The imaginary abelian number fields with class numbers equal to their genus class numbers

J

OURNAL DE

T

HÉORIE DES

N

OMBRES DE

B

ORDEAUX

K

U

-Y

OUNG

C

HANG

S

OUN

-H

I

K

WON

The imaginary abelian number fields with class numbers

equal to their genus class numbers

Journal de Théorie des Nombres de Bordeaux, tome

12, n

o

_{2 (2000),}

p. 349-365

<http://www.numdam.org/item?id=JTNB_2000__12_2_349_0>

© Université Bordeaux 1, 2000, tous droits réservés.

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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques

The

_imaginary

abelian number fields with class

numbers

_equal

to

their

_genus

class numbers

par KU-YOUNG CHANG et SOUN-HI KWON

RÉSUMÉ. Titre

_{français :}

Sur les _corps abéliens dont le nombre de classes est

_égal

au nombre de _genres.

Nous savons

_qu’il

n’existe

_qu’un

nombre fini de corps abéliens

imaginaires

pour

lesquels

le nombre de classes est

_égal

au nombre de genres. Ceux de ces _{corps qui} sont

_cycliques

et non

quadra-tiques ont été classés dans

_[Lou2,4]

et dans

_[CK].

Dans cet

arti-cle,

nous déterminons tous les corps abéliens non

cycliques

dont le nombre de classes est

_égal

au nombre de _genres. Cela achève la classification des _corps abéliens

_possédant

une classe par genre,

sauf dans le cas des _corps

_quadratiques

_imaginaires.

ABSTRACT. We know that there exist

_only

_finitely

_many imagi-nary abelian number fields with class numbers

equal

to their _genus class numbers. Such

_{non-quadratic cyclic}

number fields are

com-pletely

determined in

_[Lou2,4]

and

_[CK].

In this paper we de-termine all

_non-cyclic

abelian number fields with class numbers

equal

to their _genus class

_numbers,

thus the one class in each

genus

problem

is

solved, except

for the imaginary

quadratic

num-ber fields.

1. Introduction

For an abelian number field N the narrow _genus field

_GN

of N is the maximal abelian number field

_containing

N such that the extension

_GN/N

is unramified at all finite

_places.

The

_degree

_{~GN : N]}

is called the _genus class number of N and is denoted

_by

_{gN .} The _genus class number of N is easy to

_determine,

_namely,

_gN = where ep is the

ramifica-tion index of _p in N. In

_particular,

when N is

_imaginary,

the _genus field of N is contained in the Hilbert class field of

_N,

whence _gN divides the class number of N. In

_[Loul],

Louboutin has

_proved

that there exist

_only

finitely

many

imaginary

abelian number fields with class numbers

_equal

to

with class numbers

_equal

to their _genus numbers have been

_recently

classi-fied ([Lou2, 4], [CK]).

It is known that under a suitable

generalized

Riemann

Hypothesis,

there are

_exactly

65

_imaginary

_quadratic

number fields with class numbers

_equal

to their _genus

_numbers([We]

and

_[Lou5]).

These fields

are listed in Table I. The aim of this _paper is to _prove the

_following

result: Theorem 1. There are

exactly

_{424 imaginary}

non-quadratic

abelian

num-ber

_fields

with class numbers

_equal

to their _genus class numbers:77 out

_of

them are

_cycdic

and

₃₄₇

are

non-cyclic.

Their

degrees

are less than or

_equal

to

_24,

their class numbers are

_{1, 2}

or

_4.

The conductors

_of

these

fields

are

less than or

equal

to 65689. These

fields

are listed in tables at the end

of

this paper.

We note that in

_[M]

under the

_assumption

of the Generalized Riemann

Hypothesis, Miyada

has determined all

_imaginary

abelian number fields

N with class numbers

_equal

to their _genus class numbers such that the Galois _groups are

_elementary

_2-groups.

(There

seems to be one

_misprint

in

_[M,

Table

_2]:

the field K

_{= Q}

_(~/~2,

_V-37)

_appears

_twice,

i.e. for

9K

= hK

= 1 and for

gK

= hK

= 2. In

fact,

gK

= hK

=

2.)

However,

we do not need _any

_assumption

for the

_proof

of Theorem 1. This _paper

is

_organized

in the

_following

_way. Section 2

_presents

some well-known

facts on

_imaginary

abelian number fields. In Section 3 we illustrate our

computations.

