THE CLASS NUMBER ONE PROBLEM FOR SOME NON-ABELIAN NORMAL CM-FIELDS ...

22 Pages·1997·0.39 MB·English

TRANSACTIONSOFTHE
AMERICANMATHEMATICALSOCIETY
Volume349,Number9,September1997,Pages3657–3678
S0002-9947(97)01768-6
THE CLASS NUMBER ONE PROBLEM FOR SOME
NON-ABELIAN NORMAL CM-FIELDS
STE´PHANE LOUBOUTIN, RYOTARO OKAZAKI, AND MICHEL OLIVIER
Abstract. LetNbeanon-abeliannormalCM-fieldofdegree4p,panyodd
prime. NotethattheGaloisgroupofNiseitherthedicyclicgroupoforder4p,
orthedihedralgroupoforder4p.Weprovethatthe(relative)classnumberof
adicyclicCM-fieldofdegree4pisalwaysgreaterthenone. Then,wedetermine
allthedihedralCM-fieldsofdegree12withclassnumberone: thereareexactly
ninesuchCM-fields.
Introduction
A.M. Odlyzko proved that there are only (cid:12)nitely many normal CM-(cid:12)elds with
class number one (see [Odl]), J. Ho(cid:11)stein proved that they have degree less than
orequalto 436(see [Hof]),andK.Yamamuradeterminedallthe abelianCM-(cid:12)elds
withclassnumberone(see[Yam]). Wewillnotethatthereisnonon-abeliannormal
CM-(cid:12)eldofdegree4,6or10.Then,S.LouboutinandR.Okazakideterminedallthe
non-abelian normal CM-(cid:12)elds of degree 8 with class number one (see [Lou-Oka]).
Here,thankstogoodupperboundsonresiduesofDedekindzeta-functionsats=1
and a technique for computing relative class numbers of non-abelian CM-(cid:12)elds, we
willdetermineallthe non-abeliannormalCM-(cid:12)eldsofdegree12withclassnumber
one:
Theorem 1. There are exactly 16 non-abelian normal CM-fields N of degree 12
with relative class number one, namely the 16 dihedral CM-fields N = KM of
degree 12 with relative class number equal to 1 which are listed in Table 1, where
K=Q(αK)(withPK(αK)=p0)isatpotallyrealnon-normalcubicnumberfieldwith
discriminantdK andM=Q( −D0, −D1)(withD0 (cid:21)1andD1 (cid:21)1square-free)
is an imaginary bicycplic biquadratic number field with relative class number equal
to1 and such that Q( dK)(cid:18)M.Exactly 9 out of these fields N have class number
one: those which appear in Table 1 and are such that the class numberhN+ of their
maximal totally real subfields N+ is equal to 1.
Throughout this paper we let p denote any odd prime.
The proof of this theorem will be divided into four steps:
In section 1, we prove some useful results on CM-(cid:12)elds. We will make use of
some of them in a forthcoming paper when we solve the class number one problem
fornon-abeliannormalCM-(cid:12)eldsofdegree20.WealsoprovethattheGaloisgroup
G of a normal CM-(cid:12)eld of degree 4p is isomorphic to either the dicyclic group T4p
or the dihedral group D4p.
ReceivedbytheeditorsJuly16,1995and,inrevisedform,March21,1996.
1991 Mathematics Subject Classi(cid:12)cation. Primary 11R29; Secondary 11R21, 11R42, 11M20,
11Y40.
Key words and phrases. CM-field,dihedralfield,relativeclassnumber.
3657
(cid:13)c1997 American Mathematical Society
3658 S. LOUBOUTIN, R. OKAZAKI, AND M. OLIVIER
Table 1
(d ,P (X)) (D ,D ) h
K K 0 1 N+
(148,X3+X2−3X−1) (1,37) 1
(469,X3+X2−5X−4) (7,67) 1
(473,X3−5X−1) (11,43) 1
(756,X3−6X−2) (3,7) 1
(940,X3−7X−4) (1,235) 2
(1304,X3−X2−11X−1) (2,163) 1
(1425,X3−X2−8X−3) (3,19) 1
(1620,X3−12X−14) (3,15 1
(1708,X3−X2−8X−2) (1,427) 2
(1944,X3−9X−6) (2,3) 1
(2700,X3−15X−20) (1,3) 1
(4104,X3−18X−16) (6,19) 2
(4312,X3+X2−16X−8) (2,11) 3
(4860,X3−18X−12) (3,5) 2
(8505,X3−27X−51) (7,35) 2
(14520,X3−33X−22) (3,10) 2
In section 2, we prove that dicyclic CM-(cid:12)elds of degree 4p with p (cid:17) 3 (mod 4)
have even relative class numbers, and therefore even class numbers (see Theorem
6). Moreover,we wil prove that for any odd prime p the relative class number of a
dicyclic CM-(cid:12)eld of degree 4p is always greater than one (see Theorem 7). Hence,
for the remainder of the paper we will focus on dihedral CM-(cid:12)elds of degree 4p.
