Frobenius endomorphism - Imported Knowlpedia Content

Let $F$ be a field of characteristic $p>0$ . The Frobenius endomorphism of $F$ is the map

\mathrm{Fr}_F: F\to F,\qquad x\mapsto x^p.

It is a ring endomorphism because in characteristic $p$ one has $(x+y)^p=x^p+y^p$ (all intermediate binomial coefficients are divisible by $p$ ), and $(xy)^p=x^p y^p$ .

Remarks

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).

Examples

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

$\mathbb{F}_4$ . 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 $2$ .

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.