On Tate’s refinement for
aconjecture of Gross
and its generalization
par NOBORU AOKI
RÉSUMÉ. Nous étudions un raffinement dù à Tate de la conjecture
de Gross sur les valeurs de fonctions L abéliennes en s = 0 et for- mulons sa généralisation à une extension cyclique abitraire. Nous prouvons que notre conjecture généralisée est vraie dans le cas des
corps de nombres. Cela entraine en particulier que le raffinement de Tate est vrai pour tout corps de nombres.
ABSTRACT. We study Tate’s refinement for a conjecture of Gross
on the values of abelian L-function at s = 0 and formulate its
generalization to arbitrary cyclic extensions. We prove that our
generalized conjecture is true in the case of number fields. This in
particular implies that Tate’s refinement is true for any number field.
1. Introduction
In
[9]
Grossproposed
aconjecture
whichpredicts
a relation between twoarithmetic
objects,
theStickelberger
element and the Grossregulator,
bothof which are defined for any data
(K/k, S, T),
where is an abelian ex-tension of
global
fields andS,
T arefinite,
non-empty subsets of theplaces
of k
satisfying
certain conditions. In the same paper, heproved
the con-jecture
in the case of unramified extensions of numberfields,
and obtaineda
partial
result when the extension iscyclic
ofprime degree.
Since thenthe
conjecture
has been verified to be true in several cases(see Proposition
2.2
below),
but yet it remains to beproved
ingeneral. Taking
a close lookat the
conjecture, however,
one caneasily
notice that it becomes trivial insome cases. For
example,
if S contains aplace
whichcompletely splits
inK,
then both theStickelberger
element and the Grossregulator
are zero, and theconjecture
istrivially
true. Besides such a trivial case, there arestill some cases where the
conjecture
becomes trivial. As observedby
Tate[22],
this is the case if is acyclic
extension whosedegree
is a power ofa
prime
number l and if at least oneplace
of S "almostsplits completely"
Manuscrit reçu le 6 juin 2003.
in
(see
Section 3 for thedefinition)
and anotherplace
in Ssplits
inHe then
proposed
a refinedconjecture
in that case.The purpose of this paper is to
study
Tate’s refinedconjecture
from acohomological
viewpoint
and togeneralize
it toarbitrary cyclic
extensions.(In
aforthcoming
paper[2],
a furthergeneralization
of theconjecture
willbe
given.)
We will prove that a weak congruence holds for anycyclic
l-extension
(Theorem 3.3),
whichimplies
Tate’s refinedconjecture
when kis a number field
(Theorem 3.4). Piecing
the congruencestogether
forall
primes 1,
we will also obtain a weak congruence forarbitrary cyclic
extensions
(Theorem 4.2),
which is apartial
result in the direction of ourconjecture.
Inparticular
it shows that thegeneralized conjecture (and
hence the Gross
conjecture)
is true forarbitrary cyclic
extensions of number fields(Theorem
4.3 andCorollary 4.4).
In the lastsection, using
the resultsabove,
we willgive
a newproof
of the Grossconjecture
forarbitrary
abelianextensions
K/Q (Theorem 10.1),
whichsimplifies
ourprevious proof
in[1].
The main idea of the
proof
consists of twoingredients:
one is an inter-pretation
of the Grossregulator
map in terms of Galoiscohomology,
andthe other is genus
theory
forcyclic
extensionsK/k.
Hereby
genustheory
we mean a formula
(Theorem 7.1)
for the(S, T)-ambiguous
class number of and it willplay
animportant
role when we relate theStickelberger
element to the Gross
regulator
in theproof
of Theorem 4.2. The idea touse genus
theory
can bealready
found in the paper of Gross[9],
wherehe
implicitly
used it to prove a weak congruence in the case ofcyclic
ex-tensions of
prime degree.
Thus ourproof
may beregarded
as a naturalgeneralization
of his.Acknowledgements
I would like to thankJoongul
Lee andKi-Seng
Tanfor
reading
themanuscript
verycarefully
andmaking
a number ofhelpful suggestions. Especially
I amconsiderably
indebted to Lee for the treatmentof the case
"mo
> 0" in Theorem 9.2. I am verygrateful
to David Burns forletting
me know of recent workby
himself[3]
andby Anthony Hayward [11]
both of which areclosely
related to this article. I also wish to express mygratitude
to John Tate for many valuable comments andsuggestions.
