Motivation behind the definition of ideal class group

Solution 1:

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.

Exercise 1: Check that the "class" of all principal ideals is actually a class. Namely if $I$ is an ideal such that $\alpha I = (\beta)$ then show that $I$ is actually principal. Hint: $\mathcal{O}_K$ is an integral domain.

Exercise 2: Show that this definition of an ideal class group is actually equivalent to the one given in Neukirch. Hint: Use the first isomorphism theorem.

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:

Exercise 3: Let $I$ be an ideal of $\mathcal{O}_K$ show that there is a finite extension $L/K$ so that $I\mathcal{O}_L$ is principal. Hint: By finiteness of the class group there is an $n$ so that $I^n = \alpha$ for some $\alpha \in \mathcal{O}_K$. Now consider $L = K(\sqrt[n]{\alpha})$.

Exercise 4: Show that there is a finite extension of $L$ in which every ideal of $\mathcal{O}_K$ becomes principal. Wowowowowow! Hint: Use exercise 3 and the definition of the ideal class group given in the beginning of my answer, not the one in Neukirch.

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.

Solution 2:

At a crucial point in the proof that the equation $y^2=x^3-31$ has no solutions in the integers, one has an equation relating ideals, $$((y+\sqrt{-31})/2)=(2,(3+\sqrt{-31})/2)A^3$$ for a certain ideal $A$ in the ring of integers in ${\bf Q}(\sqrt{-31})$. But it turns out that the class number of that ring of integers is $3$, so $A^3$ is a principal ideal, and you get a contradiction, since you can prove that $(2,(3+\sqrt{-31})/2)$ is not a principal ideal.

Needless to say, this is just one example among many where class group properties help solve Diophantine equations.

Solution 3:

We have the short exact sequence of groups

$$1\to k^\times\to J_k\to \text{cl}(k)\to 1$$

The issue with this is that we have been forced to go to "group land" because we don't want to deal with monoids (e.g. the monoid of $\mathcal{O}_k$ under multiplication). So, let's forget that we have taken monoids at each step and done some kind of fractional process to get groups. What we end up with is

$$1\to \mathcal{O}_k\to \left\{\text{ideals of }\mathcal{O}_k\right\}\to \text{cl}(k)\to 1$$

So, we see that the class group can be roughly though about as the ratio of $\{\text{ideals of }\mathcal{O}_k\}$ to $\mathcal{O}_k$ itself. In other words, in our quest for unique factorization we have passed from elements of $\mathcal{O}_k$ (which may not be a UFD) to ideals of $\mathcal{O}_k$ (which always do have unique factorization). In a perfect world of sunshine and lollipops the ideals of $\mathcal{O}_k$ would be the same as $\mathcal{O}_k$ because $\mathcal{O}_k$ would be a PID. Alas, we live in no such sweet world and in general we are going to have to "add in extra" to get the unique factorization we need. The ratio of $\left\{\text{ideals of }\mathcal{O}_k\right\}$ to $\mathcal{O}_k$ (i.e the class group) measures just how much extra we had to add in.