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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.

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    $\begingroup$ 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. $\endgroup$ – Alex J Best 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
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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):

http://www.math.ucsd.edu/~kedlaya/ants10/angelakis/paper.pdf

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 $K$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ There are links here $\endgroup$ – Mathmo123 May 8 '19 at 17:34

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