Short exact sequence in the ideal class group

Question

In the answer to the Motivation behind the definition of ideal class group, I have seen one by user Alex Youcis and he/she claimed that the following is a short exat sequence: $$1\to \mathcal{O}_k\to \left\{\text{ideals of }\mathcal{O}_k\right\}\to \text{cl}(k)\to 1$$

But why is that? I couldn't figure it out. Since $\mathcal{O}_k$ is a Dedekind Domain, each ideal is either principal or generated by two elements. Then, how to construct the map between $ \mathcal{O}_k\to \left\{\text{ideals of }\mathcal{O}_k\right\}$ such that it is injective? I have tried to construct $f(x)=\langle x \rangle$, but if $\langle x \rangle=\langle y \rangle$, I can not deduce $x=y$ since $x$ and $y$ may associate. I also wonder the map between $\left\{\text{ideals of }\mathcal{O}_k\right\}\to \text{cl}(k)$. I am totally new to Algebraic Number Theory, thank you for your help.

Answer 1

Possibly the idea of "fractional ideal" is not explicit enough in what you're reading. Namely, to make "ideals" (or ideal classes...) into a group, in the first place, and to have $k^\times$ map to it at all, we need multiplicative inverses to ideals, under ideal multiplication.

The thing about needing at most two generators is misleading. Rather, it turns out that a good characterization of "fractional ideal" is as a (ok, non-zero) finitely-generated $\mathfrak o$-submodule of the field, where $\mathfrak o$ is the ring of integers. Then, for Dedekind domains (maybe as a definition/characterization) we prove that the collection of (non-zero) fractional ideals really is a group.

Then it makes more sense to talk about a/the map from $k^\times$ to the group of fractional ideals. Yes, send $\alpha\not=0$ to the fractional ideal generated by it, namely, $\alpha\cdot \mathfrak o$.

No, this is NOT injective, as the questioner suspects, because two elements of $k^\times$ differing by a unit (in $\mathfrak o^\times$) hit the same fractional ideal.

But I'd wager that any accidental claim that it does inject is irrelevant to the rest of the development. :)

EDIT: if we allow ourselves to step away from a tooo-classical viewpoint: instead of "ideal classes" as the middle joint in a purported short exact sequence, we might/should have the ideles of the global field. And, perhaps, even the smaller group of ideles of idele-norm $1$, since $k^\times$ injects to that, by the product formula. :) And then the product is the (far more universal) "idele class group", ... maybe with some extra archimedean components, which we could manage to kill off if we really wanted.

Point is: the injectivity is (of course) correct with "idele class group"... and/but not for the (in-effect, level-one) specific class group.

Answer 2

The precise version of the exact sequence is the four-term sequence $$ 1 \to \mathcal{O}_k^\times \to k^\times \to I(k) \to \text{Cl}(k) \to 1. $$

Here $I(k)$ is the group of fractional ideals and the map from $k^\times$ is the one taking a number to the principal ideal generated by it.

Answer 3

The ideal class group of a number field $K$ is defined as $$ {\cal C}_K= \{\text{non-zero fractional ideal in $K$}\}/\{\text{non-zero principal fractional ideals}\} $$ Since any two principal fractional ideals $I=\alpha{\cal O}_K$ and $J=\beta{\cal O}_K$ are equal if and only if $\alpha$ and $\beta$ are associated we can write an exact sequence $$ 1\rightarrow {\cal O}_K^\times\rightarrow K^\times\rightarrow {\cal I}_K\rightarrow {\cal C}_K\rightarrow 1 $$ where ${\cal I}_K$ denotes the group of non-zero fractional ideals.

A somewhat different and in some sense more explicit description can be given in terms of the group of ideles of $K$