The relative Frobenius $x\mapsto x^q$ of an extension of finite fields

Let ${\mathbb{F}}_q$ be a finite field of order $q$ (Finite fields and their order) and let $E$ be a finite field having ${\mathbb{F}}_q$ as a subfield. The relative Frobenius of the extension $E\mathrm{/}{\mathbb{F}}_q$ is the map

${\sigma }_q\text{\hspace{0.17em}⁣}:E⟶E,{\sigma }_q(x)=x^q\mathrm{.}$

It is an ${\mathbb{F}}_q$-automorphism of $E$, and no separate result is needed for that. Let $p$ be the characteristic of $E$; the subfield ${\mathbb{F}}_q$ has the same identity element and hence the same characteristic, so $q=p^k$ with $k=[{\mathbb{F}}_q:{\mathbb{F}}_p]$ by Every finite field has order $p^n$ for a unique prime characteristic $p$ and positive integer $n$. The Frobenius map ${\mathrm{Fr}}_E\text{\hspace{0.17em}⁣}:x\mapsto x^p$ is an injective field endomorphism of $E$, is an automorphism because $E$ is finite, and has $k$-fold iterate $x\mapsto x^{p^k}$ (Frobenius $x\mapsto x^p$ is an injective endomorphism in characteristic $p$, and an automorphism for finite fields); that iterate is ${\sigma }_q$. Every $a\in {\mathbb{F}}_q$ satisfies $a^q=a$ (A field with $q$ elements is the splitting field of $x^q-x$ over its prime subfield), so ${\sigma }_q$ fixes ${\mathbb{F}}_q$ pointwise. Hence

${\sigma }_q\in \mathrm{Aut}(E\mathrm{/}{\mathbb{F}}_q)$

(Relative field automorphisms and $\mathrm{Aut}(K\mathrm{/}F)$). Its iterates are ${\sigma }_q^i(x)=x^{q^i}$ for $i\in \mathbb{N}$, with ${\sigma }_q^0$ the identity.

• The letter. The relative Frobenius is written ${\sigma }_q$ rather than ${\phi }_q$ because $\phi$ is Euler's totient (The unit group $(\mathbb{Z}\mathrm{/}n{)}^{\times }$ and Euler's totient $\phi (n)=\mid (\mathbb{Z}\mathrm{/}n{)}^{\times }\mid$ for $n\ge 1$) everywhere below, and the two symbols would otherwise stand side by side in the same formula.

• Relative, not absolute. ${\mathrm{Fr}}_E$ is intrinsic to $E$; ${\sigma }_q$ depends on the chosen base field ${\mathbb{F}}_q$, and it is the identity exactly when $E={\mathbb{F}}_q$. Taking ${\mathbb{F}}_q$ to be the prime field returns ${\mathrm{Fr}}_E$ itself.