Let

be irreducible in

.
If

for
some

, then

.
This is a contradiction to the fact that

is irreducible.

Let

and suppose

is reducible and

where

are monic of
degree

and

respectively. Let

be the splitting field of

over

and

be a root of

in

. Then

and

. Thus

is the only root of

.
Since

. Let

.
Let

so that

.
Hence

But

which is a contradiction.

2. Show that

is the largest perfect subfield of

3. Identify the finite fields

and

More generally we show that if

,then

Since

. Hence

Hence

is surjective. If

then. Thus

and

Thus

. Now put

and

to see that

and

4. Find a necessary and sufficient condition on

and

so
that

is a subfield of

5. Draw the poset of subfields of

Divisors of

are

Thus the poset of lattices of the subfields are:

6. Show that the order of the Frobenius automorphism

is

Note that

for all

Let

denote the order in the group of all

-automorphisms
of

Then

.
Suppose

.Then

for all . But

has only

roots.
This is a contradiction. Hence

.

is a splitting field of

over

any embedding

is an

-automorphism of

But

is perfect and so

is
a separable extension of

Therefore the number of

-automorphisms of

is

But

Hence
all the roots of

are obtained by applying

to

successively. Hence the roots are

where

8. Let

and

be subfields of

having

and

elements respectively. How many elements
does the field

have ?

As

for some

and. Thus

On the
other hand

.
Hence

Hence we have the following
diagram of subfields.