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Solutions Chapters 6–10
Section 6.1
1. We have
r
1
=2
,
r
2
=1
,
r
3
=1so
t
1
,
t
2
=0
,
t
3
= 1. The algorithm terminates in
one step after after subtraction of (
X
1
+
X
2
+
X
3
)(
X
1
X
2
X
3
). The given polynomial
can be expressed as
e
1
e
3
.
2. We have
r
1
,r
2
3
=0s
o
t
1
,t
2
3
= 0. At step 1, subtract
(
X
1
+
X
2
+
X
3
)(
X
1
X
2
+
X
1
X
3
+
X
2
X
3
). The result is
−
3
X
1
X
2
X
3
+4
X
1
X
2
X
3
=
X
1
X
2
X
3
. By inspection (or by a second step of the algorithm), the given polynomial
can be expressed as
e
1
e
2
+
e
3
.
3. Equation (1) follows upon taking
σ
1
(
h
) outside the summation and using the linear
dependence. Equation (2) is also a consequence of the linear dependence, because
σ
i
(
h
)
σ
i
(
g
)=
σ
i
(
hg
).
4. By hypothesis, the characters are distinct, so for some
h
∈
G
we have
σ
1
(
h
)
6
=
σ
2
(
h
).
Thus in (3), each
a
i
is nonzero and
σ
1
(
h
)
−
σ
i
(
h
)
(
=0 if
i
=1;
6
i
.
This contradicts the minimality of
r
. (Note that the
i
= 2 case is important, since
there is no contradiction if
σ
1
(
h
)
−
σ
i
(
h
) = 0 for all
i
.)
5. By (3.5.10), the Galois group consists of the identity alone. Since the identity Fxes all
elements, the Fxed Feld of
G
is
Q
(
3
√
2).
6. Since
C
=
R
[
i
], an
R
automorphism
σ
of
C
is determined by its action on
i
. Since
σ
must permute the roots of
X
2
+ 1 by (3.5.1), we have
σ
(
i
i
or
−
i
. Thus the Galois
group has two elements, the identity automorphism and complex conjugation.
7. The complex number
z
is Fxed by complex conjugation if and only if
z
is real, so the
Fxed Feld is
R
.
Section 6.2
1. The right side is a subset of the left since both
E
i
and
E
p
i
+1
are contained in
E
i
+1
. Since
E
i
is contained in the set on the right, it is enough to show that
α
i
+1
∈
E
i
(
E
p
i
+1
). By
1
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hypothesis,
α
i
+1
is separable over
F
, hence over
E
i
(
α
p
i
+1
). By Section 3.4, Problem 3,
α
i
+1
∈
E
i
(
α
p
i
+1
)
⊆
E
i
(
E
p
i
+1
).
2. Apply Section 3.4, Problem 7, with
E
=
F
(
E
p
) replaced by
E
i
+1
=
E
i
(
E
p
i
+1
), to
conclude that
E
i
+1
is separable over
E
i
. By the induction hypothesis,
E
i
is separable
over
F
. By transitivity of separable extensions (Section 3.4, Problem 8),
E
i
+1
is
separable over
F
. By induction,
E/F
is separable.
3. Let
f
i
be the minimal polynomial of
α
i
over
F
. Then
E
is a splitting Feld for
f
=
f
1
···
f
n
over
F
, and the result follows.
4. This is a corollary of part 2 of the fundamental theorem, with
F
replaced by
K
i
−
1
and
G
replaced by Gal(
E/K
i
−
1
)=
H
i
−
1
.
5.
E
(
A
) is a Feld containing
E
≥
F
and
A
, hence
E
(
A
) contains
E
and
K
, so that by
deFnition of composite,
EK
≤
E
(
A
). But any Feld (in particular
) that contains
E
and
K
contains
E
and
A
, hence contains
E
(
A
). Thus
E
(
A
)
≤
.
6. If
σ
∈
G
, deFne Ψ(
σ
)(
τ
(
x
)) =
τσ
(
x
),
x
∈
E
. Then
ψ
(
σ
)
∈
G
0
. [If
y
=
τ
(
x
)
∈
F
0
with
x
∈
F
, then Ψ(
σ
)
y
=Ψ(
σ
)
τx
=
(
x
τ
(
x
y
.] Now Ψ(
σ
1
σ
2
)
τ
(
x
1
σ
2
(
x
) and
Ψ(
σ
1
)Ψ(
σ
2
)
τ
(
x
)=Ψ
(
σ
1
)
2
(
x
1
σ
2
(
x
), so Ψ is a group homomorphism. The
inverse of Ψ is given by Ψ
0
(
σ
0
)
τ
−
1
y
=
τ
−
1
σ
0
(
y
),
σ
0
∈
G
0
,
y
∈
E
0
. To see this, we
compute
Ψ
0
(Ψ(
σ
))
τ
−
1
y
=
τ
−
1
Ψ(
σ
)
y
=
τ
−
1
Ψ(
σ
)
=
τ
−
1
(
x
σ
(
x
σ
(
τ
−
1
y
)
.
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This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.
 Spring '11
 sdd
 Subtraction

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