![]() F24 with Polynomial
Representation The elements of F24 are the 16
vectors:
The irreducible polynomial used will be f(x) =
x4 + x + 1. The following are sample
calculations. Addition (0110) + (0101) = (0011). Multiplication (1101) (1001) = (x3 + x2 + 1) (x3 + 1) mod f(x) = x6 + x5 + 2x3 + x2 + 1 mod f(x) = x6 + x5 + x2 + 1 mod f(x) (coefficients are reduced modulo 2) = ( x4 + x + 1)(x2 + x) + (x3 + x2 + x + 1) mod f(x) = x3 + x2 + x + 1 = (1111). Exponentiation To compute (0010)5, first find (0010)2 = (0010) (0010) = x x mod f(x) = ( x4 + x + 1)(0) + (x2) mod f(x) = x2 = (0100). Then (0010)4 = (0010)2 (0010)2 = (0100) (0100) = x2 x2 mod f(x) = ( x4 + x + 1)(1) + (x + 1) mod f(x) = x + 1 = (0011). Finally, (0010)5 = (0010)4 (0010) = (0011) (0010) = (x + 1) (x) mod f(x) = (x2 + x) mod f(x) = ( x4 + x + 1)(0) + (x2 + x) mod f(x) = x2 + x = (0110). Multiplicative Inversion The element g = (0010) is a generator for the field. The powers of g are:
The multiplicative identity for the field is
g0 = (0001). The multiplicative inverse of g7 = (1011) is
g-7 mod 15 = g8 mod 15 =
(0101). To verify this, see that (1011) (0101) = (x3 + x + 1) (x2 + 1) mod f(x) = x5 + x2 + x + 1 mod f(x) = ( x4 + x + 1)(x) + (1) mod f(x) = 1 = (0001), which is the multiplicative identity. |