There may be some technical issues with the question, but hopefully what I mean is clear...

Let $k$ be a number field (or maybe any finitely generated field over $\mathbb{Q}$ of characteristic 0)

Let $k(\!(t)\!)$ be the field of Laurent series in $t$ with coefficients in $k$, and let $\Omega$ denote an algebraic closure of $k(\!(t)\!)$, and let $\overline{k}$ denote the algebraic closure of $k$ inside $\Omega$.

There is a natural map $$\rho : \mathrm{Gal}(\Omega/k(\!(t)\!))\rightarrow \mathrm{Gal}(\overline{k}/k)$$ given by restriction.

If $\{t_n\}_{n\ge 1}\subset \Omega$ satisfies $t_1 = t$, $t_n^d = t_{n/d}$ for all $d\mid n$, then we say that it is a compatible system of roots of $t$.

Given a compatible system of roots $\{t_n\}$, and a filtration $\overline{k} = \bigcup L$ by finite extensions of $k$, the tensor product $\left(\varinjlim_L L(\!(t)\!)\right)\otimes_{k(\!(t)\!)} \left(\varinjlim_n k(\!(t)\!)(t_n)\right)$ is a field isomorphic to $\Omega$, and thus we obtain an section of the map $\rho$ by having $\sigma\in \mathrm{Gal}(\overline{k}/k)$ act on the tensor product in the obvious way in the first factor, and trivially on the second factor. In particular, this action stabilizes the compatible system of roots $\{t_n\}$.

Does every section of $\rho$ stabilize some compatible system of roots $\{t_n\}$?

Note that the kernel of $\rho$ is the subgroup $\mathrm{Gal}(\Omega/E)$ where $E = \varinjlim_L L(\!(t)\!)$ as before, and the Galois group is isomorphic to $\widehat{\mathbb{Z}}$. The group $\mathrm{Gal}(\overline{k}/k)$ acts on $\mathrm{Gal}(\Omega/E) \cong \widehat{\mathbb{Z}}$ via the cyclotomic character, and if $k$ is finitely generated over $\mathbb{Q}$ then the only element of $\widehat{\mathbb{Z}}$ fixed by all of $\mathrm{Gal}(\Omega/E)$ is the identity. This implies that if a section of $\rho$ comes from a compatible system, then it must be unique.


$\newcommand{\Gal}{\mathrm{Gal}}\newcommand{\Z}{\mathbb{Z}}$Fix a compatible system $(t_n)$ of roots of $t$. It provides us with a section of $\rho$ thus giving an isomorphism between $\Gal(\overline{k((t))}/k((t)))$ and the semi-direct product $\Gal(\overline{k}/k)\ltimes \hat{\Z}(1)$ where $\hat{\Z}(1)$ denotes $\lim\limits_{\leftarrow}\mu_n(\bar{k})$ as a $\Gal(\bar{k}/k)$-module. An element $(g,m)$ acts on $\overline{k((t))}$ as follows: in terms of your decomposition $$\overline{k((t))}=\left(\varinjlim_L L(\!(t)\!)\right)\otimes_{k(\!(t)\!)} \left(\varinjlim_n k(\!(t)\!)(t_n)\right)$$ it acts on the first factor as prescrivbed by $g$ and, on the second factor sends $t_n$ to $m_nt_n$ where $m_n$ is the projection of $m$ to $\mu_n(\bar{k})$.

If $(t'_n)$ is another compatible system of roots of $t$, it provides another section $\Gal(\bar{k}/k)\to \Gal(\overline{k((t))}/k((t)))$ which is characterized by the fact that it preserves all the elements $t'_n$. In terms of the semi-direct product, this section then looks like $g\mapsto (g,g(m)-m)$ where $m\in\hat{\Z}(1)$ is the element whose components $m_n\in\mu_n(\bar{k})$ are given by $t_n^{-1}t'_n$

In general, if $G\ltimes M$ is the semidirect product of a profinite group $G$ with a continuous module $M$, a giving a continuous section $s:G\to G\ltimes M$ is equivalent to giving a map $f:G\to M$ such that $f(g_1g_2)=g_1f(g_2)+f(g_1)$.

Coming back to our situation we see that section are in bijection with continuous 1-cocycles of the group $\Gal(\bar{k}/k)$ with coefficients in $\hat{\Z}(1)$ and a section comes from a system of roots if and only if this cocycle is a coboundary. In other words, we get

Lemma. Every continuous section comes from a compatible system of roots of $t$ if and only if $H^1_{cont}(\Gal(\bar{k}/k),\hat{\Z}(1))=0$.

In your setting, this group is never zero, for example, from the exact sequence $0\to\hat{\Z}(1)\xrightarrow{n}\hat{\Z}(1)\to \mu_n\to 0$ we see that $n$-torsion in this cohomology group is given by $$H^1_{cont}(\Gal(\bar{k}/k),\hat{\Z}(1))[n]=\mathrm{coker}(\hat{\Z}(1)^{\Gal(\bar{k}/k)}\to \mu_n(k))$$ Since the order of roots of unity contained in $k$ is bounded, $\hat{\Z}(1)^{\Gal(\bar{k}/k)}$ is zero, but, for instance, $\mu_2(k)$ is equal to $\Z/2$ for every $k$.

So, there are always other sections but you can control them by this Galois cohomology group.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.