Counting cyclic quartic extensions
of
anumber field
par HENRI
COHEN,
FRANCISCO DIAZ Y DIAZ et MICHEL OLIVIERRÉSUMÉ. Nous donnons des formules asymptotiques pour le nombre d’extensions
cycliques
quartiques d’un corps de nombresgénéral.
ABSTRACT. In this paper, we give
asymptotic
formulas for the number ofcyclic
quartic extensions of a number field.1.
Galois, Kummer,
and HeckeTheory
1.1. Introduction. Let K be a number
field,
fixed once and forall,
andlet G be a transitive
subgroup
of thesymmetric
groupSn
on n letters. The2reverse
problem of
Galoistheory
asks whether there exists an extensionL/K
ofdegree n
such that the Galois group of the Galois closure of L isisomorphic
to G. Thisproblem
is far frombeing solved, although great
progress has been made
by
Matzat and hisschool,
andhopes
have beenraised
by
Grothendieck’stheory
of dessins d’enfants. Forspecific
groups Gwe can even ask for the numbers
NK,n (G, X )
of such extensionsL /K
up toK-isomorphism,
such that the norm of the discriminant ofL/K
is at mostequal
toX,
at least in anasymptotic
sense. Ageneral conjecture
due toMalle
(see [12]
and[13])
states that there should exist constantsaK(G), bK(G),
andcK(G)
such thatand
gives
formulas for the constantsaK(G)
and(in [9],
it is shownthat the formula for
bK(G)
cannot be correct, but theconjecture
is stillbelieved to hold with corrected
values).
For ageneral
base field K thisconjecture
is known to be true in a number of cases, and inparticular
thanks to the work of
Wright ~14~,
in the case of Abelianextensions,
andthe constants
aK (G)
andbK (G)
agree with thepredictions. Unfortunately
it is very difficult to deduce from
[14]
theexplicit
value of the constantcK(G) (which
is notgiven by
Malle’sconjecture),
so all thesubsequent
work on the
subject
has been doneindependently
ofWright’s.
When thebase field is .K =
Q,
the result is known ingeneral:
after work of manyauthors,
Maki in[10]
and[11] gives
the valueofcQ(G)
for all Abelian groups G when the base field isQ.
On the otherhand,
for ageneral
base field K and Abelian groupG,
theonly
known results are due to theauthors, except
for G =C2
for which the result can be deduced from[8]:
G= C~
forprime
Ê
(see [4]
and[5]),
G= Y4
=C2
xC2 (see [6]),
and G =C4,
which isthe
object
of thepresent
paper. Note that groups such as G =Cn
withsquarefree n
can be treatedquite easily using
the methods of[4]
and[5],
butthe formulas are so
complicated
that there is not muchpoint
indoing
so.Thus the main
difficulty
in theC4
case is the fact that 4 is notsquarefree.
As in the
C~
case, the main tool that we need issimple
Galois and Kummertheory,
but we will also need a little local class fieldtheory.
1.2. Galois and Kummer
Theory.
We first consider the Galois situa- tion. LetL/K
be aC4-extension.
Then there exists aunique quadratic
subextension We will write =
K( VD)
where for the moment Dis an
arbitrary
element of K*generating
Then L =k(JZ)
for somea e
1~*,
and it is well known and easy to see that a necessary and sufficient condition forL/K
to be aC4-extension
is thatDz2
for some z E .K.Writing
a = x +yVD,
it is clear that thisimmediately implies
that D is a sum of two squares in K.
Conversely,
if D =m2 + n2 in K,
VD(m
+ is such that c~ E k* and =Dn2,
henceL = defines a
C4-extension
of K. In otherwords,
we have shownthe well known result that a
quadratic
extensionk / K
can be embedded ina
C4-extension
if andonly
if its discriminant is a sum of two squares.In a
large
part of this paper, we fix the intermediatequadratic
extensionkl K
with k = We denoteby
T thegenerator
of the Galois group ofby
d =D(klK)
the relative idealdiscriminant,
the different of and
by
T the set ofprime
ideals of Kdividing
d(i.e.,
which
ramify
inFinally
we fix once and for all an élément such that =Dn2
for some n E K(not necessarily
the onegiven above) .