2. The Gross
conjecture
In this section we
briefly
recall the Grossconjecture.
Let k be aglobal
field,
and let S be anyfinite,
non-empty set ofplaces
of l~ which contains all the archimedeanplaces
if k is a number field. Letes
be thering
ofS-integers
of k and consider the S-zeta functionwhere the summation is over the ideals a of
Os.
It is well known that this series converges forR(s)
> 1 and has ameromorphic
continuation to thes-plane,
withonly
asimple pole
at s = 1. Theanalytic
class number formula for k says that theTaylor expansion
of(s( s)
at s = 0begins:
where
hs
is the class number of®s, Rs
is theS-regulator,
ws is the numberof root of
unity
in the S-unit groupUs
=U;
and n =IS I - 1.
To achievea formula with no
denominator, following Gross,
we introduce aslightly
modified zeta function. Let T be a finite set of
places
of 1~ which isdisjoint
from
S,
and define the(S, T)-zeta
functionwhere Nv is the
cardinality
of the residue field of v. To describe the corre-sponding
formula for(S,T(S),
we define the(S, T)-unit
group of kby
Clearly
the index(Us : Us,T)
isfinite,
and is a divisorDefine the
(S, T)-class
number h =hs,T by
the formulaThen the
Taylor expansion
of(s,T(s)
at s = 0begins:
where
Rs,T
is the(S, T)-regulator (see
below for thedefinition)
and w isthe number of root of
unity
inUs,T.
We henceforth assume that T is chosenso that
Us,T
is torsion-free. Then from(2)
we obtain a formula without adenominator:
To define the
(S, T)-regulator,
letYs
be the free abelian groupgenerated by
theplaces
of S andthe
subgroup
of elements ofdegree
zero inYs.
ThenXs ^--’
ZI. Since weare
assuming
thatUS,T
istorsion-free,
we have anisomorphism
Let
(1,... un)
and(xl, ... , rn)
be l2-bases of andXs respectively.
Let
detR(A)
be the determinant of the maptaken with respect to Z-bases of
UST
andXS
chosen above. The(S, T)- regulator Rs,T
isby
definition the absolute value ofdetr (A) - By (3)
wehave
where the
sign
of coursedepends
on the choice of bases ofUS,T
andXs.
The Gross
conjecture predicts
that there is ananalogous
congruence relation if wereplace
anddetr (A)
in(4) by
theStickelberger
elementand the Gross
regulator respectively.
Wegive
a brief review of the definition of these twoobjects.
First we define the
Stickelberger
element. Let be a finite abelian extension which is unramified outside S and G the Galois group of For anycomplex
valuedcharacter X
ofG,
we define the(S, T)-L
functionassociated to X to be the infinite
product
where
Frv
E G denotes the Frobenius element at v. Then there exists aunique
elementBG
=C[G]
such thatX(OG)
= for anyX. Gross
[9]
showed thatBG
is inZ[G] using integrality properties proved independently by Deligne-Ribet [7]
andBarsky-Cassou-Nogu6s [5].
Next,
to define the Grossregulator,
consider the mapwhere rv :
k;
2013~ G denotes the localreciprocity
map. Let be then x n matrix
representing
A with respect to the bases and(rj )
chosenabove, namely:
Let
Z[G]
be theintegral
groupring
of G andIG
theaugmentation
ideal ofZ[G].
Then there is anisomorphism
G ~IG/IÕ, 9 1---+ 9 - 1. Using
thisisomorphism
we can view the matrix with entries inIGII2G
as amatrix
representing
A. Define the Grossregulator by
where the sum is over the
permutations
of~l, ... ,
We are now in a
position
to state theconjecture
of Gross.Conjecture
2.1. Let the notation be as above. ThenOG
E13
and thefollowing
congruence holds:where the
sign
is chosen in a way consistent with(4).
In the
following
we have a list of the cases whereConjecture
2.1 isproved.
Proposition
2.2.Conjecture
2.1 is true in thefollowing
cases:(i)
n=0.(ii) k
is a numberfield
and S is the setof
the archimedeanplaces.
