Finite field: Difference between revisions

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hence there is some minimal natural number N such that <math>\sum_{i=1}^N 1_F = 0</math>. Since F is a field, it has no 0 divisors, and hence N is prime.
hence there is some minimal natural number N such that <math>\sum_{i=1}^N 1_F = 0</math>. Since F is a field, it has no 0 divisors, and hence N is prime.
===Existence and uniqueness of F<sub>p</sub>===
===Existence and uniqueness of F<sub>p</sub>===
To begin with it is follows by inspection that <math>\mathbb{F}_p</math> is a field. Furthermore, given any other field F' with p elements, one immidiately get an isomorphism <math>F\to F' </math> by mapping <math>\sum_{i=1}^N 1_F \to \sum_{i=1}^N 1_{F'}</math>.
To begin with it is follows by inspection that <math>\mathbb{F}_p</math> is a field. Furthermore, given any other field F' with p elements, one immediately get an isomorphism <math>F\to F' </math> by mapping <math>\sum_{i=1}^N 1_F \to \sum_{i=1}^N 1_{F'}</math>.
===Existence - general case===
===Existence - general case===
working over <math>\mathbb{F}_p</math>, let <math>f(x) := x^{p^n}-x</math>. Let F be the splitting field of f over <math>\mathbb{F}_p</math>.
working over <math>\mathbb{F}_p</math>, let <math>f(x) := x^{p^n}-x</math>. Let F be the splitting field of f over <math>\mathbb{F}_p</math>.

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A finite field is a field with a finite number of elements; e,g, the fields (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 or (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 . hence there is some minimal natural number N such that . 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 is a field. Furthermore, given any other field F' with p elements, one immediately get an isomorphism by mapping .

Existence - general case

working over , let . Let F be the splitting field of f over . 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 ; i.e. it contains a copy of . Hence, F is a vector field of finite dimension over . Moreover since the non 0 elements of F form a group, they are all roots of the polynomial ; hence the elements of F are all roots of f.

The Frobenius map

Let F be a field of characteritic p, then the map is the generator of the Galois group .