We will
constantly
use thefollowing
lemma whose trivialproof (for
ex-ample using
Hilbert90)
is left to the reader.Lemma 1.1. Let a E k*.
( 1 )
is a square irc Kif
andonly if
there exists z E K E ksuch that a =
z-y 2
(2)
isequal
to D times a square in Kif
andonly if
there existst E K
and q
OE k such that a =wt-y 2
Thus a defines a
C4-extension
if andonly
if a =wt-y 2
for somet E
K*;
since a isonly
defined up to squares, we may in fact assume thatcx = Wt. Furthermore c~t and wt’ define
K-isomorphic C4-extensions
if andonly
ifthey
definek-isomorphic quadratic extensions,
hence if andonly
ifE
k* 2.
Since EK*,
this means that EK*2 U DK*2.
Thus,
if we consider extensions L =k( VWt)
for t EK* 1 K*2 (note
that the unit class must not be
excluded),
we will obtainexactly
twice allC4-extensions
of K up toisomorphism.
We have thus shown:Proposition
1.2. Let k = be aquadratic
extensionof
K which isembeddable in a in other words such that there exists w E k*
such =
Dn2 for
some n E K* . Then the setof isomorphism
classes
of C4-extensions L/K containing kl K
is in a noncanonical one-to-two
correspondence
with the groupK*IK * 2.
Theisomorphism
classof C4-extensions corresponding
to t EK* 1 K*2 is k( VWt),
and this is the sameextension as the one
corresponding
to tD.Definition. Recall that T is the set of
prime
ideals of K ramified inkl K.
We denote
by
T > thesubgroup
of the group of fractional ideals of Kgenerated by
the elements of T.(1)
The T-class groupClT(K)
is thequotient
group of the group of frac- tional idealsby
thesubgroup
of ideals of the formOb,
wherej3
E K*and b E T >. In other words
CLT(K)
=Cl(K)1
T >, the quo- tient of theordinary
class groupby
thesubgroup generated by
theideal classes of elements of T.
(2)
The T-Selmer groupST(K)
of K is the set of classes of elementsu E K* such that =
q2b
for some ideal q of K and some idealb E T >, modulo squares of elements of K*
We will also use this definition either for T =
0,
in which case we re- cover the notions ofordinary
class and Selmer groups, which we will denoterespectively by Cl(K)
andS(K).
We will need the
following lemma,
whose easyproof
is left to the reader:Lemma 1.3. The natural map
f rom S(K)
toS(k)
induces anisomorphism from ST(K)
to where as usual is thesubgroup of
elementsof
S(k)
stableby
T.The
following
lemma is well known in the case T= ~l (see
forexample
[2]),
and itsproof
is the same.Lemma 1.4. There exists a noncanonical one-to-one
correspondence
be-tween
K* /K*2
andpairs (a, u),
where a is asquare f reel
idealof
K co-prime
to () whose ideal class is a square in the T -class groupClT(K),
and~c E
ST(K).
= some b OE T > and to EK*,
the elementof K* 1 K*2 corresponding to (a, ù) is
the class modulo squaresof
tou.lWe will only use the term "squarefree" to mean integral and squarefree
Note for future reference that
by
theapproximation
theorem qo and u(but
of course not to ingeneral)
can be chosencoprime
to1.3. Kummer and Hecke
Theory. Important
notation. In this pa- per, the notation will be used to denote ageneric
fractional ideal of k which need not be the same from one line to another. Inaddition,
we recallthe
following
notation introduced above:~ K is a number
field,
the base field fixed in the whole paper.~ l~ is a
quadratic
extension ofK,
which will be fixed in the first three sections of the paper, D E K is an element such that k = d and D are the relative discriminant and different of T is thegenerator
of and w is an element of 1~~ such that =Dn2 for
somenEK.
~ T is the set of
prime
ideals of K whichramify
ink/K,
in other wordsdividing
d.~ L is a
quartic C4-extension
of Kcontaining
k.Definition. If VJ1 is any fractional ideal of
l~,
we denoteby s(~)
the square-free
part
of9Jt, i.e.,
theunique squarefree
ideal such that 9M = forsome
(fractional)
ideal fl. If 9Jt =aZk
for some element a, we writes(a)
instead of
s(aZk) .