(iii)
K is aquadratic
extensionof
k.(iv) k
is afunction field
and n = 1.(v) k
=Q.
(vi)
is an abelianp-extension of function fields of
characteristic p.(vii) K/k
is an abelian1-extension.,
where l is aprime
numberdifferent from
the characteristicof
k and divides neither the class numberof
k nor the numbers
of
rootsof unity
in K.(viii) K/k
is an abelian extensionof a
rationalfunctions field
k over afinite field
and 3.Proof.
In the case of(i) Conjecture
2.1 is an immediate consequence of(4).
In both cases
(ii)
and(iii)
theconjecture
wasproved by
Gross in[9].
Case(iv)
is a consequence of the work ofHayes [10] proving
a refined version ofthe Stark
conjecture.
Case(v)
was treated in ourprevious
paper[1].
Inthe cases
(vi), (vii)
and(viii)
theconjecture
wasproved by
Tan[19],
Lee[15]
and Reid[16], respectively.
0As mentioned in the
introduction,
Gross[9] proved
a weak congruence whenK/k
is acyclic
extension of aprime degree.
Here wegive
theprecise
statement because our main results
(Theorem
3.3 and Theorem4.2)
may be viewed as ageneralization
of it toarbitrary cyclic
extensions.Proposition
2.3.Suppose K/k
is acyclic
extensionof
primedegree
l.Then there is a constant c E
(7G/d7G)"
such thatProof.
See[9, Proposition 6.15].
03. Tate’s refinement for the Gross
conjecture
In this section we
give
theprecise
statement of Tate’s refinement for the Grossconjecture.
First,
we assume that G is anarbitrary
finite abelian group. Choose and fix aplace
vo E S and setS’1= S ~ ~vl, ... , vn~. Then,
as a Z-basisof
Xs,
we can take{~i 2013
vo, ... , vn - In this case we haveChoosing
a Z-basis~ul, ... , 7 Un I
ofUS,T,
we defineIt is clear from the definition that
?ZG
E13
anddetG(A) (mod ~+1).
Now,
suppose that G is acyclic
group ofdegree lm,
a power of aprime
number l. For each v E
S,
letG~,
denote thedecomposition
group of v in G. We fix theordering
of the elements of S = vi , ... , so thatClearly
N > n, and N = n if andonly
if mo = ... = 0.If mn = m, that
is,
Vnsplits completely
in then0G
=RG
= 0.Hence
Conjecture
2.1trivially
holds. Let us consider the secondsimplest
case mn = m -1.
Following Tate,
we say that theplace
vn almostsplits completely
in K if mn = m -1. Tate[22] proved
thefollowing.
Theorem 3.1. Assume that mn = m - 1. Then
8G
EIg
andRG
EMoreover, the image ofRG in
sign, independent of the
choiceof the
basis and the choiceof va.
Since
N > n,
thistheorem,
inparticular,
shows that8G
EIä.
Let usconsider the case where mo > 0. In this case we have N > n and hence Theorem 3.1 also shows that
8G
E~+1. Moreover,
one can show thathRG
= 0(mod ~+1) (see
Theorem9.2, (i)).
ThereforeConjecture
2.1 istrivially
true if mo > 0. On the otherhand,
if mo =0,
then both8G
and7ZG
are in the same idealIg.
Therefore it ismeaningful
to compare them in thequotient
group Based on thesefacts,
Tate[22] proposed
a refinement for the Gross
conjecture.
Conjecture
3.2. Assume that mo = 0 and mn = m -1. Thenwhere the
sign is
chosen in a way consistent with(4).
Obviously Conjecture
3.2implies Conjecture
2.1 since N > n andNow,
we can state one of our main results.Theorem 3.3. Let the notation and
assumptions
be as inConjecture
3.2.Then there exists an
integer
cprime
to l such thatIn
particular, if l
=2,
thenConjecture
3.2 is true.We will
give
theproof
of this theorem(more precisely,
of Theorem 9.1 which isequivalent
to Theorem3.3)
in Section 9. Here weonly
note thatthe last assertion
immediately
follows from the first one. To see this wehave
only
to observe thatIG /IG +1 ^-’
since G is acyclic
group andthat both
0G
andRG
are killedby
l in under the condition thatmn = m -1
(see Proposition 5.4) . Therefore,
if l =2,
then =~7~
(mod
ThusConjecture
3.2 is true.As a
special
case of Theorem 3.3 we have thefollowing.
