Frobenius endomorphism

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

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 $(xy)^p=x^p y^p$.

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

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

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

3. 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.