Showing $F_{64}/F_2$ is Normal I need to show that $F_{64}/F_2$ is normal. So far I generated $F_{64}$ by doing $F_2[x]/(x^6+x+1)$. But I'm not sure where to go from here. I think what I need to do is show that $F_{64}$ is a splitting field for $x^6+x+1$? Though I'm not sure why just showing it for this one polynomial is enough to show that the extension is normal. And I've been trying to factor it but I don't see how to go about that?
 A: During my qualifying oral exam, I was asked to construct a field of order $27$; so I wrote one polynomial and invoked a certain result, but was asked to do a more constructive construction that did not invoke the existence of splitting fields. I found an irreducible polynomial over $\mathbb{F}_3$ and did the usual construction. Great. Then I was given a different irreducible polynomial over $\mathbb{F}_3$ and was asked to prove the resulting extension was isomorphic to the one I had initially given. After 5 minutes of flailing to try to construct an explicit isomorphism, I was told to step back and look at the polynomial I had originally written down: $x^{27}-x$.

Basically: since a finite subgroup of the multiplicative group of a field (in fact, of an integral domain) must be cyclic, the multiplicative subgroup of $\mathbb{F}_{p^n}$ is cyclic of order $p^{n}-1$, and therefore every nonzero element satisfies the polynomial $x^{p^n-1}-1$. Therefore, the elements of $\mathbb{F}_{p^n}$ are all roots of $x^{p^n}-x$, and these are all the roots in an algebraic closure of $\mathbb{F}_{p}$. And no strictly smaller field has all the roots, because this polynomial is separable (its derivative is $-1$). Thus, $\mathbb{F}_{p^n}$ is the splitting field of $x^{p^n}-x$ over $\mathbb{F}_p$, and therefore it is normal (and unique up to isomorphism over $\mathbb{F}_p$).
