Are class numbers encoded in the absolute Galois group of ${\mathbb Q}$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:32:18Z http://mathoverflow.net/feeds/question/49960 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49960/are-class-numbers-encoded-in-the-absolute-galois-group-of-mathbb-q Are class numbers encoded in the absolute Galois group of ${\mathbb Q}$? Tim Dokchitser 2010-12-20T13:27:49Z 2010-12-20T21:33:58Z <p>The absolute Galois group $G_{\mathbb Q}=\text{Gal}(\bar{\mathbb Q}/\mathbb Q)$, as a profinite group, encodes a lot of things: the whole lattice of number fields (closed subgroups of finite index), Galois extensions (normal subgroups), abelian extensions etc. Is it possible to recognize <em>unramified</em> abelian extensions from the group-theoretic data? So,</p> <blockquote> <p>Question: Given a closed subgroup $H\subset G_{\mathbb Q}$ of finite index, is there an explicit group-theoretic way to recover the class group or the class number of the fixed field $K=\overline{\mathbb Q}^H$?</p> </blockquote> <p>(Because all continuous automorphisms of $G_{\mathbb Q}$ are inner, in theory $K$ itself can be recovered from $H$ and its class group computed, but this is not what I'd like to call "group-theoretic" or "explicit".)</p> http://mathoverflow.net/questions/49960/are-class-numbers-encoded-in-the-absolute-galois-group-of-mathbb-q/49967#49967 Answer by Minhyong Kim for Are class numbers encoded in the absolute Galois group of ${\mathbb Q}$? Minhyong Kim 2010-12-20T15:05:28Z 2010-12-20T17:51:39Z <p>Dear Tim,</p> <p>As you're probably aware, this is part of the 'anabelian' etcetera. </p> <p>It suffices to recover all intertia subgroups $I_v\subset H$, because their union will then be a normal subgroup $N$ such that $H/N$ is the Galois group of the maximal extension of $K$ unramified everywhere. We can get the ideal class group then by (topological) abelianization. The fact that we can get all the decomposition groups $D_v\subset G$ is Neukirch's theorem (together with Artin-Schreier at infinity). This says the maximal subgroups isomorphic to a local Galois group are exactly the decomposition groups. If you want to make this purely group-theoretic for the finite places, you invoke the theorem of Jannsen-Wingberg that lays out a presentation for all local Galois groups and consider maximal elements in the lattice of subgroups isomorphic to such an explicit presentation. Once you have the $D_v$, there is a standard group-theoretic recipe for $I_v$, which escapes me for the moment. But I'll get back to you with it, if you don't figure it out in the meanwhile.</p> <p>Added:</p> <p>OK, so here is the easy part. Now let $F$ be a finite extension of $\mathbb{Q}_p$ and $D=Gal(\bar{F}/F)$. We know that $D^{ab}$ fits into an exact sequence $$0\rightarrow U_F\rightarrow D^{ab}\rightarrow \hat{\mathbb{Z}}\rightarrow 0,$$ so we recover $p$ as the unique prime such that the topological $\mathbb{Z}_p$-rank of $D^{ab}$ is $r_D\geq 2$. The order $q_D$ of the residue field is 1 greater than the order of the prime-to-$p$ torsion subgroup of $D^{ab}$. Also, we know $r_D=1+[F:\mathbb{Q}_p]$. Now we apply the same reasoning to the subgroups of finite index in $D$ to figure out those corresponding to unramified extensions. That is, consider the subgroups $E$ of finite index such that $q^{r_D-1}_E=q_D^{r_E-1}$. Then the inertia subgroup of $D$ is the intersection of all of these.</p>