what can we say about center of rational absolute galois group? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T07:36:16Zhttp://mathoverflow.net/feeds/question/23711http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/23711/what-can-we-say-about-center-of-rational-absolute-galois-groupwhat can we say about center of rational absolute galois group?krolik2010-05-06T14:01:23Z2010-05-06T16:04:19Z
<p>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..</p>
http://mathoverflow.net/questions/23711/what-can-we-say-about-center-of-rational-absolute-galois-group/23733#23733Answer by Minhyong Kim for what can we say about center of rational absolute galois group?Minhyong Kim2010-05-06T15:24:29Z2010-05-06T15:24:29Z<p>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:</p>
<p>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</p>
<p>$G(P_1)\cap G(P_2)=1.$</p>
<p>Once this is stated for you, it's essentially an exercise to prove. </p>
<p>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.</p>
<p>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.</p>