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Yeah, that is how I got that. | Yeah, that is how I got that. | ||
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Latest revision as of 04:34, 11 December 2008
I think this is right...
Say $ F = Z_2[x]/<x^5 + x^3 + 1> $
Then F = {$ ax^4 + bx^3 + cx^2 + dx + e $ | a, b, c, d, e in $ Z_2 $}
So |F| = 32 and |F*| = 31
This means F* is isomorphic to $ Z_{31} $. Since 31 is prime, all non-identity elements in $ Z_{31} $ are generators. This means that all non-identity elements in F* are generators. So, x is a generator of F*
Okay, but how do you know that |F|=32? is it because we are talking about $ Z_2 $ and our polynomial is degree 5, so $ |F|=2^5 $?
Yeah, that is how I got that.
|F| = 32 b/c 2^5
