group. For any
subgroup H
ofG,
we denoteby
1~
the inducedcharacter of G from the
principal
character of H. S. Kuroda[27]
and R. Brauer[6]
proved
independently
thefollowing
relation among the classnumbers
h(M),
theregulators
R(M),
and the numbersw2(M)
of roots ofunity
of2-power
orders of intermediate fields M ofL/K:
If we have a linearrelation among
1 H
then we have the
following
relationHere,
LH
denotes the intermediate field ofL / K
corresponding
to Hby
Galoistheory.
We want to
get
h(L)
from theknowledge
ofh(.K)
andh(M)
forinter-mediate fields M. When an
arbitrary
finite group G isgiven,
there doesnot exist
necessarily
such a nontrivial(that
is,
aH 54
0 for someH)
lin-ear relation. Even if there exists a nontrivial linearrelation,
it is diflicultto use
(t)
in this form because it containsregulators.
However,
in some cases theproduct
can besimplified
into the form[EL : E]/q
H
by elementary calculations,
whereEL
is the group of units inL,
E is itssubgroup
generated
by
E~
for some subfields M ofL,
and q
is some powerof a
prime,
and moreover[EL : E]
is also some power of theprime
dividing
q. Such a
simplified
relation is often very useful for calculation of classnumbers of fields of
high degrees.
Let G =
( a,
bI
a2n
= b2
=1,
b-lab
= be the dihedralgroup of
order 2n
(n > 2)
(the
four group if n =2).
Then we have thefollowing
relation:
From this and some
elementary calculations,
weget
thefollowing
lemmas.LEMMA 2.
(~27, 28))
LetL/K
be aV4-extension
of algebraic
numberfields
and
F,
M and N its three intermediatefields.
Then we have thefollowing
class number relation:
where r is the Z-rank
of
EK,
t is the numberof infinite
primes
ramified
inL/K,
and v E10, 11
(for
thedefinition of
v, see[28]).
Moreover,
the indexThe
special
case K =Q
and L DQ( I)
of this is a classical resultdue to
Dirichlet,
which we can consider as aprototype.
By
this lemma we can calculate class numbers of some fields ofdegree
24.LEMMA 3.
([18,
19,
36])
Let K be animaginary quadratic
numberfield
or the rational numberfield
Q,
and p an oddprime
number,. Let L be aDp-extension
o f
K. Let M and M’ be any two intermediatefields o f
L/K
with[M :
K]
=[M’ :
K]
= p and N theunique
quadratic
subextensionof
L/K.
Then we have thef oldowzng
class number relation:where b =1
i f
K =G~
and L isimaginary,
and b = 2 otherwise.Moreover,
the index
[EL : EMEM,EN]
=pa
with 0 - a - b.Furthermore,
if
L/N
isunramifced,
then a =b -1,
thatis,
Let K be an
imaginary quadratic
number field whosep-class
group iscyclic
of order p. Then the Hilbertp-class
field of .K is aDp-extension
ofQ.
If p =
3,
or5,
by
Lemma 3 we cancompute
its class numberby calculating
the class number of its subfield of
degree
p. For this purpose, we use datafor cubic fields and
quintic
fields. Data for number fields ofdegrees n
with3 __
n _ 7 and small discriminant(in
absolutevalue)
are availableby
anonymous
ftp
frommegrez.math.u-bordeaux.fr
(147.210.16.17).
LEMMA 4.
([9])
Let L be animaginary D4-extension
of Q.
Let M and Nbe the two
nonisomorphic
nonnormalquartic
sub fi elds
and K itsquadratic
subfield
such thatL/K
iscyclic.
Therc we have thef ollowing
class numberrelation:
r ~ T""I T""I 1
Here, v
= 3if
L is aCM-field
and v = 2 otherwise.Moreover,
the index~EL : EMENEK]
is anonnegative
powerof
2.If K is an
imaginary quadratic
number field withcyclic
2-class group oforder
4,
thenby
this lemma the class number of theunique
unramifiedcyclic quartic
extension L of K can becomputed
from the class numbers ofand all of its nonnormal
quartic
subfields have odd class number. Thereforewe have
because the exact
2-power
dividing
h(L)
ish (2)
(K)/4,
whereh~2~ (K)
is the 2-class number of K.4. THE ACTION OF GALOIS GROUPS ON CLASS GROUPS
The action of Galois groups on class groups can often be used to obtain useful information on the structure of class groups.
We first review the
following,
often calledp-rank
theorem.LEMMA 5.