Throughout

this paper the

following

notation will be used.

For an

_imaginary

abelian number field

_K,

let

K+,

_fK,

_hK,

_Wx,

GK,

_9K,

hK,

QK

be the maximal real

_subfield,

the

_conductor,

the class

_number,

the number of roots of

_unity

in

_K,

the _genus

_field,

the _genus class

_number,

the relative class number and the Hasse unit index of

_K,

_{respectively.}

For an

odd

_prime

p let Xp be an odd Dirichlet character of conductor _p, of

_degree

p - 1. When p &#x3E;

2,

let

qbpp

be an even

_primitive

Dirichlet character of

conductor

_pP,

of order with = For the

prime

p =

2,

let X4

be the odd Dirichlet

_quadratic

character of conductor 4. When

_{p &#x3E; 3,}

let

02P

be an even

_primitive

Dirichlet character of conductor

_2P,

of order

2p-2

with

_9)p

=

2. Preliminaries

In this section we review some well-known facts

concerning imaginary

abelian number fields.

Proposition

1.

₍₁₎

Let F be an abelian number

field. If

_hF

=

gF, then

hM

=

gM

for

any

subfield

M

of

F.

(2)

Let F be an

_imaginary

abelian number

field,

_XF the _group

_of

_primitive

Dirichlet characters associated to F and

_XF

the subset

_of

_X E _XF such that

_{X( -1) =}

-1. For _{a X} E

_{XF let us}

denote the conductor

_of

_X

_by

fx.

We have

(3)

If F is

an

_imaginary

cyclic

number

field,

then

QF =

1. Let F c K

be the two

_CM-fields.

_If

_{(K : F]}

is

_odd,

then

_QK

=

QF.

(4)

Let F be an

_imaginary

cyclic

number

field of degree

2m.

If

hF

= gF,

then

Proof.

(1)

See Lemma 1 in

_{[M]. (2)}

See Theorem 4.17 in

_{[W]. (3)}

For the first statement see Satz 24 in

[H].

For the second statement see Lemma 2

in

_{[HY1]. (4)}

See

_[Lou2

and

_4]

and

_{(CK~ .}

D

3. Proof of Theorem 1

Let N be an

imaginary

abelian number field with Galois _group G. When

say N is of

type

(pP, 1, - - - ,

p2 i, - - - ).

In addition we

put *

when the

_{corresponding}

subfield is

_imaginary.

For

_example,

if N is

of

_type

_{(4*, 2),}

then N is the

_compositum

of an

imaginary

cyclic

quartic

field

_Ml

and a real

quadratic

field

M2

with

Ml

n

M2 = Q.

Then the

maximal real subfield N+ is of

_type

_(2,2).

If N is of

_type

_{(4*, 2*),}

then N is

the

_compositum

of an

_imaginary

_cyclic

_quartic

field

Ml

and an

_imaginary

quadratic

field

M2 .

In this case,

N+

is of

type

(4).

A field N of

_type

(2*, 2, 2)

can be either of

_type

(2*, 2*, 2)

or

(2*,

2*,

2*).

However we

prefer

to say that N is of

type

(2*, 2*, 2*)

with as _many as

_{possible *}

for a technical

reason to be seen later.

Let N be an

imaginary

abelian number field of

_type

(2m*,

,2m:,

_{nl ,} I

nr)

with ml 2: ... 2 ms. We may assume 2ms for _n2 which

is a

_2-power.

In order to describe N we

give

the associated _group of

Dirich-let characters

_{(71,... ,}

Ts , ~p 1, - - - , Here -rl, - - - , Ts are odd

primitive

Dirichlet characters of order for 1 i s and _{Wl, - - - , cpr} are the even

primitive

Dirichlet characters of order nj for

1 ~ j ~

r. For each

_prime

_p

let

G(P)

be the

_{p-Sylow subgroup}

of G. Let

N(2)

be the maximal subfield of

N of

_2-power

_degree.

The scheme of our

proof

of Theorem 1 is as follows.

In 3.1 we determine all

_imaginary

abelian number fields N with

hN

_{= gN}

such that

G~2~

is an

elementary 2-group.

