Show that \(x^{p^n} - x = \prod f_i(x)\) over all irreducible monic \(f_i\) of degree \(d\) dividing \(n\).
Solution
Work in an algebraic closure \(\overline{\FF}_p\).
1. The roots of \(x^{p^n}-x\) are exactly the elements fixed by the \(n\)th power of Frobenius.
For \(\alpha\in\overline{\FF}_p\), \[ \alpha^{p^n}-\alpha=0 \quad\Longleftrightarrow\quad \alpha^{p^n}=\alpha. \] These fixed points form the finite field \(\FF_{p^n}\).
2. The polynomial \(x^{p^n}-x\) is squarefree.
Its derivative in characteristic \(p\) is \[ \frac{d}{dx}(x^{p^n}-x)=p^n x^{p^n-1}-1=-1, \] so it has no repeated roots.
3. Let \(f\in\FF_p[x]\) be monic irreducible of degree \(d\), and let \(\alpha\) be a root. Then \(f\mid x^{p^n}-x\) if and only if \(d\mid n\).
The roots of \(f\) are the Frobenius conjugates \[ \alpha,\alpha^p,\ldots,\alpha^{p^{d-1}}, \] and \(d\) is the least positive integer with \(\alpha^{p^d}=\alpha\). Hence \[ f\mid x^{p^n}-x \quad\Longleftrightarrow\quad \alpha^{p^n}=\alpha \quad\Longleftrightarrow\quad d\mid n. \] Indeed, writing \(n=qd+r\) with \(0\le r<d\), the equality \(\alpha^{p^n}=\alpha\) and \(\alpha^{p^d}=\alpha\) imply \(\alpha^{p^r}=\alpha\); minimality of \(d\) forces \(r=0\).
4. Therefore\[ x^{p^n}-x=\prod_{\substack{f\text{ monic irreducible in }\FF_p[x]\\ \deg f\mid n}} f(x). \]
By 3, the irreducible divisors are exactly the displayed factors. By 2, each occurs with multiplicity one. Since both sides are monic, their irreducible factorizations agree exactly.