- Absolute Galois group
In

mathematics , the**absolute Galois group**"G_{K}" of a field "K" is theGalois group of "K"^{sep}over "K", where "K"^{sep}is aseparable closure of "K". Alternatively it is the group of all automorphisms of thealgebraic closure of "K" that fix "K". The absolute Galois group is uniqueup to isomorphism. It is aprofinite group .(When "K" is a

perfect field , "K"^{sep}is the same as analgebraic closure "K"^{alg}of "K". This holds e.g. for "K" ofcharacteristic zero , or "K" afinite field .)**Examples*** The absolute Galois group of an algebraic closed field is trivial.

* The absolute Galois group of thereal number s is a cyclic group of two elements (complex conjugation and the identity map), since $Bbb\{C\}$ is the separable closure of $Bbb\{R\}$ and $[\; Bbb\{C\}\; :\; Bbb\{R\}\; ]\; =2$.

* The absolute Galois group of afinite field "K" is isomorphic to the group ::$hat\{mathbb\{Z\; =\; lim\_\{leftarrow\}mathbb\{Z\}/nmathbb\{Z\}$. TheFrobenius automorphism Fr is a canonical generator of "G_{K}". (Recall that Fr("x") = "x^{q}" for all "x" in "K"^{alg}, where "q" is the number of elements in "K".)

* The absolute Galois group of the field of rational functions with complex coefficients is free (as a profinite group). This result is due toAdrien Douady and has its origins inRiemann's existence theorem .

* More generally, Let "C" be an algebraically closed field and "x" a variable. Then the absolute Galois group of "K"="C"("x") is free of rank equal to the cardinality of "C". This result is due toDavid Harbater andFlorian Pop , and was also proved later byDan Haran andMoshe Jarden .

* Let "K" be a "p"-adic field. Then its absolute Galois group is finitely generated and has an explicit description by generators and relations.**Problems*** No direct description is known for the absolute Galois group of the

rational number s. In this case, it follows fromBelyi's theorem that the absolute Galois group has a faithful action on the "dessins d'enfants " ofGrothendieck (maps on surfaces), enabling us to "see" the Galois theory of algebraic number fields.* Let "K" be the maximal

abelian extension of the rational numbers. Then**Shafarevich's conjecture**asserts that the absolute Galois group of "G_{K}" is the free profinite group of countable rank.**Some general results*** Every profinite group occurs as a Galois group of some Galois extension, however not every profinite group occurs as an absolute Galois group. For example Artin-Schreier Theorem asserts that the only finite absolute Galois groups are the trivial one and the cyclic group of order 2.

* Every projective profinite group can be realized as a absolute Galois group of a

Pseudo algebraically closed field . This result is due toAlexander Lubotzky andLou van den Dries .**References*** M. D. Fried and M. Jarden, Field Arithmetic, Second Edition, revised and enlarged by Moshe Jarden, Ergebnisse der Mathematik (3) 11, Springer, Heidelberg, 2004.

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