9.5. THE GALOIS CORRESPONDENCE
437
determined by
.˛/
, which is necessarily a root of
f .x/
. But by Propo
sition
9.4.1
, there is a
0
2
Aut
K
.L/
such that
0
.˛/
D
.˛/
; therefore,
D
0
j
M
. Now, the homomorphism theorem for groups gives
Aut
K
.M/
Š
Aut
K
.L/=
Aut
M
.L/:
This completes the proof of part (b).
n
We shall require the following variant of Proposition
9.5.3
.
Proposition 9.5.5.
Let
L
be a field,
H
a finite subgroup of
Aut
.L/
, and
F
D
Fix
.H/
. Then
(a)
L
is a finite–dimensional Galois field extension of
F
.
(b)
H
D
Aut
F
.L/
and
dim
F
.L/
D j
H
j
.
Proof.
We cannot apply Proposition
9.5.3
because it is not given that
L
is
finite–dimensional over
F
. Let
ˇ
be any element of
L
. We can adapt the
argument of Proposition
9.4.13
to show that
ˇ
is algebraic and separable
over
F
, and that the minimal polynomial for
ˇ
over
F
splits in
L
. Namely,
let
ˇ
D
ˇ
1
; : : : ; ˇ
r
be the distinct elements of
f
.ˇ/
W
2
H
g
. Define
g.x/
D
.x
ˇ
1
/.: : : /.x
ˇ
r
/
2
LOExŁ
. Every
2
H
leaves
g.x/
invariant,
so the coefficients of
g.x/
lie in Fix
.H/
D
F
.
Let
p.x/
denote the minimal polynomial of
ˇ
over
F
. Since
ˇ
is a
root of
g.x/
, it follows that
p.x/
divides
g.x/
. On the other hand, every
root of
g.x/
is of the form
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 Fall '08
 EVERAGE
 Logic, Algebra, Proposition, Galois theory, Cyclic group, Algebraic geometry

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