J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
D
AVID
S. D
UMMIT
J
ONATHAN
W. S
ANDS
B
RETT
T
ANGEDAL
Stark’s conjecture in multi-quadratic extensions, revisited
Journal de Théorie des Nombres de Bordeaux, tome
15, n
o1 (2003),
p. 83-97
<http://www.numdam.org/item?id=JTNB_2003__15_1_83_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Stark’s
conjecture
in
multi-quadratic
extensions,
revisited
par DAVID S.
DUMMIT*,
JONATHAN W. SANDS* et BRETT TANGEDALRÉSUMÉ. Les
conjectures
de Stark relient les unitésspéciales
dans les corps de nombres à certaines valeurs des fonctions L attachéesà ces corps. Nous considérons le cas d’une extension
abélienne,
et nous établissons la relation fondamentale de la
conjecture
deStark
lorsque
son groupe de Galois estd’exposant
2. Nousmon-trons que la
conjecture
est entièrement vérifiée pour les extensionsbiquadratiques
ainsi que dans de nombreux autres cas.ABSTRACT. Stark’s
conjectures
connectspecial
units in number fields withspecial
values of L-functions attached to these fields. We consider the fundamentalequality
of Stark’s refinedconjec-ture for the case of an abelian Galois group, and prove it when
this group has exponent 2. For
biquadratic
extensions and mostothers,
we prove more,establishing
theconjecture
in full.1. The elements of Stark’s refined abelian
conjecture
Units. Let:o
L/F
be an abelian extension of number fields in which adistinguished
(finite
orinfinite)
prime
of F denotedby
tosplits completely.
These and all number fields will be assumed to lie in a fixedalgebraic
closureof
the fieldQ
of rational numbers..
11m
be the normalized absolute value at a fixedprime
tv of L above t3..
wL be the order of the group p L of roots of
unity
in L..
uiu)
be thegroup of elements of L
having
absolute valueequal
to 1 at each(finite
orinfinite)
absolute value of Lexcept
for those associatedwith
primes
above b, in otherwords,
those which areconjugates
ofI IM.
We sometimes refer to
U(")
as the o-units of L.Manuscrit regu le 3 decembre 2001.
*
L-functions. Let:
o G be the abelian Galois group of the extension
L/F.
.G
be the character group of G.. S’ be a fixed finite set of
primes
of F ofcardinality
>3,
andas-sume that ~5‘ contains n, all finite
primes
whichramify
inL/F,
and allinfinite
primes.
The Starkconjecture
we are concerned with must be formulateddifferently
when =2,
and is known to be true in thiscase
by
[4]
and[5].
so
= S -ftil.
.
Sfin
= the set of finiteprimes
in S.· ~ run
through
the finiteprimes
of F not in S.. a run
through integral
ideals ofF, prime
to the elements of S. . Na denote the absolute norm of the ideal a.. a a E G be the well-defined
automorphism
attached to a via the Artinmap.
For
each X
EG,
we have the Artin L-function with Euler factors at theprimes
in S’ removed:It is known that has an
analytic
continuation and a functionalequation relating
it tos, 7).
The order of its zero at s = 0 isq
splits completely
in the fieldfixed
by
the kernel ofI
depending
on whether ornot X
is the trivial character xo . See[5]
for furtherbackground
and references.The
conjecture.
We firstsingle
out thekey equality
in Stark’s refined abelianconjecture
for first derivatives of L-functions whichposits
theexis-tence of a
special to-unit
eserving
as an "L-function evaluator."Conjecture
There exists an elerraent(often
called a "Starkunit ")
E Euin)
such thatRemark 1. The conditions on E
specify
all of its absolute values and thuswe
impose
Stark’s additional condition below.Nevertheless,
we sometimesrefer to - as "the" Stark unit.
The full Stark
conjecture
in thissetting
(cf.
[4],
[5])
says more.Conjecture
St(L/F,S). St’ (L/ F,S)
holds,
andfurthermore
L(e1/WL)/ F
is an abelian Galois extension.
2. Statements of the results
We assume from now on that there are at least 2 infinite
primes
ool, 002in S. Otherwise
St(L/F,S)
is known to be trueby
[4] (see
also[5,IV.3.9]).
