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!
 A: 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.
