Finite field

In abstract algebra, a finite field or Galois field is a field which contains only finitely many elements. Finite fields are important in cryptography and [coding theory]?. The finite fields are completely known, as will be described below.

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). Ever other field with p elements is isomorphic to this one.

If q = pⁿ 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.

Examples

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 by t² + t + 1 = 0. 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 Z₄ 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) = {at² + bt + c : a, b, c in GF(3)}, where the multiplication is defined by t ³ + t ² + t + 1 = 0.

Properties and facts

If F is a finite field with q = pⁿ 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.

The field GF(p^m) contains a copy of GF(pⁿ) if and only if n divides m. The reason for this is that there exist irreducible polynomials of every degree over GF(pⁿ).

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, x², ..., x^q-2 }.

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