I had an exam question which was calculate the splitting field of $X^{40}-1$ over $F_5$ and then give its' Galois Group.
Now I simplified the polynomial to products of polynomials of degree 1 to the power of 5 and times $X^4 + 1$, so it is sufficient to find the splitting field over $X^4 +1 $ right?
So we had a similiar question once and what we did is we took an element $\theta$ which satisfies $\theta^4 = -1$ (I'm not too sure why we can even find something like this) and then looked at $F_5(\theta) = {a+b \theta \; a,b \in F_5} $ (I'd be grateful if someone could also explain why $F_5(\theta)$ looks like this ).
So now we have a finite field with 25 elements so it is isomorphic to $F_{25}$ and its' Galois group should be of order 2 and it is generated by the Frobenius-Automorphism, so it is isomorphic to $C_2$ and maps $\theta$ to $-\theta$.
I'd be thankful if someone could check if what I did is right and answer my questions (why $\theta$ exists and why the extension is equal to $ a +b\theta$.
For the simplification of the polynomial if someone cares:
$ X^{40} -1 \\ =(X^{20} +1) \cdot (X^{20} -1)\\ =(X^4 +1)^5 \cdot (X^4 -1)^5 \\= (X^4 +1)^5 \cdot (X -1)^5 \cdot (X -2)^5 \cdot (X -3)^5 \cdot (X -4)^5$
The polynomial $X^4 + 1\in \mathbb{F}_5[X]$ is not irreducible: $(X^2 + 2)(X^2 + 3) = X^4 + 1\in \mathbb{F}_5[X]$, and both $X^2 + 2$ and $X^2 + 3$ are irreducible by the fact that $2, 3\in \mathbb{F}_5$ are not squares. Thus $K = \mathbb{F}_5[\theta]/(\theta^2 + 2)$ is a field extension of $\mathbb{F}_5$ of degree $\deg (X^2 + 2) = 2$, and $K$ is spanned as an $\mathbb{F}_5$-vector space by $1, \theta$. (For the latter, note that it's clearly spanned by $1, \theta, \theta^2, \dots$, but $\theta^2 = -2$ in $K$.) Furthermore, $\theta^4 = (-2)^2 = -1\in \mathbb{F}_5$. There's only one finite field of degree $q$ for each prime power $q$, so $K$ is isomorphic to $\mathbb{F}_{25}$.