Show that over \(\FF_p\), \[ x^{p^n}-x \] is the product of all monic irreducible polynomials whose degrees divide \(n\), each occurring once.
Solution
The roots of \[ x^{p^n}-x \] in an algebraic closure of \(\FF_p\) are exactly the elements of the finite field \[ \FF_{p^n}. \] Moreover, \[ \frac{d}{dx}(x^{p^n}-x)=-1, \] so the polynomial is squarefree.
Let \(f\in\FF_p[x]\) be monic irreducible of degree \(d\), and let \(\alpha\) be one of its roots. Then \[ \FF_p(\alpha)\cong\FF_{p^d}. \] Therefore \[ f\mid x^{p^n}-x \] if and only if \(\alpha\in\FF_{p^n}\), equivalently \[ \FF_{p^d}\subseteq\FF_{p^n}. \] The subfield theorem for finite fields gives \[ \FF_{p^d}\subseteq\FF_{p^n} \iff d\mid n. \]
Hence the irreducible factors of \(x^{p^n}-x\) are exactly the monic irreducibles whose degrees divide \(n\). Since the polynomial is squarefree, each occurs with multiplicity one. Thus \[ x^{p^n}-x =\prod_{\substack{f\text{ monic irreducible}\\ \deg f\mid n}} f(x). \]