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.