16
MICHAEL D. FRIED, DAN HARAN, AND HELMUT VOLKLEIN
Recall that (cf)(cr) =
f("Y)·
Combine (6) and (7) with (3) and (4) to get
cf("Yj)
=
f(crj)
=
aj =(to
f)("Yj), for each 1 :::; j:::;
r.
o
PROPOSITION
4.4. Let (K, P) be an ordered field and tan involution in G(K)
inducing P on K. Assume K
~ C.
Let
A
E
G(Q)
and p
E
1-l(K) with
(8) liP- A(q)llt 2 v 2 in K(t),
and
p
= A(p) is K-rational. Let h1:G(FIK(x))--+ Aut(G) be the embedding
corresponding top over K with Fl K(x) Galois as given in
(2.17).
Put L = K(p)
and let H be the image of
h1
.
The following hold:
(a) 8,(p) = t(p);
(b) P does not extend to L; in particular, P does not extend to F;
(c)
GX1(t) :::; H, and therefore
I~
H;
(d) h1(Ip(FIK(x)))
=
ConH(I).
PROOF.
By (2.14), 8,(p)
'I
p.
Therefore (a) implies t(p)
'I
p.
Hence
L
~
K(t), and this implies (b). Furthermore, the criterion of (2.17) implies that
t E
H.
Since G:::;
H,
GXI(t):::;
H.
So it suffices to prove (a) and (d).
Part
I. Reduction to K with archimedian orderings dense in X(K). Let K
0
be a finitely generated subfield of
K,
containing the finitely generated subfield
Q(p)
of K. Let Po be the restriction of P to K
0
.
This ordering is induced
from the restriction to
E
G ( K
o) of
t.
Let
F0
I
K
0
(
x)
be the Galois extension and
(hi)o: G(Fol Ko(x)) --+Aut( G) the embedding corresponding top over K
0
.
We
may assume that F = Fo · K from (2.18). If K
0
is sufficiently large, then the
restriction map resp0
:
G(F I K(x)) --+ G(Fol Ko(x)) is an isomorphism. If we can
show that the assertions hold for Ko, Po, to, (hi)o, then, by (2.19) and since
to(P) = t(p), they also hold for
K, P,
t,
h1. Lemma 1.6(a) shows the set of
archimedian orderings on K
0
is dense in X(K0
).
So we may assume that K
enjoys this property.
Part
II. Reduction toP archimedian. By Remark 1.8, if P' is an (archime-
dian) ordering of K sufficiently near to P, then lp(FI K(x)) = Jp,(FI K(x)).
We may assume an involution
t1
E
G(K), so near
tot,
induces
P'
that
liP- A(q)llt 2 =liP- A(q)llt' 2 and t(p) = t'(p).
Thus we may replace
P
by
P'
and
t
by
t
1

Part
III. Reduction to K = lR and
A
= 1. Assume that P is archimedian.
Extend A-
1
to an automorphism f3 of C, and let t
1
= /3t/3-
1.
Then f3(K(t)) =
(f3(K))(t') is a real closure of (f3(K),/3(P)). Hence it is also archimedian. Thus
we may assume that (f3(K))(t')
~ JR.
Hence /3t/3-
1
= t
1
= resf3(.K)c, where cis
complex conjugation on C.
Since
q
is algebraic over
Q,
/3A(q) =
q.
So, application of f3 to (8) yields
(8')
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