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

  1. $\Bbb F_q $, and
  2. 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.

Spectre
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Reader Manifold
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  • Factorize, maybe ? – Spectre Jul 24 '20 at 07:51
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    For even '$k$'s, you'll see that $(x- 1)$ is a factor . Plus, using the identity $(a + b) \times (a-b) = a^2 - b^2 $ may help. – Spectre Jul 24 '20 at 07:53
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    https://en.wikipedia.org/wiki/Cyclotomic_polynomial – Jaap Scherphuis Jul 24 '20 at 09:28
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    To start: The polynomial $x^k-1$ always can have a factor of $x-1$ factored out.

    If k is divisible by $2^n$, then the polynomial will also have factors $(x+1)(x^2+1)(x^4+1)(...)(x^{2^{n-1}}+1)$.

    If k is divisible by an odd prime n, then the polynomial will also have a factor $1+x+x^2+...+x^{n-1}$.

    I don't know how to check for any given k if the polynomial that is left from taking out these factors is a factoriazable polynomial.

     – Moko19 Jul 24 '20 at 09:38
  • I have answered this question (the part about finite fields) more times than I care to admit. I collected most of the links here. The most highly voted is this dealing with the case when cyclotomic polynomials remain irreducible over $\Bbb{F}_q$. Anyway, chenrk's answer tells the story :-) – Jyrki Lahtonen Jul 25 '20 at 09:41

2 Answers2

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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 $\alpha^m\ne1$ 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 $1\le i\le n$ and $\gcd(i,n)=1$. In fact, $\Phi_n(x)$ is a polynomial over $K$. It is not hard to see $$x^k-1=\prod_{n\mid k}\Phi_n(x).$$ 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 $q^d\equiv1\pmod n$.

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

  • Slightly more explicitly. If $\zeta$ is of order $n$ as above, $\gcd(n,q)=1$, then its minimal polynomial is $$\prod_{i=0}^{d-1}(x-\zeta^{q^i}).$$ – Jyrki Lahtonen Jul 25 '20 at 14:39
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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!

A learner
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