K. W. G
RUENBERG
A. W
EISS
Capitulation and transfer kernels
Journal de Théorie des Nombres de Bordeaux, tome
12, n
o1 (2000),
p. 219-226
<http://www.numdam.org/item?id=JTNB_2000__12_1_219_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Capitulation
and
Transfer Kernels
par K. W. GRUENBERG et A. WEISSRÉSUMÉ. On sait que pour une extension
galoisienne
finieK/k
d’un corps de
nombres,
le noyau dumorphisme
d’extension Clk ~ClK s’identifie au noyau
X (H)
du transfertH/H’ ~
A,
où H =Gal(K/k),
A =Gal(K/K)
etK
est le corps de classes de Hilbertde K.
Lorsque
le groupe G =Gal(K/k)
estabélien,
H. Suzuki amontré
que
|G| divise
|X(H)|.
Nous
appelons
noyau de transfert pour G tout groupe abélienfini X qui s’écrit
X(H)
pour un certain groupe H tel que A ~H ~ G.
Après
avoir caractérisé les noyaux de transfert en termesde
représentations
entières deG,
nous montrons que X est unnoyau de transfert pour le groupe abélien G si et seulement si on
a
|G| X =
0et
|G| divise |X|,
ce qui fournit une nouvelledémonstration du résultat de Suzuki.
ABSTRACT. If
K/k
is a finite Galois extension of number fieldswith Galois group
G,
then the kernel of thecapitulation
mapClk
~ ClK of ideal class groups isisomorphic
to the kernelX(H)
of the transfer map
H/H’ ~
A,
where H =Gal(K/k),
A =Gal(K/K)
and K
is the Hilbert class field of K. H. Suzukiproved
that when G isabelian,
|G|
divides|X(H)|.
We call a finiteabelian group X a transfer kernel for G if X ~
X(H)
for somegroup extension A ~ H ~ G.
After
characterizing
transfer kernels in terms ofintegral
rep-resentations of
G,
we show that X is a transfer kernel for theabelian group G if and
only
if |G|X
= 0 and|G| divides |X|.
Ourarguments
give
a newproof
of Suzuki’s result.Let
K/k
be a finite unramified Galois extension of number fields withGalois group G. The
capitulation
kernel for is the kernel of thenat-ural
homomorphism
of ideal class groupsClk
eCIK.
Suzuki[S]
proved
that when G is
abelian,
its orderG~
divides the order of thecapitulation
kernel. This remarkable resultencapsulates
much of the informationpre-viously
available aboutcapitulation.
We refer to the surveys[J]
and[M]
Manuscrit requ le 23 septembre 1999.
for relevant
background.
Our aim here is toexplain
a newapproach
toSuzuki’s theorem.
The transition to group
theory
(reviewed
in §
1)
allows one tointerpret
capitulation
kernels as transferkernels,
by
which we mean thefollowing:
given
a finite groupG,
then a finite abelian group X is atransfer
kernelfor
G if there exists a group extension A - H ~ G with A finite abelian so that
X is
isomorphic
to the kernel of the transferhomomorphism
H] -4
A. We shall prove thefollowing
result.Theorem 1.
If
G is afinite
abelian group, then thefinite
additive groupX is a
transfer
kernelfor
Gif,
andonly if,
= 0 andI G I
dividesWe outline what follows. In
§1
we translate theproblem
into anequiv-alent one on G-module extensions over
AG,
theaugmentation
ideal of theintegral
groupring
ZG. Then§2,
the core of the paper, is ananalysis
of thecommon structural
properties
of transfer kernels for G. This makespossi-ble the
proof
of Theorem 1 in§3.
In our final§4
we collect some commentsand
questions.
1. TRANSLATIONS
We
begin
with the classical result of E. Artin.Proposition
1. Thecapitulation
kernelfor
Klk is
atransfer
kernelfor
G.Here is a sketch of the
proof.
Let K be the Hilbert class field of K andA =
Gal( K / K).
If H =Gal(K/k)
then there is a commutative squarefrom which the
proposition
followsby
taking
kernels.Proposition
2. Thefinite
additive group X is atransfer
kernelfor
Gif,
andonly if,
there exists a G-module extension A ~ B --~ AG with Afinite
and X -H-1(G,
B).
Proof.
This result is clear from the functorialrelationship
between group extensions over G and G-module extensions over OG(cf.
[G]
§10.5).
Asthis is not the usual
approach
in the literature we sketch it here. A group extensionyields
the G-module extensionwhere B = T is induced from 7r: AH --t A(3 and
j(a)
is theappropriate
coset ofi(a) -
1.Conversely, given
(2),
letThen H is a group with
multiplication
x.y -T(x)y
+ x + y, itsidentity
element is 0 and the inverse of x is
x-1
=-g-lx,
whereT(x) - g -
1.The module
homomorphism
Tgives
the grouphomomorphism
H -+ G viaz e
r(x)
+ 1 with kernel A.If B arises from the group extension
(1),
then the hidden group H in Bgives
an extensionequivalent
to H:where
u(h) = (h - 1)
+ AA.AH. The G-coinvariants onB,
namely
BG
=B/(dG)B,
arenaturally
isomorphic
toAH/(AH)2,
whence toH/(H, H~
and so to
!Il
Notice thatH/~H, H~ ~
BG
isjust
~(H, H~ H x
+(AG)B.