We may then assume thato02 ~
b. Also assume from now on that G =Gal(L/F)
isisomorphic
to(Z/2z)m
for somepositive
integer
m. We then callL/F
amultiquadratic
extension of rank m.Our aim in this paper is to prove the
following
theorems.Theorem 1. Let C S consist
of
thefinite primes
inS,
and letrF(S)
denote the 2-rankof
theSfin-class
groupof
F.If
ISI
> m + 1 -rF(S),
then
St(L/F,S)
holdsfor
themultiquadratic
extensionL/F.
Theorem 2.
St’(L/F,S)
holdsfor
themultiquadratic
extensionLIF,
hence
for
anarbitrary multiquadratic
extension.Theorem 3.
St(L/F, S)
holdsfor
themultiquadratic
extensionL/F
if
0 is a realinfinite prime
or afinite prime,
except
possibly
when L is themaximal
multiquadratic
extensionof F
which isunramified
outsideof
Sand in which to
splits
completely.
Theorem 4.
St(L/F, S)
holdsfor
themultiquadratic
extercsionL/F
when the rankof
this extension is m =2,
i. e.L/F
isbiquadratic.
Thus it holdsfor
anarbitrary biquadratic
extensions.Remark 2.
St(L/F, S)
wasproved
for themultiquadratic
extensionL/F
in
[2]
and[3]
under theassumption
that either~S~
> m +1,
or that noprime
above 2(i.e.
nodyadic
prime)
is ramified inL/F.
3. The relative
quadratic
caseAssume
K/F
is a relativequadratic
extension. This section summarizessome basic results from
[3]
and[5]
onSt(K/F,S).
We set:o
Gal(K/ F) = (T)
of order 2.* 77K = 1 if S contains two
split
primes
ofK/ F.
. 77K =
generator
of the infinitecyclic
group with
1,
otherwise. Note that since 002 does not
split
in this case, we may alsodescribe as the S-units u of K such that
ul+T=l.
If ro is a realm. Since
17k+’T =
1,
this thenimplies
that 17K ispositive
at both of theprimes
above 0..
CLF(S)
= =Sfin-ideal
class group ofF,
thequotient
of theideal class group
Cl(F)
of Fby
thesubgroup generated
by
the idealclasses of the
primes
in.
SK
= the set ofprimes
of Klying
above those in S. .ClK(S)
=SK-ideal
class group of K.~
HK
=HK(S)
=ClK(S)/c(C1F(S)),
the cokernel of the map c inducedby
extension of ideals..
MK
=MK(S) _
IHKI,
the order of this group.Theorem
(Stark-Tate,
cf.[Ta, IV.5.4]).
St(K/F,S)
holds with Stark unitand abelian.
Remark 3. The extra factor
e+
in[5, IV.5.4]
equals
1 when 1 asthis
implies
that the infiniteprime
002 of F does notsplit
in K.4.
Passage
to themultiquadratic
case via L-functionproperties
We have assumed that
L/F
ismultiquadratic
with thedistinguished
prime
b of Fsplitting completely
inL,
and thato02 ~
b is an infiniteprime
of F. From now on, we also assume that:. 002 does not
split completely
inL/F. (Otherwise St (L/F)
istrivially
true with 6; =
1.)
Let:
o T =
complex
conjugation
at 002 inL/F.
.
Ki
for i =1, 2, ... ,
2~"2-1~
be the relativequadratic
extensions of F in L which are not fixedby T.
(These
generate
L.)
- ! Ka a
.
Mi
= MKi.
. WiProposition
1.If
lies in
L,
then it is theStark
unit eLsatisfying
St/(L/F,S).
Proof.
(This
is astraightforward
adaptation
of theproof
of Theorem 2.6of
[3].)
Clearly e
EUib)
L becauseeach rli
Eui;l
- Inparticular,
Ki L
= 1 because this
represents
the absolute value of e above 002. We now show that E is an L-function evaluator.Fix an
arbitrary character X
EG.
IfX(T)
=1,
thenrs(X)
> 1by
the formula for thisquantity,
and therefore0)
= 0. At the sametime,
by
the observation in the lastparagraph.
