# what can we say about center of rational absolute galois group?

well the question is in the title. I asked myself this question while thinking about something in grothendieck-teichmüller theory. I guess class field theory gives some insight into this, or i am missing something absolutely obvious..

-
Class Field Theory seems like the wrong way to go since it tells you about the abelianization of the absolute Galois group and not the center. – stankewicz May 6 '10 at 14:12
While I mull over how to prove it, let me just say that the centre is surely trivial, because otherwise it would correspond to a "famous" extension of Q in Q-bar. I don't know a proof yet but probably it shouldn't be hard to find. – Kevin Buzzard May 6 '10 at 14:20

The proof of triviality is a step in the famous Neukirch-Uchida theorem of anabelian geometry, which says a number field is characterized by its absolute Galois group, even functorially, in an appropriate sense. The key elementary fact is the following:

Let $k$ be a number field, $K$ an algebraic closure, and $G=Gal(K/k)$. Let $P_1$ and $P_2$ be two distinct primes of $K$ with corresponding decomposition subgroups $G(P_i)\subset G$. Then

$G(P_1)\cap G(P_2)=1.$

Once this is stated for you, it's essentially an exercise to prove.

Determining the center of $G$ becomes then completely straightforward: Suppose $g$ commutes with everything. Then for any prime $P$, $G(gP)=gG(P)g^{-1}=G(P)$. So $g$ must fix every prime, implying that it's trivial.

I think this is spelled out in the book Cohomology of Number Fields, by Neukirch, Schmidt, and Wingberg. Unfortunately, I left my copy on the plane last year.

-
Indeed it is. Corollary 12.1.3 on page 663 is the disjointness of the decomposition groups, and Corollary 12.1.6 is the triviality of the center of the absolute Galois group. – Cam McLeman May 6 '10 at 17:47