Absolute Galois group of $\Bbb Q_p$ while varying $p$

$\newcommand{\Q}{\Bbb Q} \newcommand{\N}{\Bbb N} \newcommand{\R}{\Bbb R} \newcommand{\Z}{\Bbb Z} \newcommand{\C}{\Bbb C} \newcommand{\A}{\Bbb A} \newcommand{\ab}{\mathrm{ab}} \newcommand{\Gal}{\mathrm{Gal}} \newcommand{\prolim}{\varprojlim}$ Let $p, q$ be two distinct prime numbers.

Are the groups $\Gal(\overline{\Q_p} / \Q_p)$ and $\Gal(\overline{\Q_q} / \Q_q)$ isomorphic (as abstract groups)? Are they isomorphic as topological groups?

Thoughts:

1) The absolute Galois group of a number field $K / \Q$ determines the number field, by Neukirch–Uchida theorem. But this does not apply to local fields.

2) This reference explains that the absolute Galois group of a $p$-adic number field (with $p$ fixed) only determines the field up to elementarily equivalence. But here I'm asking for various $p$.

3) I think these two profinite groups are homeomorphic, by Brouwer theorem (every non-empty second countable totally disconnected compact Hausdorff topological space without isolated points is homeomorphic to the Cantor middle-third set). But I'm asking about (topological) group isomorphisms.

Thank you!

• math.stackexchange.com/questions/495779/… here is complete description of this absolute Galois group (short answer: no, they are different even abstractly) – xsnl Jul 8 '18 at 20:37
• Possible duplicate of Galois group over $p$-adic numbers – xsnl Jul 8 '18 at 20:37
• @xsnl : how is it a duplicate? My question is much more precise, and did not find at all the answer in your link. – Watson Jul 8 '18 at 20:41
• @xsnl : you could maybe post your ideas as an answer, instead. Thank you very much! – Watson Jul 8 '18 at 20:52

Watson, the answer to your question is that those groups are nonisomorphic as topological groups, and by being a bit more careful they are nonisomorphic as abstract groups also. We will pass to the abelianization of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$, which is ${\rm Gal}(\mathbf Q_p^{\rm ab}/\mathbf Q_p)$, and show this group knows what $p$ is. First we will check that the abelianization of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ is a purely algebraic construct.

For a topological group, its abelianization is the quotient by its commutator subgroup, just like for abstract groups, but there is a catch: the definition of the commutator subgroup of a topological group is the closure of the ordinary commutator subgroup (viewing the group as an abstract group). Fortunately, it turns out that in the case of $\mathbf Q_p$, or finite extensions of $\mathbf Q_p$, the ordinary commutator subgroup of the absolute Galois group is closed. This is because of the following two facts.

1. Jannsen and Wingberg proved that the absolute Galois group of a finite extension $K/\mathbf Q_p$, for $p \not= 2$, is topologically finitely generated (in fact finitely presented): this is in a few articles in Inventiones vol. 70 (1982/83). Diekert settled the case $p=2$ shortly afterwards. Diekert gives an explicit finite presentation when $\sqrt{-1} \in K$ for a $2$-adic field $K$, but if $\sqrt{-1} \not\in K$ then by applying Diekert's result to $K(\sqrt{-1})$, an open subgroup of index $2$ in ${\rm Gal}(\overline{K}/K)$ is topologically finitely generated (in fact topologically finitely presented) so ${\rm Gal}(\overline{K}/K)$ is topologically finitely generated with one more generator.

2. Novikov and Segal (https://arxiv.org/abs/math/0604399) showed that for every topologically finitely generated profinite group, its commutator subgroup as an abstract group is closed. (Serre had proved this earlier when the group is a topologically finitely generated pro-$p$ group.)

Putting these together, the abelianization of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ as an abstract group is the same as its abelianization as a topological group, which is ${\rm Gal}(\mathbf Q_p^{\rm ab}/\mathbf Q_p)$. Local class field theory tells us that the group ${\rm Gal}(\mathbf Q_p^{\rm ab}/\mathbf Q_p)$ is isomorphic to $\widehat{\mathbf Z} \times \mathbf Z_p^\times$ as topological groups, and hence as abstract groups. We will show the abstract group $\widehat{\mathbf Z} \times \mathbf Z_p^\times$ knows what $p$ is.

