# Finite field

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A finite field is a field with a finite number of elements; e,g, the fields ${\displaystyle \mathbb {F} _{p}:=\mathbb {Z} /(p)}$ (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 ${\displaystyle \mathbb {F} _{p^{n}}}$ or ${\displaystyle GF_{p^{n}}}$ (where GF stands for "Galois field").

## Proofs of basic properties:

### Finite characteristic:

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

### Existence and uniqueness of Fp

To begin with it is follows by inspection that ${\displaystyle \mathbb {F} _{p}}$ is a field. Furthermore, given any other field F' with p elements, one immediately get an isomorphism ${\displaystyle F\to F'}$ by mapping ${\displaystyle \sum _{i=1}^{N}1_{F}\to \sum _{i=1}^{N}1_{F'}}$.

### Existence - general case

working over ${\displaystyle \mathbb {F} _{p}}$, let ${\displaystyle f(x):=x^{p^{n}}-x}$. Let F be the splitting field of f over ${\displaystyle \mathbb {F} _{p}}$. 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 ${\displaystyle 0_{F},1_{F}....\sum _{i=1}^{p-1}1_{F}}$; i.e. it contains a copy of ${\displaystyle \mathbb {F} _{p}}$. Hence, F is a vector field of finite dimension over ${\displaystyle \mathbb {F} _{p}}$. Moreover since the non 0 elements of F form a group, they are all roots of the polynomial ${\displaystyle x^{p^{n}-1}-1}$; hence the elements of F are all roots of f.

## The Frobenius map

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