In section 3, we quote results from the (cid:12)rst author which yield very good lower
−
bounds on relative class numbers h of CM-(cid:12)elds N of (cid:12)xed degree provided that
N
their Dedekind zeta-functions ζN satisfy ζN(s0) (cid:20) 0, where s0 = 1−(2/logdN).
−
Such lower bounds on h yield upper bounds on the absolute values of the dis-
N
−
criminants dN of N when hN = 1. We note that by using Stark and Ho(cid:11)stein’s
results instead of the ones we developed in this section, we would not get as good
an upper bound (on discriminants of dihedral CM-(cid:12)elds of degree 12 with relative
classnumbersequaltoone)astheonewegetinsection4.3. Indeed,ourtechniques
yield a better lower bound on the residue of the Dedekind zeta-function of N at
s = 1, and a better upper bound on the residue of the Dedekind zeta-function at
s=1 of its maximal totally real sub(cid:12)eld N+.
In section4,we (cid:12)nally solvethe relativeclassnumber one andclass number one
problems for dihedral CM-(cid:12)elds N of degree 4p = 12. This section will be divided
into sevensteps. First, in section4.1we draw a lattice ofsub(cid:12)elds ofN and notice
that N contains an imaginary bicyclic biquadratic sub(cid:12)eld M whose relative class
− − − −
numberh dividesh (weusesection1). Henceh =1yieldsh =1.Second,in
M N N M
section 4.2 we thus determine all the imaginary bicyclic biquadratic number (cid:12)elds
with relative class number one: there are 147 such number (cid:12)elds. We notice that
the Dedekind zeta-functions ζM of these 147 numbers (cid:12)elds satisfy ζM(s) (cid:20) 0,
s 2]0,1[. This will enable us to prove that if h−N = 1 then ζN(s) (cid:20) 0, s 2]0,1[.
Hence, in the third place, according to section 3, we manage to get in section 4.3
the upper bound dN (cid:20) 4(cid:1)1044 on the discriminants of the dihedral CM-(cid:12)elds N
of degree 12 with relative class number one. Moreover, with the notations of the
NORMAL CM-FIELDS OF DEGREE 4p 3659
previous Theorem 1, we prove that h−N = 1 implies dK (cid:20) 5 (cid:1) 108. Fourthly, in
section 4.4 we remind the reader of the results of the (cid:12)rst author which enable us
to compute the relative class number of any CM-(cid:12)eld. Here, we note that it would
require too much computation to complete the determination of all the dihedral
CM-(cid:12)elds N with relative class numbers equal to 1, but working with CM-(cid:12)elds
of smaller degree would drastically alleviate the computations. Hence, in the (cid:12)fth
− −
place, we give in section 4.5 a factorization of h which shows that h = 1 if
N N
− −
and only if hM = hN0 = 1, where N0 is some non-normal sextic CM-sub(cid:12)eld of
N. Moreover, in order to drastically reduce the amount of relative class number
computation, in the sixth place, we provide in section 4.6 necessary conditions for
the relative class number of N to be odd. These necessary conditions enable us to
get rid of many of the occurrences of N and N0. Finally, in the seventh place, by
usingcomputers,wedetermineinsepction4.7allthedihedralCM-(cid:12)eldsofdegree12
suchthatdK (cid:20)5(cid:1)108 withL=Q( dK)arealquadraticsub(cid:12)eldofanyoneofthe
−
147 imaginary bicyclic biquadratic number (cid:12)elds M with h = 1, and such that
M
−
the previous necessary conditions for h to be odd are ful(cid:12)lled. In this way, we
N
get a table of 11761 pairs (K,M). The computation of the relative class numbers
of the corresponding 11761 sextic CM-(cid:12)elds N0 yields the (cid:12)rst part of Theorem 1.
Its second part is an easy consequence.
