# Surjectivity of frobenius

I have a question about Faltings' paper "Crystalline cohomology and p-adic Galois representations". Suppose $R$ is a smooth $\mathbb{Z}_p$-algebra, of relative dimension $1$, such that there is an etale map $\mathbb{Z}_{p}[T,T^{-1}]\to R$.

By $\bar{R}$ we denote the maximal extension of $R$ which is etale in characteristic zero. That is if $R$ is geometrically integral we take the maximal field extension of its fraction-field such that the normalisation of $R[1/p]$ in this field is unramified over $R[1/p]$. Then $\bar{R}$ is the normalisation of R in this field. In general $R\otimes_{\mathbb{Z}_p}\bar{\mathbb{Q}}_p$ is a product of integral domains, and $\bar{R}$ is the product of the corresponding normalisations.

The paper states that the Frobenius map on $\bar{R}/p\bar{R}$ is surjective. I wonder why. Is there any reference for the proof? Thank you!

• Probably $R$ is meant to have connected spectrum (so it is a 2-dimensional regular excellent domain) and you only permit extensions that have connected Spec or else "maximal extension..." would make no sense. But you must intend more conditions left unstated, since otherwise $R[1/p]$ is such an extension and so $p$ would a unit in the "maximal" extension. Do you want the normalization of $R$ in the "universal cover" (direct limit of connected finite etale covers, controlled by geometric generic point) of Spec$(R[1/p])$? Please rewrite the question so we don't have to guess what you're asking. Jul 19, 2013 at 7:03
• I have edited the qustion to make it more clear.
– Lan
Jul 19, 2013 at 8:29
• This is better, but still unclear. The way you use "geometrically integral" is meaningless -- do you mean Spec($R$) has geometrically integral fibers over Spec($\mathbf{Z}_p$), or perhaps that $R[1/p]$ is geometrically integral over $\mathbf{Q}_p$? Also, the end of the 2nd paragraph rules out extracting roots of $T$, yet that does provide an etale extension over $\mathbf{Q}_p$. So you give inconsistent descriptions of what you want. Why not focus on $R$ a domain, since your initial $R$ of interest is a product of domains anyway, so the domain case should be all that matters. Jul 19, 2013 at 12:30
• I think "geometric integral" means that $R\otimes_{\mathbb{Z}_p}\bar{\mathbb{Q}}_p$ is integral ring.
– Lan
Jul 19, 2013 at 12:44
• Keerthi: That will introduce ramification in char. 0, so something like $t^p - ph(t) - r$ is more suitable, but not clear what to take for $h$ to avoid char-0 ramification. Jul 19, 2013 at 13:18

One can prove this using the almost purity theorem. The argument is presented below (essentially extracted from Scholze's paper on perfectoid spaces, but is presumably also in Faltings' papers), and it works in any dimension. I do not know an elementary proof.

Note first that $\overline{R}$ is an algebra over $\overline{\mathbf{Z}_p}$ (as has been clarified in the question). Hence, we can talk about almost mathematics over $\overline{R}$ with respect to the maximal ideal of $\overline{\mathbf{Z}_p}$. The almost purity theorem tells you that the Frobenius map $\Phi:\overline{R}/p^{\frac{1}{p}} \to \overline{R}/p$ is an almost isomorphism. We want to show it is actually surjective, not just almost so. So pick some $x \in \overline{R}$. By almost surjectivity, we can write

$$p^{\frac{1}{p}} x = y^p + pz$$

for suitable $z \in \overline{R}$. Define

$$w := y \cdot p^{-\frac{1}{p^2}} \in \overline{R}[\frac{1}{p}].$$

Then the first formula shows

$$w^p = y^p p^{-\frac{1}{p}} = x - p^{1 - \frac{1}{p}}z \in \overline{R}.$$

As $\overline{R}$ is integrally closed, it follows that $w \in \overline{R}$, so $y \in p^{\frac{1}{p^2}} \overline{R}$. Plugging this into the first formula and dividing by $p^{\frac{1}{p}}$ then proves the claim.

• Very good. Strictly speaking, this gives $x = {y'}^p + p^{1-1/p}z$, but surjectivity of Frobenius on $\overline{R}/(p^{\varepsilon})$ for some $0 < \varepsilon \le 1$ implies the same for $\overline{R}/(p)$ after finitely many iterations (depending on $\varepsilon$). Jul 20, 2013 at 13:58