Corollary
3.4.If K/k
is acyclic
l-extensionof number fields,
then Con-jecture
3.2 is true.Proof. Actually, Conjecture
3.2 is non-trivialonly
when 1 = 2 in the numberfield case since is an
l-group
and bothBG
and RG are killedby
2 inThus
Corollary
3.4 is a direct consequence of Theorem 3.3. ll Remark 3.5. As remarkedby
Lee[14],
one can notdrop
the conditionmo = 0 from the conditions in
Conjecture
3.2.Indeed,
he showed that there areinfinitely
manycyclic
l-extensionsK/k
for which buthdetG(A)
E In the nextsection,
we willgeneralize Conjecture
3.2 toarbitrary cyclic
extensions in order to remove the restriction on mo.4. A
generalization
of Tate’sconjecture
In this section G will be an
arbitrary cyclic
group. For any v ES,
letbe the kernel of the canonical
surjection Z[G]
-- Let us chooseand fix a
place
vo E S. LetSl
=S ~ Ivol
and consider the idealin
Z[G].
If Q is a generator ofG,
thenGVi
isgenerated byavi
:=(J(G:Gvi)
and is a
principal
idealgenerated by (Qvl -1 ) ~ ~ ~ (ow,~ -1 ) .
Since theentries of the i-th row of the matrix are in the ideal
IG(vi),
its determinant RG
belongs
to the idealIG(Si). Moreover,
theimage
ofRG
in thequotient
groupis,
up to thesign, independent
of the choice of the basis of
Us,T.
Given a finite abelian extension
K/k
ofglobal
fields and finite setsS,
T ofplaces
of k such that S fl T =0,
we call thetriple S, T)
an admissibledata if the
following
conditions are satisfied:(i)
S contains theplaces
of 1~ramifying
in K and the archimedeanplaces
of k.
(ii) U K,S,T
is torsion-free.In
[9]
Gross gave a sufficient condition forUs,T
to be torsion-free:. In the function field case
UST
is torsion-free if T isnon-empty.
. In the number field case
Us,T
is torsion-free if T contains eitherprimes
of different residue characteristics or a
prime v
whose absolute ramifi- cation index ev isstrictly
less thanp -1,
where p is the characteristicof Fv .
It is worthwhile to note that each condition above also ensures that
UK,S,T
is torsion-free for any finite abelian extension unramified outside S.
We propose the
following conjecture
which will turn out to beequivalent
to Tate’s refinement when is a
cyclic
1-extension such that mo = 0 and mConjecture
4.1. Let(K/k, S, T)
be an admissible data such that G =Gal(K/k)
is acyclic
group. Thenwhere the
sign
is chosen in a way consistent with(4).
It is very
likely
that the congruence in theconjecture
holds for any finite abelian extension In aforthcoming
paper[2],
this will be discussed in some detail and some evidence will begiven.
If l is a
prime
numberdividing
we denoteby GI
the1-Sylow subgroup
of G. For each v E
S,
letGv
be thedecomposition
group of v inG,
and put=
Gv
nGl.
Thus is the1-Sylow subgroup
ofGv. Now,
considerthe
following
condition:(5) There
exists aplace
vn Esl
such that5 for any prime
divi s or l ofsuch that l
lIn other
words,
thisrequires
that there exists aplace
vn in S which eithersplits completely
or almostsplits completely
in the maximal 1-extension of k contained in K for eachprime
divisor 1 ofIGI. Although
this condition is very restrictive in the function field case, it isalways
satisfied in the numberfield case since
=1
or 2 for any archimedeanplace v
of k.Now,
we can state our mainresult,
which may be viewed as apartial
answer to
Conjecture
4.1.Theorem 4.2.
Suppose
condition(5)
holds. ThenBG belongs
toIG(SL)
and there exists an
integer
cprime
toIGI
such thatIn the next section we will see that in order to prove Theorem 4.2 it suffices to prove it when G is a
cyclic
group of aprime
power order.In the case of number
fields,
Theorem 4.2gives
acomplete
answer toConjecture
4.1.Theorem 4.3.
