Show that every irreducible polynomial in $\mathbb Z_p$ is a divisor of $x^{p^n}-x$ for some $n$. Let p(x) be irreducible of degree m in $\mathbb Z_p[x]$. Let $K$ be the finite extension of $\mathbb Z_p$ obtained by adjoining all the zeros of $p(x)$ in $\mathbb {\bar Z_p}$. Then $K$ is a finite field of order $p^n $ for some positive integer $n$, and consists precisely of all zeros of $x^{p^n}− x$ in $\mathbb {\bar Z_p}$. Now $p(x)$ factors into linear factors in $K[x]$, and these linear factors are among the linear factors of $x^{p^n}− x$ in $K[x]$. Thus $p(x)$ is a divisor of $x^{p^n}− x$.
It's Ok?
Thanks.
 A: That's correct, but requires a small remark: the linear factors of $p$ cannot be repeated, because finite fields are perfect (that is, irreducible polynomials cannot have multiple root in any extension). This result is usually proved in every course on field extensions.
This remark is important, because $x^{p^n}-x$ has no multiple roots, but, at least in principle, $p(x)$ might.
An example of an irreducible polynomial having repeated roots is $x^2-t\in \mathbb{F}_2(t)[x]$ (but, of course $\mathbb{F}_2(t)$ is not a finite field).
A: That is correct but even too complicated: Let $f$ (wlog we can take $f$ to be monic) be an irreducible polynomial of degree $n$ over $\mathbb F_q$. Then in $F_q^n$ there exists an $\alpha$ with $f$ as minimal polynomial (up to a scalar factor if $f$ is not monic). Then we know that $\alpha^{q^n} = \alpha$. Thus by the properties of the minimal polynomial $f$ must be a divisor of $x^{q^n} - x$.
This is true because if $f$ is MP of $\alpha$ and $g(\alpha) \neq 0$ and $g = a\cdot f + b$ with $\deg b < \deg f$ then $b(\alpha) = g(\alpha) - a(\alpha)f(\alpha) = 0$ and this $b=0$. Thus $f|g$.
