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

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 $$\mathbb{Q}$$.

• 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. – Mathmo123 Apr 22 at 19:11
• 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. – Mathmo123 May 8 at 9:42
• @Mathmo123, interesting! Thank you. All right, I will write a separate post. – Malkoun May 8 at 12:42
• Following a suggestion by @Mathmo123, I opened another post: math.stackexchange.com/questions/3218399/…. – Malkoun May 8 at 12:52

## 1 Answer

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 $$\mathbb{Z}[\sqrt{-5}]$$.

• Nice. Thank you! I should study many examples. This is my plan. – Malkoun Apr 22 at 16:00
• 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 $\mathbb{Z}[\sqrt{-5}]$. – Malkoun Apr 22 at 20:07
• 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.) – Ricardo Buring Apr 22 at 20:30
• 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. – Malkoun Apr 23 at 10:30