# How to find open subgroups of finite index in $\mathbb{Q}_{3}^{\times}$?

For purposes of illustrating Local Class Field Theory, let us play with the $3$-adic numbers.

I'd like to find some open subgroups of finite index in $\mathbb{Q}_{3}^{\times}$.

I know about the decomposition $$\mathbb{Q}_{3}^{\times}\cong\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}_{3},$$ which (I think) is supposed to help us in this task of finding such subgroups, though I'm not sure.

So, even if this is the case, I'm simply an undergraduate and I'm not used to dealing with subgroups of direct products (say I'd like to find them all, parametrizing them somehow), let alone with topological structure involved. How can I do this?

Any help or reference will be greatly appreciated!

• are you sure about that $\Bbb Z_2$ ??? May 6 '16 at 8:33
• Oops; that a typo. Thanks May 6 '16 at 10:54

You’re right, counter-examples are easy to find (e.g. lines in a plane). I don’t know why, I was so convinced that your question was on the subgroups of $\mathbf Z_p$ that I was careless about the direct product. Let me try to repair this. First I prefer to write the structural expression of $\mathbf Q_p^{*}$ for $p$ odd as $p^\mathbf Z$ x $U$, where $\mathbf Q_p$ has a filtration $U > U_1 > … > U_n > …$ , with $U = \mathbf Z_p^{*}$ and $U_n = 1 + p^n\mathbf Z_p$ , a fundamental system of neighbourhoods of 1. The successive quotients are $U/U_1 \cong \mathbf F_p^{*}$ (which admits a lift to $U$ by Hensel's lemma) and $U_n/U_{n+1} \cong \mathbf F_p$ (see e.g. Cassels-Fröhlich, chap. I, propos. 4). An open subgroup $H$ of $\mathbf Q_p^{*}$ must contain one of the $U_n$’s, and then $H.\mathbf F_p^{*}$ /$U_n$ injects into $\mathbf Q_p^{*}$ /$U_n$ = $p^\mathbf Z$ x $U/U_n$. It follows that an open subgroup $H$ of $\mathbf Q_p^{*}$ with finite index is one which contains (in mixed notation)because $p^k\mathbf Z$ x $U_n$, as in Mercio's answer.

The above filtration also shows that the subgroups $(\mathbf Q_p^{*})^n$ form also a fundamental system of neighbourhoods for 1, because $U_n^p \subset U_{n+1}$ and $U_n^m = U_n$ if $m$ is prime to $p$ (op. cit. propos. 5). You ask whether one could parametrize in this way the abelian extensions of $\mathbf Q_p$ via local CFT. In principle you can, if the abelian Galois group of your extension is given precisely enough. You can think of 2 immediate types of parametrizations :

1). Your extension is the maximal abelian pro-$q$-extension of $\mathbf Q_p$, where $q$ is a prime. In principle it suffices to describe the group $G$ = pro-$q$-completion of $\mathbf Q_p^{*}$ = $\varprojlim \mathbf Q_p^{*}$ /$(\mathbf Q_p^{*})^n$ where $n$ runs through the powers of an odd prime $q \neq p$ (the « tame » case) or $n$ runs through the powers of $p$ (« wild » case). This is derived e.g. from H. Koch’s « Galois theory of p-extensions », chap. X . The tame case if $q$ does not divide $p - 1$ is actually the uramified case, precisely $G \cong \mathbf Z_q$ (op. cit. thm. 10.1); more generally, if $q^a$ is the maximal power of $q$ dividing $p - 1$, $G \cong \mathbf Z/q^a$ x $\mathbf Z_q$ (op. cit. thm. 10.2). The wild case is more interesting : $G$ is isomorphic the free abelian pro-$p$-group on 2 generators, i.e. $\mathbf Z_p^{2}$ (op. cit. thm. 10.4). Explicit generators are given in op. cit.

NB: the quotient $\mathbf Q_p^{*}$ /$\mathbf (\mathbf Q_p^{*})^2$ is also known (Serre op. cit. chap. VI, §2)

2). Your extension is squarely the maximal abelian extension $\mathbf Q_p^{ab}$ of $\mathbf Q_p$. This is the « local Kronecker-Weber theorem » : $\mathbf Q_p^{ab}$ is the field generated over $\mathbf Q_p$ by all the roots of unity, and it’s the compositum of two linearly disjoint extensions of $\mathbf Q_p$, the maximal unramified extension, whose Galois group is isomorphic to the profinite completion of $\mathbf Z$, and the $p$-primary cyclotomic extension, generated by all the $p$-primary roots of unity, whose Galois group is isomorphic to $\mathbf Z_p^{*}$ (Cassels-Fröhlich, chap. VI, thm. 1) .

Any open subgroup of $\Bbb Q_p^*$ contains an open neighbourhood of $1$, which means it has to contain $U_k = 1 + p^k \Bbb Z_p$ for some integer $k > 0$.

$\Bbb Z_p^*/U_k = (\Bbb Z/p^k\Bbb Z)^*$, so the quotient of $\Bbb Q_p^*$ by $U_k$ is isomorphic to $p^\Bbb Z \times (\Bbb Z / p^k \Bbb Z)^*$.

