TABLE OF CONTENTS We list the following result without proof.
Suppose that F is a finite extension field of GF( p) that contains all the zeros of 
. Then these zeros form a finite field of order p n .
Let F be the finite field with q = p n elements, where p is prime.
F = GF( p n) contains a subfield GF( p m) if and only if m is a positive divisor of n.
If ? ? GF( p n) then ? ? GF( p m) if and only if 
.
To prove Theorem 3.10, we need the following lemma.
If a, s, t are integers with a ? 2, s, t ? 1, then
(Recall that the vertical bar means divides. )
Proof. We write t = qs + r, where 0 ? r < s. Then
Since a qs 1 = ( a s 1)( a ( q 1) s + + a s + 1), a qs 1 is always divisible by a s 1. The last term is less than 1, and so it is an integer if and only if r = 0.
Proof of Theorem 3.10 . (a) If m n, from Fact 3.5, then GF(
