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If p is a prime, the integers modulo p form a field with p elements, denoted by Zp or Fp. Ever other field with p elements is isomorphic to this one. |
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If p is a prime, the integers modulo p form a field with p elements, denoted by Zp, Fp or GF(p). Ever other field with p elements is isomorphic to this one. |
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If q = pn is a prime power, then there exists up to isomorphism exactly one field with q elements, written as Fq or GF(q). It can be constructed as follows: find an [irreducible polynomial]? f(T) of degree n with coefficients in Fp, then define Fq = Fp[T] / (f(T)). Here, Fp[T] denotes the ring of all polynomials with coefficients in Fp, 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 Fp. The field Fq contains Fp as a subfield. |
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If q = pn is a prime power, then there exists up to isomorphism exactly one field with q elements, written as Fq 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. |
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The polynomial f(T) = T 2 + T + 1 is irreducible over F2, and F4 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 F4 is completely unrelated to the ring Z4 of integers modulo 4. |
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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. |
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To construct the field F27, we start with the irreducible polynomial T 3 + T 2 + T + 1 over F3. We then have F27 = {at2 + bt + c : a, b, c in F3}, where the multiplication is defined by t 3 + t 2 + t + 1. |
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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. |
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The field Fpm contains a copy of Fpn if and only if n divides m. The reason for this is that there exist irreducible polynomials of every degree over Fpn. |
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The field GF(pm) 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). |
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 Zp, Fp 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 Fq 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 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.
If F is a finite field with q = pn elements (where p is prime), then xq = x for all x in F. Furthermore, the Frobenius homomorphism f : F -> F defined by f(x) = xp is bijective.
The field GF(pm) 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, ..., xq-2 }.
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