If K/k
is acyclic
extensionsof
numberfields,
then we haveProof.
As we have remarkedabove,
condition(5)
isalways
satisfied in thenumber field case. As is well
known,
if k is nottotally
real or K is nota
totally imaginary,
thenConjecture
2.1 is trivial.Suppose
K is atotally imaginary
extension of atotally
real field l~.Then,
as we will see later(see Proposition 5.4),
thequotient
groupIG/IGIG(SI)
is acyclic
group of order 2. Thus Theorem 4.3immediately
follows from Theorem 4.2. 0 SinceConjecture
3.2implies Conjecture 2.1,
Theorem 4.3 proves thefollowing.
Corollary
4.4.If
is acyclic
extensionof
numberfields,
then Con-jecture
true.5.
Stickelberger
elements forcyclic
I-extensionsThroughout
this section we will assume that G is acyclic 1-group
andthat
(K/k, S, T)
is an admissible data. Inparticular UK,s,T
is torsion-free.In order to state Theorem 5.1
below,
let F be the intermediate field ofK/k
with
[K : F]
= 1 and putIf at least one
place
in ~S’ does notsplit completely
in thenhK,S,T is
an
integer by [17,
Lemma3.4].
Inparticular,
if mn = m -1,
thenhx,S,T
isan
integer.
’ ,
Theorem 5.1.
Suppose
rrzn = m - 1. ThenMoreover,
thefollowing
assertions hold.(i) If mo
>0,
0(mod
(ii) If
mo =0,
then 0(mod IGIG(Sl)) if
andonly if
=0
(mod l).
~ ’
Although
both the first statement and(i)
of the second statement follows from Tate’s theorem(Theorem 3.1)
in view ofProposition 5.4,
we willgive
a
proof
for thecompleteness.
Webegin
with a lemma.Lemma 5.2. Let M be an intermediate
field of K/k
such that(S(K)) _ IS(M) I .
ThenUK,S,T
=UM,S,T
·Proof.
Let u be any element ofUK,s,T.
For any a EG, uO"-1
is also an ele-ment of
UK,s,T
sinceUK,s,T
is G-stable. Moreover thefollowing
argument shows that is a root ofunity. Indeed,
theassumption
of the lemma im-plies
thatU K,S,T /U M,S,T
is a finite group. It follows that there exists a pos- itiveinteger
m such that um EUM,S,T.
Therefore(uO"-I)m
= 1.Thus
ul-1
is an m-th root ofunity. However,
sinceUK,S,T
istorsion-free,
this shows that
u~-1=1
for any a EGal(K~M),
hence ~c E M~. The asser-tion of the lemma then follows from the fact that =
UM,s,T.
0Proposition
5.3. Assume that mn = m - 1. Thenwhere X runs
through
thefaithful
charactersof
G.Proof.
Let F be as above. Then theassumption
onGv’s implies
thatIS(K)I = IS (F) I. This,
inparticular, implies
thatFurther we have
where X
runsthrough
the faithful characters of G. On the otherhand, by
(3),
we haveGross
[9, (6.4), (6.5)]
showed thatThe denominator of the
right
hand side is 1by
Lemma5.2,
henceThe assertion of the
proposition
thenimmediately
follows from(6)
and(7).
0Proposition
5.4. Assume that mn = m - 1. Let p be an elemerctof
Gof
order 1. Then we have an
equality
and an
isomorophism
Proof.
See[22].
DCorollary
5.5.If K/k is
acyclic
1-extension such that mo = 0 and mn =m -
1,
thenConjecture
3.2 isequivalent to Conjecture ,~.1 Proof.
If mo =0,
then from(8)
we haveSince both
8G
andRG belong
to( p -1 )7~ ~G~,
this shows thatConjecture
3.2 is
equivalent
toConjecture
4.1. DProof of
Theorem 5.1. Let denote the norm map from toQ.
