4
$\begingroup$

Let $X$ be a scheme defined over $\mathbb{F}_p$ and denote by $X_{et}$ its étale topos . Associated to $X,$ we can consider the absolute Frobenius map $F_X: X \rightarrow X$ which gives an associated geometric morphism of étale topoi $F_X: X_{et} \rightarrow X_{et}.$

Is it true that $F_X$ induces the same geometric morphism as the identity $X \rightarrow X$?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
7
$\begingroup$

Yes.

Let $f\colon U \to X$ be an étale morphism. The absolute Frobenius $U \to U$, the absolute Frobenius $F \colon X \to X$, and the map $f\colon U \to X$ (twice) form a commutative square. This commutative square induces a morphism from $U$ to the fiber product of $U$ and $X$ over $X$ taken with respect to the maps $f$ and $F$, in other words to $F^{-1}(U)$. This morphism is natural and so gives a natural transform from the identity functor from the étale site of $X$ to itself to the absolute Frobenius from the étale site of $X$ to itself.

Because $f$ is étale, this commutative square is in fact Cartesian, so the induced morphism is an isomorphism, and the natural transformation is an isomorphism of functors.

Since $F^{-1}$ is isomorphic as a functor to the identity, their induced geometric morphisms on the topos of sheaves are equivalent.

Improve this answer
$\endgroup$
8
  • $\begingroup$ This seems to show that the Frobenius is naturally equivalent to the identity. Is this natural equivalence also "relative" to the base? For example, if $X \rightarrow \mathbb{F}_p,$ I have a relative topos $X_{et} \rightarrow (\mathbb{F}_p)_{et}.$ Is then the absolute Frobenius "relatively" isomorphic to the identity? $\endgroup$
    – Crystallineperiodic
    Commented Mar 21 at 9:54
  • 1
    $\begingroup$ Here the only property that is using about étale maps is that they are relatively perfect, thus the same statement holds for the pro-étale site. $\endgroup$
    – Z. M
    Commented Mar 21 at 12:29
  • 1
    $\begingroup$ @Crystallineperiodic For $V$ over $\mathbb F_p$ étale, the formula given for $\alpha$ sends $V \times_{\mathbb F_p} X \to (V \times_{\mathbb F_p} X) \times_{X,F} X$ is the obvious map dropping the middle $X$. To make this natural transformation I wrote down give a 2-morphism, the diagram that has to commute is a pair of morphisms $V \times_{\mathbb F_p } X \to (V \times_{\mathbb F_p} X) \times_{X, F} X$, one of which is this and the other is the composition of the identity and the map from my answer taking $U = V \times_{\mathbb F_p} X$. $\endgroup$
    – Will Sawin
    Commented Mar 21 at 14:40
  • 1
    $\begingroup$ @Crystallineperiodic I think this diagram does not commute. It suffices to show the two maps do not agree after composition on the left with projection to $V$. For the $\alpha$ map in your link, the composition with projection to $V$ is again the projection to $V$. For my map, this is the composition with the projection to $U$ followed by projection from $U$ to $V$. The composition with the projection to $U$ is the absolute Frobenius $U\to U$, and projection to $V$ composed with absolute Frobenius is the same as absolute Frobenius composed with projection to $V$. $\endgroup$
    – Will Sawin
    Commented Mar 21 at 14:43
  • 1
    $\begingroup$ So the maps differ by an absolute Frobenius, which is nontrivial if $V =\operatorname{Spec} \mathbb F_{p^n}$ for $n>1$. $\endgroup$
    – Will Sawin
    Commented Mar 21 at 14:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .