How to factor $x^k -1$?

Question

Is there any way to find the degrees of the irreducible factors of a polynomial $x^k - 1$ over the field

$Bbb F_q $, and
over $Bbb Q$, in general,

for any k? The reason why I am asking this is because I have been trying to factor them, and of course, I have been able to do it when the $k$ is given but had never found a way to do it in general.
Edit 1: I just wanted to clarify that this exercise was suggested, when I was learning field theory and Galois theory, and we were asked to do it in as much generality as possible. Even though I tried it for quite sometime, I wasn't able to progress beyond specific cases. Now that the course is over, I realised that I perhaps haven't solved this completely even now, and hence the question.
Edit 2: Also, the question I mentioned doesn't need the polynomial to be factored, just to find the degrees of irreducible factors. I have hence edited the question. Sorry for the confusion.

CHENRK · Answer

Let $K$ be a field and $n$ be a positive integer. An element $zeta$ in the algebraic closure of $K$ satisfying $alpha^n=1$ but $alpha^mne1$ for $0<m<n$ is called an $n$-th primitive root of unity, and the $n$-th cyclotomic polynomial is defined to be
$$Phi_n(x)=prod_i(x-zeta^i),$$
where $1le ile n$ and $gcd(i,n)=1$. In fact, $Phi_n(x)$ is a polynomial over $K$. It is not hard to see
$$x^k-1=prod_{nmid k}Phi_n(x).$$
Then it remains to factor the cyclotomic polynomials. For the finite field $mathbb F_q$ and $mathbb Q$, we have two results as follows.

Theorem. If $gcd(q, n)=1$, then $Phi_n$ factors into $phi(n)/d$ distinct monic irreducible polynomials in $mathbb F_q[x]$ of the same degree $d$, where $phi$ is the Euler function and $d$ is the least positive integer such that $q^dequiv1pmod n$.

Theorem. Cyclotomic polynomials over $mathbb Q$ are irreducible.

A learner · Answer

For the field $mathbb{Q} $, the highest degree of irreducible factor(which is a cyclotomic polynomial) of $ x^k - 1$ is $phi(k)$, here $phi $ is Euler-$phi $ function.
For finite field $mathbb{F}_{q} $, this can't be applied !
For example , $ x^3 - 1 $ has two irreducible factors $(x-1)$ ,$( x^2 + x + 1 ) $ over the field $mathbb{Q} $ , But over the field $mathbb{F}_{3} $, $( x^2 + x + 1 ) $ is not irreducible!

How to factor $x^k -1$?

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