This is intended to follow in the style of my note on vectors. I find the proofs for the interesting theorems are curious, interesting and not too difficult.

The Field

A field is a algebraic structure with two binary operators, + and juxtaposition. Addition (x+y) is an Abelian (commutative) group over all the field elements and the additive identity is 0. Multiplication (xy) is an Abelian (commutative) group over all the field elements except 0 and the multiplicative identity is 1.

Recapitulating axiomatically a field is defined by the following: 

1  x+y = y+x
2  x+(y+z) = (x+y)+z
3  ∀x∀y∃z (x+z = y)
4  x+0 = x
5  xy = yx
6  x(yz) = (xy)z
7  ∀x(x = 0 ∨ ∀y∃z (xz = y))
8  1x = x
9  x(y+z) = xy+xz
If n is a non-negative integer we define n*x as x+x+ … x with n terms. Axiomatically: 
0*x = 0
(n+1)*x = (n*x) + x
We take −x to denote the additive inverse of x, (as in ordinary arithmetic) and extend the definition of * thus: 
(−n)*x = −(n*x)
n*x is thus defined for all integers n.

The Finite Field

Suppose we have a field F known to be finite. What can we deduce?

Theorem: There is an integer n>0 such that n*1 = 0.
Proof: Suppose that z is the first field element to repeat in the sequence 0*1, 1*1, 2*1, … (Since F is finite there must be repetitions.) Suppose that n*1 is that element. We want to show that n*1 = 0. By hypothesis n*1 is a repeat and we thus have for some m, 0≤m<n and m*1 = n*1. Suppose that m > 0. The mth and nth elements of the sequence are the same but they must have the same predecessors since the predecessor of z may be computed as z−1. This violates the assumption that n*1 was the first repeat. This is a contradiction and thus a reduction ad absurdum of the negation of the hypothesis m > 0. We conclude that m=0. Element n of the sequence is n*1 and is the same as element 0 which is 0. Thus n*1 = 0. QED.

This proof is tedious and depends on several simple rules that can be inductively derived from the axioms such as (m+n)*x = m*x + m*x.

We now show that n is prime. Otherwise n = pq and p > 1 and q > 1 and (p*1)(q*1) = (pq*1) = n*1 = 0. But in fields xy = 0 implies that x=0 or y=0 which would entail p*1 = 0 or q*1 = 0 but either of those would have violated the hypothesis that n*1 was the first repeat in the sequence. We conclude that n is prime.

n from this discovery is called the characteristic of the field. At this point we have discovered a sub-field of F that is isomorphic to the field integers modulo n, written GF(n). n is prime and we rename it p henceforth.

Theorem: For some prime p and some positive integer q, F has pq elements.

We invoke some vector space theory. Here are the pieces of the vector space:

field over which the vector space is defined:
GF(p) (embodied as integers n, 0≤n<p).
vectors of the vector space:
elements of F.
addition of vectors:
addition of F.
multiplication between GF(p) element and vector:
*, as defined above.
Origin of vector space:
0 in F
Since F obeys the vector space axioms the elements of F can be seen to form a vector space over the field GF(p).

We build a basis for the vector space. 1 from F is the first basis element. {1} spans a sub vector space isomorphic to GF(p). We continue to choose basis elements while we can and since the elements of F are a finite set this terminates yielding a basis set {b0 … bq−1} where b0 = 1. Every combination of q values from GF(p): ai for 0≤i<q yields a distinct value for Σ[0≤i<q]aibi. We have pq elements in the vector space and thus in F. QED.

We have not considered the behavior of the multiplication in F.