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 $a^{-p}$ 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 $X_0 = \operatorname{Spec}(A_0)$ is (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 $H^1(E,\mathbf{Z}_\ell)$.

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 $H^1(E,\mathbf{Z}_\ell)$.

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 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 $a^{-p}$ 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 $\mathbf{F}_p$ here. And I'll even suppose $X$ arises by base change to $k$ from the affine $k_0 = \mathbf{F}_p$-scheme $X_0 = \operatorname{Spec}(A_0)$ is (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 then 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 $H^1(E,\mathbf{Z}_\ell)$.

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 $a^{-p}$ 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 $X_0 = \operatorname{Spec}(A_0)$ is (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 $H^1(E,\mathbf{Z}_\ell)$.