According

to Theorem 2

below,

if

G~2~

is an

_elementary

_2-group,

then the 2-rank of

G (2)

is less than or

equal

to 3. In 3.2 we determine all

_imaginary

abelian number fields N with

hN

=

gN such that

N (2)

is of

type

(2’ni ,

2 M2)

with mi &#x3E; m2 2: 1. Moreover

we _prove that if N is an

_imaginary

abelian number field with

hN -

_gN,

then 2-rank of

G (2)

is less than or

equal

to 3. If it is

_equal

to

_3,

then

G~2~

is

number fields N with such that

N(2)

is of

_type

_{(2’~i , 2’~l )}

with

ml 2: 2 and ml 2: m2.

Finally

it remains to determine all fields N with

hN

= 9N such that

N~2~ _

(4*, 2*, 2*).

These fields are treated in 3.4. Our

determinations

_rely

on the

computations

of the relative class numbers

h jo

of N. The class numbers

_hN+

of

N+

which we need are known in

~G~ , ~L~ ,

[Ma]

and

_[Mä].

In the

_remaining

_part

of this _paper we _say

briefly

"character

of order m" instead of

_"primitive

Dirichlet character of order m" .

3.1.

N(2)

is of

_type

_{(2~,... 2*).}

Proposition

2. Let N be an

_imaginary

abelian

number field,

G the Galois

group

of

N /Q.

Assume that G is not

cyclic

and that the

2-Sylow subgroup

of G is cyclic. If hN

=

gN, then N is

of type

(2*, 3, 3)

and N is associated

with

_(X3~9,~)’

·

Proof.

First,

we claim that if

_{hN -}

_9N, then N is of

_type

_{(2~3, -"}

_~3).

According

to

_[CK]

there are 4

_imaginary

cyclic

number fields of

degree

10

with class numbers

_equal

to their _genus class numbers.

_{By Proposition}

_{I . ( I )}

we cannot have a field N with

hN

_{= 9N} such that N is of

_type

(2*, 5, 5).

_By

the same

_argument

if N is of

_type

(2*, 7, 7),

(2*, 9, 9),

(4*, 3, 3)

or

_{(4*, 5, 5),}

then _{gN .}

_According

to

_Proposition

_{1. ( 1 )}

it follows that if

hN

=

9N,

then N is of

_type

_{(2*, 3,’-’ ,3).}

Let us now consider with the fields of

type

(2*, 3, 3).

Let x

be an odd

quadratic

character,

two cubic

characters,

N the field associated with

_(x,

_~2)-

Let

_{Mi, M2, M3}

and

_M4

be the subfields associated with and

_{(X, ’P2),}

respectively.

The fact that

hN

implies

that

_hMi

= gM; for 1 ~ i _~ 4.

According

to

_{(Lou4) ,}

there is

_only

one N such that N = with

9Mi for

1 ~

i ~ 4. That

is,

N =

MiM2M3M4 = MiA/2,

where

Ml, M2, M3

and

_M4

are associated with

(x3~9~(X3,X~:B:3~~9~

and

(X3?X~)

respectively.

We

_verify

that this field has

hN

=

9N = 1. Since

there is

_only

one field of

_type

(2*,

_{3, 3)}

with class number

_equal

to its _genus

class

_number,

there is no field of

_type

(2*,3,3...3),

of

_{degree 2:}

54 with

class number

_equal

to _{its genus} class number. D

Theorem 2. Let N be an

_imaginary

abelian number

field

such that the

Galois _group G is an

_{elementary 2-group.}

Assume that _9N =

hN .

Then _we

have

, , __ - , - o

Proof.

See Theorem 1 in

_[M].

D

Proposition

3.

₍₁₎

Let N be an

_imaginary

bicyclic biquadratic

number

field

and let

kl

and

k2

be two

_imaginary

_{quadratic subfields. If}

_hN

=

(2)

There are

_exactly

219

bicyclic biquadratic

number

_fields

with class

numbers

_equal

to their _genus class numbers. These

_fields

are listed in

Table III.

(3)

There exist

_exactly

_{42 imaginary}

abelian number

_fields

N

_{of degree}

&#x3E; 4 with class numbers

_{equdl to}

their _genus numbers such that

N (2)

is

_of

_type

_(2*,

_2*).

These

_fields

are listed in Table IV.

Proof.

(1)

Suppose

that

_hN

_{= gN} and 8. Since

_{hN =}

_{9.¡-hkl hk2’}

we must have

_4,

_{QN =}

_1,

_8,

_{hk2 = 1}

and

_{h~,+ =}

1.