Thenby Proposition
5.3 we havewhere v = Since l
completely
ramifies in it followsfrom this that
’ ’
Since
I S (K) I -
1 = N +lm-l -
1 and0G
E(p - 1)Z[G’],
fromProposition
5.3 and the lemma of
[22]
we deduce thatThus
OG
Eby Proposition
5.4. This proves the first statement.If mo >
0,
thenlm° > l,
whenceby Proposition
5.4. Therefore8G
EIGIG(Sl) by
Theorem 3.1. If mo =0,
then
by
the sameproposition
we haveTherefore from
(10)
and(11)
we deduce that6G
EIGIG(SL)
if andonly
if(mod 1).
Thiscompletes
theproof.
D6. Reduction to
cyclic
I-extensionsWe wish to reduce Theorem 4.2 to the case of
cyclic
1-extensions. For that purpose we consider thering homomorphism Wl :
-induced from the canonical
surjection
G -Gl.
Letwhere 1 runs
through
theprime
numbersdividing
For anyplace v
ESl,
we have
*1 (IG (v))
=IGl(v)
and hence = If * denoteseither v or
Sl,
then we define the mapto be the restriction of W to
IG(*).
Proposition
6.1. Thefollowing
assertions holdfor
the mapsW v
andW SI . (i) Both Wv and Ts,
aresurjective.
(ii) Ker(Wv)
=Ker(ws1).
(iii)
Themaps Wv
andrespectively
induceisomorphisms
where l runs
through
theprime
numbersdividing ~G~.
Before
giving
theproof
of thisproposition,
we show how theproof
ofTheorem 4.2 can be reduced to the case of
cyclic
1-extensions. We continue to use the notation above and assume that condition(5)
holds.Then 9G belongs
toIG(vn). Suppose
that Theorem 6.1 is true for anyGl,
thatis,
foreach
prime
divisor 1 ofI G 1, 8Gl belongs
toIG~ (Sl)
and we have a congruence(12) (mod IGlIGl(SI))
for some
integer
clprime
to l. Since =Proposition 6.1, (iii)
shows that
8G belongs
toIG (Si).
Let c be aninteger
such that c - cl(mod l)
for anyprime 1 dividing I G 1. By Proposition 5.4, IG~ (Sl )
is trivial or
isomorphic
toaccording
asIGv,zI
=1 or l. It follows thatfor all l
dividing IGI.
ThenProposition 6.1, (iv)
and(12)
shows thatas desired. D
To state the
following lemma,
for anysubgroup
H ofG,
we denoteby IH
the kernel of the naturalsurjection 7G~G~ ~
Lemma 6.2. Let G a
finite
abelian group andGl, G2 subgroups of
G suchthat
]G2 [)
= 1. Thenfor
anypositive integer
t.Proof.
We haveonly
to show thatfor any 9i E
Gi
and anypositive integer
t. First we note that Efor i =
l, 2. By
induction on t one caneasily
see that EIG21
for any
positive integer
t. SinceGCD(IGII, IG21)
=1,
there existintegers
al, a2 such that
allGllt
+a21G21t
= 1. Then theidentity
shows that
equation (13) holds, completing
theproof.
0Proof of Proposition
6.1. It is clear from the definition thatfor any v E s and for any
prime
ldividing IGI.
SinceIG(v)
isgenerated by
as 1 ranges over the
prime
divisors of~G~,
it follows from(14)
thatWv
is
surjective.
From(14)
and the definition ofWl
we also deduce thatSince
IG(S’1)
containsfIvESlIGv,l
forall l,
we conclude that the mapBII Sl
is also
surjective.
To prove
(ii)
note that the idealKer(BIIv)
isgenerated by
theproducts
of two ideals
IGv,l
and asl, l’
runsthrough
distinctprime
divisors of
I GI. By
Lemma 6.2 we have an inclusionfor any
positive integer
t. Since CIG (S’1 )
for any t > n, this shows thatIav7lIav7l1
9IG(Sl).
ThereforeKer(wv)
gIG (Si),
whenceKer
To prove the first
isomorphism
of(iii),
for * = v orSl
letSince
PG(v)IPG(Si)
isisomorphic toED, IGl(v)/IGl(SI), Wv
induces a mapWe have to show that Wv is an
isomorphism.
To thisend,
consider the commutativediagram
Since both
W Sl
and~v
aresurjective,
thisdiagram
shows that~v
is alsosurjective.
