$x^{q^n}-x$ is a product of all monic irreducible polynomial in $\mathbb{F}_q[x]$ of degree dividing $n$?

Question

I don't know whether the following statement is true or not:

$x^{q^n}-x$ is a product of all monic irreducible polynomial in $\mathbb{F}_q[x]$ of degree dividing $n$

Note that we are not assuming $q$ is prime (so it's some power of some prime). I already know that if $q$ is a prime then the statement is true. But I don't know if this general version is also true. Could you explain this to me? Thanks in advance.

Answer 1

Let $p(X)=p_1p_2\dots p_N$ be the product of all monic irreducible polynomials in $\mathbb F_q[X]$ with degree dividing $n$. Since $\mathbb F_q$ is perfect, $p(X)$ does not have any repeated root in $\bar {\mathbb {F}}_q$, the algebraic closure.
Let $\alpha$ be a root of $p(x)$. Then $\alpha$ is a root of some $p_i$. So $\alpha \in \mathbb F_{q^{\deg p_i}}\subset \mathbb F_{q^{n}}$ since $\deg p_i | n$. In other words, $\alpha $ is a root of $X^{q^n}-X$.
Conversely, assume $\alpha$ is a root of $X^{q^n}-X$ so it is an element of $\mathbb F_{q^n}$. Let $m(x)$ be the minimal polynomial of $\alpha$. Then we have $\mathbb F_q\subset \mathbb F_q(\alpha)\subset \mathbb F_{q^n} $. Then the tower formula says $\deg m=[\mathbb F_q(\alpha): \mathbb F_q] \ | \ [\mathbb F_{q^n}: \mathbb F_q] =n$. Thus $\alpha$ is a root of $p(x)$.
Since both the polynomials have only simple roots, $p(x)=X^{q^n}-X$.

Answer 2

This is also true if $q$ is not a prime number, but the power of a prime. Let say $q=p^m$. Let's denote $K$ the splitting field of $s(x) = x^{q^n}-x$ over $\mathbb F_q$. $K$ is the finite field $\mathbb F_{p^{mn}}$. The elements of $K$ are exactly the roots of $s$.

Suppose that $r \in \mathbb F_q[x]$ is a monic irreducible polynomial of degree $a$ dividing $n$. The splitting field $k$ of $r$ over $\mathbb F_q$ is a field extension $k/\mathbb F_q$ of degree $a$. As $a \mid n$, we have $s(y) = 0$ for any $y \in k$. Therefore $r$ divides $s$.