TITLE: Galois Group over Finite Field
QUESTION [10 upvotes]: I am having a bit of difficulty trying to answer the following question:

What is the Galois group of $X^8-1$ over $\mathbb{F}_{11}$?

So far I have factored $X^8-1$ as 
$$X^8-1=(X+10)(X+1)(X^2+1)(X^4+1).$$ 
I know $X^2+1$ is irreducible over $\mathbb{F}_{11}$ since $10$ is not a square modulo $11$. Also, $X^4+1$ is irreducible over $\mathbb{F}_{11}$. The roots of $X^2+1$ and $X^4+1$ over $\mathbb{Q}$ are $\pm i$ and $\pm \frac{\sqrt{2}}{2} \pm \frac{\sqrt{2}}{2} i$, respectively. We also see that $\sqrt{2} \not \in \mathbb{F}_{11}$ since no element squared is equal to $2$. I would then think that $\mathbb{F}_{11}(i, \sqrt{2})$ is a splitting field for $x^8-1$ over $\mathbb{F}_{11}$, which is clearly Galois. If all this were true, I would then venture that the Galois group is $V_4$. I have the feeling, however, that I have made many mistakes in my reasoning. How should one approach a problem like this?

REPLY [3 votes]: Let $E$ be a splitting field of $x^8 - 1$ over $F_{11}$. Then $E = F_{11}(\alpha)$, where $\alpha$ is a primitive $8$th root of unity. 
It follows that $G = \operatorname{Gal}(E/F)$ is isomorphic to a subgroup of $(\mathbb{Z}/8\mathbb{Z})^*$, which is isomorphic to the Klein four-group. Now $G$ cannot have order $1$ because $x^8 - 1$ does not split over $F_{11}$, it cannot have order $4$ because then it wouldn't be cyclic (see this question). Therefore $G$ must be cyclic of order $2$.
In general, suppose $F$ is a field and $E = F(\alpha)$, where $\alpha$ is a primitive $n$th root of unity. Then you can show that $\operatorname{Gal}(E/F)$ is isomorphic to a subgroup of $(\mathbb{Z}/n\mathbb{Z})^*$.