automorphisms of finite fields

Let K be a finite field. Recall that K has pⁿ 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 Fⁿ=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.