Well known problem about irreducibles I think this is well known, but I can't find a proof anywhere. Any help is appreciated. 

If $f(x)$ is monic and irreducible in $\mathbb{F}_p[x]$ with degree $m$, then $f(x)\mid x^{p^m}-x$.

 A: If you consider the quotient ring $K:=\Bbb F_p[x]/(f)$, this is a field extension of $\Bbb F_p$ of degree $m$. 
Each element $\alpha$ of a degree $m$ extension of $\Bbb F_p$ satisfies $\alpha^{p^m}=\alpha$. 
In other words, $x^{p^m}-x$ is $0$ in $K$, but that means that $x^{p^m}-x\,\in\,(f)$.
A: Hints: suppose that $f(x)$ has degree $m$, and let $\alpha$ be a root of $f(x)$. 


*

*What is $[\mathbb{F}_{p}(\alpha):\mathbb{F}_{p}]$? (this denotes the dimension of $\mathbb{F}_{p}(\alpha)$ as a vector space over $\mathbb{F}_{p}$) In more explicit terms, how many elements does $\mathbb{F}_{p}(\alpha)$ have?

*Recall that any two finite fields of the same size are isomorphic. 

*What is true about every element of the field $\mathbb{F}_{p^{m}}$? Can you think of a polynomial for which every element of $\mathbb{F}_{p^{m}}$ is a root?

*Finally, since $f(x)$ is the minimal polynomial for $\alpha$ (it is monic and irreducible and has $\alpha$ as a root), and the minimal polynomial must divide any other polynomial which has $\alpha$ as a root, what must be true? 
(Note that by following this logic with a slight variation at the second step, you prove a stronger statement: that if $f(x)$ is monic and irreducible of degree $n|m$, then $f(x)|x^{p^{m}}-x$. Instead, you note that $\mathbb{F}_{p}(\alpha) \cong \mathbb{F}_{p^{n}} \subset \mathbb{F}_{p^{m}}$, and the rest follows.)
