22
MICHAEL D. FRIED, DAN HARAN, AND HELMUT VOLKLEIN
Theorem 5.2 tells about the structure of the absolute Galois group of
K(x).
For instance, every finite group is realizable over
K(x).
This isn't new
[DF2,
Theorem 5. 7]. Still, the precise information about real involutions gives more.
THEOREM 5.3.
Let K be a formally real PRC field. Let G be a finite group,
and
let G0 be
a
normal subgroup of G
generated
by involutions. There is
a
Galois extension N/K(x) with Galois group G such that the fixed field ofG0 in
N is the maximal totally real extension of K ( x) in N.
PROOF. Let
(E)
be a group of order 2. Put
H
=
G x
(E)
and
Ho =Go x
(E),
and let
7!':
H _...... G(K(
H)/
K)
be the epimorphism with kernel
G.
The set
Io
of
involutions in
Go
generates
Go.
Therefore
h
=
(Io U {1}) x { E}
generates
Ho.
It
is a conjugacy domain in H. Theorem 5.2 (with m =
1)
gives a Galois extension
F
of
E
=
K(x)
that contains
A.
such that
G(F/E)
=
H, G(F/E(H))
=
G,
and
J1
is the set of real involutions in
G(F/E).
Let
N
be the fixed field of
E
and
N'
the fixed field of
Ho
=
(11
).
The last condition means that
N'
is the
maximal totally real extension of
E
in
F,
and, thererefore, also in
N.
Clearly
G(N /E)
~
G
and
N'
is the fixed field of
G0
in
N.
o
6. Totally real Hilbertian fields
As in the preceding sections, all fields are of characteristic
0.
Let
S / R
be
a Galois cover of rings
[F
J, Definition
5.4]
with
F / E
the corresponding Galois
extension of quotient fields. Thus
R
is an integrally closed domain and there is
z
E
S integral over R such that S
=
R[z] and the discriminant
dE(z)
of
z
over
E
is a unit of
R.
We call such
z
a
primitive element
for
S / R.
Assume
S / R
is
real [HL,
Definition 4.2]:
R
is a regular ring and
F
is not formally real.
LEMMA
6.1. The integral closure S' of R in each intermediate extension F'
ofF/ E is
also a
regular ring.
PROOF. Observe that
S/S'
is also a Galois cover. By
[R,
p. 75] it suffices to
show that
S'
I
R
is etale. i.e.' flat and unramified. We have
s
=
tiJt;:~
Rzi'
where
d =
[F : E],
and so
Sf R
is faithfully flat. Similarly
Sf S'
is faithfully flat. By
the descent property
[Ma,
(4.B)],
S' / R
is (faithfully) flat.
To show that
S' / R
is unramified, let
q
be a prime of
S,
and let
p
=
q
n
S'
and
m
=
q
n
R.
Replace
R, S',
and
S
by their localizations at these primes to
assume that they are local rings. Then
S / S'
and
S' / R
are still faithfully flat.
As
SjS'
is unramified, the field extension
(Sjq)j(Rjm)
is separable and finite.
Hence so is its subextension
(S'fp)j(R/m).
AsS/Rand
SjS'
are unramified,
mS
=
q
and
pS
=
q.
Thus
(mS')S
n
S'
=
pS
n
S'.
But
S/S'
is faithfully
flat, hence
[Ma,
(4.C)],
mS'
=
p.
o
Let M be a field. Every homomorphism
¢:
R _...... M extends to a homomor-
phism
'1/J: S _...... M,
and
'1/J
induces a group homomorphism¢*:
G(M) _...... G(F/ E):
(1)
'1/J('l/J*(a)(x))
=
a(x)
for xES.
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