R
ADAN
K
U ˇ
CERA
A note on circular units in Z
p-extensions
Journal de Théorie des Nombres de Bordeaux, tome
15, n
o1 (2003),
p. 223-229
<http://www.numdam.org/item?id=JTNB_2003__15_1_223_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
A
note
oncircular
units
in
Zp-extensions
par RADAN
KU010CERA
RÉSUMÉ. Nous nous intéressons aux limites
projectives
desgrou-pes de Sinnott et des groupes de
Washington
des unités circulaires dans laZp-extension
d’un corps abélien. Nous montrons par unexemple qu’en général
ces deux limites ne coincident pas.ABSTRACT. In this note we consider
projective
limits of Sinnott andWashington
groups of circular units in thecyclotomic
Zp-extension of an abelian field. A concrete
example
isgiven
to show that these two limits do not coincide ingeneral.
1. Introduction
For any
positive integer
m, let~’~.,.L
=e21ri/m
and letQ(m) =
bethe mth
cyclotomic
field. Let K be an abelian field of conductor m,i.e.,
m is the smallest
positive integer
such that K is a subfield of LetC(K)
be the Sinnott group of circular units of K(see
[S]).
It is well known(see
[L])
that this group can be defined as the intersection of the groupE(K)
of all units in K and thesubgroup
of themultiplicative
group of Kgenerated by -1
andby
all norms(~),
where 1rim
and(a, r)
= 1. LetC(K)
be the group ofcyclotomic
units of .~ mentioned in[W,
page143],
i.e.,
C(K) - E(K)
nC{~’’’2~ ).
We shall call the latter group theWashington
group ofcyclotomic
units.Now,
let p be an oddprime
which does notramify
in K. LetB/Q
be thecyclotomic
Zp-extension
of the rationalnumbers, i.e.,
Q
=Bo
cB1
CB2
C... are abelian fields ramified
only
at p,[Bn : Q]
=pn,
and BU’
Bn.
Hence KB =
KBn
is thecyclotomic
Zp-extension
of K.We shall consider the
following projective
limits(with
respect
tonorms)
C’ =
Zp),
and C =Zp).
It has beenproved
in
[KN]
that C is of finite index in C. But there has remained an openquestion
whether C = ~’ ingeneral
or not. This note is devoted to thenegative
answer to thisquestion.
Moreprecisely, by
anexplicit
constructionwe shall prove the
following
Manuscrit requ le 3 d6cembre 2001.
C(KBN)
n
Let us mention that
by
an easy andstraightforward
modification of ourconstruction one can
get,
for anygiven
oddprime
p, anexample
of anabelian field of
degree
p2,
ramified at2p
+ 1primes,
whosecyclotomic
Zp-extension
satisfies pI
[C : C].
A similar
example
has beengiven independently by
J.-R.Belliard,
who describes in[B]
an abelian field K ofdegree
p2,
ramified at p + 1primes,
and shows
by
means of anexplicit
construction inZp-extension
KB/K
thatC is not a A-free
A-module,
where A =Zp[[Gal(KB/K)]].
Moreover,
he proves here that there exists - E C such that EP E Cand - 0
C if andonly
if C is not a A-free A-module. To compare these two constructions: on theone hand our
example
needs more ramifiedprimes,
but on the other hand we do not need to tensorby
Zp
to be able to define units rln Egiving
the norm-coherent sequence(in)
EC,
so - in some sense - ourexample
is moreexplicit.
Notation. Let ql, q2, .. - , q7 be different
primes,
allcongruent
to 1mod-ulo
9,
chosen in such a way thatql ~
1(mod 27)
and that for differenti, j
E{I,
2, ... ,
71
theprime
qi is a cubic residue modulo qj. For each iE {I, 2, ... ,7},
let x2 be a cubic Dirichlet character modulo qi. Let K be the abelian fieldcorresponding
to the group of Dirichlet charactersgener-ated
by
X1X2X3X4X5 andx2x3x4x5xsx7.
SoK :
iQ]
= 9 and K has thefollowing
four subfields ofdegree
3:For the sake of
brevity,
let us writeKo
=K,
f o
= M2M4M6?7?K5
=Q,
is == 1.
