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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.

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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.

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  • $\begingroup$ Thanks so much both for correcting my wrong question and giving the right answer! $\endgroup$ – HLEE Jul 27 '19 at 22:23
  • $\begingroup$ How to show $E(u)$ is invertible in $W(k)[[u]][\frac{1}{u}]$ ? $\endgroup$ – M. A. SARKAR Jan 22 at 14:02
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    $\begingroup$ @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? $\endgroup$ – SashaP Jan 22 at 15:34
  • $\begingroup$ @SashaP, sorry I meant to say $E(u)$ is unit in $\widehat{W(k)[[u]][\frac{1}{u}]}$. How to show it ? $\endgroup$ – M. A. SARKAR Jan 22 at 15:46
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    $\begingroup$ @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, $\endgroup$ – SashaP Jan 22 at 15:49

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