# maximal unramified extension of Breuil ring in $A_{cris}$

Here is the notation. Let $$k$$ be a perfect field of characteristic $$p$$, $$W=W(k)$$ the ring of Witt vector and $$\mathfrak{S}:=W[[u]]$$, $$\mathcal{O}_{\mathcal{E}}$$ is the $$p$$-adic completion of $$W[1/u]$$ which is DVR. Denote $$R=\lim_{\leftarrow}\mathcal{O}_{ \bar{K}}/p$$ where the transition map is Frobenius. The ring of Witt vector $$W(R)$$ has a canonical surjection $$W(R) \xrightarrow{\theta} \mathcal{O}_{C_K}$$ where $$C_K$$ is the completion of $$\bar{K}$$. The ring $$A_{cris}$$ is a divided power envelope of $$W(R)$$ with respect to $$\ker \theta$$.

One can embed $$\mathcal{O}_{\mathcal{E}}$$ into $$W(R)\subset A_{cris}$$ by sending $$u\mapsto [\underline{\pi}]$$ ($$\underline{\pi}=(\pi,\pi^{1/p}, \cdots)$$). In this paper, Kisin defines $$\mathcal{E}^{un}$$ as the maximal unramified extension of $$\mathcal{E}=Frac(\mathcal{O}_{\mathcal{E}})$$ contained in $$W(Frac(R))[1/p]$$ and $$\mathfrak{S}^{un}:=\mathcal{O}_{\mathcal{E}^{un}}\cap W(R)$$ (which must be done by Breuil earlier).

Here is my question. Consider $$\mathfrak{S}^{un}$$ as a subring of $$A_{cris}$$ by composing the Frobenius on $$\mathfrak{S}^{un}$$ and the inclusion to $$W(R)$$ as above.(In particular, $$u\mapsto [\underline{\pi}]^p$$) In the same paper of Kisin, it says that $$\mathfrak{S}^{un}\cap pA_{cris}=p\mathfrak{S}^{un}$$ when $$p>2$$.(proof of Theorem 2.2.7.) Why is this true? He mentions a paper of Brueil(proof of 3.3.2.), but I don't see why this helps the assertion. Also, the structure of $$\mathfrak{S}^{un}$$ is somewhat mysterious to me. Naively it should be a subring of $$\mathcal{O}_{\mathcal{E}^{un}}$$ consists of elements "without $$u$$ in the denominator", as $$\mathcal{O}_{\mathcal{E}^{un}}$$ is a DVR with residue field $$k((u))^{sep}$$. However, as the residue field is not perfect, this description is not concrete.

It is false that $$\mathfrak{S}^{un}\cap pA_{cris}=p\mathfrak{S}^{un}$$. For example, $$E(u)\in\mathfrak{S}\subset\mathfrak{S}^{un}$$ is not divisible by $$p$$ in $$\mathfrak{S}^{un}$$ but gets mapped to $$\varphi(\xi)$$ where $$\xi$$ is a generator of $$\theta:W(R)\to\mathcal{O}_C$$ which is divisible by $$p$$ in $$A_{cris}$$ because $$\varphi(\xi)-\xi^p\in pW(R)$$.

The claim in the article is weaker: Kisin says that if the image of a Frobenius-equivariant map $$f:\mathfrak{M}\to \mathfrak{S}^{nr}$$ from a height $$1$$ Breuil-Kisin module becomes divisible by $$p$$ in $$A_{cris}$$, then it was divisible by $$p$$ already in $$\mathfrak{S}^{nr}$$. Here is how we can prove such statement(the proof of course uses the assumption on the height of $$\mathfrak{M}$$):

