be a finite field
elements, for a prime p
and a positive integer n
and that there is exactly one such field
up to isomorphism
We are going to compute the automorphism
s of K
and as a corollary
we will use Galois theory
to compute the subfields of K
Theorem The automorphism group of K
is cyclic of order n generated by the Frobenius endomorphism.
Corollary The subfields of K are exactly the fields with
pr elements for r a divisor of n.
Proof of the theorem:
First let's recall a few facts about K that we'll need in the
proof. Firstly K has Zp
as a subfield. Notice that if f:K-->K is an automorphism
then it must be that f is the identity on
Zp. For f(1)=1 and f(0)=0 and the
elements of Zp are obtained by adding 1s together:
0,1,1+1,1+1+1,...,p-1 Thus any automorphism of K is
actually a Zp-automorphism. It follows that
the automorphism group of K is the same as
I claim that K is a Galois extension of Zp.
Well it was shown in All finite fields are isomorphic to GF(p^n)
that it is a splitting field of the polynomial
h(x)=xpn-x over Zp.
This polynomial is separable
because its derivative is -1.
Thus we can apply Galois theory. It follows that
Gal(K/Zp) has order n since
this is the dimension of K as a vector space over
Zp. Now consider the Frobenius endomorphism
F:K-->K. We know that this is an automorphism of F.
We have to show that it has order n. Since every element of
K is a zero of h(x) it follows that Fn=1.
Suppose that F has an order strictly smaller than this d, say. Then
apd=a for each a in K.
But this means that the elements of K are all zeroes of a polynomial
with degree pd. In particular K can have
at most pd elements, a contradiction. Thus
the Frobenius has the correct order and the group is cyclic as required.
Proof of the corollary: From Galois theory we know that
the subfields of K are exactly KG for the
subgroups G of Gal(K/Zp)=<F>.
and that [K:KG]=|G| (*).
The subroups of <F> are precisely <Fd>
for d a divisor of n. By (*) the corresponding fixed field
has dimension r over Zp and hence
has pr elements.