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Let $A$ be an $p$-adically complete (isomorphic to the inverse limit system) integral domain and $I$ be a prime ideal such that both $A$ and $A/I$ has no $p$-torsion. Is it possible that $A/I^2$ (or higher power $A/I^n$) has $p$-torsion? Basically, I want to show that each $A/I^n$ is $p$-adically complete

CO2
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1 Answers1

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Consider the finite field with $p$ elements $\mathbb{F}_{p}$. The kernel $I$ of the canonical map of Witt vectors $W\left(\mathbb{F}_{p}\left[X\right]\right)\to W\left(\mathbb{F}_{p}\right)$ is a prime ideal of the $p$-adically complete integral domain $W\left(\mathbb{F}_{p}\left[X\right]\right)$. The quotient $W\left(\mathbb{F}_{p}\right)$ has no $p$-torsion.

The ideal $I$ consists of Witt vectors $\left(w_{0},w_{1},\ldots\right)\in W\left(\mathbb{F}_{p}\left[X\right]\right)$ such that $X$ divides all $w_{n}$ for $n\in\mathbb{N}$. In particular, the Teichmüller representative $\left[X\right]$ is an element of $I$, as well as the elements $V^{n}\left(\left[X^{p-1}\right]\right)$ and $V^{n}\left(\left[X\right]\right)$ for any $n\in\mathbb{N}^{*}$. This implies that this is an element of $I^{2}$: \begin{equation*} V^{n}\left(\left[X^{p-1}\right]\right)V^{n}\left(\left[X\right]\right)=V^{2n}\left(\left[X^{p-1}\right]^{p^{n}}\left[X\right]^{p^{n}}\right)=V^{2n}\left(\left[X^{p^{n+1}}\right]\right)=p^{n+1}V^{n-1}\left(\left[X\right]\right)\text{.} \end{equation*}

For the second equality, see [Bourbaki, Algèbre commutative, chapitre IX, Proposition 5].

But as $\left[X\right]$ is not an element of $I^{2}$, we get that $W\left(\mathbb{F}_{p}\left[X\right]\right)/I^{2}$ has $p$-torsion.

In fact, $W\left(\mathbb{F}_{p}\left[X\right]\right)/I^{2}$ might not $p$-adically complete. Indeed, according to [The Stacks project, 031A], the ring $A/I^{2}$ is $p$-adically complete if and only if $I^{2}=\bigcap_{n\in\mathbb{N}^{*}}\left(I^{2}+p^{n}A\right)$.

But it is uncertain that series such as $\sum_{n\in\mathbb{N}^{*}}p^{n+1}V^{n-1}\left(\left[X\right]\right)$ are in $I^{2}$, that is it is unclear that it can be written as a finite sum of products of elements of $I$. This seems like a difficult question to me.

Nevertheless, this would have worked if $I$ was principal. Indeed, let $w,g\in W\left(\mathbb{F}_{p}\left[X\right]\right)$. If for all $n\in\mathbb{N}^{*}$ there is $a_{n}\in W\left(\mathbb{F}_{p}\left[X\right]\right)$ such that $w\equiv ga_{n}\pmod{p^{n}}$, then $w=g\left(a_{1}-\sum_{n\in\mathbb{N}^{*}}\left(a_{n}-a_{n+1}\right)\right)$.

In general, $p$-torsion and $p$-completeness are unrelated, as $\mathbb{Z}/p^{2}\mathbb{Z}$ is $p$-complete for instance.

user242429
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  • You are right $p$-torsion-free do not imply $p$-adically separted. In fact, I want to show that the $A_{\mathrm{inf}}/\ker(\theta)^i$ is $p$-adically complete without knowing $\ker(\theta)$ is principal. However, it seems it is not possible? Cf. https://mathoverflow.net/a/394795?noredirect=1. – CO2 Jun 10 '21 at 12:51
  • Actually the answer by SashaP on your link explains why $A_{\mathrm{inf}}/\operatorname{ker}\left(\phi\right)^{i}$ is $p$-adically complete. They mention $p$-adic separation, which is satisfied here. In case you don't know it, $p$-adic separation is equivalent to having $\bigcap_{n\in\mathbb{N}^{*}}p^{n}A=\left{0_{A}\right}$. – user242429 Jun 10 '21 at 13:07
  • Thank you. We were discussing about how to show it without using the $\ker(\theta)=(\xi)$ because the paper I was reading uses the conclusion $A_{\mathrm{inf}}$ being $\ker(\theta)$-adically complete to show its principality. But as SashaP said: he could only see $\ker(\theta)$-adically completeness by using $\ker(\theta)$ being principal. (Cf. https://arxiv.org/pdf/1606.01921.pdf on page 31 ) – CO2 Jun 10 '21 at 13:16
  • I see! Well, here, if I understood the notations of the paper right, $R^{\mathrm{perf}}$ is a perfect ring of characteristic $p$, and $A_{\mathrm{inf}}$ is the ring of Witt vectors of $R^{\mathrm{perf}}$, it is therefore $p$-adically separated and complete (see Bourbaki, but Proposition 6). So we can use the above proposition of the Stacks project to conclude that $A_{\mathrm{inf}}/I$ is $p$-adically complete, for every ideal $I$ of $A_{\mathrm{inf}}$. – user242429 Jun 10 '21 at 13:21
  • $I^{n}=\bigcap_{m\in\mathbb{N}^{*}}\left(I^{n}+p^{m}A\right)$ this condition is equivalently to saying $I^n$ is closed under $p$-adic topology I guess? Suppose there is an element $a$ in each $\left(J+p^{m}A\right)$. Topologically, say any nonclosed ideal $J$ and $x\in\bar{J}\backslash J$, then the equality might not hold. So it means every ideal in $A_{\mathrm{inf}}$ is closed? – CO2 Jun 10 '21 at 13:58
  • Thank you very much for your comment, it made me realise a mistake in my previous comments and answer. I updated it, but I might remove this later as this was wrong. In short, it is not clear that all ideals are closed and that their quotient is $p$-adically complete. – user242429 Jun 11 '21 at 11:13