Is Frobenius on $R^\circ/p$ surjective for general perfectoid rings $R$?

Question

In [1], Propisition 6.1.9(2), it said that if $R$ is a perfectoid ring such that $pR^\circ$ is closed in $R^\circ$ (this includes the case if $R$ is of character $p$, or if $p$ is invertible in $R$, in particular if $R$ is a perfectoid $K$-algebra, $K$ a field), then the Frobenius $\Phi:R^\circ/p\to R^\circ/p$ is surjective (it is easy to see that this is equivalent to $R^{\flat\circ}\to R^\circ/p$ is surjective because $R^{\flat\circ}=\varprojlim R^\circ/p$). In [1], Remark 6.1.10 it said that probably $pR^\circ$ is not closed in general.

However, in the proof of [1], Lemma 6.2.7, it makes the extensive use of the surjectivity of thie map $R^{\flat\circ}\to R^\circ/p$ (say, by choosing $R^+$ to be $R^\circ$), and all subsequent development of the theory relies on this lemma.

So my problem is: is this map surjective in general (i.e. when $pR^\circ$ is not closed)? I noticed that in the video of Scholze's fall 2014 course he said that this is surjective without giving details. Is it wrong, or still open, or I missed something obvious? Thanks in advance.

[1] Peter Scholze's lectures on $p$-adic geometry, fall 2014

Answer 1

[Probably this question is no longer interesting to the author. But since I faced the same problem while trying to learn basics of perfectoid spaces I decided to write down an argument here]

We start with a perfectoid ring $R$ and a pseudo-uniformizer $\varpi$ s.t. $\varphi:R^{\circ}/\varpi \to R^{\circ}/\varpi^p$ is surjective. This means that for any element $x\in R^{\circ}$ there is a pair of power-bounded elements $x_0, y_1$ s.t. $x=x_0^p+\varpi^py_1$. Apply the same procedure to $y_1$ to get $x_1,y_2\in R^{\circ}$ s.t. $$ x=x_0^p+\varpi^p(x_1^p+\varpi^py_2)=x_0^p+\varpi^px_1^p+\varpi^{2p}y_2. $$

Keep going to obtain $x_1,\dots, x_n, y_{n+1}\in R^{\circ}$ s.t. $$ x=x_0^p+\varpi^px_1^p+\varpi^{2p}x_2^p+\dots+\varpi^{np}x_n^p+\varpi^{(n+1)p}y_{n+1}. $$ This implies that $$ x-(x_0^p+\varpi^px_1^p+\varpi^{2p}x_2^p+\dots+\varpi^{np}x_n^p)\in \varpi^{(n+1)p}R^{\circ}. $$ Now note that since $R$ is a uniform Tate ring, so the induced topology on $R^{\circ}$ is just $\varpi$-adic topology and $R^{\circ}$ is complete with respect to this topology. Therefore, we can conclude that $$ x=\sum_{i=0}^{\infty}\varpi^{ip}x_i^p. $$ The last thing to observe is that $$ x-(\sum_{i=0}^{\infty} \varpi^{i}x_i)^p= \sum_M \varpi^M \Big(\sum_N \sum_{k_0+\dots k_N=p, k_N\geq 1, k_n\neq p, k_1+2k_2+\dots+Nk_N=M} \prod_{t=1}^N \frac{p!}{k_0!\dots k_N!}x_t^{k_t}\Big)\in pR^{\circ}. $$ So we are done.

[To rigorously prove that this sum makes sense and lies inside $pR^{\circ}$ note that each sum in parentheses is actually finite and each of $\frac{p!}{k_0!\dots k_N!}$ is divisible by $p$ in $\mathbf Z$ (under the assumption that $k_N\geq 1$ and $k_N\neq p$). Let's say that $\frac{p!}{k_0!\dots k_N!}=pc_{k_0,\dots, k_N}$ for some integer $c_{k_0,\dots, k_N}$. Then we have $$ \sum_M \varpi^M \Big(\sum_N \sum_{k_0+\dots k_N=p, k_N\geq 1, k_n\neq p, k_1+2k_2+\dots+Nk_N=M} \prod_{t=1}^N \frac{p!}{k_0!\dots k_N!}x_t^{k_t}\Big)= $$ $$ x-(\sum_{i=0}^{\infty} \varpi^{i}x_i)^p=p\bigg(\sum_M \varpi^M \Big(\sum_N \sum_{k_0+\dots k_N=p, k_N\geq 1, k_n\neq p, k_1+2k_2+\dots+Nk_N=M} \prod_{t=1}^N c_{k_0,\dots, k_N}x_t^{k_t}\Big)\bigg) \in pR^{\circ}.] $$