$(\varphi,\Gamma)$-modules and valuation in Zp

Question

The theory of $(\varphi,\Gamma)$-modules and a reciprocity law due to Cherbonnier and Colmez involve the ring $A_K$, that is described below.

The question is why for the valuation defined as $l_ν(a):=min\{i| p^ν - does- not- divide - a_i\}$ of $a=\sum_i a_i(\pi_k)^i$ $\in$ $A_K$, the $l_1(a)$ is independent of a choice of uniformizer for $A_K$ but $l_v(a)$ for $v\geq2$ is not?

Can someone help me here, please? Is there an example of where one can see this?

The ring $A_K$: Set $E$ to be the set of sequences $(x^{(0)},x^{(1)}, . . . )$ of elements of $C_p$ satisfying $(x^{(n+1)})^p=x^{(n)}$, with addition given by $(x + y)^{(n)} = \lim_m (x^{(n+m)} + y^{(n+m)})^{p^m}$ and $(xy)^{(n)} = x^{(n)}y^{(n)}$. Then $E$ is a complete, algebraically closed field of characteristic $p$. Fix a compatible system of roots of unity $(\zeta_p^m),\forall m\in\mathbb N$ where $(\zeta_p^{m+1})^p= \zeta_p^m$ , and write $K_n=K(\zeta_p^n)$. We set $\epsilon=(1,\zeta_p,\zeta_p^2,...)$ and denote by $E_{\mathbb Q_p}$ the subfield of $E$ given by $F_p((\epsilon− 1))$. We write $E_s$ for its separable closure and note that $E$ is the completion of the algebraic closure. We write $E_K=E^{H_K}$ for the subfield of $E$ fixed by $H_K=\ker\chi^{cyclo}$.

We take $A=W(E)$ the ring of Witt vectors over $E$, and $B=A[1/p]=Frac(A)$. This is a complete discrete valuation field with residue field $E$. We can write $x\in A$ as $\sum_{k=0}^\infty p^k[x_k]$ where $x_k\in E$ and $[\cdot]$ is the Teichmüller lift.

We write $\pi=[\epsilon]−1$, and $A_{\mathbb Q_p}$ for the closure of $\mathbb Z_p[\pi,\pi^{−1}]$ in $A$; it is a complete discrete valuation ring with residue field $E_{\mathbb Q_p}$. We write $B_{\mathbb Q_p}=Frac(A_{\mathbb Q_p})=A_{\mathbb Q_p}[1/p]$. We have actions of $\varphi$ and $G$ on $B$ given by $\varphi(\pi)=(1 + \pi)^p−1$ and $g(\pi)=(1+\pi)^{\chi^{cyclo}(g)}−1$.

We write $B_c$ for the closure of the maximal unramified extension of $B_{\mathbb Q_p}$ in $B$ and $A_c=B\cap A$ such that $A_c[1/p]=B$ We have $B_K=B^{H_K}$ and $A_K=A^{H_K}$. We observe that $$ B_K=\{\sum_{n \in\mathbb Z} a_n \pi_K^n : a_n \in F, \lim_{n \rightarrow -\infty} a_n=0\}$$

where $\pi_K$ is $e-$th root of $\pi$ and a uniformizer of $B_K$ . We likewise have

$$ A_K=\{\sum_{n \in\mathbb Z} a_n \pi_K^n \in B_K : a_n \in\mathcal O_F, \forall n\in\mathbb Z\}$$

Answer 1

After clarifying terminology in the comments, I think this is what's going on here:

Let $F$ be a local field with ring of integers $O_F$ whose maximal ideal is $(p)$, and residue field $k := O_F/(p)$. Let $A$ be the set of Laurent series

$\displaystyle \sum_{i\in \Bbb Z} a_i \pi^i: a_i \in O_F$ such that $\lim_{i\to -\infty} a_i=0$

(i.e. the series can be infinite on both sides, but to the left, the coefficients go to $0$). Then one can show that $A$ is a complete DVR with maximal ideal $(p)$, whose residue field $A/p$ (that's $E_K$ in your question) is of the form $k((t))$ (usual Laurent series, with finite negative part, over the field $k$). Notice that $k((t))$ is itself a complete field with a discrete valuation, whose ring of integers is $k[[t]]$ (power series over $k$). The uniformisers of this are precisely the elements of the form $\bar a_1 t + \sum_{i\ge 2}\bar a_i t^i$, where $\bar a_i \in k$ and $\bar a_1 \neq 0$.

Next, let $\varpi \in A$ be any lift of any such uniformiser. It is of the form

$\varpi = \displaystyle \sum_{i\in \Bbb Z} a_i \pi^i$ with $a_i \in (p)$ for $i\le 0$, and $a_1\in O_F^*$

(a special case being $\varpi = \pi$ itself).

Claim: Every element of $A$ can also be written uniquely in the form $$\displaystyle \sum_{i\in \Bbb Z} b_i \varpi^i: b_i \in O_F \text{ such that } \lim_{i\to -\infty} b_i=0.$$

Now in the article they define functions $l_\nu: A \rightarrow \Bbb Z$ by

$l_\nu(\sum_{i\in \Bbb Z} a_i \pi^i) := \min\lbrace i: p^\nu \text{does not divide } a_i\rbrace$.

but one could also define, for each $\varpi$, the variant

$l_{\nu, \varpi}(\sum_{i\in \Bbb Z} b_i \varpi^i) :=\min\lbrace i: p^\nu \text{does not divide } b_i\rbrace$

(so that $l_\nu$ is the choice $l_{\nu, \pi}$).

Now:

• For $\nu \ge 2$, in the examples $\varpi = p^{\nu-1} + \pi$ we have $l_\nu(\varpi) = 0$ but $l_{\nu, \varpi}(\varpi) = 1$.

• For $\nu=1$ however, a little consideration shows that both $l_{\nu}$ and $l_{\nu, \varpi}$ can actually be described as

$l_\nu(a)= l_{\nu, \varpi}(a) = \text{ the } t-\text{adic valuation of } \bar a \in A/(p) = k((t))$

which does not depend on the choice of $\varpi$.