_By

[Corollary

1,

M],

there exist at most three rational

_primes

ramified

in N. The fact that

_hN

=

9N = 4

= ~

IT

ep

implies

that there exist

exactly

three ramified

_primes

_including

2 with e2 = 4. This leads _a

contradiction. Since the fact that e2 =

4,

hk2

= 1 and

hN+

= 1

implies

k2 E

and N+ E

with _{p == 1} mod

_4,

there exist at most two ramified

_primes

in N =

k2N+.

(2)

There are 81

_{imaginary quadratic}

number fields with class numbers

_1,

2 or

4(~51,2~

and

[A]).

_Among

these 81 fields there are 51 fields with

class numbers

_equal

to their _genus class numbers. In order to

_get

all

bicyclic

biquadratic

number fields with class numbers

_equal

to their

genus class numbers it is sufficient to examine the

_composita

of two of those 51

_quadratic

fields. Note that we do not assume the Generalized

Riemann

_Hypothesis.

(3)

Let N be an

_imaginary

abelian number fields of

_{degree &#x3E;}

4. Assume

that

_hN

=

gN and is of

type

(2*, 2*).

Bearing

in mind the

imaginary

cyclic

fields with class numbers

_equal

to their _genus class numbers and

_Proposition

2 it remains to consider the fields N of

_type

(2*, 2*, 3)

or

(2*, 2*, 5).

i)

Let Tl, T2 be two odd

quadratic characters,

cp a cubic

characters,

N

the field associated with

_{(Tl, T2, p) .}

Let

_Ml

and

_M2

be the

imag-inary

sextic subfields associated with and

_{(T2, W),}

respec-tively.

Let E be the

_imaginary

_bicyclic

_biquadratic

subfield

asso-ciated with

_{(Tl, T2),}

kl

and

k2

the

_{imaginary quadratic}

subfields associated

_{with 71}

and T2,

respectively. Using

[Lou4],

we make a

list of

_(Ml,

_M2)’s

such that

_hMl

=

gMl,

hM2

=

g,yl2 and

hE

= gE.

There are 40

_pairs

of

_{(Ml, M2)}

_satisfying

_hm,

=

gMl,

h,yl2

=

9M2 and

_hE

=

9E. For these 40 fields we determine

_using

the fact

h-

h-that

_hN=

and

_{QN = QE.}

The class numbers of WM1 WM2

real sextic number fields are known

by

[Mä].

There are 39 fields

N of

_type

_{(2*, 2*, 3)}

with

_hN

=

gnr.

(The

remaining

field is

as-sociated with

_{X51, x7).}

This field has class number 3 and

ii)

By

the same method as

i)

we

verify

that there are three

_imaginary

abelian number fields N of

_type

_(2*,

_2*,

₅₎

with

hN

= gN.

D

Proposition

4. There are

_exactly

19

_imaginary

abelian number

_fields

N

with class numbers

_equal

to their _genus class numbers such that the

_subfield

N (2) is

_type

_of

_{(2*, 2*,}

_2*).

There are

exactly

17

fields of

_type

(2*, 2*, 2*),

and two

_{fields of}

_type

_(2*,

_{2*, 2*,}

_3).

These

_fields

are listed in Table V.

Proof.

Let N be an

_imaginary

abelian number field N with class

num-ber

_equal

to its _genus class number such that the subfield

N (2)

is

_type

of

(2*,

2*,

2*).

According

to

_[Lou4]

and

_[CK],

_{if p &#x3E;}

_11,

then

G(P)

is trivial. Let us

_begin

with the fields of

_type

(2*, 2*, 2*).

If N is of

_type

_{(2*, 2*, 2*)}

and if

_hN

=

9N, then

hN

= 1 _or 2.

Let

_{~2, k3}

and

k4

be the four

_imaginary

_quadratic

subfields of N. Since

hN =

where

hi

=

hz

and wi

is the number of roots of

N i= Wi

unity in ki

for i

_{=1, 2, 3}

and

_4,

there are at least three

_{imaginary quadratic}

subfields ki

with _{4 among} the four

_imaginary

_quadratic

subfields of N. It remains us to consider the

_{composita klk2k3}

of three

_imaginary

quadratic

subfields ki

with class numbers

_1,

2 or

_4,

i =

1, 2

and 3. There

are 18

_composita

N =

ki k2k3

satisfying

the

following:

i)

hki

=1, 2

or 4 for i

_{=1, 2, 3.}

ii) hki

for i

_{= 1,2,3.}

iii)

For the six

_imaginary

_{bicyclic biquadratic}

subfields M of N we have

hM

=

9M.