(See
[53,
Theorem10.8])
LetL/K
be afinite cyclic
extensionof degree
nof algebraic
numberfields.
Let p be aprime
nurriberwith p f
nand assume that all
fields
E withK C E §
L have trivialp-class
group.Then
for
eachpositive integer
a, theof
thep-class
groupof
L isa
multiple of
theorder f of
p modulo n. Inparticlar, if
pI
h(L),
thenpf
h(L),
andif h(L)
pf,
The
following
proposition gives
us more information in a sense.PROPOSITION 2. Let K be art
imaginary
abelian numbersfield
andKg
itsgenus
field,
thatis,
the maxirraalunramified
abelian extensionof
K which is abelian overQ.
Let L be the Hilbert classfield of
Kg.
Thenfor
anyprime
number p withp f
[L : Q],
thep-class
groupCl(p)(L)
of
L is trivialor
noncyclic.
Proof.
We first note that both L and its Hilbertp-class
field M are normal overQ.
The Galois group r =Gal(L/Q)
actsby conjugation
onGal(M/L),
which is
isomorphic
toCl(p)(L)
by
class fieldtheory.
This action induces agroup
homomorphism
Now we assume that
CI~P~(L)
iscyclic.
ThenAut(CI~P~(L))
is abelian andtherefore the kernel of p contains the derived group of r. Hence if we let F
be the field
corresponding
toKer(p),
F is abelian overQ
and therefore Fis contained in
Kg.
SinceKer(p)
=Gal(L/F)
acts onGal(M/L)
trivially,
by
theassumption p f [L :
Q]
=If
we conclude thatGal(M/F)
is factoredThis
implies
that the class number of F is divisibleby
IGal(Mj L)I =
and thereforeI
h(Kg) .
HenceCl(p)(L)
must betrivial.
Remark. As described in the
Introduction,
if K is animaginary quadratic
number
field,
we caneasily
calculate the class number ofK9.
Thus,
thiswill be a
powerful
tool.The group
homomorphism
p above oftengives
a useful information onCI(l) (L)
also forprime
divisors l of[L : Q].
The sameargument
works inmore
general
situation. The refereepointed
out that thisproposition
is aspecial
case of a result due to O. Grfn withproofs
(essentially
the same asabove)
by
L. Holzer and A. Scholz(Jaresber.
DMV 44(1934),
74-75).
Let
L/K
be a finite Galois extension ofalgebraic
number fields withGalois group r. Let p be a
prime
number and A thep-class
group of L.The action of r on V = induces a group
homomorphism
where r is the
pl-rank
of A.Thus,
theknowledge
of the structure of lineargroups over finite fields is often useful for the
study
of the structure of classgroups. Later we use the
following
facts.LEMMA 6.
([22,
II,
Satz7.3a])
Let p be aprime
andf
and npositive
integers.
GL(f, pn )
has acyclic subgroup
Sof
orderpin -
1.(S
is calledSinger cycle.)
For anysubgroups
Tof
S whose order is not a divisorof
for
any ef ,
its norrrialixer inGL( f, pn) is
the semi-directproduct
of
S with acyclic
group Cof
orderf
such that the actionof
onegenerator
of
C on S ispowering.
Remark. Let q be a
prime
number
p andf
the orderof p
modulo q.Then any
Sylow q-subgroup
ofGL( f, p)
is asubgroup
of aSinger
cycle
andits normalizer in
GL( f, p)
has orderf (pf -
1).
LEMMA 7.
([8])
Let q be a powerof
an oddprime
and W aSylow 2-subgroup
of
GL(2, q).
If
q - 1(mod 4),
then W isisomorphic
toC2s
1C2,
the wreathproduct of C2s by C2,
where 21 is the exact powerof
2dividing
q -1,
whileif
q - 3(mod 4),
then W is a semi-dihedral groupof
order2t+2,
zuhere2t
is the exact power
of
2dividing
q + 1.Remark.
By
this lemmaSylow 2-subgroups
ofGL(2, 3)
andGL(2,5)
arehas three maximal
subgroups isomorphic
toC4,
D4
YC4,
andSDls.
Here, D4
YC4
denotes the centralproduct
ofD4
andC4.
Later for an unramified Galois extension L of an
imaginary quadratic
number field which is normal over
Q,
we consider the action of theGa-lois group
Gal(L/Q)
on the class groupCl(L)
of L. Then animportant
problem
is to determine theimage
of the induced grouphomomorphism
Aut(Cl(L)).
