Prove that an irreducible polynomial \(\pi(x)\in \FF_p[x]\) divides \(x^{p^n}-x \iff \deg \pi(x)\) divides \(n\).
Solution
Let \(\pi\in\FF_p[x]\) be irreducible of degree \(d\), and let \(\alpha\) be a root in \(\overline{\FF}_p\). Then \[ \FF_p(\alpha)\cong\FF_{p^d}. \] Also, the roots of \(x^{p^n}-x\) in \(\overline{\FF}_p\) are exactly the elements of \(\FF_{p^n}\).
If \(\pi\mid x^{p^n}-x\), then \(\alpha\in\FF_{p^n}\), so \[ \FF_{p^d}=\FF_p(\alpha)\subseteq\FF_{p^n}. \] By the finite-field subfield criterion, \(d\mid n\).
Conversely, if \(d\mid n\), then \(\FF_{p^d}\subseteq\FF_{p^n}\), hence \(\alpha^{p^n}=\alpha\). Therefore \(\alpha\) is a root of \(x^{p^n}-x\). Since \(\pi\) is the minimal polynomial of \(\alpha\) over \(\FF_p\), \[ \pi(x)\mid x^{p^n}-x. \] Thus \[ \boxed{\pi\mid x^{p^n}-x\iff \deg\pi\mid n}. \]