For any
prime
q3/o?
letFrob(q)
mean a fixed Frobeniusautomorphism
of q on
KB /Q,
i.e.,
any chosen extension to KB of the Frobeniusautomor-phism of q
on the maximal subfield of KB unramified at q. Since qi is acube modulo
folqi
for any i =1, 2, ... , 7,
theprevious
choice can be donein such a way that E for each i =
1, 2, ... , 7.
We shall introduce
generators
of For any i =0,1, ... ,
4 andMoreover,
let = -1 andfor any
positive
integer
n. Thefollowing
lemma shows that units Ei,j with0
j
n aregenerated by
allconjugates
of the units eij,where j
= 0 ori =
n.Lemma 1. For any
integer
n >0,
the Sinnott groupof
circular unitsC(K Bn)
is the Galois modulegenerated by
Proof.
Using
[L], C(KB,,)
is the intersection ofE(KBn)
and of the Galois module Dgenerated
by -1
andby
all norms where1 r
I 3n+l fOe
We can write r =3jq,
where 0j
n +1,
and qfo.
LetSo,
denoting
B_ 1
=Q,
we have nKBn
=KiBj-1-
ThereforeBut
where q in the
product
runs over allprimes q
( r,
qf 3J fj (an
empty
product
equals
1).
Hence D isgenerated by -1, by
for all i =0,1, ... ,
4 andj
=0,1, ... , n,
andby
for j -
1, 2, ... , n ~- 1.
If 0 j
n thenand
(3’+’) =
(1 -
(3nH)(1 -
~3 +1 ) .
Therefore .D is theGalois module
generated by - l, by
Ej, j for all i =0, 1 , ... ,
4and j
E10, ?~},
and
by
{ 1-
(3n+l )(I -
~~ +1 ) .
It is well-known that all the numbers but the last one are units and that thequotient
of any twoconjugates
of the lastnumber is a unit. The lemma follows
using
the fact that theautomorphism
sending
each root ofunity
to its fourth powergenerates
the Galois groupof 0
The
generators
describedby
Lemma 1 are notindependent.
For anyintegers j
> 0 andi, l
E~0,1, ... ,
,5}
such that 1 = 0 or i = 5 we have thewhere q in the
product
runs over allprimes
dividing
filfi.
Moreover,
forany
integer j >
0 and i E10,
1, ... ,
5}
Construction. As we have
already mentioned,
the Frobeniusautomor-phism
Frob(qi)
EGal(KB/K)
for each i =1, 2, ... , 7. Moreover, qi -
1(mod 9)
implies
thatFrob(qi)
is a cube inGal(KB/K) -
(7G3, +).
There-fore there is
unique o
EGal(KB/K)
such that03
=Frob(ql)-1.
Because1
(mod
27), ~
is atopological
generator
ofGal(KB/K).
Hence,
foreach i =
2, 3, ... ,
7 there is auniquely
determined
3-adicinteger
aisatis-fying
Frob(qi)-1 =
’Ø30i.
It is easy to see that
Gal(KBn~Bn) ~ {id}
is thedisjoint
unionTherefore
Let us consider the group C(KBn)/(C(KBn))3 written
additively,
wherewe shall
identify
any unitofC(KBn)
with itsimage
inC(KBn)/(C(KBn))3.
Therefore the
identity
(4)
meansLet us fix n > 2.
By
abuse ofnotation,
we shall denote the restrictionof ’0
to
KBn
again by 9.
We shallapply
on
(5).
Since =1,
we haveIt is easy to see that
+’lj;203n-l
is the normoperator
ofK Bn/ K
Similarly,
for any i =2, 3, 4 using
(2)
we haveLet be a
positive integer
satisfying
(mod 3nZ3).
Thenwhere
and
(6)
and(1)
give
A similar
computation
gives
Putting (5), (7),
(8),
and(9)
together,
we obtainThis
equality
inC(KBn)/(C(KBn))3
means that there EC(KBn)
such that
Since
KBn
is a real abelianfield,
isuniquely
determinedby
the pre-viousidentity.
