# The kernel from $A_\mathrm{inf}$ to $\mathcal{O}_{\mathbb{C}_K}$

I tried to understand this paper 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 $$A_{\mathrm{inf}}$$?

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).

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.

• Also, how do we see the $p$-adically separatedness of each $A_{\mathrm{inf}}/\ker^n$? I could only see it when $n=1$ since it is a closed ideal. Do we need to unravel the arithmetic for Witt vector or there is a direct way to see?
– CO2
Jun 8, 2021 at 12:26
• @CO2 1)Sure, the system $A/I^{m_i}$ is cofinal inside the system $A/I^n$ so their inverse limits coincide. 2)Good point, I was too hasty at this step. I think we have to appeal to principality of $\ker$ to establish this, I've edited the answer Jun 8, 2021 at 15:03
• So then the logic of th paper is a bit flawed? Becaus they used the completeness to prove the principality.
– CO2
Jun 8, 2021 at 16:33
• @CO2 There certainly could be a direct argument for separatedness that I'm not seeing, perhaps you can ask the authors if you're interested in setting up the theory in this particular order. Jun 8, 2021 at 17:24
• Thank you. Last question: why do we only need to reduce mod $p$ in the last step? It seems that I also need to reduce mod all $p^n$.
– CO2
Jun 9, 2021 at 21:31