GCD between an irreducible polynomial of degree $n$ in $\mathbb{F}_p[x]$ and $x^{p^n} - x$ I was viewing the proof of Proposition 1.1 in some notes I found on factoring polynomials in $\mathbb{F}_p[x]$ at https://wstein.org/129-05/section/m129-section-factoring-poly-mod-p/factoring_poly_mod_p.pdf. Specifically, I'm only worried about the second sentence in the first paragraph of the proof. 
Basically, if I have an irreducible polynomial $f$ of degree $n$ in $\mathbb{F}_p[x]$, the author argues that $(f(x), x^{p^n} - x) \neq 1$ in $\mathbb{F}_{p^n}[x]$. This part is completely fine for me. 
But, the author then makes the claim that this implies that $(f(x), x^{p^n}-x) \neq 1$ in $\mathbb{F}_p[x]$. I'm struggling with seeing why this implication follows. Is there a fairly trivial reason for this that I'm not seeing ? Or is there more high-powered machinery involved ? 
Thanks! 
 A: Note that as $f$ is irreducible, $L=\mathbb{F}_p[x]/f(x)\mathbb{F}_p[x]$ is a field with $p^n$ elements. So for every nonzero element $\alpha(x) \in L$, the equation $\alpha^{p^n-1}(x) \equiv 1$ mod $f(x)$ holds. [As in every field $K$ with $q^n$ elements the equation $a^{q^n-1} = 1$ holds for each $a \in K^{\times}$.] This implies that for every $\alpha(x) \in L$ [including trivially $\alpha=0$], the resulting $\alpha^{p^n}(x)-\alpha(x)$ is 0 in $L$.
This implies that for every $\alpha(x) \in \mathbb{F}_p[x]$ [including trivially $\alpha=0$], the resulting $\alpha^{p^n}(x)-\alpha(x)$ is a multiple of $f(x)$ in $\mathbb{F}_p[x]$ i.e., divide $\alpha^{p^n}(x)-\alpha(x)$ by $f(x)$ in $\mathbb{F}_p[x]$ the remainder will be 0. This implies that for every $\alpha(x) \in \mathbb{F}_p[x]$ there is a $g_{\alpha}(x) \in \mathbb{F}_p[x]$ such that $\alpha^{p^n}(x)-\alpha(x) = g_{\alpha}(x)f(x)$.
Taking $\alpha(x) = x$ one concludes $x^{p^n}-x$ is a multiple of $f(x)$ in $\mathbb{F}_p[x]$. Which gives you your desired result.
