- #1
PsychonautQQ
- 784
- 10
Theorem: Let F be any field. If G is a finite subgroup of the multiplicative group F* of F, then G is cyclic. In particular, is F is finite, then F* is cyclic.
Corolarry 1: GF(p^n) = Z_p(u), where u is any primitive element for GF(p^n).
So <u> = GF(p^n)*, so |u| = GF(p^n) - 1.
I'm now trying to imagine what Z_p(u) would look like, maybe:
{a_0 + a_1*u + a_2*u^2 + ... a_(n-1)*u^n-1 | a_i are elements of Z_p, u^n = (0?)}
This would make sense because this field would have order n... but it would also mean that u^n = (0?) or something, when u^n should just equal u^n because |u| = GF(p^n) - 1 > n.
Anyone understand my dilemma? If anyone could drop some knowledge on this topic in general it'd be appreciated.
Corolarry 1: GF(p^n) = Z_p(u), where u is any primitive element for GF(p^n).
So <u> = GF(p^n)*, so |u| = GF(p^n) - 1.
I'm now trying to imagine what Z_p(u) would look like, maybe:
{a_0 + a_1*u + a_2*u^2 + ... a_(n-1)*u^n-1 | a_i are elements of Z_p, u^n = (0?)}
This would make sense because this field would have order n... but it would also mean that u^n = (0?) or something, when u^n should just equal u^n because |u| = GF(p^n) - 1 > n.
Anyone understand my dilemma? If anyone could drop some knowledge on this topic in general it'd be appreciated.