J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
B
ART DE
S
MIT
The cyclic subfield integer index
Journal de Théorie des Nombres de Bordeaux, tome
12, n
o1 (2000),
p. 209-218
<http://www.numdam.org/item?id=JTNB_2000__12_1_209_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
The
cyclic
subfield
integer
index
par BART DE SMIT
RÉSUMÉ. Dans cet
article,
nous nous intéressons à l’indice dans l’anneau des entiers d’une extension abélienne d’un corps denom-bres K du sous-groupe
engendré
par les entiers contenus dans dessous-corps
cycliques
sur K. Cet indice est fini et nedépend
que du groupe de Galois et dudegré
de K. Nous en donnons uneexpression combinatoire.
Lorsqu’on
considèreplus généralement
des anneaux deDedekind,
des termes correctifs apparaissent, s’ily a une extension
inséparable
du corps résiduel. Nousexplicitons
ces termes dans le cas d’une extension abélienne de type
(p,p).
ABSTRACT. In this note we consider the index in thering
ofintegers
of an abelian extension of a number field K of theaddi-tive
subgroup generated by integers
which lie in subfields that arecyclic
over K. This index isfinite,
itonly depends
on the Galoisgroup and the
degree
ofK,
and wegive
anexplicit
combinatorial formula for it. Whengeneralizing
to moregeneral
Dedekinddo-mains, a correction term can be needed if there is an
inseparable
extension of residue fields. We
identify
this correction term for abelian extensions of type(p, p).
1. INTRODUCTION
We first
give
the main result in somespecial
cases. Let A be thering
ofintegers
in an abelian extensionof Q
oftype
(p, p),
where p is aprime
number. Then the additive
subgroup generated by
allintegers
in A withdegree
pover Q
has indexpp(p-l)/2
in A. For p = 2 this seems to bewell-known,
and for p = 3 this has been shownby
Parry
[11,
Lemma5].
It wasproved
by
A.Fajardo
Mir6n[6]
that for a Galois extensionof Q
with abelian Galois group of order2 k
andexponent
2,
the index in thering
ofintegers
of thesubgroup generated by quadratic
integers
is2~"~ *~.
In this paper wegive
such anexplicit
formula for any abelian extensionof number
fields,
and we considergeneralizations
to moregeneral
abelianextensions of
quotient
fields of Dedekind domains.Manuscrit requ le 9 septembre 1999.
The author was supported by a fellowship from the Koninklijke Nederlandse Akademie van
In order to state the full
result,
we first introduce some notation. Let Gbe a finite abelian group of order n and let
Z[G]
be the groupring
of Gwith coefficients in Z. For any
Z[G]-module
M we letMcyc
be the additivesubgroup
LH
MH
ofM,
where H ranges over allsubgroups
of G for whichG/H
iscyclic,
andMH
denotes the set of H-invariants of M. We letc(G)
be the indexWe will first
compute
thisinteger explicitly.
Theorem 1. Let n = theprime
factorization of n and for
d > 1 letOd(G)
be the numberof
elementsof
Gof
exact order d. Then theprime
factorization of
c(G)
isgiven
by
It is easy to see that Cp = 0 if and
only
if thep-Sylow subgroup
of G iscyclic..
oLet A be a Dedekind domain and let B be its
integral
closure in a finiteabelian extension of the
quotient
field of A with Galois group G. Then B is a Dedekind domain as well[13,
Ch.I,
§4, Prop.
8,
9]
andBcyc
is thesub-A-module of B
generated by
allintegers
in B thatgenerate
acyclic
extension of the
quotient
field of A.The results and
arguments
willdepend
strongly
on thefollowing
condi-tion,
which may or may not hold:for
all maximal ideals qof B
theq-adic completion of B is
gen-(*)
eratedby
asingle
element as aring
extensionof
thecompletion
of A.
