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 unramified abelian extensions from the group-theoretic data? So,

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 $K=\overline{\mathbb Q}^H$?

(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".)

  • $\begingroup$ Maybe you should've typed $\overline{\mathbb{Q}}^H$ ? $\endgroup$
    – Qfwfq
    Commented Dec 20, 2010 at 21:04
  • $\begingroup$ @unknown&David: Thank you for fixing this! $\endgroup$ Commented Dec 20, 2010 at 21:44

1 Answer 1


Dear Tim,

As you're probably aware, this is part of the 'anabelian' etcetera.

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.


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.

  • 2
    $\begingroup$ Dear Minhyong, I don't know whether you would consider that as cheating, but maybe it's ok, since you are already identifying the decomposition groups via their presentation à la Neukirch: I don't have Neukirch in front of me but I seem to remember that he writes down the inertia subgroup explicitly in terms of generators of that presentation. I think, in fact, he first writes down a presentation for the inertia subgroup, and then the one for the whole absolute galois group in such a way that it is visible an extension of the inertia subgroup. $\endgroup$
    – Alex B.
    Commented Dec 20, 2010 at 15:41
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    $\begingroup$ Dear Kim, i don't know what your criterion is, but a very simple way to get the maximal unramified extension of a local field with residue characteristic $p$ "group theoretically" is the following: you obtain the maximal tamely ramified extension as the compositum of all extensions such that the interction of the p-Sylow groups of the Galois group is trivial. Then you obtain the maximal unramified taking the maximal abelian extension in that maximal tame, and quotient out the torsion from the Galois group $\endgroup$ Commented Dec 20, 2010 at 15:43
  • $\begingroup$ I agree with both of you. But if we stick just to the portion concerning the recovery of the inertia, the (well-known) argument I gave seems to be more elementary. $\endgroup$ Commented Dec 20, 2010 at 15:56
  • $\begingroup$ Dear Minhyong, really nice!! Let me think a bit more about it, but I am sure I'll accept it. Of course, computing maximal subgroups and identifying their presentation is probably impossible "in practice", but it is a theoretical question anyway. I am not really hoping for anything more explicit... $\endgroup$ Commented Dec 20, 2010 at 21:02

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