14
JURGEN RITTER AND ALFRED WEISS
L E M MA 2.5. Let K be totally real. Then
1. \[s C(K*)
and
%iS have the same Zi-rank.
2. If K satisfies Leopoldt's conjecture, then G(M/K) has Zi-rank 1.
With respect to the first claim look at the exact sequence C(UP) —• £(Kp) Zj, which arises
from /-completing the valuation sequence Up Kp Z. Because of Up Up K
p
, with
* , denoting the residue field of K„ we have C(UP) = ( f f ^ * . { ' , and therefore,
rk{C(K?))-^ ^rVTTl= _
1 if p f /
l + ikU^ l + [Kp:Qi\ ifp|/.
AS ~
I
ZS -
i
z -
£
n*1
P
x
1
G(M/K)
Summing up, and denoting the set of infinite primes of K by SQOJ we get
rktf l W ) ) = r k ( J ] W ) ) = | 5 \ 5 o o | + B ^ P
:
«M = 1^1 " l^ool + [* : Q] = |5 |
since C(Kp) is finite for infinite p, and since |5oo| = [K : Q] by assumption.
The second claim of the lemma now follows from the left column in (D2.3) and Dirichlet's
unit theorem. Indeed, rk(Z/g)£) + ik(G(M/K)) = \S\ and rk(Zj g £ ) = | 5 | - 1 . This finishes
the proof of the lemma.
We now suppose that we are given a G-equivariant homomorphism Z 5 TlpeS Kp which
maps AS injectively into E. This fits into a commutative diagram
(D2.5)
in which E Yis Kp
ls
the canonical map and \[s
Kpx
—* G(M/K) arises from Yls Kp
]\SC(KX) -+ G{M/K). Recall that E is sent to zero in G(M/K) by the reciprocity law,
since every Up, p 0 5, is in the kernel.
We will also require that Z —• G(M/K) is injective.
The /-completed diagram of (D2.5) is
Z; ® AS - Zj g E
I i
(D2.5«) z 5 ~ nA^i
PES'
i I
Zi ~ G(M/K)
with right column exact by Lemma 2.3. We stick to this notation.
Notice that the above maps are G-monomorphisms from the right ends of the exact 4-term
sequences in (D2.4) into their respective left ends. Therefore we are in a position to perform
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