The condition
(*)
seems to be the natural condition under which thetra-ditional results of ramification
theory
[13,
Ch.III,
IV]
hold. It is satisfied if all residue field extensions of B over A areseparable
[13,
Ch.III,
§6,
Prop.
12].
Inparticular,
(*)
holds forrings
ofintegers
in number fields. It also holds when G iscyclic
ofprime
order. One can show ingeneral
thatcondition
(*)
isequivalent
to the condition that the module of differentials which is a B-module of finitelength,
iscyclic
as aB-module, i.e.,
itcan be
generated
as a B-moduleby
asingle
element;
see[3].
For an inclusion M C N of
finitely generated
modules over a Dedekinddomain A we let the A-index
[N :
M~A
be theFitting
ideal of theA-module
N/M.
IfN/M
has finitelength
as anA-module,
then we canwrite
NIM
A/al ~ ~ ~ ~ ~ A/at
for non-zero ideals al, ... , at ofA,
andthen
[N : M~A
= 0. Note that the usual index of M in N isgiven
by
(N : M] = [A : [N :
MIA]-Theorem 2. Let A be a Dedekind domain and let B be its
integral
closure in afinite
Galois extensionof
thequotient
field of
A with an abelian Galois group G.If
condition(*)
holds,
then we haveIn the number field case one can deduce Theorem 2 from the theorem
of Fr6hlich
[7]
that says that B is "factorequivalent"
to the groupring
A~G~,
and a characterization of factorequivalence by
Burns[1,
Prop.
(1)].
Conversely,
theproof
of Theorem 2given
belowgives
rise to an alternativeapproach
to Eb6hlich’sresult;
see[4].
See[5]
forapplications
in aslightly
different context.
The situation is much more cumbersome if condition
(*)
does not hold.It was shown in
[3]
that in the case that G is oftype
(~, p)
there is asingle
B-ideal c1 which can be used to extend some ramification theoretic
results,
notably
[13,
Ch.IV,
§1,
Prop.
3, 4],
to the case where(*)
does not hold. This ideal D measures thedegree
to whichf2B/A
isnon-cyclic.
It isgiven
by
1 = and it is the smallest B-ideal for which there exists aB-module
epimorphism
Theorem 3. Let A be a Dedekind domain and let B be its
integral
closure in a Galois extensionof
thequotient
field
of
A with an abelian Galois group Gof
type
(p, p~
for
someprime
number p. Then we haveThe
proofs
of the Theorems aregiven
in Section 3.They
use somegen-eral
properties
of modules over abelian groups which aregiven
in the nextsection.
2. MODULES OVER ABELIAN GROUPS AND THE LEMMA OF DE
BRUIJN-REDEI
For a
positive
integer
m we let EZ[X]
be the mthcyclotomic
polyno-mial. We will need a basic lemma about these
polynomials,
which was firststated
by
R6dei[12]
andproved by
DeBruijn
[2].
Gillard[8]
and Gras[9]
attribute it to Martinet. We include a differentproof
forcompleteness.
We first introduce some notation from
[10, §2]
that will be usedthrough-out the paper. If C is a
cyclic
group of order m, then theQ-algebra
Q(C)
will be the
quotient
of the groupring
Q[C]
by
the idealgenerated
by
with g
a generator of C. Note that this ideal does notdepend
on the choiceof g
and thatQ(C)
isisomorphic
to the field of mth roots ofunity.
Lemma 4
(De Bruijn-Redei).
Let n > 1 be aninteger,.
The idealof
Z[X]
generated by
thepodynomials
(X’~ -
1)/(Xn/p -
1),
where p ranges over theprime
factors of
n, is theprincipal
idealgenerated by
Proof.
