I tried to understand [this paper](https://arxiv.org/pdf/1606.01921.pdf) on page 31.


Let $K$ be an finite extension of $\mathbb Q\_p$ and $\overline{K}$ be its algebraic closure; $\mathcal{O}\_{\overline{K}}$ is the ring of integers of $\overline{K}$; $\mathcal{O}\_{\mathbb{C}\_K}$ is its $p$-adic completion. We have $R:=\mathcal{O}\_{\overline{K}}/(p)\cong\mathcal{O}\_{\mathbb{C}\_K}/(p)$ canonically.


There is a natural projection onto the first component:
$$\phi:R^{\mathrm{perf}}\to R,$$
where the $R^{\mathrm{perf}}$ contain elements of the form $(x\_1,x\_2,...)$ with $x\_i^p=x\_{i-1}$.


Now, use the universal propety of Witt vector we get a lift


$$\theta:W(R^{\mathrm{perf}})\to\mathcal{O}\_{\mathbb{C}\_K}.$$
And the ring $W(R^{\mathrm{perf}})$ is denoted by $A\_{\mathrm{inf}}$.


I have two question about it:


1. Then the author claims $R^{\mathrm{perf}}$ is $\ker(\phi)$-adically complete, but I doubt whether it is true. If we consider the map. In fact, if we consider the system $\phi\_n:R^{\mathrm{perf}}\to R$ by projecting onto the $n$-the component. It is easy to see $R^{\mathrm{perf}}$ is isomorphic to the completion this system $\{\ker(\phi\_n)\}$. But I doubt $\ker(\phi\_n)=\ker(\phi)^{n}$.
2. Even this is true how do I get the $\ker(\theta)$-adic completeness of $A\_{\mathrm{inf}}$?



The kernel from $A_\mathrm{inf}$ to $\mathcal{O}_{\mathbb{C}_K}$
1)Pick a sequence of elements $p^{1/p^n}\in \mathcal{O}\_{\overline{K}}$ such that $(p^{1/p^{n+1}})^p=p^{1/p^n}$. The ideal $\ker\phi$ is in fact principal and is generated by the element $p^{\flat}:=(\dots, p^{1/p^2},p^{1/p},0)$ (where we view $p^{1/p^n}$ as elements in the reduction $\mathcal{O}\_{\overline{K}}/p$). Indeed, any element of $\ker(\phi)$ has the form $x=(\dots, x\_2, x\_1,0)$ and $x\_n$ has to be divisible by $p^{1/p^n}$ inside $\mathcal{O}\_{\overline{K}}/p$ because $x\_n^{p^n}=0$. We therefore can divide $x$ by $p^{\flat}$, inductively picking ration $x\_n/p^{1/n}$ in a way compatible under raising to the $p$-th power.


Having established that, we can conclude that $\ker(\phi)^{p^n}=\{x^{p^n}|x\in \ker(\phi)\}$ but the right hand side is equal to $\ker(\phi\_{n+1})$ because $\phi\_{n+1}=\phi\circ \varphi^{-n}$ where $\varphi$ is the Frobenius automorphism on $R^{\mathrm{perf}}$. Therefore completeness wrt $\{\ker(\phi\_n)\}\_n$ implies completenes wrt to the powers of the ideal $\ker(\phi)$.


2)We want to prove that the canonical map $\alpha:A\_\inf\to \lim\limits\_n A\_\inf/(\ker\theta)^n$ is an isomorphism. It is in fact true that $\ker(\theta)$ is principal and one can explicitly write down a generator (see e.g. Proposition 4.4.3 of [https://math.stanford.edu/~conrad/papers/notes.pdf](https://math.stanford.edu/%7Econrad/papers/notes.pdf)).


The ring $A\_{\inf}$ is $p$-complete by construction and the quotients $A\_{\inf}/\ker(\theta)^n$ are $p$-complete because they are $p$-adically separated (they admit filtrations with graded pieces $\ker(\theta)^i/\ker(\theta)^{i+1}\simeq \mathcal{O}\_{\mathbb{C}}$ because $\ker(\theta)$ is generated by one element that is not a zero divisor) and are quotients of a complete ring, so to prove that $\alpha$ is an isomorphism, it is enough to do so for its mod $p$ reduction. The mod $p$ reduction of $A\_{\inf}$ is $R^{\mathrm{perf}}$, by construction, and $(\lim\limits\_n A\_\inf/(\ker\theta)^n)/p=\lim\limits\_n R^{\mathrm{perf}}/\overline{(\ker\theta)^n}$ because each $A\_\inf/(\ker\theta)^n$ is $p$-torsion free. Finally the inclusion $\overline{\ker\theta}\subset\ker\phi$ is equality because $\theta$ is surjective and $\mathcal{O}\_{\mathbb{C}}$ is torsion free. Therefore the fact that $\alpha$ mod $p$ is an isomorphism amounts to the first part of the question.


