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 | 1 | initial version |
If $p$ is a prime, then GF(p^n,'x') is obtained by computing $F_p / (f(x))$ where $f$ is a monic, irreducible polynomial of degree $n$ in $F_p[x]$. For $n=1$, you just get $F_p[x] / (x) \cong F_p$.
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 $F_p[x] / (f(x))$ where $f$ is a monic, irreducible polynomial of degree $n$ in $F_p[x]$. For $n=1$, you just get $F_p[x] / (x) \cong F_p$.
So, for any prime $p$, GF(p,'x') is [0,1,2,...,p-1].
