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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?

the man
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    See also this question. – Dietrich Burde Feb 10 '19 at 19:57
  • 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 $n\mid q^t-1$. – Jyrki Lahtonen Feb 11 '19 at 04:10
  • 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. – Jyrki Lahtonen Feb 11 '19 at 04:27
  • 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. – Jyrki Lahtonen Feb 11 '19 at 04:37

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