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..
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3$\begingroup$ 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. $\endgroup$– stankewiczCommented May 6, 2010 at 14:12
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18$\begingroup$ 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. $\endgroup$– Kevin BuzzardCommented May 6, 2010 at 14:20
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1 Answer
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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.
answered May 6, 2010 at 15:24
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11$\begingroup$ 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. $\endgroup$ Commented May 6, 2010 at 17:47