Galois Group over Finite Field
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?
An extension of finite fields is always cyclic: the Galois group must be cyclic. So the Galois group certainly cannot be $V_4$.
Note that $F_{11}(i)$ does have a square root of $2$: $(3i)^2 = -9\equiv 2\pmod{11}$. So once you adjoint $i$ to $F_{11}$, you also get $\sqrt{2}$. Thus, $F_{11}(i,\sqrt{2}) = F_{11}(i)$.
Likewise, $X^2+1$ is reducible over $F_{11}(\sqrt{2})$, since $(4\sqrt{2})^2 = 32\equiv -1\pmod{11}$. That is, $F_{11}(i) = F_{11}(\sqrt{2})$.
The key to remember is that there is a unique field of order $121=11^2$; so any extension you get from $F_{11}$ by adjoining the square root of a nonquadratic residue is the same. So $F_{11}(i) = F_{11}(\sqrt{2}) = F_{11}(\sqrt{6}) = F_{11}(\sqrt{7}) = F_{11}(\sqrt{8})$.
Your other mistake is that $x^4+1$ is not irreducible over $F_{11}$: it splits as a product of two irreducible quadratics. You can figure this out by replacing $\frac{\sqrt{2}}{2}$ with $7i$ and $\frac{\sqrt{2}}{2}i$ with $-7$ (the values in $F_{121}$), or by solving a system of simple equations. Either way, you get $$x^4 + 1 = (x^2-3x-1)(x^2+3x-1)$$ in $F_{11}$. (Had $x^4+1$ been irreducible, then your extension would have been of degree $4$: you need an extension of degree $4$ to get a root for $x^4+1$, that one already contains the quadratic extension, and so you would get all the roots you need. In that case, the Galois group would have been cyclic of order $4$).
Your mistake was in assuming that $x^4 + 1$ was irreducible over $\Bbb{F}_{11}$. In fact there is something incredible about this polynomial; it is irreducible over $\Bbb{Z}$ but reducible over $\Bbb{F}_p$ for every prime $p$!
The following is the proof given in Dummit and Foote:
If $p = 2$ the polynomial is clearly reducible by the schoolboy binomial theorem. Now if $p$ is odd, notice that
$$p^2 \equiv 1 \mod 8$$
for every prime $p$. This is because mod 8, $p$ is congruent to 1,3,5, or 7 all of which square to 1 mod 8. It follows that we have the following divisibilities:
$$x^4 + 1 | x^8 - 1 | x^{p^2 - 1} - 1| x^{p^2} - x.$$
In particular we conclude that $x^4 + 1 | x^{p^2} - x$ and so all the roots of $x^4 + 1$ are roots of $x^{p^2} - x$. It follows that the extension generated by any root of $x^4 + 1$ is at most a degree 2 extension, and so $x^4 + 1$ cannot be irreducible.
$\hspace{6in} \square$
It follows that the Galois group of $x^8 - 1$ over $\Bbb{F}_{11}$ is isomorphic to the cyclic group of order 2.
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 $(\mathbb{Z}/n\mathbb{Z})^*$.