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Q1: Do we have a criterion which allows us to say when is a profinite group $G$ topologically finitely generated?

For example, if $G$ is topologically finitely generated then, for a fixed integer $N$, there are only finitely many open subgroups $H\leq G$ such that $[G:H]=N$. In general, I guess that this condition is not sufficient.

Q2: Where can I find a proof that the absolute galois of $ \mathbf{Q}_p $ is topologically finitely generated?

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For Q2, see Pete Clark's answer mathoverflow.net/questions/13733/wild-ramification/13743#13743 to a previous question. –  Rob Harron Apr 27 '11 at 1:39

4 Answers 4

up vote 6 down vote accepted

See the theorem of Mann quoted by Jaikin in this question for a nice criterion guaranteeing topological finite generation in terms of the number of open subgroups of index N. See Pete Clark's answer to that question for pointers towards verifying that criterion for Gal(Q_p).

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The following profinite group is not finitely generated but has only finitely many open subgroups of each degree.

$$G=\prod_{n=5}^\infty (Alt_n)^{(n+1)!}.$$

Note that you need at least $n+1$ elements in order to generate $(Alt_n)^{(n+1)!}$.

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Thanks Andrei for the simple example. –  Hugo Chapdelaine Apr 27 '11 at 11:47

Here is the bare outline of a proof that the absolute Galois group of a finite extension $K$ of $\mathbf{Q}_p$ is finitely generated (as a profinite group). Let $\Omega$ be an algebraic closure of $K$, and denote groups such as $\operatorname{Gal}(\Omega|K)$ by $\Gamma(\Omega|K)$.

The profinite group $\Gamma(\Omega|K)$ has the closed normal subgroup $\Gamma(\Omega|K')$ (the ramification group --- a pro-$p$-group), where $K'$ is the maximal tamely ramified extension of $K$ in $\Omega$, giving rise to the short exact sequence $$ 1\to\Gamma(\Omega|K')\to\Gamma(\Omega|K)\to\Gamma(K'|K)\to1. $$ It is well known that the profinite group $\Gamma(K'|K)$ admits the presentation $$ \langle\sigma,\tau\mid\sigma\tau\sigma^{-1}=\tau^q\rangle, $$ where $q=p^f$ is the cardinality of the residue field $k$ of $K$ (see for example a Hasse's book or a paper by Iwasawa in the Transactions). In view of this, it is sufficient to show that the pro-$p$-group $\Gamma(\Omega|K')$ is finitely generated as a closed normal subgroup of $\Gamma(\Omega|K)$.

This would follow if its abelianisation $\Delta(\Omega|K')$ were finitely generated as a $\mathbf{Z}_p[[\Gamma(K'|K)]]$-module, which would in turn follow if $\Delta(\Omega|K')/\Delta(\Omega|K')^p$ were finitely generated as an $\mathbf{F}_p[[\Gamma(K'|K)]]$-module. But such is indeed the case because the $\mathbf{F}_p[\Gamma(K'|K)]$-module $K'^\times/K'^{\times p}$ is finitely generated (by local class field theory or more simply by Kummer theory), completing the proof.

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Thanks a lot Chandan for the nice sketch of the proof. –  Hugo Chapdelaine Apr 27 '11 at 12:11
    
Hi @Chandan, I was wondering if you had a reference for the following statement that you made: If $\Delta(\Omega/K')$ is a finitely generated $Z_p[[\Gamma(K'/K)]]$-module then you may find finitely many elements of $\Gamma(\Omega/K')$ such that the normal closure of these elements generate it. –  Hugo Chapdelaine Jun 2 '12 at 16:23
    
@Hugo: Sorry for the belated reply; I was away for a few days. Perhaps a more detailed account of the above outline will get written up some time soon. –  Chandan Singh Dalawat Jun 7 '12 at 5:48

If $G_1$ has rank $r$, then all finite quotients of $G_1$ will have rank $r$.

Conversely, suppose $G$ is a profinite group such that every finite quotient has rank $\leq r$ (and some finite quotient has rank $r$). The set of finite quotients forms a lattice, since if one has epimorphisms $G\twoheadrightarrow A, G\twoheadrightarrow B$ where $A$ and $B$ are finite groups, then one has a map $G\twoheadrightarrow H\leq A\times B$, such that $G\twoheadrightarrow H\twoheadrightarrow A$ and $G\twoheadrightarrow H\twoheadrightarrow B$.

For any given finite quotient $G\twoheadrightarrow A$, consider the subset $Gen(A)\subset A^r$ consisting of $r$ elements which generate $A$. Since $A$ is finite, $Gen(A)$ is finite for all $G \twoheadrightarrow A$. By hypothesis, $Gen(A)$ is non-empty for all $G \twoheadrightarrow A$, and clearly if we have $G \twoheadrightarrow H \twoheadrightarrow A$ where $H$ is finite, then we have a map $Gen(H) \twoheadrightarrow Gen(A)$.

Thus, we may choose for all $G \twoheadrightarrow A$ an $r$-tuple $g(A)=(a_1,\ldots, a_r)\in Gen(A)$, such that if $G\twoheadrightarrow H \twoheadrightarrow A$, then $g(H)\mapsto g(A)$. This follows by the lattice property of finite quotients: we may find $G \twoheadrightarrow \cdots H_i \twoheadrightarrow \cdots \twoheadrightarrow H_2 \twoheadrightarrow H_1$ an inverse sequence of finite groups such that for any $G \twoheadrightarrow A$, there exists $i$ such that $G \twoheadrightarrow H_i \twoheadrightarrow A$. Choose $g(H_i)$ inductively so that for each $j>i$, there exists $v\in Gen(H_j)$ such that $v\mapsto g(H_i)$. Then define $g(A)=\phi(g(H_i))$ for some $i$ such that there is a factoring $G \twoheadrightarrow H_i \overset{\phi}{\twoheadrightarrow} A$. Now, the collection $\{ g(A), G \twoheadrightarrow A\}$ defines a unique $r$-tuple in $G$ which determines a topological generating set for $G$, since the cosets of the groups $ker\{G \twoheadrightarrow A\}$ form a neighborhood basis of $1\in G$, and therefore any nested sequence of cosets of $ker\{G \twoheadrightarrow A\}$ determines a unique element of $G$.

Thus, $rank(G)$ is the supremum of ranks of finite quotients of $G$.

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