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answered 12 years ago

calc314 gravatar image

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].

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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].