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").

Proofs of basic properties:

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 F_p

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 $Gal(F/\mathbb {F} _{p})$.