Set $G = \widehat{\mathbf Z} \times \mathbf Z_p^\times$, so the torsion subgroup $G_{\rm tor}$ is $(\mathbf Z_p^\times)_{\rm tor}$, which is a finite subgroup of $\mathbf Z_p^\times$: $G/G_{\rm tor} \cong \widehat{\mathbf Z} \times (\mathbf Z_p^\times)/(\mathbf Z_p^\times)_{\rm tor}$. The group $\mathbf Z_p^\times$ is a topological group, whether you want to think of it like that or not. Using the $p$-adic logarithm, we have $\mathbf Z_p^\times/(\mathbf Z_p^\times)_{\rm tor} \cong \mathbf Z_p$ as topological groups, and thus also as abstract groups. Therefore $$G/G_{\rm tor} \cong \widehat{\mathbf Z} \times \mathbf Z_p \cong \left(\prod_{\ell} \mathbf Z_\ell\right) \times \mathbf Z_p \cong \prod_{\ell \not= p} \mathbf Z_\ell \times \mathbf Z_p^2$$ as abstract groups. To show the abstract group $\widetilde{G} := G/G_{\rm tor}$ knows $p$, observe that for a prime number $\ell$, $\widetilde{G}/\ell\widetilde{G}$ is isomorphic to $\mathbf Z/\ell\mathbf Z$ if $\ell \not= p$ and is isomorphic to $(\mathbf Z/p\mathbf Z)^2$ if $\ell = p$, so $|\widetilde{G}/\ell\widetilde{G}| \not= \ell$ only at $\ell = p$. Thus $G$, as an abstract group, knows what $p$ is.

Next you will probably ask, for $K$ a finite extension of $\mathbf Q_p$ and $L$ a finite extension of $\mathbf Q_q$, where $p$ and $q$ are different primes, could ${\rm Gal}(\overline{K}/K)$ be isomorphic to ${\rm Gal}(\overline{L}/L)$ as abstract groups? If they were, then again we could pass to the abelianizations (abstract and topological abelianizations are the same thing by the two hard theorems at the start) and local class field theory then implies $\widehat{\mathbf Z} \times \mathcal O_K^\times \cong \widehat{\mathbf Z} \times \mathcal O_L^\times$ as abstract groups. We want to show the abstract group $G := \widehat{\mathbf Z} \times \mathcal O_K^\times$ knows what $p$ is, where $K$ is a finite extension of $\mathbf Q_p$.

The solution we used when $K = \mathbf Q_p$ can be made to work when $K$ is just a finite extension of $\mathbf Q_p$, with some further details needed. The torsion subgroup $G_{\rm tor}$ equals $(\mathcal O_K^\times)_{\rm tor}$ and $G/G_{\rm tor} \cong \widehat{\mathbf Z} \times (\mathcal O_K^\times)/(\mathcal O_K^\times)_{\rm tor}$ as abstract groups. Using the $p$-adic logarithm (extended from $1 + \mathfrak m_K$ to $\mathcal O_K^\times$ by killing the prime to $p$ roots of unity), $(\mathcal O_K^\times)/(\mathcal O_K^\times)_{\rm tor}$ is isomorphic to an open subgroup of $\mathcal O_K$ as topological groups. Since $\mathcal O_K \cong \mathbf Z_p^n$ as topological groups, where $n = [K:\mathbf Q_p]$, its open subgroups are isomorphic to $\mathbf Z_p^n$. Thus $G/G_{\rm tor} \cong \widehat{\mathbf Z} \times \mathbf Z_p^n$ as topological groups, and thus also as abstract groups. (I am using topology merely as a tool to understand the structure of $G/G_{\rm tor}$ as an abstract group.) Setting $\widetilde{G} := G/G_{\rm tor}$, for each prime number $\ell$ we have $\widetilde{G}/\ell\widetilde{G} \cong \mathbf Z/\ell\mathbf Z$ if $\ell \not= p$ and $\widetilde{G}/p\widetilde{G} \cong (\mathbf Z/p\mathbf Z)^{n+1}$, with $n+1 \geq 2$, so as before, $|\widetilde{G}/\ell\widetilde{G}| \not= \ell$ only at $\ell = p$. Hence the abstract group $\widehat{\mathbf Z} \times \mathcal O_K^\times$, where $K$ is a finite extension of $\mathbf Q_p$, knows what $p$ is.

