Ma
5
c
HOMEWORK
7
SOLUTION
SPRING 09
The exercises are taken from the text,
Abstract Algebra
(third edi
tion) by Dummit and Foote, unless stated otherwise.
Page 582,
17
. c)
Gal
(
K/F
) =
{
1
, σ
}
, with
σ
(
a
+
b
√
D
) =
a

b
√
D
.
Therefore
N
K/F
(
a
+
b
√
D
) = (
a
+
b
√
D
)(
a

b
√
D
) =
a
2

b
2
D
.
d) Since [
F
(
α
) :
F
] =
d
,
d

n
follows from the tower law.
α
has
d
conjugates since the minimal polynomial is of degree
d
. Suppose
H
0
is
the subgroup of
G
corresponding to
F
(
α
), then
H
⊂
H
0
of index
k
=
n
d
.
Therefore,
α
is fixed by
k
embeddings of
K
, and each conjugate
α
i
appears
k
times in
Q
σ
(
α
). Furthermore,
Q
α
i
= (

1)
d
a
o
. Therefore
N
K/F
(
α
) =
Q
σ
(
α
i
) =
Q
(
α
i
)
k
= (

1)
n
(
a
0
)
k
.
2.a) The minimal polynomial for 1

ζ
is Φ
p
r
(1

x
) =
(1

x
)
p
r

1
(1

x
)
p
r

1

1
=
1 +
t
+
· · ·
+
t
p

1
, where
t
= (1

x
)
p
r

1
.
N
(1

ζ
) = Φ
p
r
(1

0) =
p
.
b)Suppose
n
=
p
r
m
, where
p
doesn’t divide
m
,
q

m
for some other
prime
q
. then Φ
n
(
x
) =
Φ
m
(
x
p
r
)
Φ
m
(
x
p
r

1
)
, since the numberator and denomina
tor each has the same number of terms, Φ
n
(1

x
) has constant term
1. Therefore
N
(1

ζ
) = 1.
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 Spring '09
 SusamaAgarwala
 Algebra, Polynomials, Prime number, primitive dth root

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