The $k$-linear Frobenius and the Frobenius twist

(Absolute Frobenius)

Let $X$ be a scheme all of whose local rings contain $\FF_p$. The Frobenius morphism $F : X \to X$ is the identity on the underlying topological space, with \[ F^\sharp : \OO_X \to \OO_X, \qquad f \mapsto f^p . \] This is a ring homomorphism in characteristic $p$, so $F$ is a morphism of schemes.

(The twist and the $k$-linear Frobenius)

For $X$ over $k$ with structure morphism $\pi : X \to \Spec k$, let $X_p$ denote $X$ with the structure morphism twisted by Frobenius, so that $k$ acts on $\OO_{X_p}$ through $p$-th powers. Then $F$ becomes a $k$-morphism \[ F' : X_p \to X , \] the $k$-linear Frobenius. For $X$ a curve over perfect $k$ it is finite of degree $p$, and on function fields it is the inclusion \[ k(X) \subseteq k(X)^{1/p} . \]

(Effect of the twist)

The absolute $F$ is not a morphism over $k$ unless $k=\FF_p$: $F^\sharp$ sends $\lambda \in k$ to $\lambda^p$, so $\pi\circ F=\operatorname{Frob}_{\Spec k}\circ\pi$, where $\operatorname{Frob}_{\Spec k}$ is the Frobenius of $\Spec k$. Over $k=\FF_p$ the Frobenius of $\Spec k$ is the identity, so $F=F'$. In $X_p$ the $k$-structure is composed with the $p$-th power map of $k$, so $\pi\circ F'$ is the structure morphism of $X_p$ and $F'$ is a morphism over $\Spec k$.

The extension $k(X)^{1/p}/k(X)$ has degree $p$: $k(X)$ has transcendence degree $1$ over the perfect field $k$, so $k(X)^{1/p}$ is generated over $k(X)$ by the $p$-th root of a separating transcendental element. For $X=\PP^1$, this is $k(t) \subseteq k(t^{1/p})$, of degree $p$ since $t^{1/p}$ is a root of the irreducible polynomial $T^p-t$.

Twisting applies the Frobenius of $k$ to the coefficients of the defining equations, so $X_p \cong X$ over $k$ whenever $X$ is defined over $\FF_p$, for instance $X = \AA^1$. For $k$ algebraically closed and an elliptic curve $E$ over $k$, the twist replaces $j(E)$ by a $p$-th power or $p$-th root of it, so $E_p \cong E$ over $k$ if and only if $j(E)^p=j(E)$, that is, $j(E) \in \FF_p$; a curve with $j(E)\notin\FF_p$ has $E_p\not\cong E$.

(Absolute versus relative Frobenius)

The absolute Frobenius of $\PP^n$ is finite, flat, bijective, and nowhere smooth, because $d(t^p) = 0$ (FE-MORFROB). The same identity $d(t^p)=0$ shows that $k(X)^{1/p}/k(X)$ is purely inseparable, which PR-IV2INSEP uses.

Define the Frobenius morphism of a scheme of characteristic $p$.

Why is it not a morphism of $k$-schemes, and how is that fixed?

What is the degree of the $k$-linear Frobenius, and which field extension does it induce?

When is $X_p \cong X$ over $k$?

Backlinks

PR-IV2INSEPPurely inseparable morphisms of curves are composites of Frobeniususes

What links to this

Inseparable morphisms and Lüroth