It is easy to see that EC(KBn-i) ,
EKBn-1-Since
KB,,/Q
isabelian,
we have E hence EQ(3n fo).
On the other hand 1Jn-1 E
imphesq,,-,
Ef-)).
But forcyclotomic fields,
the Sinnott groups of circular unitssatisfy
the Galoisdescent
(see
[GK]),
soIt is easy to see that for any
positive integers
a and bUsing
theprevious
identities we caneasily
prove that the numbersaj (n)
canbe
replaced by
a~~
in(10).
Using
(1)
thisimplies
thatso
Denoting
1]0 = we haveproved
E
C(KBn),
sor~3
e To showthat 11 rt
itis
enough
to prove thatC(KB1).
For this purpose we shall describe abasis of
C(KB1).
Lemma 2. There is a basis
of
C(KBi)
consisting
of
(i)
4conjugates of
eachof
the units ~1,1, ê2,1, e3,i, and s4,1, and~ ~
(ii)
of
2conjugates of
eachof
the units ê1,O, £2,0, ê3,0, é:4,0, and ~5,1’Proof.
Sincewhich is the number of units mentioned in the
lemma,
it isenough
to show thatthey together
with -1generate
We have chosen theprimes
ql , .. - , q7 in such a way that all
automorphisms
Frob (q7 )
acttrivially
onKB1.
Hence relations(4)
and(2)
for n =0, 1
give -3,
=1,
soEo,o = eo,l = 1. Because
£5,0 = -1,
Lemma 1gives
that themultiplicative
group,generated by -1
andby
allconjugates
of units mentioned in thelemma, equals
Let us fix any i
== {I, 2, 3, 4}
and consider The Galoisgroups
Gal(KZ B1 /Ki )
andGal(KiB1/ B1)
are groups of order3;
let p andv be their
generators,
respectively. Then [L
and vgenerate
Gal(KiBi /Q),
so with r, s =
0, l, 2,
is thesystem
of allconjugates
of Ei,i .Using
the trivial action of Frobeniusautomorphisms,
(2)
implies
Similarly,
(3)
gives
Therefore all
conjugates
of £i 1
aregenerated by
the fourconjugates
with r, s =
0, 1,
andby
all threeconjugates
of’
Similarly,
one can show that each of the units £1,0, £2,0, £3,0, E4,0, and E5,1has
precisely
threeconjugates
and theproduct
of them(i.e.,
its absolutenorm) equals
one. The lemma follows. 0Since
1/13
actstrivially
onequality
(10)
gives
773
= But Lemma 2 shows that -,,l is not a cube in henceC(KB1),
which
gives
q~
The theorem will be
proved,
if we show that there existprimes
q1, ... , q7satisfying
allassumptions
made in the first sentence of Notation. But thiscan be done either
by
a standardtechnique using
Tchebotarevdensity
theorem
(e.g.,
see Theorem 3.1 in[R])
orjust by
checking
thatprimes
satisfy
all theseassumptions.
References
[B] J.-R. BELLIARD, Sous-modules d’unités en théorie d’Iwasawa, to appear in Publications
mathématiques de l’Université de Franche-Comté.
[GK] R. GOLD, J. KIM, Bases for cyclotomic units. Compositio Math. 71 (1989), 13-27.
[KN] R. KU010CERA, J. NEKOVÁ0158, Cyclotomic units in Zp-extensions. J. Algebra 171 (1995), 457-472.
[L] G. LETTL, A note on Thaine’s circular units. J. Number Theory 35 (1970), 224-226.
[R] K. RUBIN, The main conjecture, appendix in S. Lang, Cyclotomic Fields I and II,
Springer-Verlag, New York, 1990.
[S] W. SINNOTT, On the Stickelberger ideal and the circular units of an abelian field. Invent. Math. 62 (1980), 181-234.
[W] L. C. WASHINGTON, Introduction to cyclotomic fields. Springer-Verlag, New York, 1996. Radan KucERA Department of Mathematics Masaryk Univerzity Janackovo nam. 2a 662 95 Brno Czech Republic