Exercise 11.1.4

Proof. Write $F={\mathrm{Frob}}_p$ the Frobenius automorphism of ${𝔽}_{p^n}$, which generates $G=\mathrm{Gal}({𝔽}_{p^n}∕{𝔽}_p)=⟨F⟩$ (Theorem 11.1.7).

Let $H=⟨F^m⟩$ be the subgroup of $G$ generated by $F^m$. As $m\mid n$, $H$ is a cyclic subgroup with $n∕m$ elements, isomorphic to $\mathbb{Z}∕\frac{n}{m}\mathbb{Z}$.

By the Galois correspondence, $H$ corresponds to the fixed field $L_H$, and

So $L_H$ is the set of the roots of $x^{p^m}-x$. $L_H$ is the unique subfield of ${𝔽}_{p^n}$ with $p^m$ elements, written ${𝔽}_{p^m}$ (cf Exercise 3).

By the Fundamental Theorem of Galois Theory (section 7.3),

So

where the isomorphism sends $F^m={({\mathrm{Frob}}_p)}^m$ on the generator $[1]\in \mathbb{Z}∕\frac{n}{m}\mathbb{Z}$. □