Let $K$ be an algebraic number field, and consider the Galois group:

$G = Gal(\bar{\mathbb{Q}}, K)$.

Is knowing the Galois group $G$ alone, without other information on $K$, enough to determine the ideal class group of $K$?

A user suggested that in short the answer is "yes", via class field theory, as a comment to my other post:

Does the abelianization of the Galois group determine the ideal class group?

In that post, I was looking at the wrong Galois group.

I don't require a long answer. It would not be fair to ask for a complete explanation, because it seems like a standard result in class field theory (which is on my list of subjects to learn about). However, a brief outline with a couple of references would be great.

  • 3
    $\begingroup$ Its beyond class field theory: but you may also be interested in the Neukirch–Uchida theorem en.wikipedia.org/wiki/Neukirch%E2%80%93Uchida_theorem which loosely says that your Galois group determines the field completely, and hence all other invariants of the field. $\endgroup$ May 8 '19 at 16:04
  • $\begingroup$ @AlexJBest, thank you! This is a very nice theorem, the Neukirch-Uchida theorem, thank you! Moreover, your comment is a great addition to the discussion here. $\endgroup$
    – Malkoun
    May 8 '19 at 16:08

The answer is yes, but the explanation given by @Mathmo123 is incorrect.

While it is true that class field theory gives an adelic description of $G^{ab}$, it is not at all clear one can recover from this description what the maximal unramified extension of $K$ should be. In order to compute the class group, one has to take the quotient of the idele class group (or really the idele class group modulo a maximal connected component at the infinite place) by $\widehat{\mathcal O_K^\times}$ - but this subgroup is given in terms of $K$, and the problem is exactly about giving a description which depends only on $G$, and not on $K$.

This error is a fatal one - it turns out that the abelianization $G^{ab}$ of $G = G_K$ is not enough to determine the class group. For some references and examples of this phenomenon, see this paper (in particular, the last line of the first page):


That said, the answer by Alex J Best in the comments gives a complete positive answer to the question; the entire group $G$ determines $K$ by the Neukirch-Uchida theorem, and then knowing $K$ determines the class group of $K$.

  • $\begingroup$ Thank you so much. This discussion is beyond my current level in Algebraic Number Theory. If a consensus is reached, in particular with Mathmo123, I can reevaluate my answer. This will benefit future readers. $\endgroup$
    – Malkoun
    Jun 26 '19 at 11:27
  • 1
    $\begingroup$ Malkoun, I'm not sure Thomas Kuhn would approve of such a "truth by consensus" approach. I think you should reevaluate your answer regardless at some point! (Although I expect @Mathmo123 will agree with this answer when he reads it.) $\endgroup$
    – The Piper
    Jun 26 '19 at 17:24
  • $\begingroup$ Thanks for pointing this out. I've edited my answer to explain the problem. $\endgroup$
    – Mathmo123
    Jun 27 '19 at 9:20
  • $\begingroup$ All right. I am still trying to understand Class Field Theory, as of now, which makes me an unqualified judge for the time being. I will try to take a "crash course" on the topic to understand enough to be able to appreciate this discussion. Consensus is reached but let us not upset Thomas Kuhn! Please give me a bit more time. $\endgroup$
    – Malkoun
    Jun 27 '19 at 9:40
  • $\begingroup$ Thank you, and thank you @Mathmo123 too. This generated an interesting discussion. I kind of had to read the wikipedia page on class field theory (and a few other resources) a number of times to be able to appreciate this discussion. $\endgroup$
    – Malkoun
    Jul 8 '19 at 20:36

Edit: The answer below is incorrect. While it is true that, via class field theory, we can recover the class group as a quotient of $G^{ab}$, the problem, as @ThePiper points out, is that this quotient is by $\widehat{\mathcal O}_K^\times$, which $G^{ab}$ knows nothing about.

Given the whole of $G$, we would be able to recover $\widehat{\mathcal O}_K^\times=\prod_{v}\widehat{\mathcal O}_{K_v}^\times$ via class field theory if we could recover the inertia groups $I_v$ from $G$: by local class field theory, $I_v\cong {\mathcal O}_{K_v}^\times$.

It is possible to recover the inertia groups from $G$. However, the fact that we can do so is a key part of the Neukirch-Uchida theorem.

The answer is yes. Let $G^{ab}$ denote the abelianisation of $G$ $-$ i.e. $G^{ab} = G/\overline{[G,G]}$. By global class field theory, we have a canonical isomorphism $$K^\times\backslash\mathbb A_K^{\times}/\overline{(K_\infty^\times)^0}\cong G^{ab}.$$

Here, $\mathbb A_K^\times$ are the ideles of $K$, and $\overline{(K_\infty^\times)^0}$ is the closure of the identity connected component of $(K\otimes_\mathbb Q\mathbb R)^\times$ viewed as a subgroup of $\mathbb A_K^\times$.

This isomorphism gives a concrete connection to the class group of $K$: the class group of $K$ is canonically isomorphic to $$K^\times\backslash\mathbb A_K^{\times}/\widehat{\mathcal O_K^\times} K_\infty^\times,$$ and is therefore a quotient of $G^{ab}$.

On the Galois side, this quotient of $G^{ab}$ cuts out a finite abelian extension of $K$ -- the Hilbert class field.

  • $\begingroup$ Great! Thanks. I have to build up to reach the ideles and adeles. I am reading Markus's book, which is great as an introduction. However, what should I read at the same time, or maybe after that book, that has material on adeles and ideles please, and hopefully, an explanation of those fascinating two-sided quotients that you mention? $\endgroup$
    – Malkoun
    May 8 '19 at 15:09
  • $\begingroup$ A standard reference is Neukirch's algebraic number theory. For the bigger picture, Kevin Buzzard recently gave a lecture course at MSRI on the Langlands program, of which class field theory is a small slice. The course notes contain a brief introduction to the adeles/ideles and explain class field theory from the perspective of the absolute Galois group. $\endgroup$
    – Mathmo123
    May 8 '19 at 15:50
  • $\begingroup$ Nice. I shall look at Kevin Buzzard's lecture course. Thank you so much. Yes of course, I am interested in the Langlands program (though I am quite scared of its technical requirements hehe, but it seems like fun). $\endgroup$
    – Malkoun
    May 8 '19 at 16:05
  • $\begingroup$ @Mathmo123, I could not find the course notes of that Summer long program at MSRI organized by Kevin Buzzard. I did find the videos though but, did someone turn these lectures into notes by any chance? $\endgroup$
    – Malkoun
    May 8 '19 at 17:19
  • 1
    $\begingroup$ There are links here $\endgroup$
    – Mathmo123
    May 8 '19 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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