1. Some useful results on CM-fields
We assume that all the number (cid:12)elds we will be looking at lie in the (cid:12)eld of
complex numbers. We let c denote the complex conjugation. If N is normal, we
may think of c as being in its Galois group. When E is a number (cid:12)eld, we let UE
betheunitgroupofthe ringofalgebraicintegersofE,WE bethe groupofrootsof
unityinE,wE betheorderofWE anddE betheabsolutevalueofthediscriminant
ofE.WhenE/Fisanextensionofnumber (cid:12)elds welet jE/F be the canonicalmap
from the group of fractional ideals of F to that of E and NE/F be the norm map
fromthe groupoffractionalideals of E to that of F. When K is a CM-(cid:12)eld, we let
K+ be its maximal totally real sub(cid:12)eld (hence K is a quadratic extension of K+),
QK =[UK :WKUK+]be Hasse’sunitindex ofK andκK bethe orderofthekernel
of jK/K+. Note that QK =[UK :WKUK+]=[UKc−1 :WK2]=2/[WK :UKc−1]. Since
WK2 =WKc−1 (cid:18)UKc−1 and since WK is a cyclic group, we get κK 2f1,2g. We have
an injective homomorphism
φK : ker(jK/K+)7!WK/UKc−1
which maps an ideal class C in this kernel to the image of α(cid:22)/α in the factor group
WK/UKc−1,where(α)=jK/K+(C).Moreover,wehaveWK2 =WKc−1 (cid:18)UKc−1.Hence,
wegetκK 2f1,2g(note aslightmistakeinthe proofof[Wa,Th. 10.3]). Note also
c−1
that QK =2 implies [WK :UK ]=1 and κK =1. Finally, if K/k is an extension
of CM-(cid:12)elds, note that NK/k(cid:14)jK/K+ = jk/k+ (cid:14)NK+/k+ on fractional ideals and
onidealclasses(use the (cid:12)nitness ofthe classgroupstoreducethe prooftothe case
of principal ideals).
Lemma 2. (i). Let K be atotally imaginary numberfield and let N beany normal
closure of K. Then, K is a CM-field if and only if for all g 2 Gal(N/Q) the
commutators [c,g]=c−1g−1cg lie in Gal(N/K).
(ii). Hence, a totally imaginary normal number field is a CM-field if and only if
the complex conjugation lies in the center of its Galois group.
3660 S. LOUBOUTIN, R. OKAZAKI, AND M. OLIVIER
(iii). Any subfield of a CM-field is either a CM-field or a totally real number
field.
(iv). There is no non-abelian normal CM-field of degree 2p or 2p2, p any odd
prime.
Proof. We notice that point (i) readily implies points (ii) and (iii), and we now
prove point (i). Set G = Gal(N/Q), H = Gal(N/K), and for any sub(cid:12)eld K+
of N set H+ = Gal(N/K+). Since for any g 2 G we have Gal(N/g(K+)) =
gGal(N/K+)g−1 =gH+g−1, then K+ is totally realif andonly if for allg 2G we
havec2gH+g−1.Moreover,as K is totally imaginarythen forany g 2Gwe have
c 62 Gal(N/g(K)) = gGal(N/K)g−1 = gHg−1 and g−1cg 62 H. Therefore, as K is
a CM-(cid:12)eld if and only if we can (cid:12)nd a totally realsub(cid:12)eld K+ of N with K+ (cid:18)K
and [K:K+]=2, then K is a CM-(cid:12)eld if and only if there exists a subgroup H+
of G such that
1) H(cid:18)H+,
2) [H+ :H]=2,
3) g 2G implies g−1cg 2H+nH.
Now, 1), 2) and 3) imply [c,g] 2 H for all g 2 G. Conversely, if [c,g] 2 H for all
g 2G then H+ =hH,ci, the subgroup of G generated by H and c, satis(cid:12)es 1), 2)
and 3).
Finally, point (iv) follows from the fact that G=S2Sp, where S2 =fId,cg is a
subgroup of the center of G and Sp is any p-Sylow subgroup of G, and from the
fact that any group of order p or p2 is abelian.
Hence, the smallest possible degrees for non-abelian normal CM-(cid:12)elds are 8,
12, 16 and 20. Recently S. Louboutin and R. Okazaki solved the class number one
problemforthenon-abeliannormalocticCM-(cid:12)elds(see[Lou-Oka]). Now,weprove
that the Galois group of a non-abelian normal CM-(cid:12)eld of degree 4p (which will
copewithnormalCM-(cid:12)eldsofdegree12and20)isisomorphiceithertothedihedral
group D4p or the dicyclic group T4p:
Lemma 3. Let G be a non-abelian group of order 4p. If the center Z(G) of G has
even order, then G is isomorphic either to the dihedral group D4p or to the dicyclic
group T4p. Here, we set
D4p =ha,b: a2p =b2 =1,b−1ab=a−1i (dihedral group of order 4p),
T4p =ha,b: a2p =1,ap =b2,b−1ab=a−1i (dicyclic group of order 4p).