Thus,
we prepare thefollowing.
LEMMA 8. Let K be a
quadratic
numberfield
and L itsfinite
extensionwhich is normal over
Q. Suppose
thatL/K
isunramified
at allfinite
Prirraes.
Then the Galois groupof
L/Q
isgenerated by
elementsof
ordertwo not contained in the Galois group
of
L/K.
Inparticular,
Gal(L/Q)
isisomorphic
to noneof
the2-groups C4, Q8,
SD16,
andC4
IC2.
The former assertion of this is a
special
case of[11,
Corollary
16.31],
which is
easily
deduced fromCebotarev monodromy
theorem. None of the2-groups C4, Q8, SDls,
andC4 62
isgenerated by
elements of order two.We note that if a
quadratic
number field has an unramifiedquaternionic
ex-tension which is normal over
Q,
then its Galois group overQ
isisomorphic
to
D4
YC4
([30]).
5. RESULTS ON CLASS FIELD TOWER
We also use results on class field tower. We review here some results.
The
following
is well known.LEMMA 9.
(See [49,
TheoremI].)
Let K be analgebraic
numbersfield of
finite degree and P
anyprime
number,.If
thep-class
group,i.e.,
thep-part
of
the class groupof
K iscyclic,
then thep-class
groupof
the Hilbertp-class
field
of K is
trivial.Moreover, if
p = 2 and the 2-class groupof
K isisomorphic
toV4,
then the 2-class groupof
the second Hilbert 2-classfield,
thatis,
the Hilbert 2-classfield of
the Hilbert 2-classfield,
of
K is trivial.This can be
easily proved by using
grouptheory.
All fields we consider have
cyclic p-class
group for any oddprime
p. Infact,
anoncyclic p-class
group for an oddprime
p first occurs asC1(Q(B/-3299)),
which isisomorphic
toCg
xC3
([7]).
On the otherhand,
noncyclic
2-class groups often occur as the 2-class group of(a(f ), d
0with small But for all fields we
consider,
the 4-ranks of the class groups are at most one. The firstexample
forbigger
4-rank is d = -2379.In the rest of this
section,
we review some results on the 2-class fieldtowers of
imaginary quadratic
number fields K:K =
Ko2~ C Ki2~ c K22~ C ~ ~ ~
(Ki+i
is the Hilbert 2-class field ofK~2~.)
By
the lemmaabove,
if the 2-class groupC1~2~ (K)
of K iscyclic,
then the2-class field tower of K terminates with that
is,
K22~
= Whendoes
K22~
= hold? F.Lemmermeyer
[29]
gave a necessary condition:
If
K22~
=Ki2~,
thenCI(2) (K)
iscyclic,
orisomorphic
toC2
x1).
He also obtained sufficient criteria in terms of the factorization of d. If
C1~2~ (K) ^_-’
V4,
thenK32~
=K22~
by
the lemma above. We know thatany nonabelian finite groups of order 2’~ with n > 2 whose abelianization is
isomorphic
toV4
isisomorphic
toQ2n ,
orSD2,.. Thus,
if(K) E£
Y4,
thenV4, D 2- Q2n,
orSD2n,
where n is determinedby
2 n-2
11
h(K (2))
(and
therefore n can beeasily calculated,
becauseKi2) _
Kg).
H.Kisilevsky
[25]
characterized the occurrence of these groups interms of the factorization of the discriminant d of K.
When does
K32~
=K22~
hold? Thecomplete
answer has not beengiven
yet.
However, Lemmermeyer
[29]
determined when/ K)
ismeta-cyclic.
(Of
course, if ismetacyclic,
K32
=K2 )
by
Lemma9.)
Heproved
that if is nonabelian butmetacyclic,
thenC1~2~ (K) ’-’
V4,
or d is of the form-4pq,
where pand q
areprimes
congru-ent to 5 modulo 8:
(i)
If(plq)
=1,
thenGal(’ 2
/K)
isisomorphic
to the groupgiven by
according
as the norm of the fundamental unit of the realquadratic
numberfield is -1 or 1.
Here, n
is determinedby
2n11
h(Q(ýPq))
(in
both
cases),
and m is determinedby
11
In[5],
these groups are denoted
by
MC,~
andrespectively.
We use thesenotations in the
Appendix
2.(ii)
If(p~q) _ -1,
thenM2m+2,
where m is determinedby
and that if we
put
T = thenT2
= 1 and =a 27n+l
.
