Injectivity of Frobenius on $A_{cris}$

Question

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?

Answer 1

$\newcommand{\Z}{\mathbb{Z}}$ $\def\cO{\mathcal{O}}$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.