Frobenius endomorphism

If $F$ is a finite field, then $\mathrm{Fr}_F$ is bijective (hence a field automorphism); indeed, any injective map $F\to F$ is automatically surjective because $F$ is finite. More generally, in characteristic $p$ the Frobenius is an automorphism precisely when $F$ is perfect.

For $F=\mathbb{F}_{q^n}$ over $\mathbb{F}_q$, the $q$-power Frobenius $x\mapsto x^q$ generates the Galois group of the extension (see finite-field Galois group is cyclic), and the fixed field of $x\mapsto x^q$ is $\mathbb{F}_q$ (compare fixed field).

If $X$ is a scheme of characteristic $p$, the ring maps $a\mapsto a^p$ glue contravariantly to the absolute Frobenius morphism

It is the identity on the underlying topological space and raises local functions to their $p$th powers. It is a morphism of schemes over $\mathbb F_p$, but generally not a morphism over a larger base field $k$: on constants it acts by the Frobenius of $k$.

For a $k$-scheme one therefore also uses the relative Frobenius $F_{X/k}:X\to X^{(p)}$, where $X^{(p)}$ is obtained by base change along Frobenius on $k$. Partial Frobenius morphisms on products and shtuka spaces apply relative Frobenius in selected factors.

1. Prime field. On $\mathbb{F}_p$, Frobenius is the identity map since $a^p=a$ for all $a\in\mathbb{F}_p$.

1. $\mathbb{F}_4$. In $\mathbb{F}_4=\mathbb{F}_2(\alpha)$ with $\alpha^2=\alpha+1$, the Frobenius is $x\mapsto x^2$. It satisfies $\alpha\mapsto \alpha^2=\alpha+1$ and $\alpha+1\mapsto (\alpha+1)^2=\alpha$, so it is a nontrivial automorphism of order $2$.

1. A non-surjective Frobenius (imperfect field). Let $F=\mathbb{F}_p(t)$, the rational function field. Then Frobenius sends $t\mapsto t^p$, so $t$ is not a $p$th power in $F$; hence Frobenius is not surjective and not an automorphism.

1. The Stacks Project Authors, “Varieties,” §33.36, “Frobenii.” Stacks Project.