Note the
following
trivial butimportant
lemma:Lemma 1.5. For two
squarefree
ideals ai and a2define
the
"syrnrraetric difference" of al
and a2. Thenfor
any two idealsVJtI
and9X2
we haveS(VJtIVJt2)
=S(VR2).
Lemma 1.6. be an ideal
of k
and let 9X be afractionad
idealof
k.The
following
two conditions areequivalent:
(1)
There exists anof k
such that(VRÙ2,e:)
= 1.(2)
We have = 1.If,
inaddition,
9K comesfrom K (i.e.,
9R =MZk for
some ideal mof K),
then
in (1)
we may chooseÙ2 of
theform bqZk for
q an idealof
K andb E T >.
Proof.
Since 9N =~(9~)Û~,
we have(2)
~(1). Conversely,
assume(1).
If~3
is aprime
idealdividing C
then = 0 hence = 0(mod 2),
so =0, proving (2).
Finally,
assume that 9K = and write m =aq26
where a issquarefree
coprime
to c1 and b E T >. It is clear thats(fl)
=aZk ,
so a iscoprime
to C
by (2),
hence if we take £"2 = where~2
=bZk,
we haveVRi12
=aZk,
so i1 is suitable. DDefinition. Let C be an ideal of A’
dividing 2Zk .
(1)
We will denoteby Q(e:2)
the group of elements a E k*satisfying
thefollowing
two conditions:~
(s(a), G)
= 1 orequivalently, by
the abovelemma,
there existsan idéal of k such that
G)
= 1.~ The
multiplicative
congruence = 1(mod * (t2 )
has a solu-tion in k.
(2)
We will denoteby
the group of elements z E K*satisfying
the
following
two conditions:~ There exists an ideal b E T > such that
(zb, C)
= 1.~ The
multiplicative
congruencezlx 2 =
1(mod *e:2)
has a solu-tion in k
(not necessarily
inK).
Remarks.
(1)
When a(or z)
isalready coprime
to e: this means that a -x 2 (mod *e:2)
has a solution. We will see that it is essential to have alsothe
coprimeness
condition.(2)
Weclearly
have CK,
but we do not haveequality
in
general,
sinceby
Lemma 1.6 if z E K there exists q and b E T > such thatC)
=1,
notnecessarily
with q = 1. What is trueby
theapproximation
theorem is that if z EQ(e:2) n K,
thereexists n E K* such that
zn2
EQK((!2).
Since
Gi 1 1 2Zk implies trivially
thatQ(~2)
C thefollowing
definition is reasonable:
Definition. With the above
notations,
for a E ~* we denoteby
thelargest
ideal(for divisibility) dividing 2Zk
and such that a EQ(e:2).
With these
notations,
we recall thefollowing important special
case of atheorem of Hecke
(see
forexample [2], Chapter 10) :
Theorem 1.7. Let k be a number ,
field.
Thenfor
0: E 1~*(including
ce E
l~*2~
the relative ideal discriminant.given by
Thanks to
Proposition
1.2 and Lemma1.4,
we see that we mustapply
this theorem to a = hence must first
compute s(a).
Note thatwill
usually
denote thesquarefree part
of the ideal 9R ink,
even if 9R comesfrom an ideal of K. It will be clear from the context when the notation is used to denote the
squarefree part
in K.By
definitiontouZK
=aq 2b
for somesquarefree
ideal acoprime
to d andsome b E T >. Since the elements of T
ramify
in and a iscoprime
to D,
it follows thats(tou)
= a(this
is of course the wholepoint
ofusing
ideals
coprime
toc~).
Let us now
compute
Lemma 1.8.
Keep
the above notation. There exists asquarefree
ideal a,of K
such that =as(Z),
where 0 = thedifferent of
Furthermore,
the classof a~,
inClT(K)lClT(K)2 is independent of
w,i. e., depends only
on the extensionNotation. If
j3
isany element
such that =Dz2
with z EK*,
we will write aj3 for the
squarefree
ideal of Kcoprime
to D such that=
aj3s(1’). Moreover,
ideals of k will be denotedby capital gothic
letters and those of K
by
lowercasegothic
letters.Proof. Writing
= andtaking
norms from toK,
we see thatDq2
for some ideal q of K. Since the different satisfies= e~ and since =
()m2
for some ideal m, it follows thatNk/K(s(w)I1’)
=qi
for some ideal ql of K.By
Hilbert 90 forideals,
ortrivially directly,
this isequivalent
tofor some ideal q2 of K
(and
as usual Û an ideal ofl~).
