18.
Find
α
∈
E
that is not in
F
. Now
α
is algebraic over
F
, and must be of degree 2 because [
E
:
F
] = 2
and [
F
(
α
) :
F
] = deg(
α, F
). Thus irr(
α, F
) =
x
2
+
bx
+
c
for some
b, c
∈
F
. Because
α
∈
E
, this
polynomial factors in
E
[
x
] into a product (
x

α
)(
x

β
), so the other root
β
of irr(
α, F
) lies in
E
also. Thus
E
is the splitting field of irr(
α, F
).
19.
Let
E
be a splitting field over
F
. Let
α
be in
E
but not in
F
. By Corollary 50.6, the polynomial
irr(
α, F
) splits in
E
since it has a zero
α
in
E
. Thus
E
contains all conjugates of
α
over
F
.
Conversely, suppose that
E
contains all conjugates of
α
∈
E
over
F
, where
F
≤
E
≤
F
. Because
an automorphism
σ
of
F
leaving
F
fixed carries every element of
F
into one of its conjugates over
F
,
we see that
σ
(
α
)
∈
E
. Thus
σ
induces a onetoone map of
E
into
E
. Because the same is true of
σ

1
, we see that
σ
maps
E
onto
E
, and thus induces an automorphism of
E
leaving
F
fixed. Theorem
50.3 shows that under these conditions,
E
is a splitting field of
F
.
20.
Because
Q
(
3
√
2 ) lies in
R
and the other two conjugates of
3
√
2 do not lie in
R
, we see that no map of
3
√
2 into any conjugate other than
3
√
2 itself can give rise to an automorphism of
Q
(
3
√
2 ); the other two
maps give rise to isomorphisms of
Q
(
3
√
2 ) onto a subfield of
Q
. Because any automorphism of
Q
(
3
√
2 )
must leave the prime field
Q
fixed, we see that the identity is the only automorphism of
Q
(
3
√
2 ). [For
an alternate argument, see Exercise 39 of Section 48.]
51. Separable Extensions
167
21.
The conjugates of
3
√
2 over
Q
(
i
√
3 ) are
3
√
2
,
3
√
2

1 +
i
√
3
2
,
and
3
√
2

1

i
√
3
2
.
Maps of
3
√
2 into each of them give rise to the only three automorphisms in
G
(
Q
(
3
√
2
, i
√
3 )
/
Q
(
i
√
3 )).
Let
σ
be the automorphism such that
σ
(
3
√
2 ) =
3
√
2

1+
i
√
3
2
.
Then
σ
must be a generator of this
group of order 3, because
σ
is not the identity map, and every group of order 3 is cyclic. Thus the
automorphism group is isomorphic to
Z
3
.
22. a.
Each automorphism of
E
leaving
F
fixed is a onetoone map that carries each zero of
f
(
x
) into
one of its conjugates, which must be a zero of an irreducible factor of
f
(
x
) and hence is also a zero of
f
(
x
). Thus each automorphism gives rise to a onetoone map of the set of zeros of
f
(
x
) onto itself,
that is, it acts as a permutation on the zeros of
f
(
x
).
b.
Because
E
is the splitting field of
f
(
x
) over
F
, we know that
E
=
F
(
α
1
, α
2
,
· · ·
, α
n
) where
α
1
, α
2
,
· · ·
, α
n
are the zeros of
f
(
x
).
As Exercise 33 of Section 48 shows, an automorphism
σ
of
E
leaving
F
fixed is completely determined by the values
σ
(
α
1
)
, σ
(
α
2
)
,
· · ·
, σ
(
α
n
) that is, by the
permutation of the zeros of
f
(
x
) given by
σ
.
c.
We associate with each
σ
∈
G
(
E/F
) its permutation of the zeros of
f
(
x
) in
E
.
Part(
b
) shows
that different elements of
G
(
E/F
) produce different permutations of the zeros of
f
(
x
).
Because
multiplication
στ
in
G
(
E/F
) is function composition and because multiplication of the permutations
of zeros is again composition of these same functions, with domain restricted to the zeros of
f
(
x
), we
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 Spring '08
 OGUS
 Algebra, φ, Identity element, Inverse element