Math 5126
Monday, March 17
Seventh Homework Solutions
1. Section 14.2, Exercise 2 on page 581.
Determine the minimal polynomial over
Q
for the element 1
+
3
√
2
+
3
√
4.
Let
α
denote the above element. We find a Galois extension
K
containing
α
, and then
consider the conjugates containing
α
under the action of the Galois group Gal
(
K
/
Q
)
.
Thus here we take
K
=
Q
(
3
√
2
,
ω
)
, where
ω
is a primitive cube root of 1.
Then
Gal
(
K
/
Q
)
is generated by complex conjugation and the automorphism determined
by
3
√
2
→
ω
(
3
√
2
)
and
ω
→
ω
. Thus the conjugates of
α
under the Galois group are
1
+
3
√
2
+
3
√
4
,
1
+
ω
3
√
2
+
ω
2
3
√
4
,
1
+
ω
2
3
√
2
+
ω
3
√
4
.
This will yield a polynomial of degree 3. Since there are 3 conjugates under Gal
(
K
/
Q
)
,
the minimal polynomial must have degree at least 3, and thus the polynomial will be
the minimal polynomial. Explicitly, the polynomial is
(
x

1

3
√
2

3
√
4
)(
x

1

ω
3
√
2

ω
2
3
√
4
)(
x

1

ω
2
3
√
2

ω
3
√
4
) =
x
3

3
x
2

3
x

1
.
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 Fall '07
 PALinnell
 Math, Algebra, Group Theory, Galois group, Field extension

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