# 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 prime 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").

## Proofs of basic properties:

### Finite characteristic:

Let F be a finite field, then by the pigeonhole principle 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 divisors of 0, and hence N is prime.

### Existence 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 immediately get an isomorphism $\mathbb{F}_p\to F$ by mapping $\sum_{i=1}^N 1_{\mathbb{F}_p} \to \sum_{i=1}^N 1_F$ for all N.

### Existence - 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-zero 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 this polynomial.

## The Frobenius map

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