Note that7G~X~/(Xn -1)
is the groupring
Z [Cn]
of thecyclic
groupCn
of order ngenerated
by
theimage
of X. LetIn
be theimage
inZ [Cn]
of the idealgenerated by
thepolynomials
(Xn - 1)/(Xn/p - 1)
where p rangesover the
prime
factors of n.For every m > 1 we have
ITdlm
=1,
and sinceQ[X]
is aunique
factorization domain thisimplies
that isisomorphic
to thecyclotomic
field It suffices to show that is a torsionfree abelian group because it then follows that
In
is the kernel of the mapZ
If n is a
prime
power then this is trivial. Weproceed by
inductionon the number of
prime
factors of n.Suppose n
= mk with m andk
coprime
and smaller than n. We have and un-der thisisomorphism
In
maps to + It follows that which is torsionfree,
becauseby
the in-ductionhypothesis
it is a tensorproduct
of torsion free abelian groups. DLet G be a finite abelian group, and let C be the set of
cyclic quotients
of G. The canonical mapsQ(p)
forp E G give
rise to a canonicalisomorphism
ofQ-algebras:
See
[10, §2]
for a shortproof.
It follows that everyQ[G]-module
Vdecom-poses as a
product
V = whereV(P)
=For any
Z[G]-module
M and p =G/H
E C we letMp
=MH
be thesubmodule of H-invariants of M. For a
Q[G]-module
V wehave Vp
=V(’),
where thepartial
order on C is definedby
G/H’
G/H
~H C H’. Note that for v E V we have v E
V(P)
if andonly
if v EVp
and= 0 for some
generator g
of p.Now let M be a
Z [G]-module
which is torsion free as an abelian group.Viewing
M as asubgroup
of theQ[G]-module
welet
M(P)
be theimage
ofMp
under theprojection
on the factorLemma 5. The kernel
of
theprojection
Proof.
Note thatKer(x)
= The inclusionMu
CKer(Tr)
is clear.Suppose g
is agenerator
of p and let m =#p.
PutBy
theprevious lemma,
thepolynomials
Pp =
1),
with p aprime
divisorof m,
generate
the unit ideal inZ[X].
By writing
1 =Lplm
Qp
in
Z [X]
withQp
E we see that every x EKer(7r)
can be written asx =
Lplm
xp with xp =Qp(g) .
x EMp.
Using
thatwm(g)x
= 0 one seesthat
g"’rp
fixes zp, so that x EMeT.
D In the next lemma we follow anargument
that Gillard[8, §4]
gives
in thecontext of
cyclotomic
units.Lemma 6. Let A be a Dedekind domain
of
characteristic0,
and let N C M be an inclusionsof finitely generated
A[G]-modules,
which are torsionfree
as A-modules.If
(M : NIA 0
0 thenProo, f.
For a subset D ofC,
denote1:,CD
Mo-
by
MD .
We claim that forevery subset D of
C,
for which a E D whenever a p and p ED,
we haveTaking
D = C the Lemma will follow. We prove this claimby
inductionto
#D.
If D isempty,
then there isnothing
to prove. Assume D isnon-empty,
choose a maximal element p ED,
andput C
=DBIpl.
It is clearthat
Mg
is contained in which in turn lies in the kernel of theprojection
mapM-D 7r)(MOQ)(P).
Thisimplies
that7r(Mv) =
7r(Mp) =
M(P) . By
Lemma 5 one sees that the kernel of 7r isequal
toME.
By
applying
the sameargument
to N onegets
adiagram
with exact rows inwhich the vertical maps are
injective:
By
the snake lemma weget
a short exact sequence of the cokernels ofthe vertical maps. Over a Dedekind
domain,
taking
theFitting
ideal of amodule of finite
length
ismultiplicative
over short exact sequences, so itfollows that
3. DISCRIMINANTS AND CONDUCTORS
Let A be a Dedekind domain and let B be a commutative
A-algebra
whichis
finitely
generated
and free as anA-module,
with basis wi , ... , wn. Thediscriminant
AB/A
is the A-idealgenerated by
where is the trace of the matrix with coefficients in A definedby
zwj =
rj
If B’ is asub-A-algebra
which is also free as an A-moduleof rank ~, then
ABI/A
=(B :
When A = Z we oftenidentify
a Z-ideal such as a discriminant or an A-index with itsunique positive
generator.