Recently
E.Benjamin,
F.Lemmermeyer,
and C.Snyder
characterized K whose 2-class fields havecyclic
2-class group[5].
They proved
thatGal(K 2 (2)
isnonmetacyclic
and its derived group iscyclic,
if andonly
if d is of one of thefollowing
forms: d =-4Pp’,
wherep and
p’
areprimes
with p -
1,
p’ -
5(mod
8)
and(p/p’)
=1,
(p/p’)4(p’/p)4
=-1;
d =-rpp’,
where -r is anegative
discriminant #
-4,
and p andp’
arepositive
prime
discriminants with(plp’) = (r/p)
= 1 and(r/p’) _
=-1.
Moreover,
/ K)
isisomorphic
tor:n,t
oraccording
asd =
-4pp’ (in
this case2t
=h~2~(Q( -4p)),
ord
=-rpp’
(in
this case2t
= h,~2~(Q( -rp)),
and in both cases 2m =h (2) (K)/2.
6. DETERMINATION OF
In this
section,
we determine Ourprocedure
for each K = is as follows:(1)
We take a normal unramified extension L of K oflarge degree
for which we can obtain an upper bound forL]
less than 60by
discriminantbounds
(unconditional
foridi
~ 420 and conditional foridl
>420).
(2)
We showh(L)
= 1 and conclude = Lby Proposition
1.(For
fields withcyclic
class group of odd orderprime
to15,
we can conclude =K,
from[Kur : Ki]
168(see below).)
(3)
We determine the structure of(This
is not difficult formost
fields.)
As described in
§2,
for mostK,
we take L =(Kg),.
Theexceptional
cases are the
following
two cases:(El)
d is divisibleby
the discriminantdE
of aquartic
number field E andthe
quotient
d/dE
is a fundamentalquadratic
discriminant or 1. In this casethe normal closure M of E is an unramified
A4-extension
of thequadratic
number field
Q(J5)
and thecomposite
field KM is not contained in This case is divided into thefollowing
three subcases:(a)
d =dE
=-p, where p is a
prime -
3(mod 4):
d= -283, -331, -491,
-563, -643,
-751. In this case, K has an unramifiedA4-extension,
whichyields
aquaternionic
extension ofKi
=(Kg),
by
composition.
(b)
d =dE
is acomposite:
d = -731 = 17.(-43).
In this case, K hasan unramified
A4-extension,
whichyields
aV4-extension
ofKl
=by
composition.
d = -687 =
(-3) .
229, d
= -771 =(-3) ~
257, d
= -916 =(-4)’
229. Inthis case, K has an unramified
S4-extension,
whichyields
aV4-extension
of
(Kg),
by composition.
(E2)
Though
K does notsatisfy
the condition in(El),
we can check thatK has an unramified
S4-extension:
d =-856,
-996.These
exceptional
cases will be treated in the next section. In the restof this
section,
we treat the other fields. Let L = Then we will showK
We can treat fields with same class group
similarly.
(Of
course, for somefields,
additional consideration isneeded.)
None of K withh(K)
= 1 hasany nontrivial unramified extension. For the fields K with
h(K)
=2,
wehave determined in
[58].
We first treat fields with odd class number > 3. Note that
Kg
= K if andonly
ifh(K)
is odd. We note thatexcept
for fields listed in(a)
no fields with odd class number appear in our table as fields with1 ~ 2. For all K with odd class number
and idl
1000except
for d =-787, -827, -859, -883, -947, -967, -971, -991,
weget
Ki]
60by
discriminant bounds. Therefore our task is to proveh(Ki)
= 1. We have thefollowing.
PROPOSITION 3. Let K be an
imaginary
quadratic
numberfield
withcyclic
class group
of
odd order n, > 7. Assume thatfor
all intermediatefields
Eof K,IK,
we haveEl = Kl.
Thenif
h(Kl)
> 1(this
isequivalent
toK2 ~
Kl),
we have26
= 64.(This is
valid alsofor
realquadratic
number
fields.
In the case where K isimaginary
andh(K)
is aprime
(>_
7),
64 can be
improved
to132
= 169. See theremark after
theproof.)
Proof.
Let 1 be aprime
factor ofh(K)
= n. Thenby
theassumption
there exists a subextension E of
Kl/K
such thath(E)
= l andEl
=Kl.
Therefore
l ~ h(Kl)
by
Lemma 9.Let p
be aprime
factor ofh(Kl)
and r thep-rank
of thep-class
groupof
Kl.