Subgroups of $\Bbb Q_p^*$ containing $U_k$ with finite index correspond to subgroups of finite index of the quotient. Since the right factor is a finite group, any subgroup of finite index of $p^\Bbb Z \times (\Bbb Z / p^k \Bbb Z)^*$ must contain an element $p^l$ for some integer $l > 0$ (it has to contain some $p^l \times u$, but taking a suitable power of this annihilates the second component, leaving you with some pure element).

And so, an open subgroup of finite index of $\Bbb Q_p^*$ has to contain $p^{l\Bbb Z} \times U_k$ for some integers $l,k>0$.

Conversely, any subgroup containing any of those two is open and has finite index since the quotient $\Bbb Q_p^* / (p^{l\Bbb Z} \times U_k)$ is isomorphic to $p^{\Bbb Z/l\Bbb Z} \times (\Bbb Z/p^k \Bbb Z)^*$, which is finite.

• Cool, thanks! I also learned yesterday that the groups of powers $(\mathbb{Q}_{p}^{\times})^{n}$ are also a fundamental system of neighbourhoods for $1$. My point with this question was to find such a subgroup and to find the associated finite abelian extension of $\mathbb{Q}_{p}$ via local CFT. Maybe to parametrize such abelians in terms of the subgroups. I guess this is not very nice to compute though. For example, say I want to find the abelian extension of $\mathbb{Q}_{3}$ corresponding to the subgroup of squares $(\mathbb{Q}_{3}^{\times})^{2}$. Can we do this explicitly? Thanks. May 6 '16 at 13:26
• well the subgroup of squares has index $4$ and corresponds to the extension obtained by adjoining $\sqrt 3$ and $\sqrt 2$ if i'm not mistaken May 6 '16 at 13:59
• @Shoutre it is not true that the powers $(\mathbf Q_p^\times)^n$ are a fundamental system of neighborhoods of 1 since those powers don't eventually all get close to 1, e.g., it contains $p^n$ and $(1/p)^n$, which get very close to 0 and very far from 1. The powers $(1 + p\mathbf Z_p)^n$ for $n \geq 0$, or better $(1 + p\mathbf Z_p)^{p^k}$ for $k \geq 0$, are a fundamental system of neighborhoods of 1.
– KCd
May 23 '19 at 16:28
• @KCd Hm, you're right. But I'm confused, open subgroups of finite index $n$ should contain the $n$-th powers. And from what I understand these open subgroups of finite index form a fundamental system of neighborhoods. What am I missing? thank you May 27 '19 at 16:56
• @Shoutre in a profinite group (like $\mathbf Z_p^\times$, but not $\mathbf Q_p^\times$) the open subgroups of finite index as a basis of neighborhoods of the identity.
– KCd
May 28 '19 at 2:50

Any subgroup of a direct product A × B is a direct product of a subgroup of A and a subgroup of B. This is straightforward, so I guess that your problem is actually with the subgroups of $(\mathbf Z, +)$ and $(\mathbf Z_p, +)$ . Note that the structural expression that you give for $\mathbf Q_3^*$ can be generalized to any odd prime $p$ instead of 3 .

Even at undergruate level, you certainly know that all the subgroups of $\mathbf Z$ are of the form $m\mathbf Z$, $m \in \mathbf Z$ ; this is usually shown by using the ring structure of $\mathbf Z$, which is an euclidian domain. A similar argument works for $\mathbf Z_p$ using the $p$-adic (additive) valuation in $\mathbf Q_p$ : let $H$ be a non trivial (additive) subgroup of $\mathbf Z_p$ , and pick $x \in H$ such that $v_p (x)$ is minimal (this is possible because the set $v_p (H)$ is contained in $\mathbf N$). For any $y \in H$, $v_p (y) \ge v_p(x)$, hence $y = xz$, with $n = v_p (z)\ge 0$, i.e. $z \in \mathbf Z_p$. We can further write $z = up^n$, with $u \in \mathbf Z_p^*$. Summarizing : any non trivial subgroup of $\mathbf Z_p$ is of the form $p^n \mathbf Z_p$ ; moreover it can be easily shown that $\mathbf Z_p / p^n \mathbf Z_p$ is isomorphic to $\mathbf Z /p^n\mathbf Z$, see e.g. chapter I of Serre’s « A Course in Arithmetic » .

• Thanks for answering. I'm pretty sure it is not true that subgroups of direct product are direct products of subgroups, though. Other that that, the difficults were mainly topological and if products of subgroups of finite index have finite index in the direct product. May 6 '16 at 10:59
• I already knew about subgroups of each component. The problem is "putting things together". May 6 '16 at 11:01
• Just to emphasize what Shoutre said, your first claim is false (and already $A = B = \mathbb{Z}_2$ is a counterexample; the diagonal copy of $\mathbb{Z}_2$ is not a direct product of a subgroup of $A$ and a subgroup of $B$). The correct statement is given by Goursat's lemma (en.wikipedia.org/wiki/Goursat%27s_lemma). May 6 '16 at 17:36