So E is an L-function evaluatorfor this
type
of X.Now suppose that
x(T) _
-1. The fixed field of the kernelof X
must then be one of theKi
for some i =i(X).
Letting
Gi
=Gal(L/Ki),
weobserve that
We use the definition of c, the fact that
~(r) = 20131,
and the fact thatGi
fixesalong
with the evaluation of the last sum andfinally
theStark-Tate theorem for relative
quadratic
extensions and the inflationproperty
of Artin L-functions to see that
So c is an L-function evaluator for this
type
of X
aswell,
and theproof
iscomplete.
05. Class field
theory
Recall thatHK
=Proposition
2. Let K be anyof
the Ki
for
which ~7Z=1=
1. Thenrank2(HK)
>rank2(Clp(S))
=rF(S),
withequality holding if
ISI
=3.
Proof.
We will show that the norm map induces asurjective
homomorphism
HK/HK
eCIF(S) /CIF(S)2
which is anisomorphism
when = 3.The
assumption that "Ii =1=
1implies
that 002 and the otherprimes
ofSo
do not
split
inK/F.
Thus thecomplex conjugation
T at 002 restricts to agenerator
ofGal(K/F’).
Let
IK
denote the group of fractional ideals ofK,
PK
denote thesub-group of
principal
fractionalideals,
andIF
denote thesubgroup
of fractionalideals of K which are extended from fractional ideals of F. Also let
Sfin(K)
denote the set of ideals of K which lie above those in
Sgn.
This contains all of the ideals of K which are ramified over F. From the factorization of ideals intoprimes,
we see thatIF(Sfin(K))
=IIK+-JK(Sfi.(K)),
whereJK
denotes the group
generated
by
theprime
ideals of K which are inert overF. Then
Under the Artin map of class field
theory,
IK/PK (= CIK)
corresponds
to the maximal unramified abelian extension
(Hilbert
ClassField)
H ofK,
in the sense that this map induces anisomorphism
ofIK/PK
withFrom this it is clear that
IKIPK12K
corresponds
to the maximal abelian unramifiedelementary
2-extension of K in the same way.Similarly,
corresponds
to the maximal abelian unramifiedelementary
2-extension T of K
having
theproperty
that T acts(by conjugation)
triv-ially
onGal(.F/K).
By maximality,
:FjF
is Galois.Let
Gllp
=Gal(0/F)
andN~
= Then is normal of index2 in
Gjr,
and T actstrivially
onNr.
So and any lift of T commute with which suffices to show thatNF
lies in the center ofG~.
Nowis
cyclic
of order2,
so thatG,
modulo its center iscyclic.
Thisimplies
that
G:F
is abelian. Hence in fact is an abelian extension.We now know that
corresponds
to the maximal unramifiedelementary
2-extension .~’ of K which is abelian over F. So the extensioncorresponds
to the maximal such extensionGK
ofK in which all
primes
ofSfin(K)
split completely.
For each finite or infinite
prime
PESo,
letDp
denote itsdecomposition
group in the abelian extension
LKIF.
Under ourassumptions,
such aprime
p does notsplit
in thequadratic
extensionK/F,
while theprime q3
aboveit in K
splits completely
in£K / K.
ThusDp
=(7p)
has order 2. Let D be thesubgroup
ofGal(GK/F)
generated by
all theDp
for finite and infinite2-rank
rank2(D) 1 Sl - 2.
LetL’p
be the fixed field of D. SinceGF
eLK,
no
primes
ramify
Also, only
primes
inSo
canramify
inK/F.
Soonly
primes
inS°
canramify
inL’ F IF.
The definition ofL’p
requires
thatthe
primes
inS°
other than 002split completely
in Hencecan
ramify only
at 002. ThusL’p
is contained in the ray class field modulo002 for F. But the ray classes modulo 002 are the same as the ray classes modulo
1,
due to the presence of the unit -1. Thus£h /F
is unramifiedat 002 as
well,
and is thereforeeverywhere unramified,
with allprimes
in Ssplitting completely.
