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+xzIf 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) + xWe 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.

**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 p^{q} 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

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 {b_{0} … b_{q−1}} where b_{0} = 1.
Every combination of q values from GF(p): a_{i} for 0≤i<q yields a distinct value for Σ[0≤i<q]a_{i}b_{i}.
We have p^{q} elements in the vector space and thus in F.
QED.

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