Computing

the unit indices

_{QN by}

_[HY3]

and

_hN+

_by

_[Satz

5,

_K]

for these 18 fields

_N,

we

verify

that there are

exactly

17 octic fields N with

hN

= _gN.

Consider now the fields of

_type

(2*,

2*, 2*,

3).

Let _71,72,73 be three

odd

_{quadratic characters, cp}

a cubic

character,

N the field associated with

(71,72,73,

Let

_{Ml, M2, M3}

and

M4

be the four

_imaginary

sextic sub-fields of N associated with

_{(7i,~),(72~)?(73,~),}

and

respec-tively.

Let E be the subfield associated with

_{(Tl, T2, T3~.}

_According

to

[Lou4],

there are

only

two fields N such that

_h~2

=

gM2 for 1 i 4 and

hE

=

gE. For these two fields we

_compute

_hN

and

_{hN+ .}

The results are

summarized in Table V.

Finally,

it is sufficient to notice that there is no field N of

_type

(2*, 2*, 2*, 5)

with

_hN

_{= gN.} D

Proposition

5.

₍₁₎

There are 15

_imaginary

abelian number

fields of

_type

(4*,

2)

with class numbers

_equal

to their _genus class numbers. These

(2)

Let m be an odd

integer

with m &#x3E; 3. There is no

_imaginary

abelian

number

_{field of degree}

8m with class number

_equal

to _{its genus} class number such that the

_{subfield N (2)}

is

_of

_type

_{(4*, 2).}

(3)

There is no

field of

_type

(8*, 2)

with class number

equal

to its _genus

class number. There is no

field of

_type

(16*, 2)

with class number

equal

to _{its genus} class number.

_{Consequently, if}

N is an

_imaginary

abelian nurrzber

_field

with

_hN

= _gN such that the

subfield N (2)

is

of

type

(2’"’~i, 2’"’’2)

with ml &#x3E; m2 &#x3E;

1,

then N is

of

type

(4*, 2).

(4)

There is no

_imaginary

abelian number

field of

_type

(4*,

2,

2)

with class

number

_equal

to its genus class number.

(5)

If

N is an

imaginary

abelian number

field

with

hN

= _gN, then the

2-rank

_{of C(2)}

is less than or

equal

to 3.

Moreover,

if

the 2-rank

of

C(2)

is

_equal

to

_3,

, then

G(2) =

(2*,

2*,

2*)

or

(4*,

2*,

2*).

-Proof.

(1)

Let cp be an odd

primitive

character of order

4,

_X an even

primitive

quadratic

character,

N the associated field with

_{(rp, X).}

Let

Ml

and

_M2

be two

_imaginary

_cyclic

_quartic

subfields of N associated with

_(cp)

and

_(CPx),

_{respectively.}

The real

_quadratic

subfields of

_Mi

and

_M2

coincide. In order to obtain N with

hN

=

gN it is sufficient to

consider the

_composita

Ml M2

such that

_hMl

=

gMl and

hM2

=

There are 35 fields N such that N =

MlM2,

hMl

=

9Ml and

hM2

=

9M2. The Hasse unit indices

Qnr

are

_easily

obtained from Sätze 15

and 22 of

_[H].

The relative class number

_hN

can be

expressed

as

The class number

_hN+

can be

_computed

_following

~K~ .

It remains 15

fields N such that

hN

=

gN.

Similarly

we

verify

(3).

(2)

Let N be an

imaginary

abelian number field of

degree

8rrL, containing

a subfield of

_type

(4*, 2).

Suppose

hN

=

gN.

According

to

[CK],

we have m = 3. Let

p, x and w be an odd character of order

4,

an

even

quadratic

character and a cubic

character,

respectively.

Assume

that N is associated with

_(p,

_X,

_w).

Let

Ml

and

M2

be the subfields associated with

_(pw)

and

_(CPxw),

_{respectively.}

_Using

Table II in

_[CK],

we

verify

that there is no

_pair

of

(Ml, M2)

such that

_hMl

=

gMl and

hM2 =

9M2 - O

(4)

Let

_{cp be}

an odd

_primitive

character of order

4,

_Xl and _{x2 two} even

primitive

quadratic

character,

N the associated field with

_x2~.