We will showp’’ >__
26
= 64. Sinceby
Lemma 5 andProposition
2we have
r >
2 and 1(mod n),
it suffices to eliminate thefollowing
four
possibilities:
(a)
p= 2, r
= 3 if n =7;
(b)
p =2, r
= 4 if n =15;
(c)
p = 2, r = 5 if n = 31;
(d) p = 3, r = 3
if n = 13. Note that for all thesepossibilities,
r is the order of p mod n.Put r =
Dn
and A =Cn.
The action of rIf A n
{1},
then we would haveI h(F)
for the field Fcorres-ponding
to A nKer(p),
which contradicts theassumption.
(Cf.
theproof
ofProposition
2.)
Therefore A nKer(p) _
{1}
andp(A) =
A ££Cn.
SinceIm(p)
is a factor group of F ££Dn,
Dn.
Since A is a normalsubgroup
ofr,
Im(p)
is contained in the normalizer N ofp(A)
inGL(r, p) .
Therefore N must have asubgroup
ofisomorphic
toD~,.
Assume that n is a
prime
and that r is the orderof p
mod n. Thenp(A)
is a
subgroup
of aSinger
cycle
ofGL(r, p)
satisfying
the condition of Lemma6. Therefore N ^--’
Cpr-l
xCr.
Since N has asubgorup
ofisomorphic
tor must be even. This eliminates the
posiibilities
(a),
(c),
and(d).
We know
GL(4,2)
=PSL(4,2) ~
A8
[22,
II,
Satz6.14,
(5)].
Since thisgroup does not have a
subgroup isomorphic
toD15,
andn = 15. This
completes
theproof.
Remark.
Regrettably
the sameargument
does not work forh(K)
= 5: TheSinger cycles
ofGL(4,2) ~
A8
have asubgroup isomorphic
toDs.
However,
the class number relation(Lemma 3)
enables us tocompute
h(Kl)
by
calculating
class numbers ofquintic
subfields ofKl.
For all K withh(K)
= 5(all
such K havediscriminant ~
2683[2, 52]),
we haveh(Kl)
= 1and therefore
K2
=Kl.
We note that we can find in[20]
the class numbers of the unramifiedcyclic
quintic
extensions F of K with 51
h(K)
and~d~
1000. For all suchK,
h(F)
=h(K)/5.
By
Lemma 3 we can checkthat for all K with 3
1 h(K)
andidl
1000except
for d listed in(a),
we have
h(F)
=h(K)/3,
where F is theunique
unramifiedcyclic
cubic extension of K.F.
Hajir
checked theparities
ofh(Kl)
of K withCq
(q
an oddprime
19)
andIdi
15000by using elliptic
units[16]:
For all such Kexcept
for d = -283(q
=3),
-331(q
=3),
-643(q
=3),
and -14947(q
=17), h(Kl)
is odd(28
= 2561 h(Kl)
for d =-14947).
All K withh(K)
= 7 have discriminant 5923[2,
52]
and thereforeh(Ki)
is odd forall such K.
By
this reason in the case where(K
isimaginary
and) h(K)
= q(q
an oddprime >_
7),
64 can beimproved
toh(Kl) >_
132
= 169.(Note
thatGL(2,13)
has asubgroup isomorphic
toD7.)
Moreover,
in thecase where
h,(K) >_
11 and21,63,
64 can beimproved
toh(Kl) >_
2$ =
256(without
assuming
theimaginarity
ofK.) (Note
thatGL(8,2)
has asubgroup isomorphic
toD17.)
Suppose
Cn,
where n is an oddinteger
prime
to 15. The estimate 169 under theassumption
h(Kl)
> 1 isimportant.
By
this,
if[fur :
168,
then we can conclude =Kl
under thenormal over
Q.
(Though
there exist(probably
infinitely)
manyimaginary
quadratic
number fields withcyclic
class group of odd orderprime
to 15having
such anextension,
the maximal discriminant of such fields is -2083.(h(Q( -2083)
= 7. See§8.))
For thisconclusion,
we needonly
to showthat
Kl
does not have an unramifiedA5-extension.
(Note
that the secondminimal order of nonabelian
simple
groups is168.)
Suppose
thatKl
has such an extension M. Then the Galois groupGal(M/K)
is an extensionof
A5 by
C~,.
Since any such extension is the directproduct
A5
xCn,
K has an unramifiedA5-extension
and this extension must be normal overQ,
for otherwise its normal closure has
degree
2 ~ 60z
= 7200. This contradicts theassumption. Thus,
[Kur : Kl~
168implies Kur
=Kl.