Moreoverby
the snake lemma we have an exact sequenceThen the
equality Ker(wsl)
= shows thatKer(Wv)
=0,
whenceivv
is anisomorphism.
In order to prove the second
isomorphism
of(iii),
letThen
QG(Sl)
is asubgroup
of and isisomorphic
to Hence
iYsl
induces a mapTo show that
’Qs,
is anisomorphism,
we consider the commutativediagram
where Ws, denotes the restriction of to As we have seen
above,
issurjective. Moreover, quite similarly
as in theproof
of thesurjectivity
of one can prove that cpsl is alsosurjective. Hence, by
thesnake lemma
again, 4’s,
is alsosurjective
and we obtain an exact sequenceWe wish to show that = 0. To see this note that in the
proof
of(i)
we have
actually proved
that is contained in the idealIGIG(Si).
Therefore is also contained in the
same ideal,
whenceThus from
(15)
we deduce thatKer(ws1)
=0,
as desired. Thiscompletes
theproof
ofProposition
6.1. 07. The
(S, T)-ambiguous
class numberLet
S,
T be any finite sets ofplaces
of such that S fl T = 0.(We
do notimpose
any other condition on S andT.)
LetJ~
be the id6le group of k.If v is a
place of k,
then we denoteby kv
andZfv
thecompletion
of k andthe
integer ring
of k at v,respectively.
We define the(S’, T)-idele
groupJs,T
=JK,S,T
of k to be thesubgroup
of
Jk ,
where for v E T we put®v 1 = ~u
EU; 1 u
-1(mod v) I - Clearly
we have
’
The
(S’, T)-id6le
class groupCs,T
=CK,S,T
is defined to be thequotient
group
Let
Ck
= be the id6le class group of k. It follows from(16)
that theinclusion ~
Jk
induces aninjective
mapCs,T
~Ck.
We define the(S, T)-ideal
class group =Clk,S,T
of kby
Then
Cls,T
isisomorphic
to the ray class groupcorresponding
to the
subgroup k x JST
ofJ.
To seethis,
note thatBy (16)
we have anisomorphism
whenceLet h be the
(S, T)-class
number defined in(1).
Thisnaming
will bejustified
if we show that h = To show
this,
note that we have an exactsequence
For
simplicity
we letCls
:=Cls,g (the
S-ideal class group ofk)
andJs
=(the
S-idele group ofl~), respectively. Then,
as aspecial
case of(17),
we have an exact sequence
For
any v E T,
letIw
be the residue field of v. Then two exact sequences(17)
and(18)
fit into thefollowing
commutativediagram:
Applying
the snake lemma to thisdiagram,
we obtain an exact sequenceSince
hs
=iClSI,
it follows that, , ,
Therefore h = as desired.
Now,
let be a finite Galois extension with the Galois group G.We define
UK,S,T,
etc.similarly
as above.Then, taking
H’
(G, -)
of the exact sequenceof
G-modules,
we obtain thelong
exact sequenceThe next formula will
play a key
role in theproof
of Theorem 4.2.Theorem 7.1.
Suppose
G iscyclic.
Then.. 1.-, /
Proof.
SinceHI(G,CK)
=0, taking cohomology
of the short exact se-quence
yields
the exact sequenceNoticing
thatCK
=Ck,
we obtainFor any G-module
M,
letQ(M)
denote the Herbrandquotient:
From the exact sequence
(20)
we deduce thatBy
class fieldtheory
we know thatQ(CK) = Moreover,
we haveQ(CIK,s,T)
= 1 sinceCLK,S,T
is a finite group.Therefore,
from the exactsequence
we obtain
IGI.
SinceQ(JK,S,T)
=Q(Ilx,s,T)Q(Cx,s,T),
itfollows that
IGI.
Thereforeby (22)
we have. - - . -... - - , , .
Substituting
this into(21),
we obtainSince
ICls,TI
=h,
this proves the theorem. 0 Remark 7.2. If S is the set of archimedeanplaces
and T =0,
then his the usual class
number hk
of k and Theorem 7.1 reduces to a well known formula for theambiguous
class number for thecyclic
extension- , - - . - ,
where
hk
=e(K~k)
andEk
denote the class number ofk,
theproduct
of ramification indices in of the
places
of k and the unit group ofk, respectively.