Let

_{Ml, M2, M3}

and

M4

be four

_imaginary

_cyclic

_quartic

subfields of

N associated with

_(cp~,

_(WX1),

_(px2)

and

_{respectively.}

The fields

_{Mi,1 i 4,}

have the same real

quadratic

subfield.

Suppose

that

_{hN =}

_{gN .} Then we have

_hMi

=

gM; , for 1 i

4,hMIM2 =

and

_{hM3 M4}

= We

verify

that there is _no such

quadruple

of(Mi,M2,M3,M4).

(5)

By

(3)

and

₍₄₎

it is sufhcient to

_verify

that there is no field N of

type

(4*, 4*, 2*)

with

_{hN -}

_gN.

_Suppose

that there is a field N of

type

(4*,

4*,

2*)

with

hN

9N. Let cpl and p2 be two odd

primitive

characters of order

_{4, X}

the odd

_quadratic

character such that N is associated to the group

(~pl, ~p2, x) .

Then we would have three fields

of

_type

_{(4*, 4*)}

with class numbers

_equal

to their _genus class

_numbers,

i.e. the fields associated with

_(Wl,

_~p2),

_((Pl,

and

_~cp2,

This contradicts

_Proposition

_{8. ( 1 )}

below.

0

3.3. is of

_type

_{(2’ni , 2m2 )}

with 2 and m2 ·

Bearing

in mind

_Proposition

_5.(3)

if

hN =

gN and

N(2)

is of

type

(2’ni , 2m*)

with m2 2::

2,

then ml =

m2 = 2. We need

_only

consider N such that

N~2~

is of

_type

_{(4*, 2*), (8*, 2*), (16*, 2*)}

or

(4*, 4*).

Proposition

6. There exist 39

_imaginary

abelian numbers

_fields

N with

hN

= gN such that the

subfield N (2)

is

of

type

(4*, 2*):

36

_of

them are

of

type

(4*, 2*)

and 3

_of

them are

_of

_type

(4*,

_2*,

3).

These

_fields

are listed in

Table VI.

Proof. According

to

_[CK],

if N is an

_imaginary

abelian number field with

hN

= gN such that the subfield

N (2)

is of

type

(4*, 2*),

then

_{[N : Q]}

40. Our first

_step

is to determine all fields of

_type

_{(4*, 2*).}

Let _p be an odd

character of order

_{4, X}

an odd

_{quadratic character,}

N the field associated

with

_{~~p, x) .}

Let M be the field associated with

_(p)

and let

ki

and

k2

be the two

_{imaginary quadratic}

subfields associated with

_(X)

and

respectively. Using

[Lou2]

we make a list of N such that

hM =

_gM and

hE

=

gE, where E is the field associated

(p2,

x .

We use the formula

/.... ’B. 7-....

To determine

_Qnr

we use the tables in

~H~,

[YH1]

and

[YH2]

for the fields

of conductors

_200,

and Satze 15 and 22 in

_[H]

for the fields of conductor &#x3E; 200. For

_example

let us consider

QN

for the field N which is associated

with

_{X 3).}

We have

_{N+ = Q}

~7

(5

+ and N =

N+

(H).

In order to determine

_{Q N}

it is sufficient to know whether the

_prime

ideal of

N

_lying

above 7 is

_principal

or not.

_Using

the function

_{IdealIsPrincipal}

of

KASH([KT])

we know that this

_prime

ideal is

_principal.

Hence we conclude

that

_QN

= 2. On the other

hand,

the class number of real

quartic

fields associated with

_(cpX)

are known

by

[G].

_Using

the above results we can

easily

determine all fields of

_type

_(4*,

_2*,

₃₎

and

_verify

that there is no field

of

_type

_(4*,

_2*,

₅₎

with class number

_equal

to _{its genus} class number. The

computational

results are

compiled

in Table VI. D

Proposition

7.

₍₁₎

Let N be an

_imaginary

abedian numbers

_field

with

hN

=

gnr. If

the

subfield N (2)

is

of

type

(8*, 2*),

then N is

_of

_type

(8*, 2*).

There exist three

_imaginary

abelian number

_{fields of}

_type

(8*, 2*)

with class numbers

_{equal to}

their _genus class numbers. These

fields

are

_given

in Table VII.

(2)

There is no

_imaginary

number

field of

_type

(16*, 2*)

with class number

equal

to its genus class number.