The fact that 002
splits
inL’p/ F
but not in thequadratic
extensionK / F
implies
thatLp
fl K = F. Thus theelementary
abelian2-group
Nc
=Gal(£K/K)
and theelementary
abelian2-group
D =generate
the abelian groupGal(LK / F),
which is therefore also anelemen-tary
abelian2-group.
Hence if q is anyprime
of F which is inert inK,
wemay consider the Frobenius of the extended
prime il
of K in the extension£K / K
and use theproperties
of the Frobenius in relative extensions(see
[1, 111.2.4]):
a(q,£K/F)2
= 1. This shows thatJK
hastrivial
image
inGal(£K / K)
under the Artin map.We return now to the
isomorphism
from toN~ _
which is induced
by
the Artin map as described above. Sincethe
image
ofJK
lies in the kernel of thisisomorphism,
we conclude thatNow we observe that
GF
has an intrinsic definition in terms of F. SinceGF/F
is an unramifiedelementary
abelian 2-extension in which allprimes
of Ssplit completely,
it is contained in the maximal suchextension,
whichwe denote
by
GF.
Then K is an unramifiedelementary
abelian2-extension of K in which all
primes
ofSfin(K) (indeed S(K),
as unramifiedis the same as
split
for the infiniteprimes)
split completely,
and is abelianover F.
But GK
was defined to be the maximal suchextension,
so£K D
GF.
As allprimes
in Ssplit
completely
inLFIF,
Gp
must be fixedby
the
decomposition
groupsgenerating
D. This means thatGF
C Weconclude that
GF
=GF.
Finally
defineDo
=GF)),
which has index 2 in D =Gal(GK/GF).
ThusDo
is anelementary
abelian2-group
withrank2(Do)
181 -
3. Then we have an exact sequence:This
simply
comes from the natural restriction mapidentifying
Gal((K -
LF)IK)
withThus in terms of class groups
(using (1)
and(3)),
the exactsequence(2)
becomes
where the kernel
Co
is anelementary
abelian2-group
of 3 and the map on theright
is inducedby
the norm map on ideals. The conclusionof the theorem follows. 0
Corollary
1. Theintegers 2TF(S)
dividesMi
when 1.Proof.
This is clear sincerF(S)
=rank2(HK)
for K =0 6. Proof of Theorem 1
The
assumption
is that m + 2 -rF(S).
In view ofProposition
1,
we consider-- , , , I I
. - , ....
where the
product
mayclearly
be taken over i forwhich qj #
1. Forsuch i,
the
expression
ei =M, -
is aninteger multiple
ofby
Corollary
1,
and this in turn isintegral by
assumption.
Thus ei isintegral
and so.. ~
does in fact lie in
L,
sinceeach qi
lies inKi
C L. ThenSt’ ~L/F,,S)
holdsby Proposition
1. Furthermoreand lies in an abelian extension of
~’, by
the Stark-Tate theorem. As thecomposite
of abelian extensions isabelian,
we conclude that-l/WL
lies in an abelian extension of F. This
completes
theproof
of Theorem 1. 7. Kummertheory
Let:
· G =
GS
be thecomposite
of allquadratic
extensions of Fin Q
withrelative discriminant
dividing
4ms.
.
C~F
be thering
ofintegers
of F.. be the
ring
ofS¡in-integers
ofOF.
Lemma 1.
Suppose
(K : F]
= 2. Then K =for
some1 E F which
generates
afractional
idealof F of
theforra
(7)
=a26
with bsupported in
8fin
if
andonly if
the relative discriminants8(K/F)
of
K over F dividesProof.
First suppose that K = with(-y)
= and bsupported
in6’fin.
The relative discriminant may becomputed
locally,
so we reduce to the case of an extension of local fieldsKTIFP
by passing
to thecompletions
at a fixed
arbitrary prime p
of F and aprime T
over p in K. Thatis,
thep-part
of the relative discriminant ofK/F
equals
the relative discriminant ofKTIFP,
and it suffices to show that this divides4tns
foreach p.
Let 7r be auniformizing
parameter
for thering
ofintegers
Op
ofF~ .
Then -Y
= where u is a unit ofOp
and aequals
0 or 1. SoKT
Fp(..j7)
= We treat the twopossibilities
for aindividually.
