# Does every section of the map Gal$(\overline{k(\!(t)\!)}/k(\!(t)\!))\rightarrow$ Gal$(\overline{k}/k)$ stabilize a compatible system of roots of $t$?

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.

## 1 Answer

$$\newcommand{\Gal}{\mathrm{Gal}}\newcommand{\Z}{\mathbb{Z}}$$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.