Show that \(\mathbb{F}_{p^d} \le \mathbb{F}_{p^n} \iff d \mid n\).
Solution
Suppose first that \[ \mathbb F_{p^d}\subseteq\mathbb F_{p^n}. \] Then the tower law gives \[ n=[\mathbb F_{p^n}:\mathbb F_p] =[\mathbb F_{p^n}:\mathbb F_{p^d}]\,[\mathbb F_{p^d}:\mathbb F_p] =[\mathbb F_{p^n}:\mathbb F_{p^d}]\,d, \] so \(d\mid n\).
Conversely, assume \(d\mid n\). In an algebraic closure of \(\mathbb F_p\), \[ \mathbb F_{p^m}=\{x:x^{p^m}=x\}. \] If \(n=dk\) and \(x^{p^d}=x\), then iterating the \(p^d\)-power Frobenius \(k\) times gives \[ x^{p^n}=x. \] Hence every element of \(\mathbb F_{p^d}\) lies in \(\mathbb F_{p^n}\). Thus \[ \boxed{\mathbb F_{p^d}\subseteq\mathbb F_{p^n}\iff d\mid n.} \]