When a = 0 we have Then the relative discriminant
J(Kq3/Fp)
divides the discriminant of thepolynomial
x2 -u
which is(4u) =
(4),
and thisclearly
divides4ms.
When a =
1,
itevidently
must be the casethat p
dividesb,
and thereforep divides my - We have = where
~r’
is anotheruni-formizing
parameter.
ThusKTIFP
is an Eisenstein extension for which it is known thatOq3
= Thereforeequals
the discriminantof
~2 -
x’,
namely
(4~r’)
=4~.
Again
this divides as p divides ms.This
completes
the first half of theproof.
Next assume that the relative discriminant
J(KIF)
of K over F divides4tns .
SinceK / F
is a relativequadratic extension,
we know that K =F(..j7)
for some q E F. Write(q)
= and b a square free fractional ideal. If aprime
p appears in the factorization ofb,
letq3
be aprime
above p in K. Then we are in the situationappearing
in the first half of theproof
where a = 1 andKq3lFp
is an Eisenstein extension. In this case wesaw that
6(Kqg /Fp) =
4p.
We areassuming
that this divides so mayclearly
concludethat p
divides mis andthus p
is inS’fin.
This shows that bis
supported
in,S’fin,
and concludes theproof.
0Proposition
3. Thefield
L= ,CS
contains i = 1 ...,2~"~}).
Proof.
We show that ,C contains i = 1...2"z-1 ~ )
by showing
that ,C contains L and eachV1ii.
First,
each is aquadratic
extension,
soKi =
We may write =ab
with bsquare free.
Then
is ramified at the divisors ofb,
by
Kummertheory.
Since is unramified outside,S’,
we conclude that b issupported
inSfin.
It now follows from theLemma that
Ki
is aquadratic
extension of F with relative discriminantdividing
4ms .
But ,C was defined to be thecomposite
of all such extensions.Thus L contains the
composite
of all the which isL,
as we observed inthe
beginning
of section III.Having
shown that L containsL,
weproceed
to show that ,C contains each This is trivial if 1}i =1,
so we may assume that we are notin this situation. Then the
image
of qj
generates
the infinitecyclic
groupThus qj
is not a square inK2,
and qj
does not lie in F. Sois an extension of
degree
4 over F. We know thatqi
= sothe
conjugates
of over F Thus isa Galois extension of
degree
4. It is in fact thecomposite
of the relativequadratic
extensionKi
F(?7i)
in which 002ramifies,
and the relativequadratic
extensionKi =
+1/ r~Z)
in which 002splits.
We havealready
seen thatI~2
lies in,C,
so we now show that lies in ,~. This willimply
that thecomposite
liesinC,
as desired.Above we saw In since this
extension is unramified at v.
Since qi
lies inUJ;!,
the Lemmayields
xs Hence
’
which divides
16n28(Ki/F)2.
Similarly,
S(Ki/F)2
divides and thus divides We conclude thatWe examine this
divisibility
statement oneprime
at a time and showthat it
implies
b(Ki/F)p~4ms
for each p. Observe that is unramified outside ofS,
sob(Ki/F)p
=(1)
for p notdividing
ms.Consequently
which divides
4ms
in this case.Now for p
dividing
ms, the lemmaapplied
toKi/F
implies
thatdivides
4p
which in turn divides4ms.
This shows that6 (K§ /F)
divides4ms.
HenceKi’
lies inG, by
its very definition. 0Proposition
4.[,C,g : F]
=2TF(S)+!SI
Proof.
Let =l, ... ,
t}
be a minimal set ofgenerators
for the2-torsion
subgroup
of theSfin-class
groupCIF(Sfin)-
So t =rank2(CIF(Sfin))
=rF(S).
We viewClp(Sfin)
as the group of invertibleideals modulo
principal
fractional ideals of thering
of elements ofF which are
integral
at all finiteprimes
not inSfin. Using
Chebatorev’sden-sity theorem,
we choose therepresentatives ;
to beprime
ideals ofThe units of this
ring
are denotedUF(Sfin)
and called theSfin-units.
Nowa?
=aiOp(Sfin)
for some ai. Let A =({ai :
i =l, ... ,
tl) UF (Sfin)
-We
begin by noting
that A =({ai}) x UF(Sfin).
