If $G$ is a smooth scheme over $S$ of characteristic $p$, is the relative Frobenius morphism $F_{G/S}$ faithfully flat?

Question

Let $G$ be a smooth scheme over $S$ of characteristic $p$, do we have that the reltaive Frobenius morphism $F_{G/S}$ is faithfully flat?

There is an excersice in Liu's book saying that this is true when $S$ is the spectrum of a field, it's Exercise 3.13. I think it should be possible to generilize it, feel free to add nice properties to $S$ or $G$, e.g. Noetherian, finite type $k$-schemes.

Because in my case, $S=\DeclareMathOperator{\Spec}{Spec} \Spec A/\mathfrak{m}^n$, where $(A,\mathfrak{m},k)$ is a discrete valuation ring, and $G=\mathbb{G}_{m,S}=\Spec A/\mathfrak{m}^n[T,T^{-1}]$ the multiplicative group scheme over $S$, and it's smooth over $S$ since $\Spec \mathbb{Z}[T,T^{-1}]$ is smooth over $\Spec \mathbb{Z}$, and I want to show that $F_{G/S}:G\rightarrow G^{(p)}$ is faithfully flat.

Answer 1

It's true under the condition that $S$ is locally Noetherian. Clearly relative Frobenius morphisms are heomorphisms hence bijection hence surjective. Since faithfully flat is just surjective and flat, it suffices to show flatness.

We would like to use this lemma [Tag 039D, stacks project]. There are three conditions, which we will show one by one.

(1) $S,G,G^{(p)}$ are all locally Noetherian.

$G/S$ is smooth implies that it is locally of finite type. Since $S$ is locally Noetherian and that $G\rightarrow S$ is locally of finite type, we have that $G$ is localy Noetherian.

$G^{(p)}/S$ is the pull back of $G/S$ by $F_S:S\rightarrow S$. So it's also smooth hence locally of finite type, since $S$ is locally Noetherian, so is $G^{(p)}$.

(2) $G$ is flat over $S$.

It follows from that $G/S$ is smooth.

(3) For every $s\in S$, the morphism $F_{G/S,s}=F_{G_s/\kappa(s)}:G_s\rightarrow G^{(p)}_s$ is flat.

Since $G/S$ is smooth, $G_s/\kappa(s)$ is smooth. By Exercise 3.13 in Liu's book [Algebraic geometry and arithmetic curve], we have that the relative Frobenius morphism of $G_s/\kappa(s)$ is flat.

The result follows.

Answer 2

Here's an alternative way to think about this that contains methods useful for dealing with relative Frobenius more generally, for any $X/S$ smooth, and doesn't use locally Noetherian.

Note that if $X=\mathbb{A}^n_S$ then $F_{X/S}$ is evidently faithfully flat, in fact its locally free of rank $p^n$. Moreover, the relative Frobenius map is an isomorphism in the etale case (use the fact that a universal homeomorphism which is etale is an isomorphism). Since smooth maps are etale locally of the form $\mathbb{A}^n_S\to S$ you can actually reduce to that case. In fact, this shows that $F_{X/S}$ (when $X/S$ is smooth of relative dimension $n$) is locally free of rank $p^n$.