# Geometric and arithmetic Frobenius

I read in Serre's "Lectures on $N_X(p)$" that when $X$ is a scheme defined over $\mathbb{F}_q$ (a finite field), the geometric Frobenius $F: X \mapsto X$ is defined by fixing every element of the topological space of $X$, and acting as $f \mapsto f^q$ on the structure sheaf. As a particular case, if $X$ is affine (say $Spec(\mathbb{F}_{q^n}[x,y])$), then "$F$ is the "standard Frobenius map" (on points): $(u,v) \mapsto (u^q,v^q)$." Why is that ? Doesn't the action on the structure sheaf implies that every element of $\mathbb{F}_{q^n}[x,y]$ is raised to the $q$-th power ? Also, I have seen other definitions of geometric Frobenius. (I have noticed related questions on MO, but the answers confuse me even more.)

Secondly, it is mentioned that the geometric and arithmetic Frobenius act in the same way on $X(\overline{\mathbb{F}_q})$. Can anyone explain this in some detail ?

Thanks !

• For $P\in \mathbb{F}_q[x,y]$, $P^q=P(x^q,y^q)$.
– abx
Jun 23, 2016 at 14:48
• Sure, but what if $P \in \mathbb{F}_{q^n}[x,y]$?
– THC
Jun 24, 2016 at 7:18

Let $X=Spec R$, $R$ a finite type $F_q$-algebra,then the endomorphism $a\mapsto a^q$ induces an endomorphism $Fr$ on $X$, which is usually referred as absolute Frobenius.
Let $Y=X\times Spec F$ be the base change of $X$ to $F$, where $F$ is an algebraic closure of $F_q$. Then one calls the endomorphism $Fr\times Id$ a geometric Frobenius on $Y$. Meanwhile, the Galois conjugation $r\mapsto r^q$ gives an automorphism $fr$ of $Spec F$, and the automorphism $Id\times fr$ on $Y$ is called an arithmetic Frobenius.
In particular, the above geometric Frobenius is an endomorphism over $F$, while the arithmetic Frobenius is not defined over $F$.
On the rational points $Y(F)$ the geometric Frob is the inverse to the arithmetic Frob.