If $\mathrm{GF}(2^6) = \mathbb{Z}_2 [x]/ (x^6 + x + 1)$ and $u$ is a generator with $o(u)= 63$ how do I show $\mathrm{GF}(2^3)=\{0,1, u^9, u^{18}, u^{27}, u^{36}, u^{45}, u^{54}\}$?
Here's what I came up with: Since $\mathrm{GF}(2^3)$ is the set of roots of $x^8-x$, if the set were equal to $\mathrm{GF}(2^3)$ each element would satisfy $f(x)=x^8-x=0$.
$f(u^9) = (u^9)^{8} -u^9 = u^9-u^9 =0 $
$f(u^{18}) = (u^{18})^8 -u^{18} = u^{18}-u^9 =0$
etc. Is that sufficient for $\mathrm{GF}(2^3)=\{0,1, u^9, u^{18} ,u^{27}, u^{36}, u^{45}, u^{54}\}$ or is there something else to show? Maybe something with $\mathbb{Z}_2 [x]/ (x^6 + x + 1)$? In general, if a set contains all roots of $x^{p^q}-x$ is the set always equal to $\mathrm{GF}(p^q)$?