We apply the above results to proving that
every polynomial has a root in some (extension) field.
proof:
We may assume that
does not
have a linear factor in
since otherwise that factor
would yield the desired root (in
, which is
an extension field of itself!).
Let
denote a factor of
which is irreducible.
Let
. This is an extension field of
. The representative
of the element
satisfies
, so
.
Example 2.6.12
If
and
then
has a root
in
(which we see upon identifying
with
).
Exercise 2.6.13
Construct an extension field of
in
which
has a zero.
Exercise 2.6.14
Show that the smallest extension field of
in
which
has a zero
is isomorphic to
the smallest extension field of
in
which
has a zero.
(Hint:
is the ``reciprocal'' of
:
.)
Exercise 2.6.15
In Example 2.6.7, show that
.
Exercise 2.6.16
In Example 2.6.8,
show that
.
Exercise 2.6.17
Let
. Let
.
Write down the addition and multiplication tables for
.
Show that
is a field.
Exercise 2.6.18
Let
. Let
.
Show that
is isomorphic to the field in
Example 2.1.19.
Exercise 2.6.19
State and prove a variant of Euler's Theorem 1.8.4.
for polynomials in
, where
is in
and
is a finite field. (Hint: What's the analog for Euler's
function for polynomials?).
Exercise 2.6.20
Let
and
in
.
Compute
in
and in
.
[Hint: one of
and
is irreducible, the other isn't.
Use Exercise 2.6.19.]
Exercise 2.6.21
In high school algebra you learned how to ``rationalize the
denominator"; for instance,
is
.
Express
in terms of
.
[Hint: What does this have to do with the inverse of
in
? (and you know how to find this quickly, right?)]
David Joyner
2007-09-03