Note that in both cases Z(G)=f1,apg and G/Z(G) is the dihedral group of order
2p.
Proof. Since the product of an element of order p and of an element of order 2
which lies in Z(G) has order 2p, there exists an element a of order 2p in G. Then,
the subgrouphai of order 2p whichhas index two in G is a normalsubgroupof G.
We let S2 be a 2-Sylowsubgroupof G whichcontains ap, which has order2. Then
S2 has order 4.
First,assumethatS2 =hbiiscyclic,whereb4 =1.Then,thereexistsanintegerk
suchthat b−1ab=ak.We geta=a−paap =b−2ab2 =b−1akb=ak2.Hence, k2 (cid:17)1
(mod 2p), which implies k (cid:17) (cid:6)1 (mod 2p) and b−1ab = a or b−1ab = a−1. In the
(cid:12)rst case, we would have ab=ba and G would be abelian, a contradiction. Hence,
b−1ab = a−1 and G is somorphic to T4p. Second, assume that S2 = h1,ap,b,apbi
NORMAL CM-FIELDS OF DEGREE 4p 3661
is bicyclic, where b2 = 1. As in the previous case we get b−1ab = a−1 and G is
isomorphic to D4p.
Lemma 4. (i). (See [Lou-Oka] and [Lou 7]) Let N be a CM-field and let t be
the number of distinct prime ideals of N which are ramified in N/N+. Then 2t−1
−
divides h .
N
(ii). (See [Mar]) Let N+ be a dihedral number field of degree 2p (p any odd
prime), let L be the only quadratic subfield of N+, and let q be any prime positive
rational integer which is ramified in L/Q, i.e. (q) = Q2L in L. Then, either QL
splits completely in N+/L, or QL is totally ramified in N+/L. In the latter case,
we also have q = p. Moreover, the conductor of the extension N+/L is a positive
integer f such that for any prime q which divides f we have q (cid:17) χL(q) (mod p),
where χL dpenotes thequadratic Dirichlet character associated with L. In particular,
if L = Q( p) with p (cid:17) 1 (mod 4) prime, then q 6= p and q divides f imply q (cid:17) 1
(mod p).
(iii). Let K be any subfield of degree p of a dihedral number field N+ of degree
2p. Then, N+ = KL, dN+ = dLd2K, and there exists a rationalpinteger f (cid:21) 1psuch
that dN+ = dpLf2(p−1) and dK = d(Lp−1)/2fp−1. Hence, L = Q( dL) = Q(p−1dK)
is well determined by dK.
The last assertion of Lemma 4(ii) will be proved at the begining of the proof of
Proposition 8. We note that by using point (ii) of the following Theorem 5 and
[Hor, Corollaryon page 519]we could provepoint (iii) of Theorem5. However,we
will give a di(cid:11)erent proof of this point (iii).
Theorem 5. Let k(cid:18)K be two CM-fields. Assumethat [K:k]=m is odd. Then,
(i). Qk =QK.
(ii). jK+/k+ : ker(jk/k+)−!ker(jK/K+) is an isomorphism. Hence, κk =κK.
− −
(iii). h divides h .
k K
Proof. To start with, we notice that both the factor groups WK/WK2 and Wk/Wk2
haveorder2. Moreover,since Wkm =NK/k(Wk)(cid:18)NK/k(WK)and since m is odd,
thenNK/k induces anisomorphismfromthe factorgroupWK/WK2 ontothe factor
groupWk/Wk2.Inparticular,ifζ 2WK,thenNK/k(ζ)2Wk2 ifandonlyifζ 2WK2.
We provepoint(i). Since QK andQk areequalto 1 or2, itis su(cid:14)cientto prove
that QK = 2 if and only if Qk = 2. Assume that [UKc−1 : WK2] = QK = 2. Then,
thereexists(cid:15)2UKsuchthatζ =(cid:22)(cid:15)/(cid:15)2WKnWK2.Setη =NK/k((cid:15)).Asthecomplex
conjugationcommuteswithallthe embeddingsofK(Lemma1(i)), thenη isaunit
in k such that η(cid:22)/η =NK/k((cid:15)(cid:22)/(cid:15))=NK/k(ζ)62WK2. Hence, Qk =[Ukc−1 :Wk2]=2.