We claim there is a commutative square
where
G
is the normendomorphism
and the lowerisomorphism
isinduced
by j .
-In view of
(3)
we mayreplace
Hby
H andview j
as inclusion. Take atransversal
G,
for A in H. If x E H withT(x)
= J~ -1,
then theimage
of x under transfer is11~
which is the same as1:g
and so the transfer
image
isGx
asrequired.
Proposition
2 followsby taking
kernels. D 2. TRANSFER KERNELSLet G be a finite group, not
necessarily
abelian. Put A =ZG/(G)
andidentify
AG
with If M is aZG-module,
thendG(M)
denotes theminimum number of module
generators
of M.Theorem 2. The
following
areequivalent:
(a)
X is atransfer
kernelfor
G;
(b)
X isisomorphic
to the cokernelof
ahomomorphism
cp :Am
where m >dG(AG)
and U is afinitely generated
G-submoduleof
Q
(c)
X isisomorphic
toMG
for
sorrcefinitely generated
G-moduleM,
whereGM
= 0 andQM
contains aQG-copy
of
QA;
(d) IGIX
= 0 and there exists asurjective homomorphism
X -MG
withM as in
(c).
Proof.
(a)
~(b).
Using
Proposition
2 we may, andshall,
assume X riH-1 (G,
B),
where A - B - AG. Take a free resolution ofB,
sodeter-mining
m and S in thefollowing diagram:
Now
H-1(G, B) =
(we
use Tatecohomology
throughout)
andthe exact sequence
SG --+ S -4* U gives
where J is the
connecting homomorphism.
Since A isfinite,
QS
=QR ri
whence
SG -
and Note that U is Z-torsion-free and so U is contained inQU.
AlsoGU
= 0gives
H-1 (G,
U)
=UG
andH~(G,
U)
= 0. ThusUG 6 )Am --*
X is exact.(b)
~(c).
The exact sequence A -AG
of G-modulesstays
exact when we
apply
Hom(U, -)
because U is Z-free. So we obtain theexact sequence
where the
right
hand term is 0 because it isisomorphic
towhich is 0 because
HomG(U,7G)
= 0 asIt follows that the
given homomorphism cp :
UG --~
AG
lifts to aG-homomorphism Q :
U e A"’twith QG
giving
the commutativediagram
-
with M =
Coker,B.
This induces anepimorphism
M - X andhence,
by
taking
G-coinvariants,
anisomorphism
X.Finally, QM (1) QU
contains
(a
G-copy
of)
QA"~ ,
whenceQM
containsQA.
(c)
~ (d)
is clear sinceIGI I
annihilatesMG
=H-1 (G, M).
(d)
~ (b) .
Choose m >dG (AG),
d(X ) ~
and take a G-freepresentation
L -4 M ofM.,
SinceGM =0,
LG
=thereby giving
the exact sequence where U =L/LG,
andhence the exact sequence
UG
Am--*
GMG.
Choose a :AJ-X
G and let{3: X - MG
be thegiven
homomorphism.
ThusCoker iG ^_r
Imoa,
whichimplies,
by
the Lemmabelow,
thatImiG -
Ker,8a. Consequently
Kera isisomorphic
to asubgroup
D ofImiG.
There exists a map ofUG
onto D
(remember
we aredealing
with finite abeliangroups),
giving
thecomposite
p :UG -~
D "Am,
with Kera.Again
using
theLemma
below,
Cokerp ri
X and soUG
Am
-X is exact.Finally,
QU
eQM ~ QA"
shows thatQA
implies
QU
CLemma.
(i)
Givenepimorpltisms
fi , f 2 :
Az -
X ,
thenKer f 2 .
(ii)
Givenepimorphisms gi :
Xi ,
i =l, 2,
withKergl ^-r
Kerg2,
then
X2.
Proof.
In(i)
thehomomorphisms
f1, f2
are freepresentations
of X asSo Schanuel’s Lemma and the Krull-Schmidt
property
give
the result. For(ii),
dualise withrespect
toQ/Z
and obtainXZ
by
(i).
(b)
~(a).
Our aim is to prove that the X of(b)
is a transfer kernel inthe module-theoretic sense of
Proposition
2. We use theisomorphism
given by
integral duality: ~
~(x
H~.x).
Hence p
corresponds
to auniquely
determined extension 7~’"~ ~--> S -H U whose associatedconnecting
homomorphism
H-1(G,
U)
e is cp(e.g.
11.1 in[GW]).
ThusUG ~
H°(G,
S)
is exact and so X -H°(G,
S).
