If p is a prime, then GF(p^n,'x')
is obtained by computing Fp/(f(x)) where f is a monic, irreducible polynomial of degree n in Fp[x]. For n=1, you just get Fp[x]/(x)≅Fp.
So, for any prime p, GF(p,'x')
is [0,1,2,...,p-1]
.
![]() | 2 | No.2 Revision |
If p is a prime, then GF(p^n,'x')
is obtained by computing $F_p $F_p[x] / (f(x))wherefisamonic,irreduciblepolynomialofdegreeninF_p[x].Forn=1,youjustgetF_p[x] / (x) \cong F_p$.
So, for any prime p, GF(p,'x')
is [0,1,2,...,p-1]
.