Without loss ofgenerality
we may assume that q2 issquarefree
in K(include
all squares inÛ)
and iscoprime
to 0(include
all ramifiedprimes
inÛ).
The firstpart
of the lemma follows from the fact that is asquarefree
ideal of k.Furthermore,
since thegeneral
solution to =Dz2
is 0152 = we have aa =s ( a~, t) .
IftZK
=atqb
for b E T >,by
Lemma 1.5 this istherefore
equal
to aw6.at = Since the classof at
isclearly
asquare in
ClT(K),
the secondpart
of the lemma follows. D Sinces(tou)
= a, Lemma 1.5gives
thefollowing:
Corollary
1.9. With the abovenotation,
we haveCorollary
1.10.Keep
the above notation and let L = The relative ideal discrim2nantof by
2. The Dirichlet Series As usual in this kind of
investigation,
we setwhere the
symbol J1r
without asubscript
willalways
indicates the abso- lute norm from K toQ,
and ranges overK-isomorphism
classes ofC4-extensions.
Whenusing
the relative norm from toK,
we willalways
write
(as above)
and whenusing
the absolute norm from toQ,
wewill
always
writeBy
the Galoisdescription
ofC4-extensions
and the discriminant-conduc- tor formula7J(LIK) =
we can writewhere the sum is over
isomorphism
classes ofquadratic
extensions which are embeddable in aC4-extension,
andIn the
above,
thequadratic
extensionsL /k
are of courseonly
taken up tok-isomorphism,
and the factor1/2
in front comes from the fact that wehave a one-to-two
correspondence
inProposition
1.2.The
goal
of thefollowing
sections is tostudy
the inner Dirichlet seriesWe will come back to the
global
series~K,4(C4, s) only
in Section4.
Thus in this section and the
following,
we fix aquadratic
extension k ofK,
an element D E K* such that 1~ = which is a sum of two squares inK,
an element VJ E 1~* of relative norm D times a square, andmore
generally
wekeep
the notations of thepreceding
section.To
simplify notations,
denoteby
1 the set ofsquarefree
ideals of Kwhich are
coprime
to c~ and whose ideal class in is a square. Letn =
[K : Q]
be the absolutedegree
of the base fieldK,
so that[k : Q]
= 2n.It follows from
Proposition 1.2,
Lemma 1.4 andCorollary
1.10 that wehave:
where
(We
have used the trivial fact thatJVkIK(8(0»
=s(D),
where here ofcourse denotes the
squarefree part
of D inK).
For C
coprime
tos (1))
and to aAa~(in
other words tos(wtou)),
setIn the above notations gto and
f to,
we do not write theexplicit dependence
on a.
We
clearly
havewhere we take
only
the Dcoprime
tos(D)
and toThus, by
a formof the Mbbius inversion formula we deduce that
(same
restrictions onD),
where ILK is the Môbius function on ideals of K.It follows that
To
compute
we first need thefollowing
definitions and notations.Definition. Let C be an ideal of k
dividing 2Zk,
and set c= ~2 n
K.(1)
We define as thequotient
group of the group of fractional ideals of Kcoprime
to cby
thesubgroup
of ideals of the form z6 whichare
coprime
to c, where b E T > and z E(see
Definition1.3).
(2)
We define as the set of elements u EST (K)
such thatu E
Q~(~2)
for some lift u(hence
for all lifts u such that there exists with(ub, G) =1 ) .
Note the
important
fact thatalthough
the elements z and u are in the base fieldK,
the congruencesdefining
are still in k and not in K.Definition. Let G be an ideal
dividing 2Zk
andcoprime
to(1)
We will say that C satisfies condition(*)
if there exists/3
- 1(mod *~2)
such thatNk/K(/3)
isequal
to D times a square in K.(2)
Let a be an ideal of K. We will say that a satisfies condition(**) Q:2
ifa is
squarefree
andcoprime
to"0,
and if there exists,~
== 1(mod *~2)
withNk/K(/3) equal
to D times a square such that(or, equivalently, s(/3)
=s (w a»
Lemma 2.1.
