Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $T$ be a finite set of primes and let $K$ be the maximal extension of $\mathbf{Q}$ unramified outside $T$.

We have three Galois groups:

$G_{\mathbf{Q}} = \mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$

$G_{T} = \mathrm{Gal}(K/\mathbf{Q})$

and for any prime number $p$

$G_p = \mathrm{Gal}(\overline{\mathbf{Q}_p}/\mathbf{Q}_p)$

Are these compact topological groups?

Also, are there any canonical maps between these groups? I think $G_T$ maps to $G_p$ if $p$ is in $T$. Is that correct?

share|improve this question
4  
They're profinite groups: inverse limits of finite groups. The inverse limit is a closed subspace of the direct product, which is compact by Tychonoff (or an easier case of it). –  Dylan Moreland Apr 17 '12 at 21:00

2 Answers 2

up vote 9 down vote accepted

First of all, your question doesn't make sense if you do not specify what topology you want on your group. A group can be equipped with various topologies that yield non-isomorphic topological groups.

There is, however, a nice topology that can be defined on Galois groups called the Krull topology. Let $E/F$ be a Galois extension and $\mathrm{Gal}(E/F)$ its Galois group. The Krull topology on $\mathrm{Gal}(E/F)$ has as basis for its closed sets the subgroups of $\mathrm{Gal}(E/F)$ which fix some finite intermediate extension of $F$ in $E$ (together with all right and left cosets of such subgroups). With the Krull topology, $\mathrm{Gal}(E/F)$ is a compact topological group. In fact, it is a profinite group, i.e. it is Hausdorff, compact, and totally disconnected, or equivalently it is the inverse limit of discrete finite groups.

share|improve this answer

To answer the second part of your question: $G_T$ is obviously a quotient of $G_{\mathbf{Q}}$, and one can view $G_p$ as a subgroup of $G_{\mathbf{Q}}$, but this is only uniquely defined up to conjugation in the latter group -- to define it but one has to make a non-canonical choice (one has to choose, compatibly, a prime above $p$ in every number field).

Composing these one obtains a map $G_p \to G_T$. If $p \notin T$ then this map is rather dull (it factors through a rather small quotient of $G_p$). If $p \in T$ then the map is very interesting: it has recently been shown by Chenevier and Clozel that it is injective if $T$ is large enough (I think $|T| \ge 2$ suffices).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.