## 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 $$\rho : Gal(\Omega/k((t)))\rightarrow Gal(\overline{k}/k)$$ 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 $$\sigma\in Gal(\overline{k}/k)$$ 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 $$Gal(\Omega/E)$$ where $$E = \varinjlim_L L((t))$$ as before, and the Galois group is isomorphic to $$\widehat{\mathbb{Z}}$$. The group $$Gal(\overline{k}/k)$$ acts on $$Gal(\Omega/E) \cong \widehat{\mathbb{Z}}$$ 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 $$Gal(\Omega/E)$$ is the identity. This implies that if a section of $$\rho$$ comes from a compatible system, then it must be unique.