By
the discriminant0(K)
of a number field K one meansthe discriminant of its
ring
ofintegers
as aZ-algebra,
and the absolutediscriminant of K is the real number where d is the
degree
of K.Proof of
Theorem 1. Weidentify
Q[G]
withflp
Q(p),
so that thering
N =becomes a
subring
of theproduct
M = of therings
ofintegers
in
Q(p).
We havec(G) = ~N :
Ncyc]
and M =Mcyc,
so with Lemma 6 weget
* 1/2
is
generated
as an abelian groupby
EG/H~,
whereSx
is the formal sum of all elements in the coset x of G mod H. Under theprojection
map, such asum Sx
ismapped
to the element(#H)x
of=
Z(p),
so N(P) =(#H)M~P~.
The Z-rank ofM(P)
iscp(#p),
where cp is the Eulerphi-function.
One deduces that[M(P) :
N(P)] =
so that
By
duality
of finite abelian groups, G has the same number ofcyclic
sub-groups as
cyclic
quotients
of each order.Using
the fact that the mthcyclotomic
field hasdegree
and that acyclic subgroup
of order m isgenerated by exactly
of itselements,
weget
where
ord(g)
is the order of thecyclic
groupgenerated by g,
andd(m)
denotes the absolute discriminant of the mthcyclotomic
field.Next,
one remarks that for two abelian groupsG1
andGZ
ofcoprime
order n1 and n2 one hasc(Gi
xG2)
= and that one hasthe
corresponding
identity
for the other side of theequality
in Theorem 1. We may therefore assume that n is a power of aprime
number p. With the1
formula
d(p"’)
=pm-
p- 1 for m > 1(see
e.g.[14,
Prop.
2.1)
the formula inTheorem 1 now follows
easily.
0Proof of
Theorem 2. Let A and B be as in Theorem 2. Note first that= A ~ which in turn
implies
that[A[G] :
A[G]cyc]A
=c( G) .
A.Let us start with the easy case that A has
positive
characteristic p. Ifp f c(G)
thenA[G]
=A[G]cyc
and B =A[G]cyc.
B CBcyc,
so B =Bcyc
andwe are done. Now suppose that p ~
c(G).
By
the normal basis theorem we can choose anA[G]-module
injection
cp:A[G]
e B. Let us also choose a non-zero element x E A so that xB is contained in theimage
of cp. Then we have inclusionsand since
[A[G] :
A[G]cyc]A
= 0 it follows that~B :
xBcyc]A
= 0. But wehave
[Bcyc : XBcyc]A =1=
0,
and therefore[B :
Bcyc]A
= ~ =c( G) .
A.Now assume that A has characteristic zero and let n be the A-rank of B.
Let K and L be the
quotient
fields of A and B. The relativediscrimi-nant
AB/A
can be defined as the A-idealgenerated by
all determinantswhere ranges over all sequences of
length
n in B. It is a non-zero ideal of A.By
induction to thecardinality
of p we can define for eachp E C
an A-idealf (p),
called the conductor of p, suchthat
We can write down this definition in one stroke
by
M6bius inversion:I ... ,
where p
is the M6bius function. The conductor discriminantproduct
for-mula now says thefollowing.
Lemma 7.
If
condition(*)
holds,
thenProof. For g
E G let ag be the A-ideal which is the norm of the B-idealgenerated by
all x - gx with x in B. For anysubgroup
H of G Hilbert’s formula andtransitivity
of the differentimply
thatwhere denotes the trace of the action of an element x on a
Q-vector
space V. For each
p e C
we deduce thatThe lemma now follows from the
Q[G]-module
isomorphism
Q[G] ££ fl
Q(p).