AssumeconverselythatQk =[Ukc−1 :Wk2]=2.Then,thereexistsη 2Uk suchthat
η(cid:22)/η 62Wk.Now,Wk\WK2 =Wk2 forw =ζ2 2Wk\WK2 implieswm =NK/k(w)=
(NK/k(ζ))2 2Wk2 andw 2Wk2. Thus,η isaunitofKsuchthatη(cid:22)/η 62WK2.Hence,
QK =[UKc−1 :WK2]=2.
We prove point (ii). Since NK+/k+ (cid:14)jK+/k+(C) = Cm on ideal classes C of k+,
since kerjk/k+ has order κk (cid:20) 2 and since m is odd, then jK+/k+ induces a well
de(cid:12)ned injective homomorphism
(1) j : kerjk/k+ −!kerjK/K+.
Hence, if κk =2 then κK =2 and j of (1) is an isomorphism. Conversely, assume
that κK = 2 and let I be an ideal of K+ such that jK/K+(I) = (α) with ζ =
3662 S. LOUBOUTIN, R. OKAZAKI, AND M. OLIVIER
α(cid:22)/α2WKnUKc−1. Since κK =2, then [WK :UKc−1]=2 and QK =1. According to
point (i) we get Qk = [Ukc−1 : Wk2] = 1. Therefore, UKc−1 = WK2 and Ukc−1 = Wk2.
Then, jk/k+(cid:14)NK+/k+(I)=NK/k(cid:14)jK/K+(I)=(β) with β =NK/k(α). Hence, the
ideal class of J = NK+/k+(I) is in the kernel of jk/k+. Since β(cid:22)/β = NK/k(α(cid:22)/α) =
NK/k(ζ)2WknWk2 =WknUkc−1, then κk =2 and j of (1) is an isomorphism.
We prove point (iii). We have
− jkerNK/kj jcokerNK+/k+j
(2) h = .
K jkerNK+/k+jjcokerNK/kjh−k
LetHk andHk+ betheHilbertclass(cid:12)eldsofkandk+,andletAk/kandAk+/k+
be the maximal unrami(cid:12)ed abelian sub-extensions of K/k and K+/k+. According
to class (cid:12)eld theory we have
(3) jcokerNK+/k+j = [K+\Hk+ :k+] = [Ak+ :k+].
jcokerNK/kj [K\Hk :k] [Ak :k]
Since Ak+/k+ is an unrami(cid:12)ed abelian sub-extension of K+/k+, then Ak+k/k is
anunrami(cid:12)edabeliansub-extensionofK/k,whichyieldsAk+ (cid:18)Ak,whichinturn
yields Ak+ (cid:18) A+k. Conversely, since Ak/k is an unrami(cid:12)ed abelian sub-extension
of K/k and since Ak is a CM-(cid:12)eld (Lemma 2(iii)), then A+k/k+ is an abelian sub-
extension of K+/k+ which is clearly unrami(cid:12)ed if m = [K : k] is odd. Hence, we
get A+k (cid:18)Ak+. Therefore, we have A+k =Ak+ and [Ak+ :k+]=[Ak :k]. Hence,
(2) and (3) yield
(4) h−K/h−k =jkerNK/kj/jkerNK+/k+j.
Since NK/k(cid:14)jK/K+ = jk/k+ (cid:14)NK+/k+, then jK/K+ induces a well de(cid:12)ned homo-
morphism
(5) J : kerNK+/k+ −!kerNK/k.
According to point (ii), if C is in kerJ then C = jK+/k+(C0) with C0 2 kerjk/k+.
Hence, we have 1 = NK+/k+(C) = C0m. Since m is odd and C0 has order (cid:20) 2
(since κk 2f1,2g), we get C0 =1, which implies C =1. Thus, J of (5) is injective,
jkerNK+/k+j divides jkerNK/kj, and (4) provides us with the desired result.
2. Class number problems for dicyclic CM-fields of degree 4p
Let N be a normal CM-(cid:12)eld with Galois group T4p, p any odd prime. With the
notations of Lemma 3, we let N+ be the (cid:12)xed (cid:12)eld of the group Z(T4p)=f1,apg,
and M be the (cid:12)xed (cid:12)eld of the cyclic subgroup ha2i = f1,a2,(cid:1)(cid:1)(cid:1) ,a2(p−1)g. Ac-
cordingtoLemma 2(ii) andLemma3, wehavethat ap is the complexconjugation.
