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Polynomial representation of GF(7)?

asked 2012-06-12 03:38:30 -0600

bk322 gravatar image

updated 2012-06-12 03:41:03 -0600

Why sage would give me polynomial representation of GF(8), but not GF(7)?

sage: G = GF(8, 'x')
sage: G.list()
[0, x, x^2, x + 1, x^2 + x, x^2 + x + 1, x^2 + 1, 1]
sage: G = GF(7, 'x')
sage: G.list()
[0, 1, 2, 3, 4, 5, 6]

Maybe there's no such thing as polynomial represenation of GF(7)?

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answered 2012-06-12 05:11:47 -0600

calc314 gravatar image

updated 2012-06-12 05:13:27 -0600

If $p$ is a prime, then GF(p^n,'x') is obtained by computing $F_p[x] / (f(x))$ 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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Oh I see - so it's just because `n=1`. Thank You for Your answer.

bk322 gravatar imagebk322 ( 2012-06-12 05:47:52 -0600 )edit

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Asked: 2012-06-12 03:38:30 -0600

Seen: 63 times

Last updated: Jun 12 '12