Irreducible polynomial in $\mathbb{F}_p[x]$ divides $x^{p^n}-x$

Question

Let $p(x)=x^{p^n}-x \in \mathbb{F}_p[x]$, $n \in \mathbb{N}$. I want to prove that any irreducible polynomial in $\mathbb{F}_p[x]$ with degree $n$ divides $p(x)$.
Let $q(x)$ be this irreducible polynomial: $$q(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0 \quad , \quad a_i \in \mathbb{F}_p$$ I know that $q(x) \vert p(x)$ means that $q(x) \vert (x^{p^n-2}+x^{p^n-3}+\ldots +1)$ but I don't think this leads me anywhere. Can someone help me?

Answer 1

Let $q(x)$ be an irreducible polynomial of degree $d$ dividing $n$. Then $F=\mathbb{F}_p[x]/(q(x))=\mathbb{F}_p[\alpha]$ is a field of $p^d$ elements, where $\alpha$ is a primitive element of $F$ and $q(x)$ is the minimal polynomial of $\alpha$. The set of non-zero elements $\left( \mathbb{F}_p[x]/(q(x)) \right)^{*}$ is a multiplicative group of order $ p^d-1$ and $\alpha$ is the generator of this multiplicative group. By Lagrange's theorem, the order of any element of this multiplicative group divide $p^d-1$ and hence, we have $\alpha^{p^d-1}=1$ or $\alpha^{p^d}=\alpha$. (One can use the Fermat's Little Theorem also)

Since $d~|~n$, we have $n=dm$, for some integer $m$, we have $$\alpha^{p^n}=\alpha^{p^{dm}}=\alpha,$$ and we see $\alpha$ is a rootof $x^{p^n}-x.$

Now $q(x)$ is irreducible in $\mathbb{F}_p[x]$, so either $q(x)$ divides $x^{p^n}-x$ or by Bezout's identity, we have $$f(x)q(x)+g(x)(x^{p^n}-x)=1, ....(*)$$ for some polynomials $f(x),~ g(x) \in \mathbb{F}_p[x]$. In the second case, setting $x=\alpha$ in equation $(*)$, we get $0=1$, which is not possible. Hence the $d$-degree irreducible polynomial $q(x)$ must divide $x^{p^n}-x$. As $n ~|~n$, your question is answered.