A finite extension of a finite field of order $q$ is Galois with cyclic Galois group generated by $x\mapsto x^q$

Let ${\mathbb{F}}_q$ be a finite field of order $q$ and let $E$ be a finite field having ${\mathbb{F}}_q$ as a subfield, with $[E:{\mathbb{F}}_q]=n$ (The degree $[K:F]={\dim }_{F}K$ of a finite field extension). Then $E\mathrm{/}{\mathbb{F}}_q$ is a finite Galois extension (Finite Galois extensions and $\mathrm{Gal}(K\mathrm{/}F)$) and

$\mathrm{Gal}(E\mathrm{/}{\mathbb{F}}_q)=⟨{\sigma }_q⟩$

is cyclic of order $n$, generated by the relative Frobenius ${\sigma }_q\text{\hspace{0.17em}⁣}:x\mapsto x^q$ (The relative Frobenius $x\mapsto x^q$ of an extension of finite fields).

• Separability and normality are never argued separately. The usual route checks that $E$ is a splitting field of $t^{q^n}-t$ and that this polynomial has no repeated root. Routing through Artin's fixed-field theorem: $[K:K^G]=\mathrm{\mid }G\mathrm{\mid }$ and $\mathrm{Aut}(K\mathrm{/}{K}^G)=G$ instead replaces both checks by one count: an automorphism group of order $n$ whose fixed field is ${\mathbb{F}}_q$ already forces $\mathrm{\mid }\mathrm{Aut}\mathrm{\mid }=[E:{\mathbb{F}}_q]$, which is one of the equivalent Galois conditions.