0
We continue the
proof
of Theorem 2. Consider the tensorproduct
N =B~AB
as a module over the commutativering
M =B[G]
by letting
Bact on the left factor and G on the
right
factor. For everysubgroup
H ofG we have
NH
=BQ?)A(BH)
andMH
=B~A(A[G]H).
Thisimplies
that= and
Mcyc
= so that[N :
Ncyc]B
=[B :
B and
[M :
Mcyc]B
=c( G) .
B. Since the canonical map from theideal group of A to the ideal group of B is
injective,
it suffices to show that(N :
(M :
The
advantage
of this basechange
to B-coef&cients is that we now have a canonicalB[G]-linear
mapFor every
subgroup
H of G we claim that[MH :
ABHIA
B. To seethis,
let p be aprime
ofA,
letAp
be thep-adic completion
ofA,
put
Bp
= and choose a basisc~l, ... , wn of over
Ap.
TheBp-linear
map inducedby
cp is thengiven
by
the matrix U =(oi(wj))ij,
where~Ql, ... , an)
=G/H.
On the onehand,
itfollows that
det(U)
is thep-part
of[MH :
But on the otherhand,
it is well known thatdet(U)2
is thep-part
of the discriminantof BH
over
A;
see[13,
Ch.III,
§3,
4].
This proves the claim.For each p E C we now
get
twoproduct
expansions
of[Mp :
onefrom our definition
(**)
of the conductor and one from Lemma 6applied
to the
p-action
onMp.
By
induction to the size of p(or
M6biusinversion)
it follows that(M~p~ :
=f(p) ~ B
forall p
E C.By Lemma 6,
applied
now to the G-action onM,
we therefore have[Mcyc :
=IlPEc
f(P)~
One can summarize this with thefollowing
commutativedia-gram of
injections
ofB-modules,
where the labels of the arrows indicateNow we use condition
(*)
in order to invoke Lemma 7. We obtainSince
ABIA 4
0 thisimplies
thatProof
of
Theorem 3. Note first that indeedc(G) =
by
Theorem1,
so that the secondequality
holds.The case that A has
positive
characteristic isagain
easy: either p = 0 inA,
in which case we sawalready
that[B :
vanishes,
or 0 inA,
in which case B is
tamely
ramified over A so that condition(*)
holds and we have D = B.Thus,
we assume that A is of characteristic zero. For each maximal idealq of B we need to check that the
q-parts
of the two B-ideals are the same.Let p be a maximal ideal of A.
Suppose
that condition(*)
holds for theprimes
q of Bextending
p. Then theq-part
of l is trivial for these q, andby
localization and Theorem 2 we are done. The case that(*)
failsfor some q
extending
p remains. Since(*)
holds for extensions ofdegree
1 and p, this canonly
happen
when thecompletion
Bq
has rankp 2
overAp,
and p is the residue characteristic. For the remainder of theproof
wemay therefore assume that A and B are
complete
discrete valuationrings
of residue characteristic p. For this case the results in[13,
Ch.IV,
§1~
havebeen extended:
by
Theorem 2.2 of[3]
andtransitivity
of the different wehave
Here
1J B/A
denotes the different of B overA,
and the ideal r,according
to Theorem 5.1 of
[3],
isgiven by
t =c~p-1.
Dividing by
1JB/A
andraising
both sides to the power p, we deduce with definition(**)
thatWe now use the same
diagram
as in theprevious
proof
and findThis proves Theorem 3. D
Example.
Let us take p = 2 and let A be acomplete
discrete valuationring
of characteristic zero, whose residue field is the field
IF2 (x, y)
with xand y
algebraically independent
overF2 .
Lifting x
and y
to elements xand g
ofA,
we now consider B =j4[B/~,~/~].
Then B is Dekekind with an actionof an abelian Galois group G of
type
(2,2).
We now have B =Bcyc
andc( G) .
B = 2B =D 0
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Mathematisch Instituut Universiteit Leiden
P.O. Box 9512, 2300 RA Leiden The Netherlands