---
tag: DPFMXS71D1
kind: definition
name: Absolute Galois group
msc: [12F10, "20E18", 11R32]
defines: PGKWBJDTKX
uses: [0QA8CHH8MG, JJ91A1V8NQ, 0XVR72M51M, F3GTF5NT4M]
sources:
  - url: https://en.wikipedia.org/wiki/Absolute_Galois_group
---

## Definition

Let $K$ be a [[field|0QA8CHH8MG]] and let $K^{s}$ be a [[separable closure|JJ91A1V8NQ]] of $K$.
The *absolute Galois group* of $K$ is the [[Galois group|0XVR72M51M]]
$$G_K = \Gal(K^{s}/K)$$
of field automorphisms of $K^{s}$ that fix $K$ pointwise, with the [[Krull topology|F3GTF5NT4M]].

## Remarks

It does not depend on the choice of $K^{s}$ up to isomorphism, because two separable closures are isomorphic over $K$.
The induced isomorphism of Galois groups is unique only up to an inner automorphism.
If $K$ is perfect, for example of characteristic $0$, then $K^{s}$ is an algebraic closure of $K$.
Example: $G_\R = \Gal(\C/\R)$ has order $2$, generated by complex conjugation, since $\C$ is algebraically closed and $[\C : \R] = 2$.
Example: for a finite field $\F_q$, the group $G_{\F_q}$ is isomorphic to the profinite completion $\hat{\Z}$ of $\Z$, topologically generated by $x \mapsto x^q$.
Non-example: $\Gal(\Q(i)/\Q)$ is not the absolute Galois group of $\Q$, since $\Q(i)$ is not separably closed; $x^2 - 2$ has no root in it.
If $K$ is separably closed, then $G_K$ is trivial.