Hence, N+ is the maximal totally real sub(cid:12)eld of N and M is an imaginary cyclic
quartic (cid:12)eld. We let L be the real quadratic sub(cid:12)eld of M. Then N+/Q is dihe-
dral of degree 2p, and N/L is cyclic of degree 2p. We have the following lattice of
sub(cid:12)elds:
NORMAL CM-FIELDS OF DEGREE 4p 3663
N
C
2 C
C
N+ Cp
C
C
p C
M
(cid:2)
L (cid:2)
(cid:2)2
2 (cid:2)
(cid:2)
Q
Now, assume that h− is odd. Thus, there exists at most one prime ideal of N+
N
whichisrami(cid:12)edinthequadraticextensionN/N+ (Lemma4(i)),thusatmostone
primeidealofMwhichisrami(cid:12)edinM/L.SinceM/Qiscyclicquartic,thenthere
exists exactly one rational prime q which is rami(cid:12)ed in M/Q, and that prime q is
totallyrami(cid:12)edinthis extensionM/Q.Now,accordingto Lemma4(ii), ifq 6=por
q = p but p is not rami(cid:12)ed in N+/L, then the prime ideal QL of L lying above q
splitscompletelyinN+/L.SinceM/Qistotallyrami(cid:12)edatp,thesepprimeideals
ofN+ lyingaboveQL arerami(cid:12)edinN/N+,and2p−1dividesh−N.Acontradiction.
Thus, q =p and p is totally rami(cid:12)ed in N+/Q. Finally, as L is rami(cid:12)ed only at p,
p
we must have p (cid:17) 1 (mod 4), and L = Q( p). Thus, we have proved point (ii) of
the following Theorem
Theorem 6. Let N be a dicyclic CM-field of degree 4p, p any odd prime.
(i). If p(cid:17)3 (mod 4) then 2p−1 divides h−.
N
(ii). Ifh− isoddthenp(cid:17)1 (mod 4),Mhasconductorpandpistotallyramified
N
in N+/Q.
(iii). If h− =1 then p2f5, 13, 29, 37, 53g.
N
Proof. (i). In that case (2) = P2 is rami(cid:12)ed in L/Q, and the previous discussion
L
showsthatPL splits completelyin N+/Qandisrami(cid:12)edinM/L.Thus,atleastp
prime ideals of N are rami(cid:12)ed in N/N+ and Lemma 4(i) yields the desired result.
− −
(iii). If h = 1, then h = 1 (Theorem 5(iii)), and M has conductor p. But
N M
it is well known that there are only (cid:12)nitely many imaginary cyclic quartic number
(cid:12)eldsMwithrelativeclassnumberequaltoone,andB.Setzerdeterminedallthese
(cid:12)elds. According to [Set], we get the desired result.
The remainder of this section is devoted to proving the following result:
Theorem 7. There does not exist any dicyclic CM-field of degree 4p, p any odd
prime, with relative class number equal to one.
To get this result we will (cid:12)rst prove in Proposition 8 that if M has conductor
p and p (cid:17) 1 (mod 4) and h− is not divisible by p, then N+ must be a so-called
N
pure real dihedral number field of degree 2p, which means that p is totally rami(cid:12)ed
in N+/Q and no other rational prime is rami(cid:12)ed in N+/Q. We will then provide
in Proposition 9 a powerful necessary condition for the existence of a pure real
dihedral number (cid:12)eld of degree 2p. Hence, we will easily get that there is no pure
real dihedral number (cid:12)eld of degree 2p with p2f5, 13, 29, 37, 53g. According to
Theorem 6, we will have proved Theorem 7.
3664 S. LOUBOUTIN, R. OKAZAKI, AND M. OLIVIER
Proposition 8. Let p be any odd prime. Let N/M be a cyclic extension of degree
p of CM-fields Assume that N+/M+ also is a cyclic extension of degree p. Let T
be the number of prime ideals of M+ which split in M/M+ and are ramified in
N+/M+. Then, pT−1h− divides h−, and pTh− divides h− if p does not divide
M N M N
wM, the order of the finite group of roots of unity in M. In particular, if N is a
dicyclic CM-field of degree 4p, p(cid:17)1 (mod 4) an odd prime, if M has conductor p,
and if N+ is not a pure real dihedral number field of degree 2p, then p divides h−.
N
Proof. The last assertion follows from the previous ones when we notice that any
rationalprimeq 6=pwhichdividestheconductorf oftheextensionN+/Lmustsat-
isfy q (cid:17) χL(q) (cid:17) (cid:6)1 (mod p), where χL denotes the quadratic Dirichlet character
associatedwithL(seeLemma4(ii)). HerewehaveχL(q)=(q/p)(Legendre’ssym-
bol). Ifwehadq (cid:17)−1 (mod p)thenwewouldgetχL(q)=(q/p)=(−1/p)=16(cid:17)q
(mod p). Hence q (cid:17)1 (mod p). Since M has conductor p, then q splits completely
in M/Q. Hence, if N+ is not pure then we get T (cid:21)2, and p divides h−.