(1) If a satis,fies
condition(**)e2,
then a A aw iscoprime
to c
(or, equivalently, to (t).
(2) If a
and aosatisfy
condition(**)Q:2,
thenalao
iscoprime
to c.Proof. (1).
Letfl3
be aprime
ideal of kdividing
G. Thenby assumption
= 0 and = 0
for /3
- 1(mod *lî2) .
Thus condition(**),r2 implies
that - 0(mod 2).
Since a A a, issquarefree
andcoprime to D,
its extension to k is alsosquarefree
so that in factaw)
=0, proving ( 1 ) .
(2).
Assume that = ands(cva)
= with/3o
and/3
asabove.
Including
theimplicit
idealsû2 of k
anddividing,
thisimplies
that(,3/~0),~2.
As in(1),
ifS1J
is an idealdividing C,
we have= 0
(mod 2).
Since a and ao aresquarefree
andcoprime
toc~,
we have E{20131,0,1},
hence = 0 as claimed. DProposition
2.2. Fix an ideal (tdividing 2Zk,
let a be an idealof K
andlet to be as above.
(1)
We have 0if and only if a satisfies (**)~z.
(2)
There exists an ideal aosatisfying (**)~z if
andonly if Le satisfies
condition
(*).
- 1(mod *~2)
is such that isequal
toD times a square
in K,
we may choose ao =ap 0
aw.(3)
Let ao be some idealsatisfying (** )Q:2.
Then asatisfies (**)~Z if
andonly if
the classof a/ao
is a square inClT,Q:2 (K)
or,equivalently, if
the class is a square in
(4)
The ideal asatisfies (**)~~ if
andonly if
the classof
a is a square inClT(K).
Inparticular
this is the case when asatisfies
thestronger
condition(**) Q:2.
(5)
When 0(hence by (1)
when asatisfies (**) Q:2 ),
we haveand in
particular
this isindependent of a satisfying (**) Q)2 .
Proof. (1).
Assume that wtou eQ( cr2)
or,equivalently,
that2
=WtOU0
for some
/3
=1(mod *cr2). By
definition of a and u, there exists b OE T >and q such that
TOUZK
= so that =aü2,
henceby Corollary
1.9
In
addition, taking
norms from to K of theequality x 2
= it isclear that
Nk/K((3)
isequal
to D times a square ofK,
so that a satisfies(~~)(~2.
Conversely,
assume that a satisfies(**) (t2,
so that/3Û~.
If weset a =
wtol(3,
this means that a ES(k),
theordinary
Selmer group of k. On the otherhand, taking
relative norms we obtainNk/K(a) = y2
forsome y E
K,
so thatby
Lemma 1.1 we obtain as usual a =n,2
for someq E k and n E K. Thus in
S(k)
we have a = n, hence n E andby
Lemma 1.3 this means that n E
ST(K) . Finally, setting
u =1/n,
we seethat u is an element of
ST(K)
such that wtou EQ(e:2), proving (1).
(2).
It follows from(1)
that the existence of an ideal asatisfying (**)(t2
implies
the existence of(3
-1(mod *e:2)
such that isequal
to Dtimes a square.
Conversely,
if such a(3 exists, by
Lemma 1.8 we can writes((3)
= for asquarefree
ideal a~prime to D,
and it is clear thatsatisfies
(**) (t2.
(3).
Assume that ao satisfies(**)e2
for some l~.By construction,
we know that
(a A
=waû2,
henceby dividing,
it is clear that asatisfies
(**)e2
if andonly
if = for some a - 1(mod *e:2)
of square norm. Thus
by
Lemma 1.1 we can write a =z-y 2
for somez E K
and q
and inparticular, Z
EQK(~2).
On the other hand the fractional ideal(a/ao)/z
of K extends to the square of the idealflq in k,
hence is of the form
q2 b
for some b E T >, so thata/ao
=By
the
approximation theorem, by multiplying
q with a suitable n E K* andchanging z
intoz/n2,
we may assume that q iscoprime
to c, hence that z EQ~(~2),
and sinceby
Lemma 2.1(2) alao
iscoprime
to c, thisexactly
means that the class of
a/ao
is a square inFurthermore,
since(a A a,)/(ao A a)
isequal
to the square of an ideal timesalao
which iscoprime
to cby
Lemma 2.1(1),
the second statement of(3)
follows.(4).
