Let $O$ be a Dedekind domain and $K$ its field of fractions. The set of all fractional ideals of $K$ form a group, the ideal group $J_K$ of K. The fractional principal ideals $(a) = aO, a \in K^*$, form a subgroup of the
group of ideals $J_K$, which will be denoted $P_K$. The quotient group
$$CL_K=J_K/P_K$$ is called the ideal class group of $K$.
This definition looks completely arbitary, I fail to see why the class group or the ideal class group is worth studying. So, why is the class group or ideal class group interesting? I've read that ideal class group "measures" how much unique factorization fails in a domain, but I don't understand how it does so.
This from Algebraic Number Theory by Neukirch.
The class group $C_K$ measures the expansion that takes place when we pass from numbers to ideals, whereas the unit group $O^*$ measures the contraction in the same process. This immediately raises the problem of understanding these
groups $O^*$ and $C_K$ more thoroughly.
I don't understand the above paragraph. What does it mean by "the expansion that takes place when we pass from numbers to ideals"? And why should we even bother about such "expansion"?
Best Answer
Let $K$ be an algebraic number field and $\mathcal{O}_K$ the integral closure of $\Bbb{Z}$ in $K$. Let us recall that a Dedekind domain is a UFD iff it is a PID. Classically, I think that the questions concerning whether or not a certain Dedekind domain was a UFD were very important, see e.g. this thread here.
Perhaps from the point of view of algebra asking if something is a PID is easier to approach: we know how to factor ideals in Dedekind domains and thus there should at least be a tool to measure how far does a Dedekind domain differ from being a principal ideal domain.
Let me now give you an alternative definition of the ideal class group. We will put an equivalence relation on the set of all ideals defined as follows. We say that an ideal $I$ of $\mathcal{O}_K$ is equivalent to $J$ iff there is $\alpha,\beta \in \mathcal{O}_K$ so that
$$\alpha I = \beta J.$$
One easily checks that this is an equivalence relation. With a little bit more work, one can show that the set of all equivalence classes has a well defined multiplication law and is actually a group. The identity element being the class of all principal ideals. Now for some exercises.
Now do you see how the definition I have given you of an ideal class group actually measures nicely whether or not $\mathcal{O}_K$ is a PID? We see that $\mathcal{O}_K$ is a PID iff its ideal class group is trivial.
Now on to more interesting material. In advanced subjects such as Class Field Theory one can construct something know as the Hilbert Class Field of $K$. I don't know all the details of this construction as my algebraic number theory is not so advanced, but in the Hilbert class field every ideal of $\mathcal{O}_K$ becomes principal!! One can now ask the question: can we avoid talking of the Hilbert class field and find such an extension?
The answer is: Of course we can! This is where the ideal class group comes in. Firstly from Minkowski's bound we get that $Cl_K$ is actually a finite group. Using this, here are now two exercises which you can do:
If you are stuck with any of these exercises I can post their solutions for you to view here.
Solution to Exercise 1 (As requested by user Andrew):
Suppose that $\alpha I = (\beta)$. Then in particular there is $x \in I$ so that $\alpha x = \beta$. We claim that $I = (x)$. Now it is clear that $(x) \subseteq I$. For the reverse inclusion take any $y \in I$. Then $\alpha y = \beta \gamma$ for some $\gamma \in R$. Since $\beta = \alpha x$ we get that $$\alpha y = \alpha x \gamma.$$
But now $\mathcal{O}_K$ is an integral domain and so $y = x\gamma$, so that $I \subseteq (x)$. Hence $I = (x)$ and so $I$ is principal.