what can we say about center of rational absolute galois group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T07:36:16Z http://mathoverflow.net/feeds/question/23711 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23711/what-can-we-say-about-center-of-rational-absolute-galois-group what can we say about center of rational absolute galois group? krolik 2010-05-06T14:01:23Z 2010-05-06T16:04:19Z <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#23733 Answer by Minhyong Kim for what can we say about center of rational absolute galois group? Minhyong Kim 2010-05-06T15:24:29Z 2010-05-06T15:24:29Z <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>