Take a free
presentation R "
ZGm --* AG of AG and embed S in R with cokernel A. This can be done becauseTaking
thepushout
along
R - Agives
adiagram exactly
like(4)
except
that A
might
not be finite. In any caseH-1(G, B).
It remains to find a submodule L of A so that
A/L is
finite andH-1 (G,
B)
=
H-1(G, B/L).
First note that = 0by
the middle column of(4)
and
QSG --
HenceBG
is finite and soA/A
n AG.B is finite. Picka torsion-free G-submodule L of finite index in A n Then
A/L
isfinite;
alsoGL
=0,
whenceLG
= 0. The exactsequence L Y B -
B/L
then
gives
the exact sequencewhich finishes the
proof.
03. PROOF OF THEOREM 1
To prove Theorem 1 it
suffices,
in view ofProposition
2 and Theorem2,
to establish the
equivalence
of(i)
X is afinite
additive group such thatIGIX
= 0 andIGI
divideswith
(ii)
X isisomorphic
toMG
for
somefinitely generated
G-moduleM,
whereGM
= 0 andQM
contains aQG-copy of QA.
(i)
#(ii).
By (d)
of Theorem 2 it suffices to prove X has a transferkernel for G as a
homomorphic
image.
We shall show that anyimage
of Xof order
~G~
is a transfer kernel.Change
notation and call thisimage
X.So we have and shall use induction on when X =
0,
then G = 1 and so we can take M = 0.Now let X =
Xl
Since~X~,
so G has animage
G =GIG,
of order
p’
and thenIG11 =
By induction,
for anappropriate Ml.
Define M =A,
where A =ZG/(G)
is aG-module
by
inflation. ThenOM
= 0 sinceG1M1
=0,
whence(ii)
~(i).
Werepeat
the classicalargument.
Take a freeZG-presentation
F - M ofM,
with F = SinceMG
isfinite,
the kernel ofFG - MG
is
isomorphic
toFG
and soMG
is the cokernel of anendomorphism f
ofFG.
It follows thatdet f
=Since F -~
FG
mapsKer(F
eM)
onto theimage
off ,
there is aZG-endomorphism
f
of F such thatfc
=f
andCoker f
maps onto M.(det
/)M
= 0implies
det/
= nG for a suitableinteger
n(since
QA
9QM) .
Finally,
with edenoting
theaugmentation
onZG,
edet f
= detf
andthus
nlGI
=E(nd) =
4. REMARKS
(1)
Is the converse ofProposition
1 true? This is a fundamentalproblem.
An even
stronger
form of this is thefollowing: given
a group extension A --tH e G with A
abelian,
does there exist an unramified Galois extension L with Galois group H so that L is the Hilbert class field K of the fixed fieldK of A?
It should be noticed that any
group H
can be realised as the Galoisgroup of an unramified extension
L/k ([L],
p.121).
Then L C K and thedifficulty
lies inensuring
that L = K.(2)
Suppose
X is a finite additive group such thatIGIX
= 0. Then(a)
if X is a transfer kernel forG,
IG/[G, G]I
I
dividesI X 1;
;(b)
if~G~
dividesIXI,
then X is a transfer kernel for G.Both these facts are variations of
§3;
for(b)
one must first show thatif,
for eachprime
p, thep-primary
part of X is a transfer kernel for aSylow
p-subgroup
ofG,
then X is one for G.However,
neither(a)
nor(b)
has a converse if G is a non-abelian p-group.This is obvious for
(b) (take A = 1).
For(a),
if X isZ-cyclic
and of order , then X cannot be a transfer kernel for G: for if X -MG
with M as in(c)
of Theorem2,
thenlifting
agenerator
ofMG
to Mgives
aG-homomorphism
A - M which becomes anisomorphism
on coinvariants(by
Nakayama’s
Lemma andQA);
thenIXI =
IGI
forcing
G to be abelian.REFERENCES
[G] K.W. Gruenberg, Relation Modules of Finite Groups. CBMS Monograph 25, Amer. Math.
Soc., Providence, R.I., 1976.
[GW] K.W. Gruenberg, A. Weiss, Galois invariants for units. Proc. London Math. Soc. 70
(1995), 264-284.
[J] J.-F. Jaulent, L’état actuel du problème de la capitulation. Sem. de Théorie des Nombres de Bordeaux, 1987-1988, exposé no. 17.
[L] S. Lang, Algebraic Number Theory. Springer, New York, 1994.
[M] K. Miyake, Algebraic investigations of Hilbert’s Theorem 94, the principal ideal theorem and the capitulation problem. Expo. Math. 7 (1989), 289-346.
[S] H. Suzuki, A generalization of Hilbert’s Theorem 94. Nagoya Math. J. 121 (1991), 161-169.
K. W. GRUENBERG
School of Mathematical Sciences
Queen Mary and Westfield College
Mile End Road
London El 4NS, England
E-mail : K. W. Gruenbergtqmv. ac. uk A. WEISS
Department of Mathematics
University of Alberta Edmonton
Canada, TG6 2G1