26
MICHAEL D. FRIED, DAN HARAN, AND HELMUT VOLKLEIN
By Example 7.2,
I~
generates A. But
HE
ImD implies that IH generates H.
We have 1r(I'k)
=I~
and ConH(I'k) = IH. By an analogue of Gaschiitz' lemma
[HL,
Lemma 3.3 with n = OJ, I'k generates H.
By Example 7.2 there is an epimorphism 'lj;: D---+ H such that
7f
o 'lj; =¢and
'!j;(I~)
= I'k. Clearly '!j;(In) = IH. o
THEOREM 7.4. Let K be a totally real Hilbertian PRC field. Assume that K
has no proper totally real aJgebraic extensions and X ( K) has no isolated points.
Put G
=
(G,Ia), where G is the absolute Galois group of K and Ia is the
conjugacy domain of all involutions in G. Then
(a) A finite embedding problem
(1r:
H---+ A, ¢: G---+ A)
for G has a solution,
if(*) IH
=f. 0
is a conjugacy domain in Hand H
=
(IH)·
(b) A finite involutory structure His in ImG if and only if(*) holds.
(c)
G has the embedding property.
PROOF. The fixed field of Ia in
G
is totally real over
K.
Thus
G
=
(Ia).
PROOF OF (a). As 1
rJ_
IA = cjJ(Ia), Ker(¢) nia =
0.
Therefore the fixed field
L of Ker(¢) is not formally real. Without loss of generality A= G(L/ K) and¢
is the restriction map.
Theorem 5.2 (with m = 1,
X1
= X(K), and I
1
= IH) identifies
1r:
H ---+
A
with the restriction map resL:G(F/E)---+ G(L/K), where Eisa simple
transcendental extension of K, and F is a Galois extension of E that contains L
and is regular over L.
Therefore, G(F/E) = (I(F/E)). Choose
Q1
, ...
,Qm E X(E) with
m
I(F/E)
=
U
IQ
1
(F/E).
j=l
We may assume their restrictions P1
, ... ,
Pm E X(K) to K are distinct. Indeed,
each P1 is not isolated in X ( K), and hence there is P E X ( K) distinct from
P
1
, ... ,
P
m
and arbitrarily close to P1. By Remark 1.8(b) we may assume that
Ip(F/E) = Ip.(F/E). As Ip(F/E) = nQEX(E)IQ(F/E), there is Q E X(E)
J
Q2P
above P such that IQ(F/E) = IQ1 (F/E). We replace QJ by Q.
Let X
1
, ... ,
Xm be a partition of X ( K) into disjoint clopen sets such that
P1 E Xj. This gives the setup (2) of Section 6. As K is totally real Hilbertian,
there is a E K and an epimorphism
¢~:
G(K) ---+ G(F /E). By Remark 6.2(a),
¢~
is a solution to our embedding problem.
PROOF OF (b). Condition(*) is necessary, since Ia
=f.
0
is a conjugacy domain
in G and G =
(I0
).
Conversely, assume(*). Let A= (a)= G(K(H)/K) and
A
=
(A,
{a}), where a is the generator of
A,
and let ¢: G ---+
A
be the restriction
map. We construct below a finite involutory structure
H
that satisfies (*),
with epimorphisms
H
---+ H
and
1r:
H
---+ A.
By (a) there is an epimorphism
'lj;: G---+
H,
and hence
HE
ImG.
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