I think I have a solution for this, but I have doubts for its validity:
We immediately see that $F(\alpha^2)\subseteq F(\alpha)$. Well, in the language of Galois theory, $F(\alpha^2)$ is either an intermediate field or is equal to $F(\alpha)$.
Suppose $F(\alpha^2)$ is an intermediate field in the lattice for the splitting field of $p(x)$. This means there exists a subgroup of automorphisms in $Gal(F, F[x]/(p(x))$ such that $F(\alpha^2)$ is fixed. Let $\sigma$ be one such automorphism. Then, as $F(\alpha^2)$ is fixed, $\sigma(\alpha^2)=\sigma(\alpha)\sigma(\alpha)=\alpha^2$, so $\sigma(\alpha)=\pm \alpha$. If $\sigma(\alpha)=\alpha$, then $\alpha$ would also be fixed,
However, the roots of $p(x)$ must be $\alpha, \omega\alpha, \omega^2\alpha$, where $\omega$ is the traditional cube root of unity. Since elements of the Galois group must send roots of the polynomial to other roots, $\sigma(\alpha)\not=-\alpha$, so $\alpha$ must be fixed by all automorphisms that fix $\alpha^2$, and $F(\alpha)=F(\alpha^2)$.
My doubts are in a few parts: I did not use the characteristic 0 requirement. Is this just necessary background machinery for the Galois theorems? Also, my claim that elements of the Galois group may only permute roots of that splitting field, is that always true (in the case of the splitting field of the polynomial)?
Best Answer
You took it a bit too far, this problem doesn't require any use of automorphisms. We have $[F(\alpha):F]=3$. By the tower rule:
$3=[F(\alpha):F]=[F(\alpha):F(\alpha^2)]\cdot [F(\alpha^2):F].$
And so either $F(\alpha^2)=F(\alpha)$ or $F(\alpha^2)=F$. Assume the latter is true. Then $\alpha^2\in F$, and hence $\alpha$ is a root of the polynomial $x^2-\alpha^2\in F[x]$. But this is a contradiction, because the minimal polynomial of $\alpha$ over $F$ has degree $3$. So we must have $F(\alpha^2)=F(\alpha)$. The characteristic of $F$ doesn't matter.