Math 5126
Monday, March 24
Eighth Homework Solutions
1. Write
F

0
=
{
a
1
,...,
a
q
}
, where
q
+
1
=

F

. Then
x
q

1
= (
x

a
1
)(
x

a
2
)
...
(
x

a
q
)
. The coefficient of
x
q

1
on the right hand side is

(
a
1
+
···
+
a
q
)
, whereas the
coefficient of
x
q

1
on the left hand side is 0 (because
q
≥
3). This shows that the sum
of all elements of
F
is zero.
Next consider the constant coefficient. We obtain

1
= (

1
)
q

1
a
1
a
2
...
a
q
. Note that
(

1
)
q

1
=
1, because if not, then
q
is even which means that
F
has characteristic 2
and then

1
=
1. Thus in all cases

1
=
a
1
a
2
...
a
q
.
Finally the sum of all products of pairs of elements of
F
is the square of the sum of
the elements of
F
and hence by the first part is zero.
2. Suppose there are a finite number of subfields between
F
and
K
. Then we may write
K
=
F
(
α
)
for some
α
∈
K
. Let
f
denote the minimal polynomial of
α
over
K
, a
polynomial of degree 8, and let
E
be a splitting field for
f
over
F
containing
K
.
Let
a
1
=
α
,
a
2
,...,
a
8
be the roots of
f
in
E
(there maybe repetitions). Let
L
be an
intermediate subfield, that is
F
⊆
L
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 PALinnell
 Math, Algebra, Galois group, minimal polynomial, Galois, Field extension, abelian Galois group

Click to edit the document details