Ray Class Group 1

There are so many ways I was tempted to branch out from where I left off. Eventually I want to come back and talk some more about complex multiplication which is a fascinating topic, but before doing that it would be better to cover some other ground first. So we’re moving on to a new topic. Instead of useful algebra that could be taught in a first year course, we’re going to jump all the way to assuming knowledge of a one quarter (10 week) class in algebraic number theory.

This means we’re not assuming that much, but basic facts about number fields, discriminants, rings of integers, Dedekind domains, finiteness of class group, valuations/Hensel’s lemma, and basic elliptic curve theory will be assumed (but maybe not used).

It might seem a bit of a jump, but let’s do some class field theory. The way I saw this was via ray class groups, but it seems maybe this isn’t the most popular current viewpoint. Let’s fix a number field ${K}$ with ring of integers ${\mathcal{O}_K}$. A conductor for ${K}$ is a tuple ${\frak{m}=\{\frak{m}_f, w_1, \ldots, w_i\}}$ where ${\frak{m}_f}$ is an ideal of ${\mathcal{O}_K}$ and ${w_j: K\hookrightarrow \mathbb{R}}$ are embeddings.

A non-zero ${\alpha}$ in ${K}$ is defined to be ${\alpha\equiv 1 \mod \frak{m}}$ if it satisfies ${ord_\frak{p}(\alpha -1)\geq ord_\frak{p}(\frak{m}_f)}$ for all primes and the images under all the embeddings ${w_j(\alpha)>0}$ are positive. In the simplest case, we could take ${\frak{m}=\{\frak{m}_f\}}$ where we don’t pick any embeddings. This makes the second condition empty. Let’s unravel the first condition a little more.

To think about the first condition recall that ${ord_\frak{p}(\alpha-1)}$ can be interpreted as the first ${r}$ such that ${\alpha-1\in \frak{p}^r}$ but ${\alpha-1\notin \frak{p}^{r+1}}$ and ${ord_\frak{p}(\frak{m}_f)}$ is the power of ${\frak{p}}$ occuring when decomposing ${\frak{m}_f=\prod \frak{p}_i^{e_i}}$. Thus the inequality tells us that ${\alpha-1}$ is in all those powers occuring in ${\frak{m}}$, i.e. ${[\alpha-1]=[0]\in \mathcal{O}_K/\frak{m}_f}$. Thus ${\alpha\equiv 1\mod \frak{m}_f}$, so the notation makes sense. Another way to grapple with this definition is to set ${S=\{\frak{p}: ord_\frak{p}(\frak{m}_f)>0\}}$. Then any ${\alpha}$ such that ${\alpha\equiv 1\mod \frak{m}}$ is an ${S}$-unit in ${\mathcal{O}_K}$.

Define ${I_\frak{m}}$ to be the group of fractional ideals ${Q}$ such that ${ord_\frak{p}(Q)=0}$ if ${ord_\frak{p}(\frak{m}_f)>0}$. Unravelling this a little tells us that if we decompose ${Q=\prod \frak{p}_j^{e_j}}$, then since ${\frak{m}_f}$ is assumed an honest ideal, the decomposition involves only positive powers and hence ${ord_\frak{p} (\frak{m}_f)>0}$ just says ${\frak{p}}$ appears. So ${I_\frak{m}}$ is the group of fractional ideals such that no power (positive or negative) appears in the decomposition of the ideal if the prime appears in ${\frak{m}_f}$. Or more colloquially, it is the group of fractional ideals relatively prime to ${\frak{m}_f}$.

Now in the same way we form the ideal class group, we can form the group of principal ideals: ${Prin(\frak{m})=\{\mathcal{O}_K\cdot \alpha : \alpha \equiv 1 \mod \frak{m}\}}$. This is a subgroup of ${I_\frak{m}}$ (this is an exercise, but shouldn’t be immediately obvious that multiplication preserves congruence to ${1 \mod \frak{m}}$).

Now we can define ${Cl_\frak{m}(K)=I_\frak{m}/Prin(\frak{m})}$ to be the ray class group of ${K}$ of conductor ${\frak{m}}$. We easily recover the full ideal class group by taking ${\frak{m}}$ to be the “empty” conductor, or in other words no embeddings together with the ideal ${\frak{m}_f=\mathcal{O}_K}$. Since ${ord_p(\mathcal{O}_K)=0}$ for all primes, we get that ${I_\frak{m}}$ contains every fractional ideal and trivially all elements ${\alpha\equiv 1 \mod \mathcal{O}_K}$, so ${Prin(\mathcal{O}_K)}$ is the full group of principal fractional ideals. So ray class groups really are just generalizations of the ideal class group.

Another basic example is to take ${K=\mathbb{Q}}$ and ${\frak{m}=\{\mathbb{Z}\cdot m, i:\mathbb{Q}\hookrightarrow \mathbb{R}\}}$. The fractional ideals are principal, so ${I_\frak{m}=\{\mathbb{Z}\cdot\frac{a}{b} : (a,m)=1 \ \text{and} \ (b,m)=1\}}$ (of course WLOG assume ${(a,b)=1}$). Now for the principal ideals we finally have an embedding condition. But it turns out to be vacuous because all we need is to find some generator that maps to a positive under the embedding, and we have ${\mathbb{Z}\cdot n=\mathbb{Z}\cdot (-n)}$ so one or the other will map to a positive and we’ll assume we’ve chosen that one.

Now the other condition on principals is that ${ord_p(\frac{a}{b}-1)\geq ord_p(m)}$, and since all of these things are integers we convert this to ${m|(a-b)}$, i.e. ${a-b\equiv 0 \mod m}$. Thus we get a map ${I_\frak{m}\rightarrow \mathbb{Z}}$ by taking the positive generator ${\frac{a}{b}\mapsto a-b}$. The condition we just wrote down shows that any fractional ideal in ${Prin(\mathbb{Z}\cdot m)}$ will map to a multiple of ${m}$ and hence the map passes to the quotient ${Cl_m(\mathbb{Q})\rightarrow \mathbb{Z}/m\mathbb{Z}}$. By the earlier statement about being relatively prime to ${m}$ we see that the image of the map lands inside the units ${(\mathbb{Z}/m\mathbb{Z})^\times}$. This gives us a plausibility argument that ${Cl_m(\mathbb{Q})\simeq (\mathbb{Z}/m\mathbb{Z})^\times}$, but to get an actual isomorphism the standard argument goes through a bit of trouble involving completions and approximation theorems which we’ll skip.