For this formula we refer the reader to[12,
Lemma4.1].
Weshould also remark that Federer
[8]
obtained a similar formula forIClK,sGI [
when S is
arbitrary.
Thus our formula may be viewed as ageneralization
ofthose formulae. To see that
(23)
is aspecial
case of Theorem7.1,
it sufficesto prove the formula
in the case of s = T =
0.
To prove this note thatV 11
where v runs
through
theplaces
of k. HenceI H2(G, JK,s.) I
=Moreover,
we haveEk/NK/kEK,
whereEK
=UKS.
isthe unit group of K. Therefore
(24)
follows if we prove the formulaBut this is an easy consequence of the Hasse’s norm theorem
asserting
theinjectivity
of the natural map8.
Cohomological interpretation
of AThroughout
this section we will assume thatS, T)
is an admissibledata such that G =
Gal(K~k)
is a finitecyclic
extension. In Section 1 wehave defined A to be a map from
UST
toHowever,
in order togive
a
cohomological interpretation
ofA,
it seems natural toreplace
the targetgroup G 0
Xs
with asubgroup XG s
defined below. Tobegin with,
we letWe will
regard YG,s
as asubgroup
of G 0Ys
via the naturalinjection sending (... , gv, ...)vEs
to 0 v. Next we define asubgroup XG,s
of
YG,s by
the exact sequencewhere
Ds
is thesubgroup
of Ggenerated by Gv
for all v E s and the mapYG,s
- Ds is definedby sending (... , gv, ...)vEs
toITvEs9v.
Then theimage
of A is contained inXG~s.
We will hereafterregard
A as a mapFor
example Coker(A)
will stand for thequotient
groupXG,s/Im(A).
We now wish to reveal a connection between the Gross
regulator
map A and the(S, T)-ambiguous
class number formula(Theorem 7.1).
To thisend,
we start withstudying H2(G, J K,S,T ) .
For eachplace
v of1~, choose,
once and for
all,
aplace
w of Klying
above v. Thenby Schapiro’s
lemmawe have an
isomorphism
where v runs
through
theplaces
of k. Letbe the
projection
to the8-part
in theright
hand side of(25). Similarly
wehave an
isomorphism
Lemma 8.1. Both
®w )
and®w,1 )
vanish S.Proof. First,
for any v we have~w ) ^--’
where ev denotes theramification index of v in the extension
K/k. Therefore, if v ft S,
thenK/k
is unraxnified at v, and so
H2 (Gv, ®w )
= 0.Next,
to see thatH2 (Gv, ~w,1 )
also vanishes for
any v g S,
we consider thelong
exact sequence’
obtained from the short exact sequence
- -- -- _v
By
Hilbert’s theorem 90 we have = 0. From this and(26)
itfollows that
H2 (Gv, ®w,1 )
= 0 for any vg
S. Thiscompletes
theproof.
0By
this lemma we may viewH2 (G, JK,S,T)
as asubgroup
ofH2 (G, JK) .
Thus, restricting
the map prs toH2(G, JK,s,T),
we obtain anisomorphism
Recall that the local invariant map
is an
isomorphism
for anyplace
v. LetThen the map
is an
isomorphism.
In order tostudy
theimage
ofIm(a2) C H2 (G, Jx,s,T)
under the map
invs,
letbe the summation map. If we denote
by Ds
thesubgroup
of Ggenerated by Gv
for all v Es,
thenIm(s) =
We defineby
the exactsequence
Lemma 8.2.
invs(Im(a2))
C3XG,s.
Proof. By
class fieldtheory
we have anisomorphism
Let
H2(G, CK,S,T)
-’Z/Z
be thecomposite
mapof invs
andthe natural map
H2(G, CK,S,T)
-H2(G, CK).
Then we have a commu-tative
diagram
from which the assertion of the lemma follows. D
Now,
theconnecting homomorphism
obtained from the short exact sequence
is an
isomorphism
sinceHi (G, Q)
= 0 for i > 0. For any G-module M weconsider the cup
product
Choose and fix a faithful additive
chaxacter V)
EHom(G, Q/Z)
of G. Thendefines an
homomorphism
Lemma 8.3.