# Does this Galois group determine the class ideal class group?

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.

• 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. – Alex J Best May 8 '19 at 16:04
• @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. – 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):

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 $$K$$.

• 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. – Malkoun Jun 26 '19 at 11:27
• 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.) – The Piper Jun 26 '19 at 17:24
• Thanks for pointing this out. I've edited my answer to explain the problem. – Mathmo123 Jun 27 '19 at 9:20
• 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. – Malkoun Jun 27 '19 at 9:40
• 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. – 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.

• 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? – Malkoun May 8 '19 at 15:09
• 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. – Mathmo123 May 8 '19 at 15:50
• 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). – Malkoun May 8 '19 at 16:05
• @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? – Malkoun May 8 '19 at 17:19
• There are links here – Mathmo123 May 8 '19 at 17:34