For K =Q(~/Il827), Q( -859), Q( -967), Q(~/~991),
we do notget
Ki
60 but
get
168by
Odlyzko’s
(conditional)
discriminant bounds and therefore =Kl
(under GRH).
Now we treat fields K with even class number. For such
K,
K.First we calculate
h(Kg ) .
We have two cases:The
equality
h(Kg)
=K]
isequivalent
to =Ki.
(This
holdstrivially
in the case of odd classnumber.)
IfKur
=Kl,
thenwe have =
Kl.
We expect the converse, thatis,
if(Kg),
=Kl,
thenKur = Kl
holds for fields considered. Infact,
we can show this for mostK.
We first treat fields with nontrivial
cyclic
2-class group. For suchK,
its discriminant d is theproduct
of twoprime
discriminants: d =dld2
andKg
=Q(,/d-,,
Sinceh(K9)
buth(K) I
h(Kg)
by
Lemma9,
we have
h(Kg)
=by
Lemma 2.(Note
thatsince
d1
andd2
areprime discriminants,
both and are odd. Moreprecisely,
we have_ , F "r , , - - . F F , , r F .
where
Cl(K)
xCl(K)
butCI(K)2
= ECI(K)1,
be-cause the odd
part
of the class group of abiquadratic bicyclic
numberfield is
isomorphic
to the directproduct
of those of itsquadratic
subfields(see
[28]).
Hence in this case =h(K)/[Kg :
K]
isequivalent
toh(Q( dl))
= = 1. We taked2
asd2
> 0. How often does= 1 occur? H. Cohen and H. W.
Lenstra,
Jr.[10]
posed
heuris-tic
conjectures
(the
so-called Cohen-Lenstraheuristics)
on the distributionprobability
that a realquadratic
number field withprime
discriminant hasclass number one is about 0.75446 and this has been
numerically supported
by
A. G.Stephens
and H. C. Williams[48].
The first two discriminants of realquadratic
number fields withprime
discriminant and classnumber >
1are 229 and 257. These are also discriminants of
quartic
number fields! TheCohen-Lenstra heuristics state also that the
probability
that animaginary
quadratic
number field has class group whose oddpart
iscyclic
is about 0.97757.Thus,
forsimplicity,
we consider here fields with = 1and
Cn
(n
an oddinteger).
Thenh(Kg)
=K]
is
equivalent
to n = 1. When n =1,
under some conditionby
considering
Galois action as in
Proposition
3,
weget
an estimate forh(Ki).
When ~ >1,
we caneasily
determine the structure of We firstgive
an estimate forh(Kl)
if n = 1 and then determineGal((Kg)l/K)
ifn>1.
PROPOSITION 4. Let K be an
irraaginary quadratic
numberfield
withnon-trivial
cyclic
2-class group.(i)
AssumeC2m
(m >_ 3)
and that theunique
unrarraified cyclic
quartic
extension Eof
K has class number2m-2.
(The
latterassumption
irnplies
h(Kg)
=h(K)/2.)
If
h(Kl)
>1,
then we have72
= 49.(ii)
Assumeh(K)
=2q
(q
an odd7)
andh(Kg)
=q
(=
h(K)/2).
If h(Kl)
>1,
then we haveh(Kl) __>
26
= 64.(These
are valid alsofor
realquadratic
numberfields if
Kg
is theunique
unramified quadratic
extensionof
K.)
Proof.
Wegive only
a sketch.(i)
Put r =Gal(Ki /Q).
Then r =D2-.
By
Lemmas 7 and 8 any grouphomomorphisms
F -GL(2,3)
and F -GL(2,5)
haveimage
of order at mosteight
and thisimplies
that neitherC~
norC5
can occurby
theassumption.
(Cf.
theproof
ofProposition
2. Note thatthe
Sylow 2-subgroups
ofGL(2, 7)
areisomorphic
toSD32
andthey
have asubgroup isomorphic
toD8.)
Hence weget
the conclusion.(ii)
Put F =Q( 04).
ThenD9.
Thus,
the sameargument
as in theproof
ofProposition
3 works.Remark. Note that when K has an unramified
cyclic quartic
extensionE,
we can calculateh(E)
by using
Lemma 4. The fields withCl(K) ’-’
C2m
(m >_ 2) and
1000 are divided into twotypes:
fields K withh(E)
=2 m-2
and fields K with =2’n-lq
(q
an oddprime).
Formost of the fields of the former