TITLE: Geometric vs Arithmetic Frobenius
QUESTION [20 upvotes]: If an algebraic variety $X$ over a field characteristic p is given by equations $f_i(x_1,...,x_k) = 0$, we can consider the variety $X^{(p)}$ obtained by applying p-th powers to all the coefficients of all $f_i$'s.
Frobenius morphism, as I understand it, is a morphism $X \to X^{(p)}$,
given on points as raising all coordinates to p-th power. 
Can anyone please explain me, what is the geometric Frobenius, as opposed to the arithmetic one?
EDIT: Thanks to Florian and George for the answers. I understand the difference now.
I accepted Florian's answer because he was first and also because I found the last link 
http://www.math.mcgill.ca/goren/SeminarOnCohomology/Frobenius.pdf he provided especially helpful.

REPLY [17 votes]: Brion & Kumar ["Frobenius splitting methods in geom. and rep. thy" Birkhauser 2005] call the absolute
Frobenius endomorphism the mapping $F_{abs}:X \to X$ which is the identity on $X$ and with
comorphism given when 
$X = \operatorname{Spec}(A)$ is affine by the rule $(f \mapsto F_{abs}^*(f) = f^p):A \to A$.
This is not a morphism "over $k$" since $F^*:A \to A$ is "semilinear" for the Frobenius
endomorphism of $k$ (= Frobenius automorphism in Galois group of $k$ if $k$ is perfect).
In Jantzen ["Representations of algebraic groups", 2nd edition] 9.1 and 9.2, he describes the absolute Frobenius map - it is "the same" as the one describe by B&K, except that
the codomain is "twisted" to make $F$ a morphism over $k$. For $X = \operatorname{Spec}(A)$
this twisting amounts to: $X^{(p)} = \operatorname{Spec}(A^{(p)})$ where
the $k$-algebra $A^{(p)}$ is $A$ as a ring but an element $a \in k$ acts on $A^{(p)}$ as
$a^{p^{-1}}= a^{1/p}$ does on $A$.
Geometric and arithmetic Frobenii have meaning only (I believe) when $X$ is "defined
over" a finite field; here I'll assume $X$ is defined over $\mathbf{F}_p$. And I'll even suppose $X$ arises by base change to $k$ from the affine $k_0 = \mathbf{F}_p$-scheme $X_0 = \operatorname{Spec}(A_0)$ (otherwise, patch!). 
Then $X = \operatorname{Spec}(A)$ where $A = A_0 \otimes_{k_0} k$.
The arithmetic Frobenius map on $X$ is the $k_0$-morphism $F_{arith}:X \to X$ whose
comorphism is given by $$(f \otimes a \mapsto f \otimes a^p):A = A_0 \otimes_{k_0} k \to A = A_0 \otimes_{k_0} k$$ for $f \in A_0$ and $a \in k$.
Thus the set of $k_0$-points $X_0(k_0)$ is the set of points in $X(k)$ fixed by the
arithemtic Frobenius $F_{arith}$; i.e. the action of $F_{arith}$ on points just gives the
"usual" action of the Frobenius element of the Galois group on rational points (here I must be supposing $k$ to be perfect...)
The geometric Frobenius of $X$ is the $k$-morphism $F_{geom}:X \to X$ whose comorphism
is given by $$(f \otimes a \mapsto f^p \otimes a):A = A_0 \otimes_{k_0} k \to A = A_0 \otimes_{k_0} k.$$ If you pick an embedding $X \subset \mathbf{A}^N$ defined
over $k_0$, then $F_{geom}$ is given on $k$-points in these coordinates by
$$F_{geom}(x_1,\dots,x_N) = (x_1^p,\dots,x_N^p)$$.
The arithmetic and geometric Frobenius are defined (briefly) in Jantzen (loc. cit.). 
Note that $F_{arith} \circ F_{geom} = F_{geom} \circ F_{arith}$ is the "absolute Frobenius" of B&K mentioned above.
Also see Milne's "Lectures on Etale Cohomology" 29.11 for some discussion reconciling
the number theorists with their action of the Frobenius automorphism $\phi=(x \mapsto x^p)$ on the Tate group $T_\ell E$ of an elliptic curve defined over $k_0$ and the algebraic geometers with their action of $F_{geom}$ on $H^1(E,\mathbf{Z}_\ell)$.