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 | 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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==The Frobenius map == | ==The Frobenius map == | ||
Let F be a field of characteritic p, then the map <math>x\mapsto x^p</math> is the generator of the Galois group <math>Gal(F/\mathbb{F}_p)</math>. | Let F be a field of characteritic p, then the map <math>x\mapsto x^p</math> is the generator of the Galois group <math>Gal(F/\mathbb{F}_p)</math>.[[Category:Suggestion Bot Tag]] |
Latest revision as of 11:01, 16 August 2024
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 .