If we take G = it is clear that ao =Zk
satisfies(**)Zk
with,Q = w (since
= Thusby (3)
a satisfies(**)Zk
if andonly
ifthe class of a is a square in
ClT (K)
=ClT,Zk (K) .
Inparticular,
this is trueif a satisfies
(**) (t2.
(5).
Assume that a satisfies condition(** )(t2,
so that there existsv E
ST(K)
such thatQ(e:2).
Thus wtou EQ(e:2)
if andonly
ifu/v
EQ(e:2),
hence if andonly
ifu/v
EST,(t2.
Thus the set of suitable isequal
toVST,(t2,
whosecardinality
is of courseequal
to . DThanks to this
proposition,
we can noweasily
prove animportant
pre-liminary
formula for the Dirichlet series’I)k(8)-
Theorem 2.3. To
simplify notations,
setThen
where
and
{3(~2)
is an elementof
k suchthat ~
-1 withequal
to D times a squareof
K.Proof.
Wereplace by
its valuegiven by Proposition
2.2 in the last formula for~bk (8) given just
before Definition 2 above. Inparticular,
we canrestrict the summation to ideals C
satisfying (),
and to ideals asatisfying ( ** ) e:2. However, by
the sameproposition
if a satisfies(**) e2
then the classof a is
automatically
a square in It follows that we can removethe restriction a E Z and
replace
itby
the weaker condition that a EJ,
where
by
definitionJ
issimply
the set ofsquarefree
ideals of Kcoprime
to d. Thus
where
Note that the condition a E
,%
isequivalent
to a ~ aw EJ,
and that themap a H a ~ aw is an involution of
~%. Furthermore,
a / aw issquarefree
andcoprime
to C andd,
andby Proposition
2.2(2)
and(3),
a satisfies(**)~2
if and
only
if the class of(a ~ a,)
is a square inClT,~2(K),
with =
~(~2) Aa~,
hence if andonly
if the class of(a A
isa square in
Thus, changing
a into a ~ aw, we obtainwhere
,Q(~2) _
1(mod * e~2)
is of relative norm D times a square. Note that since(~, s(O»
=l, a,~e2>
iscoprime
to (t.Since the class of an ideal m is a square in if and
only
if for---
each x
E the group of characters of2 (K),
we havex(m)
=l,
weobtain
, l ,
since X
is of order2, proving
the theorem. As in Lemma1.8,
it iseasily
shown that the class of
a~(~2)
in isindependent
of the choice of~(~2),
so that isindependent
of0(e~2).
DIt remains to
compute ~ST~~ (K) ~ 1
and tostudy
condition(*).
3.
Computation
of]
andstudy
of condition(*)
We
keep
all the above notation. and inparticular
C willalways
be anideal
dividing 2Zk, coprime
tos(D)
andsatisfying
condition(*).
3.1. Reduction of the
problem.
Definition. As in Definition
2,
set c = K.(1)
We will say that an element z E K* isT-coprime
to c if there existsb E T > such that the ideal zb is
coprime
to c.(2)
We will denoteby Ze2
thequotient by
of the group of elements of K* which areT-coprime
to c,i.e.,
thesubgroup
of suchelements z for which the congruence
zlx 2 =
1(mod *~2)
is solublein k.
Remarks.
(1)
It is clear that if z isT-coprime
to c,then z2
EQK(~2),
hence isan abelian group killed
by 2,
like all the other groups that we consider in this section. We willcompute
itscardinality explicitly
later(see
Corollaries 3.7 and
3.13).
(2)
If we write c = crcu where cr =(c, D’)
and(cu, d)
=l,
it is clear thatz is
T-coprime
to c if andonly
if(z, cu)
=1.The basic tool which will enable us to compute
ST,ct2
is thefollowing
result. Recall that for any idéal C
dividing
we have setIn
particular, gz,(K)
=ClT(K)lClT(K)2.
Proposition
3.1. Let c== (t2
There exists a naturallong
exactsequence
where the maps will be describeâ in the
proof.
Proo f .
