J. W. S. CASSELS
Now suppose that a E R.H.S.
Then for any m, 1 I
N we have
dbi E o,
m 5 N)
B) we have d # 0, and so ldlv = 1 for almost
all u the condition
(12.2) thus implies
b, E o,
m 5 N),
This proves the lemma.
Almost all v are unramijied
in the extension K/k.
For by the results of Chapter I a necessary
and sutlicient condition for v to be un-
is that there are yl,. . ., yN E R.H.S. with ID(y,,. . ., yN)lv = 1. And for almost
all v we can put yn = ~f”-~.]
We describe here a topological
tool which will be needed later:
DEFINITION. Let sZA
(A E A) be a family of topological spaces and for almost
alIt L let On c Q, be an open subset of a,. Consider the space n whose points
are sets u =
where aA E a, for every A and cr, E On for almost all 1.
give Q a topology by taking as a basis of open sets the sets
is open for all rZ and rA = O1 for almost all 1.
topology R is the restricted topologicalproduct of the R, with respect to the O1.
COROLLARY. Let S be a finite subset of A and let Rs be the set of a E fi
with aA E O1 (2 q! S), i.e.
Then G!, is open in Sz and the topology induced in as as a subset of R is the
same as the product topology.
on the totality
but not on the individual