Let $gcd(n,q) = 1$
I'm trying to get to grips with factorising the polynomial $x^n - 1$ over $\mathbb{F}_q$. Firstly, it is a good idea to find an extension field containing all the roots of $x^n - 1$; this is always possible since splitting fields always exist.
Let $\mathbb{F}_q$ be a finite field and define $ord_n(q)$ to be the smallest positive integer $t$ such that $q^t \equiv 1 mod n$.
Then $\mathbb{F}_{q^t}$ is the splitting field for $x^n - 1$ over $\mathbb{F}_q$, which contains a primitive $n^{th}$ root of unity, $\alpha$.
So the irreducible factors of $x^n - 1$ must be the product of the distinct minimal polynomials of the $n^{th}$ roots of unities in $\mathbb{F}_{q^t}$.
I have two questions;
$(1)-$ Why is $\mathbb{F}_{q^t}$ the splitting field? What is so special about this $t = ord_n(q)$?
$(2)-$ Why are the irreducible factors of $x^n - 1$ the product of the distinct minimal polynomials?