For a non-trivial elementof
generates
an ideal which is a non-trivialproduct
of theprime
idealsa-¿, while each element of
UF(Sfin)
generates
the unit ideal. Thusby
theDirichlet-Chevalley-Hasse
unittheorem,
rank2(A)
= +rank2(UF(Sfin))
= t += rF(S)
+We will establish a one-to-one
correspondence
between the non-trivialelements of
A/A2
and the relativequadratic
extensionsK/F
contained in G. Thisimplies
thatrank2(Gal(G/F))
=rank2(A/A2),
which combinedwith the
displayed equality yields
the statement of theproposition.
Now observe that A fl(Fx)2
= A2
as follows. If_y2
E(FX)2
lies inA,
and therefore
ni a(.
The fact that this is aprincipal
idealgenerated
by
the aiimplies by
their definition that all of theexponents
areeven, ci =
2bi.
We nowhave 72
= and this shows that u= v2
is a square.Clearly v
EUF(Sfin),
so 7 = afterchoosing
the correctsign
for v. From this we see thaty2
EA2,
which was to beproved.
Given a y
representing
a non-trivial class inA/A2,
this willcorrespond
to the field K =
~(~7).
According
to the lastpaxagraph, K
will in fact bea relative
quadratic
extension of F. We check that K lies in Gby
showing
that the relative discriminant
6(K/F)
divides4ms.
The factthat y
E A meansthat 7
for someintegers
ei and some u EIJF(,Sfin).
Then theprincipal OF-ideal generated by y
is(~y)
=ab,
whereb =
(v),
andâi
is the(prime)
ideal ofOF
supported
outside ofSfin
suchthat = ai. Since b =
(v)
issupported
inSfin,
Lemma 1 allowsus to conclude that
6(K/F)
divides4ms,
as desired.Conversely,
given
a relativequadratic
extensionK/F
contained inG,
we willproduce
thecorresponding -y
E A. First we note that the relativediscriminant of a relative
quadratic
extension isequal
to the finitepart
ofits
conductor,
by
the conductor-discriminant theorem. Thus every relativequadratic
extension of F with discriminantdividing
4ms
is contained in the ray class field of F with conductorequal
to4ms
multiplied by
all ofthe infinite
primes.
Hence the field Ggenerated
by
all of these relativequadratic
extensions is also contained in this ray class field. Then anyquadratic
extension of F contained in G will have conductordividing
theproduct
of4ms
with all of the infiniteprimes,
so that its discriminant alsodivides
4ms.
We can conclude that the discriminant of ourgiven K
divides4ms.
Lemma 1 nowimplies
that K = where(Y)
=a2b
and bis
supported
inSon.
Hence(aCJp(Sfin))2,
so thataOF(Sfin)
represents
an element ofClF(Sfin)[2].
But this group isgenerated by
theimages
of the a2. Thus for some/3
E F. Then1’IOp(Sfin) =
nar,82CJp(Sfin).
Let,
= Weclearly
have K == F(~),
while Thus y =u 11
ai
for some u EUF(Sfin)
and therefore 11Corollary
2. 1. We have[L: L]
=2. Let
(i
be agenerator
of ItKi.
Wlaen is notequal
to1,
theexportent
Mi -
27G.
If it
is not inZ,
then either L = G or(G:L~=2
Proof.
1. From the fact that[L : F]
= 2’" andPropositions
3 and4,
weconclude that
(G : L]
= and thus this rational number is2. Now we can see that
[,C : LI /4
lies in4 ~.
By
Corol-lary 1,
it follows thatM, -
is in47~,
and that if it does not lie in2 ~,
we must have L = ,C. In this case, note that theambiguity
up to a root of
unity
in the choiceof qj
allows us to conclude fromProposition
3 that L = ,C contains bothV1ii
andvl-(i7-7i
andthere-fore
ae
E L. Thus is even andMi .
lies in2 7~.
Finally,
since[£ :
L]
is a power of2,
theonly
other situation inwhich =
~~C :
L]/4
is notintegral
clearly
occurs whenC,C : L] =
2 and it ishalf-integral.
