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]`

.

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