The ring $$\mathfrak{S}^{nr}/p$$ can be identified with the ring of Puiseaux series $$\{\sum c_{\alpha}u^{\alpha}|\alpha\in \mathbb{Z}[1/N]\cap\mathbb{R}_{\geq 0}\text{ for some }N\text{ coprime to }p\}$$. Suppose that for every $$x\in\mathfrak{M}$$ the image $$\varphi(f(x))$$ is inside $$pA_{cris}$$, but there exists $$x$$ such that $$f(x)$$ is not divisible by $$p$$ in $$\mathfrak{S}^{un}$$. Among such $$x$$ choose one for which $$\overline{f(x)}\in\mathfrak{S}^{un}/p$$ has the minimal lowest exponent of $$u$$ in the expansion. Denote this exponent by $$\alpha$$ so that $$\overline{f(x)}=c_{\alpha}u^{\alpha}+\sum\limits_{\beta >\alpha}c_{\beta}u^{\beta}$$.

Using that $$A_{cris}/p\to\mathcal{O}_C/p$$ is the universal characteristic $$p$$ divided power thickening of $$\mathcal{O}_C/p$$ we can identify $$A_{cris}/p$$ with the divided power envelope $$\mathcal{O}_C/p\langle x\rangle/(x-p^{1/p})$$ of the ideal $$(p^{1/p})\subset\mathcal{O}_C/p$$ via the map that sends $$x$$ to $$[p^{\flat}]$$ and an element $$a\in\mathcal{O}_C/p$$ to $$\tilde{a}^p\in A_{cris}/p$$ where $$\tilde{a}$$ is an arbitrary preimage of $$a$$ under $$\theta$$. Since $$\pi^{\flat}$$ is equal to $$(p^{\flat})^{1/e}$$ in $$R$$ up to a unit, an element of the form $$c_{\alpha}u^{\alpha}+\sum\limits_{\beta >\alpha}c_{\beta}u^{\beta}$$ dies under the map $$\mathfrak{S}^{nr}/p\xrightarrow{\varphi} A_{cris}/p$$ if and only if $$\alpha\geq e$$.

By the assumption on $$\mathfrak{M}$$ there exists $$y=\sum\limits_{i=0}^{p-1} y_i\otimes u^i\in\varphi^*\mathfrak{M}=\mathfrak{M}\otimes_{\mathfrak{S},\varphi}\mathfrak{S}$$ such that $$\varphi_{\mathfrak{M}}(y)=E(u)x$$. It means that for at least one $$i$$ the element $$\overline{f(y_i)}$$ has expansion of the form $$c'_{\gamma}u^{\gamma}+\sum\limits_{\beta>\gamma}c'_{\beta}u^{\beta}$$ with $$p\gamma\leq \alpha+e$$. It follows that $$\gamma\leq \frac{2}{p}\alpha<\alpha$$. By the minimality assumption on $$\alpha$$ and $$x$$, the reduction $$\overline{f(y)}$$ must be zero, hence $$\varphi_{\mathfrak{S}^{un}}(\overline{f(y)})=E(u)\overline{f(x)}$$ is zero which implies the vanishing of $$\overline{f(x)}$$ itself. That gives us a contradiction.

• Thanks so much both for correcting my wrong question and giving the right answer! – HLEE Jul 27 '19 at 22:23
• How to show $E(u)$ is invertible in $W(k)[[u]][\frac{1}{u}]$ ? – M. A. SARKAR Jan 22 at 14:02
• @M.A.SARKAR This is not true: if $E(u)$ was invertible in $W(k)[[u]][\frac{1}{u}]$ there would be an element $f(u)\in W(k)[[u]]$ that is not divisible by $u$ such that $f(u)E(u)=u^n$ for some $n'\geq 0$. This gives a contradiction by plugging in $u=0$. Do you think this fact is used in the above argument? – SashaP Jan 22 at 15:34
• @SashaP, sorry I meant to say $E(u)$ is unit in $\widehat{W(k)[[u]][\frac{1}{u}]}$. How to show it ? – M. A. SARKAR Jan 22 at 15:46
• @M.A.SARKAR An element of a $p$-adically complete ring $A$ is invertible if and only if its reduction in $A/p$ is invertible. In this case the reduction is a field so the claim is true because the mod $p$ reduction of $E(u)$ is non-zero, – SashaP Jan 22 at 15:49