irreducible polynomial divides $x^n-x$

Question

I want to get some hint, or solution of following question!

Let $f∈F_q[x]$. Show that if $f$ is irreducible, then $f$ divides $x^n-x$, where $n$ is $q^{deg(f)}$.

I thought if we use mathematical induction, then it can be solve, but I have no idea.. As I first, it's easy to solve, but it's hard..

Thank you for read it, and if you suggest any hint or solution, I appreciate it!

Answer 1

I just copy Milne's Corollary 4.22:

Each monic irreducible polynomial $f$ of degree $d|n$ in $\mathbb{F}_p[x]$ occurs exactly once as a factor of $X^{p^n}-X$ (hence the degree of the splitting field of $f$ is $\leq d$)

Proof. The factors of $X^{p^n}-X$ are distinct because it has no common factor with its derivative. If $f(x)$ is irreducible of degree $d$ then $f$ has a root of degree $d$ over $\mathbb{F}_p$. But the splitting field of $X^{p^n}-X$ contains a copy of every field of degree $d$ over $\mathbb{F}_p$ with $d|n$. Hence some root of $X^{p^n}-X$ is also a root $f$ and therefore $f(x)|X^{p^n}-X$. In particular $f(x)|X^{p^d}-X$ (and therefore it splits in its splitting field which has degree $d$ over $\mathbb{F}_p$).