Galois correspondence with the following field extensions I've to find the Galois correspondence between the subgroups of $Gal(E,K)$ and the intermediate fields of $E$, where  $K=\mathbb{F}_7$ and $E=K[\alpha]$ and $\alpha$ is a root of the polynomial $x^9-1$.
So far, what I've done is:
We can factorice the polynomial as
$x^9-1=(x-1)(x^2+x+1)(x^3-2)(x^3-4)$
over $\mathbb{F}_7$.
We observe that $1$ is root of $x-1$,  $2$ and $4$ are the roots of $x^2+x+1$. So, $\alpha$ is a root of $(x^3-2)(x^3-4)$. 
If $\alpha$ is a root $(x^3−2)$ then $[E:K]=3$ since $(x^3-2)$ is the minimal polynomial of $\alpha$ over $\mathbb{F}_7$.
I know that if $\sigma \in Gal(E/K)$, then $\sigma$ is completely determined by its action on $\alpha$ and that $\sigma(\alpha)$ is a root of $(x^3-2)$.
So, I would like to know how the roots of this polynomial look. I also know that 
$E=\lbrace a_2 \alpha ^2+ a_1 \alpha +a_0 | a_2,a_1,a_0 \in \mathbb{F}_7 \rbrace $
since $[E:K]=3$. I suppose that is analogous for int the case that $\alpha$ is a root of $(x^3-4)$.
Thanks in advance for any help.
 A: In this case the best thing is to note that if you have a cubic root of $2$ and a cubic root of $1$, their product is again a cubic root of $2$ (in a similar way of affine and linear associated system or in linear differential equations). So, once you ``created'' $\alpha$, it's sufficient to note that $X_3=\{1,2,4\}$ is the set of cubic root of unity. Then the roots of $x^3-2$ is precisely $\alpha X=\{\alpha,2\alpha,4\alpha\}$.
Another way (perhaps in other situations) could be to view $\alpha$ more as $\mu_\alpha:\mathbb{F}_7[\alpha]\longrightarrow\mathbb{F}_7[\alpha]$ defining $\mu_\alpha(a):=\alpha a$ (this gives you an immersion $\mathbb{F}_7[\alpha]\hookrightarrow M_{3\times 3}(\mathbb{F}_7)$
Perhaps having the more concrete matrices could be more pratical sometimes. A third, intermediate, solution coul be to use simply the relation $\alpha^3=2$ and solve the equation $x^3=2$ on elements of form (as you noticed) $a+b\alpha+c\alpha^2$. Hope this was useful.
A: I suppose you meant to write “Galois correspondence between the subgroups of the Galois group and the intermediate fields”.
But as you seem to have realized, the field extension $E\supset K=\Bbb F_7$ is of degree three and normal, so there are no fields strictly between $E$ and $K$. I guess it should be pointed out that the Galois group is cyclic of order three, generated by the Frobenius automorphism $\sigma$, for which $\forall x\in E$, $\sigma(x)=x^7$.
