Let \(p\) be prime and \(n\ge1\). Prove that \(\mathbb F_{p^n}\) is the splitting field over \(\mathbb F_p\) of \[ x^{p^n}-x. \]
Solution
Let \[ f(x)=x^{p^n}-x\in\mathbb F_p[x]. \] Its derivative is \[ f'(x)=-1, \] so all roots are simple.
In an algebraic closure of \(\mathbb F_p\), let \[ S=\{a:a^{p^n}=a\}. \] Because Frobenius powers respect addition and multiplication in characteristic \(p\), \(S\) is closed under addition, subtraction, and multiplication; if \(a\ne0\) lies in \(S\), then \[ (a^{-1})^{p^n}=a^{-1}. \] Thus \(S\) is a field. Since \(f\) has degree \(p^n\) and is separable, its splitting field contains exactly \(p^n\) roots, so \[ |S|=p^n. \] Therefore \(S\) is the finite field \(\mathbb F_{p^n}\).
Equivalently, every \(a\in\mathbb F_{p^n}\) satisfies \(a^{p^n}=a\): for \(a\ne0\) this follows from \(|\mathbb F_{p^n}^\times|=p^n-1\), and it is trivial for \(a=0\). Hence \(f\) splits completely over \(\mathbb F_{p^n}\), and its roots generate that field. Thus \(\mathbb F_{p^n}\) is its splitting field over \(\mathbb F_p\).