I am reading Brinon, Conrad "Notes on $p$-adic Hodge theory" and I can't find any reference for the proof of Theorem 9.1.8, namely the injectivity of the Frobenius endomorphism of $A\_{cris}$. Does anyone know where to find it or how to prove it?



Injectivity of Frobenius on $A_{cris}$
$\newcommand{\Z}{\mathbb{Z}}$
$\def\cO{\mathcal{O}}$There seems to be a suspiciously straightforward proof by analyzing the Witt coordinates of the elements of $A\_{cris}$ so there might well be a mistake here.


The ring $A\_{cris}$ is the $p$-adic completion of the divided power envelope of the ideal $(\xi)$ in the ring $A\_{inf}=W(\cO^{\flat}\_{C})$. The ideal $\xi$ is defined as the kernel of the map $A\_{inf}\to \cO\_C$ but for our purposes it is only important that we can pick $\xi=p-[\pi]$ where $\pi$ is an element of $\cO^{\flat}\_{C}=\lim\limits\_{x\mapsto x^p}\cO\_C/p$ of the form $(0,p^{1/p},p^{1/p^2},\dots)$ where $p^{1/p},p^{1/p^2}$ is some compatible system of roots of $p$ in $\cO\_C$. We fix such $\pi$ once and for all. $\pi$ has valuation $1$ with respect to the canonical valuation on $\cO^{\flat}\_{C}$. For this reason in what follows I will denote this valuation by $v\_{\pi}$ though it does not depend on the choice of $\pi$.


Following Brinon and Conrad, let's denote the divided power envelope of $(\xi)$ by $A^0\_{cris}$ and consider it as a subring of $A\_{inf}[1/p]$. Note that the divided power envelope of $\xi$ coincides with that of $[\pi]$ since $p$ already has divided powers in $A\_{inf}$. Every element of $A\_{inf}[1/p]$ can be written in a unique way as $\sum\limits\_{i=-N}^{\infty}p^i[x\_i]$ for some elements $x\_i\in\cO\_C^{\flat}$. Let's characterize elements of $A^0\_{cris}$ in terms of this expansion (we make sense of this infinite sum as follows -- $\sum\limits\_{i=0}^{\infty}p^i[x\_i]$ means the sum in the $p$-adic topology of $A\_{inf}$ and then we add a finite sum $\sum\limits\_{i=-N}^{-1}p^i[x\_i]$).


**Lemma.** An element $\sum\limits\_{i=-N}^{\infty}p^i[x\_i]$ belongs to $A^0\_{cris}$ if and only if $$i+f\_p(x\_i)\geq 0\qquad\qquad (\*)$$ for every $i< 0$. Here $f\_p(x)=v\_p(([v\_{\pi}(x)])!)=\frac{[v\_{\pi}(x)]-S\_{[v\_{\pi}(x)]}}{p-1}$ where $S\_a$ is the sum of digits in the $p$-adic expansion of a natural number $a$ (square brackets here mean the integral part and not the Teichmuller representative). We won't actually use this explicit formula for $v\_p(n!)$ in what follows.


*Proof.* Any element of $A^0\_{cris}$ has this form. Indeed, elements satisfying $(\*)$ form a subring of $A\_{inf}[1/p]$(concerning closedness under addition, sums of the form $p^k[x\_k]+p^k[y\_k]$ contribute to coefficients near higher powers of $p$ in a complicated way involving Witt polynomials, but if both $x\_k$ and $y\_k$ are divisible by $\pi^m$, then in the expansion $p^k[x\_k]+p^k[y\_k]=\sum\limits\_{i=k}^{\infty}p^i[a\_i]$ all $a\_i$ are also divisible by $\pi^m$ so these contributions will not spoil the condition $(\*)$ for $i>k$) so it is enough to check the condition for the elements $\frac{[\pi]^i}{i!}$. Rewrite such element as $p^{-v\_p(i!)}[\pi^i]\cdot u$ where $u\in W(\mathbb{F}\_p)$ is a $p$-adic unit. The Witt expansion of $\frac{[\pi]^i}{i!}$ thus looks like $\sum\limits\_{k=-v\_p(i!)}^{\infty}p^k[\pi^i\cdot a\_k]$ for $a\_k\in \mathbb{F}\_p$ and this expansion indeed satisfies $(\*)$. For the converse, it is enough to prove that for any $i<0$ a single term $p^i[x\_i]$ lies in $A^0\_{cris}$ provided that $i+f\_p(x\_i)\geq 0$. In this case we write $p^i[x\_i]=\frac{[\pi]^j}{j!}p^{i+v\_p(j!)}[\frac{x\_i}{\pi^j}]$ where $j=[v\_{\pi}(x\_i)]$. This product lies in $A^0\_{cris}$ since $p^{i+v\_p(j!)}[\frac{x\_i}{\pi^j}]$ lies in $A\_{inf}$. $\Box$