N
Now, let us prove the (cid:12)rst part of Proposition 8. Let a be the ambiguous
N/M
classnumberofthecyclicofextensionN/Mofdegreep,lett bethenumberof
N/M
primeideals ofM whicharerami(cid:12)edinthe cyclicextension N/Mofdegreep,and
letUMbetheunitgroupoftheringofalgebraicintegersofN.LetaN+/M+,tN+/M+
and U be de(cid:12)ned accordingly. We thus have t = (t −T)+2T =
N+ N/M N+/M+
t +T. The ambiguous class number formula yields
N+/M+
aN/M [UM+ :UM+ \NN+/M+(N+)] hM t −t
= p N/M N+/M+
aN+/M+ [UM :UM\NN/M(N)] hM+
[U :U \N (N+)]
M+ M+ N+/M+ − T
= h p .
[UM :UM\NN/M(N)] M
To get the desired results, we prove the two following assertions :
1). [UM :UM\NN/M(N)] divides (cid:15)p[UM+ :UM+ \NN+/M+(N+)] with (cid:15)p =1
if p does not divide wM, and (cid:15)p =p if p divides wM.
−
2). a /a divides h .
N/M N+/M+ N
Proof of assertion 1). We set U~M = WMUM+. Since M is a CM-(cid:12)eld, then the
Hasse index QM = [UM : U~M] is equal to 1 or 2 (see [Has], [Wa]). We (cid:12)rst note
that
[UM :UM\NN/M(N)]= [U[UMM\:NU~NM/]M[U~(MN)::U~UM~M\\NNNN/M/M(N(N)])]
=[U~M :U~M\NN/M(N)].
Indeed, if QM =1, then U~M =UM and
[UM :U~M]=[UM\NN/M(N):U~M\NN/M(N)].
IfQM =2,thenw(cid:14)ehaveaone-to-onecanonicalmorphismfromth(cid:14)e quotientgroup
UM\NN/M(N) U~M\NN/M(N) into the quotient group UM U~M of order 2,
and this morphism is onto, for (cid:15)2UMnU~M implies
(cid:15)p 2UM\NN/M(N)nU~M\NN/M(N).
NORMAL CM-FIELDS OF DEGREE 4p 3665
Now,
U~M\WNpNN+/M+(UN+)=WMUM+ \WNpNN+/M+(UN+)
(cid:18)WMUM+ \NN/M(WN)NN+/M+(UN+)
=U~M\NN/M(U~N)(cid:18)U~M\NN/M(N).
Thus, [U~M :U~M\NN/M(N)] divides [U~M :U~M\WNpNN+/M+(UN+)], and
[U~M :U~M\WNpNN+/M+(UN+)]=[WMUM+ :WMUM+ \WMp NN+/M+(UN+)]
=[WMUM+ :WMp UM+ \WMp NN+/M+(UN+)]
=[WM :WMp ][UM+ :UM+ \NN+/M+(N+)]
=(cid:15)p[UM+ :UM+ \NN+/M+(N+)].
Proof of assertion 2). Let A denote the group of ambiguous classes of the
N/M
cyclic extension N/M of degree p, and de(cid:12)ne A accordingly. Let j be
N+/M+ N/N+
the canonical map from the group C of fractional ideals of N+ to C , which is
N+ N
thesameforN.LetσbeoforderpintheGaloisgroupofthecyclicextensionN/M
of degree p. Then σ+, the restriction of σ to N+, has order p in the Galois group
of the cyclic extension N+/M+ of degree p. Now, it is easily seen that we have
j (A )(cid:18)A . Hence, we have a morphism between quotient groups
N/N+ N+/M+ N/M
~j :C /A −!C /A .
