Problem Set #2 Solutions
Chapter III.
2.2.
1.
If
F
⊆
E
⊆
K
and
E
is stable, then
G
E
G
=
G
(
K
:
F
)
.
Proof.
Let
θ
∈ G
E
,
φ
∈
G
, and
a
∈
E
. Since
E
is stable,
φ
(
a
)
∈
E
, and so
θφ
(
a
) =
φ
(
a
). Thus
φ

1
θφ
(
a
) =
φ

1
φ
(
a
) =
a
, so
φ

1
θφ
∈ G
E
.
2.
If
H
G
, then
F
H
is stable.
Proof.
Let
φ
∈
G
=
G
(
K
:
F
) and
θ
∈
H
. Since
H
is a normal subgroup
of
G
we have
φ

1
θφ
∈
H
, and so
φ

1
θφ
(
a
) =
a
whenever
a
∈ F
H
. But then
θφ
(
a
) =
φ
(
a
), so
φ
(
a
)
∈ F
H
(since
θ
∈
H
was abitrary), and hence
F
H
is
stable (since
a
∈ F
H
was abitrary).
3.1.
Suppose
F
is a field of characteristic
p >
0
.
1.
(
a
+
b
)
p
=
a
p
+
b
p
and
(
a

b
)
p
=
a
p

b
p
for all
a, b
∈
F
.
Proof.
We have
(
a
±
b
)
p
=
p
i
=0
p
i
a
i
(
±
b
)
p

i
.
Furthermore,
p

p
!
i
!(
p

i
)!
=
p
i
whenever
i
= 0
, p,
since
p
is prime and 0
< i, p

i < p
(that is,
p
divides the numerator but not
the denominator).
But char(
F
) =
p
so
p
·
1
F
= 0, and hence all the middle
terms of (
a
±
b
)
p
vanish. If
p
is odd, (

1)
p
=

1; otherwise
p
= 2, in which case
(

1)
p
= (

1)
2
= 1 =

1. Thus either way (
a
±
b
)
p
=
a
p
+ (
±
b
)
p
=
a
p
±
b
p
.
2.
Let
K
=
F
(
t
)
for an indeterminate
t
and let
f
(
x
) =
x
p

t
∈
K
[
x
]
. Then
f
(
x
)
is irreducible with just one root with multiplicity
p
in any splitting field.
Proof.
First note that
F
[
t
] is a PID,
t
∈
F
is a nonzero nonunit in
F
[
t
] and
hence prime, and
f
(
x
) is primitive in (
F
[
t
])[
x
]. Thus by Eisensteins criterion,
f
(
x
) is irreducible in (
F
[
t
])[
x
]. But
F
[
t
] is also a UFD (since PID’s are always
UFD’s, by prop II.5.10), so by proposition II.5.15
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 Spring '08
 MORRISON
 Algebra, Complex number, Galois theory, finite field, Perfect field

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