Finite field

A finite field is a field with a finite number of elements; e,g, the fields $$\mathbb{F}_p := \mathbb{Z}/(p)$$ (with the addition and multiplication induced from the same operations on the integers). For any primes number p, and natural number n, there exists a unique finite field with pn elements; this field is denoted by $$\mathbb{F}_{p^n}$$ or $$GF_{p^n}$$ (where GF stands for "Galois field").

Finite characteristic:
Let F be a finite field, then by the piegonhole peinciple there are two different natural numbers number n,m such that $$\sum_{i=1}^n 1_F = \sum_{i=1}^m 1_F$$. hence there is some minimal natural number N such that $$\sum_{i=1}^N 1_F = 0$$. Since F is a field, it has no 0 divisors, and hence N is prime.

Existance and uniqueness of Fp
To begin with it is follows by inspection that $$\mathbb{F}_p$$ is a field. Furthermore, given any other field F' with p elements, one immidiately get an isomorphism $$F\to F' $$ by mapping $$\sum_{i=1}^N 1_F \to \sum_{i=1}^N 1_{F'}$$.

Existeance - general case
working over $$\mathbb{F}_p$$, let $$f(x) := x^{p^n}-x$$. Let F be the splitting field of f over $$\mathbb{F}_p$$. Note that f' = -1, and hence the gcd of f,f' is 1, and all the roots of f in F are distinct. Furthermore, note that the set of roots of f is closed under addition and multiplication; hence F is simply the set of roots of f.

Uniqueness - general case
Let F be a finite field of characteristic p, then it contains $$0_F,1_F....\sum_{i=1}^{p-1}1_F$$; i.e. it contains a copy of $$\mathbb{F}_p$$. Hence, F is a vector field of finite dimension over $$\mathbb{F}_p$$. Moreover since the non 0 elements of F form a group, they are all roots of the polynomial $$x^{p^n-1}-1$$; hence the elements of F are all roots of f.

The Frobenius map
Let F be a field of characteritic p, then the map $$x\mapsto x^p$$ is the generator of the Galois group $$Gal(F/\mathbb{F}_p)$$.