3
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ See also this question. $\endgroup$ – Dietrich Burde Feb 10 at 19:57
  • $\begingroup$ You want the smallest extension field containing $\alpha$, i.e. $\Bbb{F}_q(\alpha)$. This has to be one of the fields $\Bbb{F}_{q^t}$ for some $t$, because those are the only (finite) extension fields that exist. The multiplicative group of $\Bbb{F}_{q^t}$ is cyclic of order $q^t-1$. Hence it contains an element of order $n$ if and only if $n\mid q^t-1$. $\endgroup$ – Jyrki Lahtonen Feb 11 at 4:10
  • $\begingroup$ You may be expected to know about the factors of $x^n-1$ over $\Bbb{Q}$. Those are called cyclotomic polynomials. They may or may not factor further ovet $\Bbb{F}_q$, see here. Even if you don't know about the cyclotomic polynomials, you can try the process described here. I'm afraid the different context may make it a bit harder to follow. $\endgroup$ – Jyrki Lahtonen Feb 11 at 4:27
  • $\begingroup$ Anyway, a general fact at play is that if a polynomial $p(x)$ with coefficients in $\Bbb{F}_q$ has $\alpha$ as a root, then $\alpha^q$ will also be a root of $p(x)$. When $\alpha$ is a root of unity of a prescribed order, this leads to a complete description of the roots of the minimal polynomial of $\alpha$. For example, if $q=4$, $n=17$, Then the roots will be $\alpha,\alpha^4,\alpha^{16}=\alpha^{-1},\alpha^{-4}$. The list stops there, because $\alpha^{-16}=\alpha$, and that was already included. $\endgroup$ – Jyrki Lahtonen Feb 11 at 4:37

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.