ThenMi -
is in2 7~,
soMi -
2ISB-m-2(WL/Wi)
isintegral
unlessWL/Wi
isodd,
i.e. L.0
8. Proofs of Theorems 2 and 3
Under our
standing
assumptions
thatL/F’
ismultiquadratic,
and thatin order to avoid
special
cases of theconjecture
which havealready
beenproved,
S’ contains at least two infiniteprimes
and one other finite or infiniteprime,
we now have:by Proposition
1.* The
exponent
Mi
is eitherintegral
orhalf-integral
when
1Ji =1=
1,
by
Corollary
2.o If it is
half-integral
forsome i,
then either ,C = L or we have both[£ : L]
= 2 andae ~
L ,
alsoby
Corollary
2.If L =
L,
then E L for alli,
since E,C, by Proposition
3. If[L : L]
= 2 andJG E
L for somei,
notice that both andÝ(i1Ji
lie in £by Proposition
3again
and theambiguity
in 1Ji. If neither of them lie inL,
then = jC =L ( ~) .
Thisimplies
thatae
EL,
which is not thecase. Hence either E L or
Ý(i’’Ii
E L.By renaming
we mayagain
assume
JK
E L.Thus in all cases, Theorem 2 follows from
Proposition
1.Turning
to theproof
of Theorem3,
we now assume that D is either real or finite. When L is not the maximalmultiquadratic
extension M of Fwhich is unramified outside of ,S’ and in which v
splits completely,
we claimthat
4 [ ~.C : L] .
Thenby Corollary
2,
rF ( S)
+[ S[ - m - 2 >
0,
and the result will follow from Theorem 1.To establish the
claim,
we first show that v is notsplit
completely
in,CS/F.
When tJ isreal,
this is clear since the definition of£s implies
that E~CS
and thus£s
istotally
imaginary.
When t) isfinite,
weproceed by
contradiction.Suppose
tosplits completely
inCSIF.
Then l1is unramified
in .cs,
soclearly
£s
_ FromCorollary
2,
we thenget
so that
of l1 must be non-trivial in this group.
By
class fieldtheory,
l1 is thennot
split completely
in the maximal unramifiedmultiquadratic
extensionof F in which every finite
prime
ofS° splits completely. However,
this extension is contained in£s, by
Lemma1,
and we have assumed that nsplits completely
inGs,
a contradiction.Let
£’5
denote thesplitting
field of n in£s. By
the claim we havejust
established, 21[£8 : ,Cs .
Since tosplits
completely
in L C,rCs,
we also haveL and
2)[jC~ : L]
unless L = Thus4~ ~,~5 : L]
unless L =£:8.
From the definitions and Lemma 1
again,
it follows that L C M c£’5.
Thus in the
exceptional
case of L = we have L =,/l~t,
the maximalmultiquadratic
extension of F which is unramified outside of ,S and inwhich to
splits completely.
9. The
biquadratic
case:proof
of Theorem 4We now assume that m = 2 and turn to the
proof
of Theorem 4. Since3,
Theorem 1 reduces us to the case where181
= 3 andrF ( S) -
0.By
Remark2,
we may assume that someprime
P2 over 2 ramifies inL/F,
so that S =
tOO1,
002,
P2},
and we must have v =ool
splitting
inL/F.
(This
is theonly
time we will make use of[3].)
Thenby Proposition 2,
wehave that
Mi
is oddfor ni ~
1. Thuswith both
Mi
odd.By
Theorem2, e
E L satisfiesSt’(L/F)
and indeed theproof
shows thatwe may take
/-i)
E L for i =1, 2.
Temporarily
fix i = 1 or 2. Notice that 002 ramifies inK2,
andonly
one other
prime
(namely p2)
is allowed toramify
over F. But some otherprime
mustramify,
for otherwiseKi
is contained in the ray class field forF modulo 002. But the ray class group modulo 002 is the same as the ray
class group modulo
1,
due to the presence of the unlit -1. This wouldimply
thatKi
is unramified at 002, a contradiction.Thusqi =,4
1.It remains to check that
L(e1/WL)
is abelian over F. For this we use astandard lemma
(see [5,
p.83, Prop.
1.2]).
Lemma 2.