Proof.

(1)

According

to

_[CK]

we know that for an odd

_integer

m &#x3E; 3 there is no

imaginary

cyclic

number field of

degree

8m with class

number

_equal

to _{its genus} class number. Let _cp be an odd

char-acter of order

_{8, X}

an odd

quadratic

character,

N the field

asso-ciated with

_(W,X).

Let

_{M, L, F, E, kl,}

and

k2

be the subfields

as-sociated with

_(W),

_{(W2X), (W2,}

_X),

_(cp4,

_X),

_(X)

and

_{respectively.}

By

[Lou2],

Propositions

3 and 6 we have three fields N such that

him

=

= 9L, and

hE

=

9E. These fields are listed in

Table VII.

_Using

the formula

we obtain

immediately

The class numbers of real

cyclic

fields of

conductor 100 are known

by

[Ma].

The class number of the real

abelian number fields of conductor 160 in Table VII is

_equal

to 2

according

to Theorem 3 of

_[L].

(2)

It is clear from the fact that there is

_only

one

imaginary

cyclic

num-ber field of

_degree

16 whose class number is

_equal

to _{its genus} class

number ( ~Lou2~ ~ .

D

Proposition

8.

₍₁₎

There are two

_imaginary

abelian number

_{fields of}

type

(4*, 4*)

with class numbers

_equal

to their _genus class number.

These

_fields

are

_given

in Table IX.

(2)

Let m be an odd

_integer

with m &#x3E; 3. There is no

_imaginary

abelian

number

_{field of degree}

l6rrc with class number

_equal

to its _genus class number such that the

_{subfield N(2)}

is

_of

_type

_{(4*, 4*).}

Proof.

(1)

Let _cpl and _p2 be two odd

_quartic

_characters,

N the field

asso-ciated with

_{(Wl, W2).}

Let

_{Ml, M2, M3}

and

_M4

be the fields associated with

_{((Pl), ((P2),}

and

_(cpicp2),

_{respectively.}

We have

_only

two

fields N such that

_hMz

=

gmi for 1 i

4,

hJB11M3

=

9MIM3 and

hMzM4

=

(2)

It follows

_immediately

from

_Proposition

_{5. (2) .}

D 3.4.

N~2~

is of

_type

_{(4*, 2*, 2*).}

Proposition

9.

₍₁₎

There are 7

_imaginary

abelian

number fields of

_type

(4*, 2*, 2*)

with class numbers

_equal

to their _genus class numbers. These

_fields

are

_given

in Table X.

(2)

Let m be an odd

_{integer &#x3E;}

3. There is no

_imaginary

number

field of

degree

16m with class number

_equal

to _{its genus} class number such that the

_{subfield N(2)}

is

_of

_type

_{(4*, 2*, 2*).}

Proof.

(1)

Let cp

be an odd character of order

4,

_X, and _{x2 two} odd

quadratic

characters,

N the field associated with Let

Ml, M2, Fl, F2, L and

E be the

_imaginary

subfields associated with

~~P~ ~

xl&#x3E;

&#x3E;

x2&#x3E;

&#x3E;

XlX2)

and

~~2,

Xl,

X2),

respectively.

Using

[Lou2],

Propositions

4 and 6 we

verify

that there are 7 fields

N with

_{hm, =}

_gmi &#x3E;

hm2 = 9M2’ hFl =

9F2’ hL =

9L and

hE

=

gE. For these 7 fields we

_compute

_hN

and

hN+ .

Note that

(2)

The

_proof

of

₍₂₎

is similar to that of

_Proposition

_5.(2).

0

To

_conclude,

the

_proof

of Theorem 1 is

_{completed by}

_Propositions

2-9. All

_computations

were carried out

_using

_PARI-GP[P]

and KANT

_V4~KT~.

Acknowledgements.

The authors wish _{to express} their

_gratitude

to

S. Louboutin and K. Yamamura for

_suggesting

the

_problem

and for _many

stimulating

conversations.

4. Tables

TABLE I continued.

TABLE V.

TABLE VI continued.

TABLE VI I.

_{(8*, 2* )}

TABLE VIII.

_{(4*, 2)}

TABLE IX.

_{(4*, 4* )}

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Ku-Young CHANG, Soun-Hi KwoN

Department of Mathematics Education Korea _University

136-701, Seoul Korea