Exactness at is trivial. If u EST(K),
thenUZK
=q 2b
withb OE T >, and
by
theapproximation theorem, changing u
inton2u
if necessary, we may assume that q iscoprime
to c, hence u isT-coprime
to c.We then send such a u to its class in
Zr2.
It is clear that it does notdepend
on the
representative
of the class u chosenT-coprime
to c.Furthermore, u
is sent to the unit element of
Zr2
if andonly
if u EQ~(C~2),
hence if andonly
if u Eproving
exactness atST(K).
Let z E
Zt2.
Since z isT-coprime
to c, there exists a b OE T > such that(zb, c)
=1,
hence also(zb, C~)
= 1. We will send z to the class of zb.By
definition of this class isindependent
of the choice of b.Furthermore,
it isindependent
of therepresentative z
of the class:indeed,
if z’ is another
representative,
then z’ = zn where n is such that thereexists b’
E T > withc)
= 1 andn/x2
=1(mod ~2)
soluble in 1~. Itfollows that z’bb’ = and the class of nb’ is trivial since
n/x2
= 1(mod Le2)
is soluble in l~. This shows that the map is well defined.To show exactness at
Zî2,
let z beT-coprime
to c, let b E T > with(z b, c)
=l,
and assume that the class of z~ is trivial in Thismeans that there exists an ideal q
coprime
to c, an ideal b’ OE T > and anelement n such that
(nb’, c)
= 1 andnlx 2
-1(mod Q:2)
soluble ink,
withzb = in other words
(z/n)ZK
= Since n E @ theclass of
z/n
inZî2
isequal
to the class of z. Thuschoosing z/n
insteadof z as
representative
of itsclass,
andnoting
thatz/n
EST(K),
we haveshown exactness at
Zî2.
Let a be an ideal
coprime
to c, and assume that its class as an elementof
QZk (K)
is trivial. This means that a =zq 2 b
for some z EK*,
someideal q and some b E T >.
Using
as usual theapproximation theorem,
wemay assume that we have chosen q
coprime
to c, so that(zb, c)
=1. Since(q, c) =1,
the class of a inÇe2 (K)
isequal
to that ofa/q2
=zb,
and sincez is
T-coprime
to c, this proves exactness at(K) .
Note that~~ (~2)
isof course sent to the unit element of
~~2 (K) .
Finally,
if a is an ideal ofK, by
theapproximation
theorem we canfind
{3
such that(0a, c)
=1,
and the class of,Qa
inQZk (K)
isequal
tothat of a,
proving
exactness at that group andfinishing
theproof
of theproposition.
DThe second much
simpler
exact sequence that we need is thefollowing.
Proposition
3.2. There ex2sts a natural short exact sequencewhere
UT(K)
denotes the groupof
T-unitsof
K(i.e.,
elements E E K*such that
vp(E)
= 0for
allprime ideals p e T),
andfor
any groupG, G[2]
denotes the
subgroup of
elementsof
G whose square is the unit elementof
G.
Proof.
Let u EST(K),
so thatUZK
=q26
for some b E T >. We sendu to the class of q in
ClT (K). By definition,
this class indeedbelongs
toClT(K)~2~. Conversely, if q
EClT(K)(2~
thenq2
= zb for some b E T >,so
clearly z
EST(K) proving
exactness atFinally,
if u EST(K)
is sent to the unit element of
ClT(K),
this means thatu7GK
=q 2b
andthat q = zb’ for some b’ E T >. Hence
UZK
=z26"
for still another b" E T >, sou/z2 (whose
class inST(K)
is the same as that ofu)
is anT-unit, proving
theproposition.
DCorollary
3.3. Let(ri, r2)
be thesignature of K.
We haveProof. By Proposition 3.1,
we haveand
by Proposition
3.2 we haveIt is well known that
UT(K)
isisomorphic
to the group of roots ofunity
of K times a free abelian group of rank ri + r2 - 1 +
ITI.
Since the groupof roots of
unity
iscyclic
of evenorder,
it follows thatIUT(K)IUT(K)21 =
On the other
hand,
for any finite abelian group G we have~G~2~) ] (look
at the kernel and cokernel ofsquaring),
so that1 ClT (K) ~2~ ~ 1
= .Putting everything together,
we obtain thegiven
formula. D