Let

*K* be a

finite field.

Recall
that

*K* has

*p*^{n}
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

automorphisms 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
*p*^{r} 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 **Z**_{p}
as a subfield. Notice that if *f:K-->K* is an automorphism
then it must be that *f* is the identity on
**Z**_{p}. For *f(1)=1* and *f(0)=0* and the
elements of **Z**_{p} are obtained by adding 1s together:
*0,1,1+1,1+1+1,...,p-1* Thus any automorphism of *K* is
actually a **Z**_{p}-automorphism. It follows that
the automorphism group of *K* is the same as
Gal(*K*/**Z**_{p}).

I claim that *K* is a *Galois* extension of **Z**_{p}.
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)=x*^{pn}-x over **Z**_{p}.
This polynomial is separable
because its derivative is -1.

Thus we can apply Galois theory. It follows that
Gal(*K*/**Z**_{p}) has order *n* since
this is the dimension of *K* as a vector space over
**Z**_{p}. 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 *F*^{n}=1.
Suppose that *F* has an order strictly smaller than this *d*, say. Then
*a*^{pd}=a for each *a* in *K*.
But this means that the elements of *K* are all zeroes of a polynomial
with degree *p*^{d}. In particular *K* can have
at most *p*^{d} 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 *K*^{G} for the
subgroups *G* of Gal(*K*/**Z**_{p})=*<F>*.
and that *[K:**K*^{G}]=|G| (*).

The subroups of *<F>* are precisely *<F*^{d}>
for *d* a divisor of *n*. By (*) the corresponding fixed field
has dimension *r* over **Z**_{p} and hence
has *p*^{r} elements.