N+ N+/M+ N N/M
If we prove that ~j is injective then we have the desired result. Now, let C+ 2CN+
beinthekernelof~j.ThenwehavejN/N+(C+)2AN/M,whichyieldsjN/N+(C+)=
σ(jN/N+(C+)) = jN/N+(σ(C+)) = jN/N+(σ+(C+)). Hence, we get C0 d=ef C+1−σ+ 2
kerjN/N+,whichobviouslyimplies σ+(C0)2kerjN/N+.Sincethis kernelhasorder
(cid:20)2, we get σ+(C0)=C0. Hence, we have
C0 =C0p =C01+σ++(cid:1)(cid:1)(cid:1)+σ+p−1 =C+1−σ+p =1,
which amounts to saying that C+ =σ+(C+), i.e., that C+ is in AN+/M+.
p
Proposition 9. Let p(cid:17)1 (modp4) be a prime and let (cid:15)p =(up+vp p)/2>1 be
the fundamental unit of Lp =Q( p). If p does not divides vp, then there does not
exist any real dihedral number field N+ of degree 2p such that p is totally ramified
in N+/Q and p is the only rational prime which is ramified in N+/Q.
Proof. Assumethatsuchanumber(cid:12)eldN+ exists. ThentheextensionN+/Lp has
conductor (p) (see [Mar, Prop. III.4 and Lemma III.1]). According to class (cid:12)eld
theoryN+ isasub(cid:12)eldofL1,whereL1 denotestheunitrayclass(cid:12)eldofconductor
p p
(p) of Lp. Hence, p divides the degree [L1p :Lp]. Since p is rami(cid:12)ed in Lp/Q, then
[L1p :Lp]=p(p−1)hp/(Up :Up1),
where hp is the class number of Lp, where Up is the unit group of the ring of
algebraic integers Rp of Lp and where
Up1 =fα; α2Up and α(cid:17)1 (mod (p))g.
3666 S. LOUBOUTIN, R. OKAZAKI, AND M. OLIVIER
Since hp < p, then p does not divide the index (Up : Up1). Since (Up : Up1) divides
p(p−1) = jRp/(p)j(cid:3), then (Up : Up1) divides p−1, hence (cid:15)pp−1 (cid:17) 1 (mod (p)).
Finally, since
(cid:15)p−1 (cid:17)2p−1(cid:15)p−1
p p
p
(cid:17)(up+vp p)p−1
p
(cid:17)upp−1+(p−1)upp−2vp p
p
(cid:17)1−upp−2vp p (mod (p))
and since up is prime to p, we get that p divides vp.
Remark. Proposition9is strongerthanthe followingresultin[JY,TheoremI.2.2]:
If p (cid:17) 1 (mod 4) is regular then the discriminant of a real dihedral field of degree
2p is not a power of p. Indeed, p is regular if and only if p does not divides Bj for
j 2f2,4,(cid:1)(cid:1)(cid:1) ,p−3g, while p divides Bp−1 if and only if p divides vp (see [Wa, Th.
2
5.37]).
3. Explicit lower bounds for relative class numbers
of CM-number fields
Wesetupsomeofthenotationwewilluseandremindthereaderofanumberof
resultsofthe(cid:12)rstauthorwewilluse. WeletNbeatotallyimaginarynumber(cid:12)eld
ofdegree2n that is a quadraticextensionof a totally realsub(cid:12)eld N+ ofdegree n,
i.e., N is a CM-(cid:12)eld with maximal totally real sub(cid:12)eld N+. In that situation, it is
well known that the class number hN+ of N+ divides the class number hN of N.
−
We set hN = hN/hN+, which is a positive integer called the relative class number
of N. We have
s (cid:0) (cid:1)
(6) h− = QNwN dN Ress=1(cid:0)ζN (cid:1),
N (2π)n dN+ Ress=1 ζN+
where QN 2f1,2g is the Hasse unit index of N, wN (cid:21)2 is the number of roots of
unity in N, dN, ζN, dN+ and ζN+ are the absolute value of the discriminant and
the Dedekind zeta-function of N and N+, respectively (see [Wa, Chapter 4]). We
now provide lower bounds on relative class numbers.
Theorem 10. (See [Lou 5, Th. 2]) Let N be a totally imaginary number field of
(cid:0) (cid:1)
degree 2n(cid:21)2, and set (cid:15)N =2πne1/n/d1N/2n. Assume that ζN 1−(2/logdN) (cid:20)0.
Then,
p
(7) h− (cid:21) 1−(cid:0)(cid:15)N (cid:1)2QNwN dN/dN+.
N Ress=1 ζN+ e(2π)n logdN
Theorem 11. (See [Lou 6]) Let L be a real quadratic number field, K be a totally
real number field, and assume that K/L is an abelian extension. Then, there exists
an explicit constant µL such that for any Artin character χ 6= 1 (of conductor F)
of such an extension we have
(cid:18) (cid:19)
(cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
1 1
(8) jL(1,χ,K/L)j(cid:20)Ress=1 ζL 1− logf + 1+ µL
f f