Note that Frobenius is an automorphism of $A\_{inf}[1/p]$ and hence induces an injection on $A^0\_{cris}\subset A\_{inf}[1/p]$. But it might well be non-injective on the $p$-completion. E. g. for $R=\Z[x\_1,x\_2,\dots]$ with injective endomorphism $f(x\_1)=px\_1, f(x\_{i+1})=px\_{i+1}-x\_{i}$ there is a non-zero element $\sum p^i x\_i$ in the kernel of $f:R^{\wedge}\_p\to R^{\wedge}\_p$. Note also that $\varphi$ can increase the $p$-adic valuation of an element of $A^0\_{cris}$ by an arbitrarily big number, e.g. $\varphi(\frac{[\pi]^i}{i!})=\frac{[\pi]^{pi}}{i!}=\frac{(pi!)}{i!}\frac{[\pi]^{pi}}{(pi)!}$ where $v\_p(\frac{(pi)!}{i!})=i$ but we should see that there is no $p$-adicaly convergent sequence in $A^0\_{cris}$ whose $p$-adic valuation increases by bigger and bigger number after applying $\varphi$.


Assume that $\varphi$ on $A^0\_{cris}$ fails to be an injection after the $p$-completion. Namely, there is a sequence $a\_1,a\_2,\dots$ of elements of $A^0\_{cris}$ such that $a\_{i+1}-a\_i$ is divisible by $p^i$ and $\varphi(a\_i)$ is divisible by $p^i$ for every $i$, but for some $i$ the element $a\_i$ is not divisible by $p^i$. Dividing by a suitable power of $p$, we can assume that $a\_1$, and consequently all other $a\_k$ are not divisible by $p$. So, consider the expansion of $a\_1=\sum\limits\_{i=-N}^{\infty}p^i[x\_i]$ with $x\_i\in\cO\_{C}^{\flat}$. There exists $i\leq 0$ with $i+f\_p(x\_i)=0$ since otherwise $a\_1$ would be divisible by $p$ in $A^0\_{cris}$. Moreover, the same is true for every $a\_k$ -- in its Witt expansion there must be a term $p^i[x\_i]$ such that $i+f\_p(x\_i)=0$. I claim that actually for any $a\_k$ there is such $i$ in the interval $[-N,0]$. Suppose this is not the case. Namely, in $a\_k=\sum\limits\_{i=-M}^{\infty}p^i[y\_i]$ we have $i+f\_p(y\_i)>0$ for every $i\in[-N,0]$. By our assumption $a\_k-a\_1$ is divisible by $p$. Since an element is divisible by $p$ iff every term of its Witt expansion is, we know that for each $i$ in the tail $[-M,-N-1]$(expansions of $a\_k$ and $a\_k-a\_1$ coincide in this range) we have $i+f\_p(y\_i)>0$ so actually $a\_k$ is divisible by $p$ which is a contradiction.


I claim that the requirement for all $a\_k$ to have at least one $p$-adically big term in a fixed bounded interval actually prohibits $\varphi(a\_k)$ from becoming arbitrarily $p$-adically small. Indeed, $\varphi(a\_k)=\sum\limits\_{i=-M}^{\infty}p^i[y^p\_i]$ so if $a\_k$ is divisible by $p^n$ then $i-n+f\_p(y\_i^p)\geq 0$. We have $$[v\_{\pi}(y\_i^p)]<pv\_{\pi}(y\_i)+1< p([v\_{\pi}(y\_i)]+1)+1$$ For each $i\in[-N,0]$ denote by $r\_i$ the biggest number such that $v\_p(r\_i!)=-i$. Then put $n=\max\limits\_{i\in[-N,0]}(i+v\_p((p(r\_i+1))!))$. Take big enough $k$ such that $\varphi(a\_k)=\sum\limits\_{i=-M}^{\infty}p^i[y^p\_i]$ is divisible by $p^{n+1}$. By the above, there is $i\in[-N,0]$ such that $i+v\_p([v\_{\pi}(y\_i)]!)=0$. It means that $[v\_{\pi}(y\_i)]\leq r\_i$ so $[v\_{\pi}(y\_i^p)]\leq p([v\_{\pi}(y\_i)]+1)\leq p(r\_i+1)$ hence $i+f\_p(y\_i^p)\leq n$ so $\varphi(a\_k)$ cannot be divisible by $p^{n+1}$. Hence, $\varphi$ stays injective after the $p$-completion.


