Ma
5
c
HOMEWORK
1
SOLUTION
SPRING 09
The exercises are taken from the text,
Abstract Algebra
(third edi
tion) by Dummit and Foote, unless stated otherwise.
Page 519,
4
. Deﬁne
φ
:
Q
[
√
2]
→
Q
[
√
2] via
a
+
b
√
2
7→
a

b
√
2, where
a,b
∈
Q
. Then
φ
(
a
+
b
√
2) +
φ
(
c
+
d
√
2) =
a

b
√
2 +
c

d
√
2 =
a
+
c

(
b
+
d
)
√
2 =
φ
((
a
+
c
) + (
b
+
d
)
√
2) =
φ
(
a
+
b
√
2 +
c
+
d
√
2).
φ
(
a
+
b
√
2)
φ
(
c
+
d
√
2) = (
a

b
√
2)(
c

d
√
2) = (
ac
+2
bd
)

(
ad
+
bc
)
√
2 =
φ
(
ac
+ 2
bd
+ (
ad
+
bc
)
√
2) =
φ
((
a
+
b
√
2)(
c
+
d
√
2)). Therefore
φ
is a
homomorhpism. It’s obvious that
φ
is both injective and surjective.
Page 530,
12
. We know that [
E
:
F
] divides [
K
:
F
] =
p
. Therefore
[
E
:
F
] = 1, in which case
E
=
F
since
E
contains
F
, or [
E
:
F
] =
p
=
[
K
:
F
], in which case
E
=
K
, since
E
is a subﬁeld of
K
.
Page 530,
13
. Note that [
Q
(
α
1
) :
Q
]

2 since
α
1
is a root of
x
2

α
2
1
∈
Q
[
x
].
Similarly, for any
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 Spring '09
 SusamaAgarwala
 Algebra, Prime number, k2, deg mβ,F

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