Since every field of characteristic 0 contains the rationals and is therefore infinite, all finite fields have prime characteristic.
If p is a prime, the integers modulo p form a field with p elements, denoted by Z_{p}, F_{p} or GF(p). Every other field with p elements is isomorphic to this one.
If q = p^{n} is a prime power, then there exists up to isomorphism exactly one field with q elements, written as F_{q} or GF(q). It can be constructed as follows: find an irreducible polynomial f(T) of degree n with coefficients in GF(p), then define GF(q) = GF(p)[T] / (f(T)). Here, GF(p)[T] denotes the ring of all polynomials with coefficients in GF(p), and the quotient is meant in the sense of factor rings. The polynomial f(T) can be found by factoring the polynomial T^{ q}T over GF(p). The field GF(q) contains GF(p) as a subfield.
There are no other finite fields.
The polynomial f(T) = T^{ 2} + T + 1 is irreducible over GF(2), and GF(4) can therefore be written as the set {0, 1, t, t+1} where the multiplication is defined (modularly) by t^{2} + t + 1 = 0. For example, to determine t^{3}, note that t(t^{2} + t + 1) = 0; so t^{3} + t^{2} + t = 0, and thus t^{3} + t^{2} + t + 1 = 1, so t^{3} = 1. Similarly, since the characteristic of the field is 2, t^{2} = t + 1.
In order to find the multiplicative inverse of t in this field, we have to find a polynomial p(T) such that T * p(T) = 1 modulo T^{ 2} + T + 1. The polynomial p(T) = T + 1 works, and hence 1/t = t + 1. Note that the field GF(4) is completely unrelated to the ring Z_{4} of integers modulo 4.
To construct the field GF(27), we start with the irreducible polynomial T^{ 3} + T^{ 2} + T  1 over GF(3). We then have GF(27) = {at^{2} + bt + c : a, b, c in GF(3)}, where the multiplication is defined by t^{ 3} + t^{ 2} + t  1 = 0.
If F is a finite field with q = p^{n} elements (where p is prime), then x^{q} = x for all x in F. Furthermore, the Frobenius homomorphism f : F > F defined by f(x) = x^{p} is bijective, and is therefore an automorphism. The Frobenius homomorphism has order n, and the cyclic group it generates is the full group of automorphisms of the field.
The field GF(p^{m}) contains a copy of GF(p^{n}) if and only if n divides m. The reason for this is that there exist irreducible polynomials of every degree over GF(p^{n}).
The multiplicative group of every finite field is cyclic, a special case of a theorem mentioned in the article about fields. This means that if F is a finite field with q elements, then there always exists an element x in F such that F = { 0, 1, x, x^{2}, ..., x^{q2} }.
The element x is not unique. If we fix one, then for any nonzero element a in F_{q}, there is a unique integer n in {0, ..., q  2} such that a = x^{n}. The value of n for a given a is called the discrete log of a (in the given field, to base x). In practice, although calculating x^{n} is relatively trivial given n, finding n for a given a is (under current theories) a computationally difficult process, and so has many applications in cryptography.
Finite fields also find applications in coding theory[?]: many codes are constructed as subspaces of Vector spaces over finite fields.
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