2
$\begingroup$

Let $K$ be a finite extension of $\mathbb{Q}_p$ with ring of integers $\mathcal{O}$, maximal ideal $\mathfrak{m}$ and uniformizer $\pi$. Let $\bar K$ be the algebraic closure of $K$ and $\bar{\mathfrak{m}}$ be the integral closure of $\mathfrak{m}$.

Let $f(x) \in x \mathcal{O}[[x]]$ be a noninvertible power series with finite Weirstrass-degree and let $f^{\circ n}(x)$ be the $n^\text{th}$ iterate of $f(x)$. We denote the set of roots of $f^{\circ n}(x)$ by $S_n=\{x \in \bar{\mathfrak{m}}~|~f^{\circ n}(x)=0 \}$. Also denote $S=\bigcup_n S_n$.

Since $f$ has finite Weierstrass degree, $\#S_n$ is finite. Assume that the field extension $K(S_n)$ is Galois such that \begin{align}\operatorname{Gal}(K(S_n)/K) \cong \left( \mathcal{O}/\pi^n \mathcal{O} \right)^{\times}.\tag{1}\label{1}\end{align} Taking the inverse limit of the relation \eqref{1}, we get \begin{align} \operatorname{Gal}(\bar K/K) \cong \mathcal{O}^{\times}.\tag{2}\label{2} \end{align} There is a continuous surjection $\operatorname{Gal}(\bar K/K) \twoheadrightarrow \operatorname{Gal}(K(S_n)/K)$ by $\sigma \mapsto \sigma\rvert_{K(S_n)}$.

My question—

Suppose $g(x) \in x\mathcal{O}[[x]]$ be another power series such that there exists a $\sigma \in \text{Gal}(K(S_n)/K)$ satisfying \begin{align}\sigma(\alpha)=g(\alpha) ~~\text{for all $\alpha \in S_n$} \tag{3} \label{3} \end{align} How can I extend the relation \eqref{3} to $$\sigma(\alpha)=g(\alpha)~~ \text{for all $\alpha \in S$ and $ \sigma \in \text{Gal}(\bar K/K)$}$$

Improve this question
$\endgroup$
9
  • 5
    $\begingroup$ Under this assumption it seems that the answer is no since the Galois group of $\overline{K}$ is nonabelian but the Galois group of $K(S)$ is (by assumption) abelian. $\endgroup$
    – Will Sawin
    Commented Dec 13, 2021 at 18:51
  • 1
    $\begingroup$ Am I missing something obvious? Isn't the answer that the fibres of $\operatorname{Gal}(\bar K/K) \to \operatorname{Gal}(K(S)/K)$ are $\operatorname{Gal}(\bar K/K(S))$-torsors, so (regardless of what $K$ and $S$ are) cannot be trivial unless $K(S) = \bar K$? Or do you mean whether there is some isomorphism other than the one given by restriction? $\endgroup$
    – LSpice
    Commented Dec 13, 2021 at 19:10
  • 3
    $\begingroup$ Why do you write ${\rm Gal}(\overline{K}/K)$ on the left side of (2)? It would amount to saying $\overline{K} = K(S)$. $\endgroup$
    – KConrad
    Commented Dec 13, 2021 at 20:50
  • 1
    $\begingroup$ Can you simply edit this question to ask only what you intend to ask, rather than leaving the old material up, which makes it a bit confusing? If you ask for all roots of all iterates of all power series, that would be enough, since every finite extension is generated by the roots of some polynomial, thus of some power series. $\endgroup$
    – Will Sawin
    Commented Dec 14, 2021 at 3:09
  • 1
    $\begingroup$ I don't find reason why to close this question. When I said there exists $\sigma \in \text{Gal}(K(S_n)/K)$ such that $\sigma(\alpha)=g(\alpha)$ for all $\alpha \in S_n$ is quite possible. I find no reason to disagree. In fact this is an established fact in a published paper in a reputed journal. $\endgroup$
    – MAS
    Commented Dec 15, 2021 at 3:05

0

Reset to default

You must log in to answer this question.