# Absolute Galois group of the field of Puiseux series over $\overline{\mathbb{F}}_p$

Let $K$ be the field of Puiseux series with coefficients in $\overline{\mathbb{F}}_p$ (the algebraic closure of the field with $p$ elements). What is the absolute Galois group of $K$?

Thank you to anyone who could help!

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One can compute the set of continuous homomorphisms from the absolute Galois group to $\mathbb F_p$ using the Artin-Schreier exact sequence. A basis, as a vector space over $\bar{\mathbb F}_p$ consists of functions of the form $T^{-a/b}$ with $a$, $b$, and $p$ pairwise relatively prime. This is a weird-looking countable-dimensional vector space, which makes it seem unlikely that one can find a nice description for the whole group. In particular, it cannot be finitely topologically generated. –  Will Sawin Nov 16 '12 at 6:11
Is this known if we replace $\bar{\mathbb{F}}_p$ with $\mathbb{C}$? Can one reason by the analogy? –  Spice the Bird Nov 16 '12 at 7:58
@Spice: if you replace $\overline{\mathbb{F}}_p$ iwth $\mathbb{C}$, then the field you get is algebraically closed. –  Laurent Berger Nov 16 '12 at 9:25
On a webpage entitled "Questions I'm thinking about", Kiran Kedlaya wrote "I have a method for computing in the algebraic closure of the rational function field over a finite field, using finite automata and generalized power series. Does it actually work in practice? I can't tell. (There has been a tiny bit of experimental work on this; contact me for details.)" math.ucsd.edu/~kedlaya/questions.shtml , last updated Dec. 2009. –  David Speyer Nov 16 '12 at 14:28
I think there is a 2001 paper by Kedlaya where he described the algebraic closure of $\overline{\mathbb{F}}_p((x))$ building on other's work, it is like the Puiseaux series together with towers of Artin-Schreier extensions. –  36min Nov 20 '12 at 2:15

Let $E$ be the field $\overline{\mathbb{F}}_p((X))$. The field of Puiseux series whose exponents have denominators prime to $p$ is a subfield of $E^{sep}$, so the group you're asking about would then be the wild inertia subgroup of $Gal(E^{sep}/E)$. The group $Gal(E^{sep}/E)$ is quite complicated, and it comes up in arithmetic geometry, for example when studying the $\pi_1$ of curves. It also occurs as a closed subgroup of $Gal(\overline{\mathbb{Q}}_p / \mathbb{Q}_p)$ by the theory of the field of norms of Fontaine and Wintenberger. Its representations on $\mathbb{Z}_p$-modules are described by $\varphi$-modules'' (like $(\varphi,\Gamma)$-modules without the $\Gamma$). If you want to include Puiseux series whose exponents have denominators divisible by $p$, then you're looking at the perfection of $E$. The group does not change, as $E^{sep}$ is dense in $E^{alg}$ by a theorem of Ax.
Thank you for the answers ! Would it be possible to explain in a few words what $(\phi, \Gamma)$-modules are ? Thanks again ! –  user29243 Nov 20 '12 at 0:09