• Not quite: the torsion subgroup of $\widehat{\mathbf Z} \times \mathbf Z_2^\times$ has order $2$, which is not $p-1$ when $p=2$. So if you wanted to count the torsion subgroup to answer your question then you would need to create a separate argument to distinguish $p=2$ and $p=3$. My answer originally had such a solution, but it breaks down badly if you want to treat general finite extensions (in case $K$ and $L$ have the same number of roots of unity). That is why I wrote the solution that is posted: it can be used to cover the general case of finite extensions. – KCd Jul 9 '18 at 12:13
• You run into a similar issue trying to show a finite extension $K$ of $\mathbf Q_p$ and a finite extension $L$ of $\mathbf Q_q$, for different primes $p$ and $q$, are not isomorphic as abstract fields. In the special case $K=\mathbf Q_p$ and $L=\mathbf Q_q$ you can just count roots of unity to tell them apart, except when the primes are $2$ and $3$ you need a different argument. To handle the general case (which can be done -- finite extensions of $\mathbf Q_p$ and $\mathbf Q_q$ are never isomorphic as abstract fields) you really need to come up with a more robust approach. – KCd Jul 9 '18 at 12:23
• @KennyLau I have rewritten my solution so that the only genuine topological issue is at the first step where abelianizations are computed (in a topological group, the commutator subgroup is the closure of the commutator subgroup as an abstract group). My proof that $G := \widehat{\mathbf Z} \times \mathcal O_K^\times$ knows $p$ is valid by thinking of $G$ as an abstract group (the torsion subgroup and the subgroup of $n$-th powers for $n \in \mathbf Z^+$ in an abelian profinite group are the same as when thinking of it as an abstract group). Therefore to address your concern [contd.] – KCd Jul 9 '18 at 13:01
• what would be needed is to show that the abelianization of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ as a topological group could be recovered from the structure of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ as an abstract group. I suspect this can be done, perhaps even easily, but at the time I write this I don't see it. For example, if the abstract commutator subgroup is closed then the abelianization as a topological group is the abelianization as an abstract group. – KCd Jul 9 '18 at 13:03
• Okay, it is true that the abstract commutator subgroup of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ is closed: Theorem 1.4 of Nikolov and Segal's article annals.math.princeton.edu/wp-content/uploads/… implies that the abstract commutator subgroup $[G,G]$ is closed in $G$ when $G$ is a (topologically) finitely generated profinite group, and the absolute Galois group of $\mathbf Q_p$ is finitely generated by work of Jannsen and Wingberg. I will update my reply to include this information. – KCd Jul 9 '18 at 14:10

Cohomological methods (such as in the book of Neukirch-Schmidt-Wingberg) give the most direct proof that the two groups cannot be isomorphic as profinite groups. For convenience, let us call $K$ a local $p$-adic field if $K$ is a finite extension of $\mathbf Q_p$, say of degree $N$, and denote its absolute Galois group by $G_K$. Let $\delta =1$ if $K$ contains a primitive $p$-th root of unity, $0$ otherwise. Then for any prime $l, dim_{\mathbf F_l} H^1(G_K,\mathbf Z/l\mathbf Z)=1+N+\delta$ if $l=p,1+\delta$ otherwise, which already shows what we want.

Another reason (also of a cohomological nature) lies in the structure of the maximal pro-$p$-quotient of $G_K$, which is pro-$p$-free on $N+1$ generators if $\delta=0$ (Chafarevitch), minimally generated by $N+2$ generators and $1$ explicit relation if $\delta=1$ (Demushkin for $p$ odd, Labute for $p=2$).

Actually, an almost complete description of $G_K$ by generators and relations is known : $G_K$ can be generated by $N+3$ generators $\sigma, \tau, x_0 ,..., x_N$ satisfying a "tame" relation $\sigma \tau \sigma^{-1}= \tau^q$, where $q$ is the cardinal of the residual field of $K$, as well as an explicit "wild" relation whose expression depends on the parity of $N$ (Jannsen-Wingberg for $p$ odd, Diekert for $p=2$ and $K(\zeta_4)/K$ unramified). Curiously, the remaining case is still unknown. All these results , and many more, can be found in chapter VII, §5, op. cit.

• Dear Nguyen Quang Do, thank you for your answer! (+1) – Watson Jul 9 '18 at 18:14