Suppose
L/F
is afinite
abelian extensionof
numberfields
with Galois group G. Let A be the annihilator idealof
the groupof
rootsof unity
iLL considered as a module over the groupring
7G(G.
Let T be a setof
generators
for
A. Then an element u in themultiplicative
group L* has theproperty
thatL(ul/WL)/F
isabelian,
if
andonly if
there exists a collectionof
aa EL*,
indexedby
a E T such that bothof
thefollowing
conditionsRecall that T E G is the
complex conjugation
in L over 002. Also let T1be the element of order 2 in G which fixes
Kl.
Thus T and Tlgenerate
G.First consider the number of roots of
unity
z,vL in L.Suppose
WL > 2.Then L has no real
embeddings,
and thesplit
prime
ool of F must becomplex,
while 002 is real. Hence F is a non-Galois cubic extensionof Q
and
[L : Q]
= 12.If L contains a
pth
root ofunity
(p
for some oddprime
p, thenL/F
must
ramify
at someprime
over p, becauseF«(p)/ F
does. But theonly
finite
prime
which canramify
inL/F
is P2 E s. If L contains a 16th rootof
unity,
then[L : Q] =
12 must be divisibleby
cP(16) =
8,
a contradiction.Thus the number of roots of
unity
in L is=2,
4,
or 8.Case 1: wL - 2. When WL =
2,
the abovearguments
show that we maytake = L =
K(vr¡2).
Since T is acomplex conjugation
andr~Z +T =~
1 for i =
1, 2;
we conclude that = 1 for i= 1, 2.
Using this,
one canverify
that the conditions of Lemma 2 hold for E =vfijïMl vr¡2M2
E L uponsetting
e, a1+T =1,
and ThusSt (L/F, S)
holds inthis case.
Now if
4~~vL,
then L containsF(B/~I),
which must be eitherK1
or.K2.
By
renumbering,
we may assume that it isK2.
Thus wl = 2eCase 2: 4. Now w, =
2,
W2 =4,
and - =r¡1 Ml vr¡2M2
E L. Wehave noticed above that ,~ contains the square roots of all the roots of
unity
in theKi.
Thus ,~C contains an 8th root ofunity (8.
Put Ela,+l =
1,
and Theargument
above showsthat and
(8
lie in£,
but notL,
although
their squares lie in L. Thus(8vfijïM¡ E
L,
sinceM~1
is odd. Alsoý9i2 E
L,
so we have confirmed thataTl-3 E L.
Again
the conditions of Lemma 2 hold andSt(L/F,,S’)
isproved
in this case.Case 3: 8. Now
2,
W2 =4,
and e =1112Ml172M2
E L. PutaWL e, aT+1 -
1,
and aTl-3 = We know that1
and
~/~2
lie in~C,
but not inL,
although
their squares lie in L. Thus aTl -3 EL,
sinceM1
andM2
are odd.Again
the conditions of Lemma 2hold and
St(L/ F,S)
isproved
in this case.Acknowledgements
The second author thanks the Mathematics
Department
of theUniversity
of Texas at Austin for its
hospitality
while much of the work of this paperwas
being
carried out. Thanks also go to CristianPopescu
for severalReferences
[1] G. JANUSZ, Algebraic number fields. Academic Press, New York, 1973.
[2] J. W. SANDS, Galois groups of exponent two and the Brumer-Stark conjecture. J. Reine
Angew. Math. 349 (1984), 129-135.
[3] J. W. SANDS, Two cases of Stark’s conjecture. Math. Ann. 272 (1985), 349-359.
[4] H. M. STARK, L-functions at s = 1 IV. First derivatives at s = 0. Advances in Math. 35
(1980), 197-235.
[5] J. T. TATE, Les conjectures de Stark sur les fonctions L d’Artin en s = 0. Birkhäuser,
Boston, 1984.
David S. DUMMIT, Jonathan W. SANDS
Dept. of Math. and Stat.
University of Vermont
Burlington, VT 05401 USA
E-mail : David . Dummitmuvm . edu, Jonathan . Sandsouvm.edu Brett TANGEDAL
Dept. of Math.
University of Charleston
Charleston, SC 29424 USA