Why is Fontaine's infinitesimal period ring $A_{\text{inf}}$ complete?

Question

Fix a perfectoid field $K$ in mixed characteristic with ring of integers $\mathcal{O}$ and pseudo-uniformizer $\varpi$. Its tilt is the fraction field of $\mathcal{O}^{\flat}=\varprojlim_{x\mapsto x^{p}}\mathcal{O}/\varpi$. Pick an element $\pi\in\mathcal{O}^{\flat}$ such that $\pi^{\sharp}/p\in\mathcal{O}^{\times}$.

I want to understand Fontaine's infinitesimal period ring $$ A_{\text{inf}}:=W(\mathcal{O}^{\flat}). $$ Many references claim that it is complete with respect to the $(p,[\pi])$-adic topology. However, I was not able to find a reference for this statement by myself. I would be grateful if someone could provide a full proof here.

Answer 1

I am not familiar with perfectoid fields, so hopefully my argument is not circular.

We first remark that $(p,[\pi])$ is a regular sequence in $\newcommand\Ainf{A_{\operatorname{inf}}}\Ainf$, since $p$ is a non-zero-divisor and $\pi$ does not vanish in the integral domain $\Ainf/p=\mathcal O^\flat$. Thus the ring $\Ainf$ is $(p,[\pi])$-adically complete if and only if it is derived $(p,[\pi])$-complete, and since the ring $\Ainf$ is already derived $p$-complete, it suffices to check that it is derived $[\pi]$-complete.

Let $\theta\colon\Ainf\to\mathcal O$ denote the Fontaine's map, whose kernel $I$ is principal. Since the ring $\Ainf$ is derived $I$-complete (and in fact, $I$-adically complete), cf. [Hesselholt–Nikolaus, Topological Cyclic Homology, Prop 1.3.4], thus we are reduced to check that the ring $\mathcal O$ is derived $\theta([\pi])$-complete, but $\theta([\pi])=\pi^\sharp$ by definition, and the result follows from the derived $p$-completeness of $\mathcal O$.