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](https://link.springer.com/chapter/10.1007/978-0-8176-4532-8_7), 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](https://www.math.u-psud.fr/~breuil/PUBLICATIONS/Hodgeentier.pdf)(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.



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


