Let $K$ be an algebraic number field, assumed to be Galois, with Galois group

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

Is knowing the abelianization of $G$ alone, without other information on $K$, enough to determine the ideal class group of $K$? Or can we have two different Galois ANF $K_1$, $K_2$, having the same abelianization of their Galois groups, but non-isomorphic ideal class group?

I have just started learning the subject, so forgive the naiveness of my question.

Edit 1: this was answered below. I wonder if the answer would be any different if instead of

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

one replaces it with

$H = Gal(\bar{\mathbb{Q}} / K)$,

where $\bar{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$.

  • $\begingroup$ The key point here in @RicardoBuring's counterexample is that both fields have the same Galois group, as well as the same abelianisation. The isomorphism class of $\mathrm{Gal}(L/K)$ as an abstract group has nothing to do with the class group. $\endgroup$ – Mathmo123 Apr 22 at 19:11
  • $\begingroup$ You should ask your edit as a new question. In short, the answer is that $H$ does determine the class group, via class field theory. $\endgroup$ – Mathmo123 May 8 at 9:42
  • $\begingroup$ @Mathmo123, interesting! Thank you. All right, I will write a separate post. $\endgroup$ – Malkoun May 8 at 12:42
  • $\begingroup$ Following a suggestion by @Mathmo123, I opened another post: math.stackexchange.com/questions/3218399/…. $\endgroup$ – Malkoun May 8 at 12:52

No, take e.g. $\mathbb{Q}(\sqrt{-1})$ and $\mathbb{Q}(\sqrt{-5})$. Both have Galois group $\mathbb{Z}/2\mathbb{Z}$ but $\mathbb{Z}[i]$ is a PID whereas $(2,1+\sqrt{-5})$ is not principal in $\mathbb{Z}[\sqrt{-5}]$.

  • 1
    $\begingroup$ Nice. Thank you! I should study many examples. This is my plan. $\endgroup$ – Malkoun Apr 22 at 16:00
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    $\begingroup$ I see. This has to do with the fact that $6$ can be written as $2$ times $3$, or as $1+\sqrt{5}$ times $1-\sqrt{5}$, in $\mathbb{Z}[\sqrt{-5}]$. $\endgroup$ – Malkoun Apr 22 at 20:07
  • $\begingroup$ Indeed. The explicit relation is that $(6) = (2,1+\sqrt{-5})^2(3,1+\sqrt{-5})(3,1-\sqrt{-5})$ as ideals, and you can get factorizations of $6$ as an element by multiplying out some of the ideals to get principal ideals. (See my answer to How to factorise a number in $\mathbb{Z}[\sqrt{-5}]$? for a more elaborate example.) $\endgroup$ – Ricardo Buring Apr 22 at 20:30
  • $\begingroup$ Is there a nice algorithmic way of getting all possible factorizations of an integer, knowing the corresponding factorization of the corresponding ideal as a product of prime ideals? The converse question is interesting too. $\endgroup$ – Malkoun Apr 23 at 10:30

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