Ray Class Groups 2

Today let’s relate the ray class groups to the ideal class group. Fix {K} a number field and choose a conductor {\frak{m}=\{\frak{m}_f, w_1, \ldots, w_i\}}. There is certainly always a map {Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)}. This is just because given some fractional ideal {\frak{a}} that is relatively prime to {\frak{m}} it is a fractional ideal, so it maps to its class in the ideal class group (well-definedness is just because we mod out by less principal ideals, but they’re all principal).

It turns out that under this map we get an exact sequence

\displaystyle 1\rightarrow \left(\frac{(\mathcal{O}_K/\frak{m}_f)^*\times \prod_{j=1}^i\{\pm 1\}}{\text{im} \Delta}\right)\rightarrow Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)\rightarrow 1,

where the diagonal map {\Delta: \mathcal{O}_K^* \rightarrow (\mathcal{O}_K/\frak{m}_f)^*\times\prod_{j=1}^i\{\pm 1\}} is given by {\Delta(u)=([u], \text{sgn}(w_i(u)))}.

This might seem scary, but let’s go back to our example of {Cl_m(\mathbb{Q})}. Something fishy was going on with having to pick the positive generator, and in general when we have lots more units floating around the above tells us exactly how that is affecting everything. For one thing we have {Cl(\mathbb{Z})=1}, and so this theorem tells us we have an isomorphism from the big quotient to {Cl_m(\mathbb{Q})}.

Our units are just {\{\pm 1\}}, and our embedding is {w_1:\mathbb{Q}\hookrightarrow \mathbb{R}}. Thus our diagonal map is {1\mapsto ([1], 1)} and {-1\mapsto ([-1], -1)}. Thus {\displaystyle Cl_m(\mathbb{Q})\simeq \frac{(\mathbb{Z}/m\mathbb{Z})^*\times\{\pm 1\}}{\text{im}(\Delta)}\simeq (\mathbb{Z}/m\mathbb{Z})^*} which is what we got last time.

What this shows is that if {\frak{m}} doesn’t involve the real place, i.e. no embedding is part of the conductor, the map given last time is not well-defined because the two generators of the same fractional ideal {\frac{a}{b}} and {-\frac{a}{b}} must get mapped to the same place. But this exact sequence tells us exactly what gets changed, and we have to mod out by {\{\pm 1\}} (the image of the diagonal) so that they really do go to the same place and hence we get a smaller ray class group {Cl_\frak{m}(\mathbb{Q})\simeq \frac{(\mathbb{Z}/m\mathbb{Z})^*}{\{\pm 1\}}}.

Now let’s prove exactness of that sequence in general. We’ll start with surjectivity of the right map. Let {[\frak{b}]\in Cl(\mathcal{O}_K)}. We want to find some fractional ideal {\frak{a}} relatively prime to {\frak{m}} such that {[\frak{a}]=[\frak{b}]}. We can use the same approximation theorem trick as in this post (see it for more details). We have {\frak{b}=\prod \frak{p}_i^{e_i}}, but for any prime appearing that also appears in {\frak{m}} choose some {a_i} with {ord_{p_i}(a_i)=-e_i}. Multiplying these together we get some element {a} and {\frak{a}=\frak{b}\cdot (a)} is altered by a principal and hence in the same ideal class, but is by construction relatively prime to {\frak{m}}.

Just by definition the kernel of that map is exactly the principal ideals that are relatively prime to {\frak{m}} modulo the principal ideals with the conductor condition, i.e. {\displaystyle \frac{Prin(\mathcal{O}_K)\cap I_\frak{m}(K)}{Prin_\frak{m}(K)}}. We’ll establish an isomorphism between this and the first term of the sequence.

Suppose {\mathcal{O}_K\cdot \beta} is a class in this group, so that {\beta} is relatively prime to {\frak{m}}. Thus our map will be {\beta\mapsto \overline{\beta}:=((\beta \mod \frak{p}^{ord_{\frak{p}}(\frak{m}_f)}\mathcal{O}_K), \text{sgn}(w_j(\beta)))} in

\displaystyle \prod_{\frak{p}|\frak{m}_f} \mathcal{O}_{K, \frak{p}}^*/(1+\frak{p}^{ord_{\frak{p}}(\frak{m}_f)}\mathcal{O}_K)^*\times\prod_{j=1}^i\{\pm 1\}

Now it is not clear this is well-defined because there is ambiguity in the choice of generator {\beta}. Thus we must check that {\mathcal{O}_K \beta=\mathcal{O}_K \beta'} (as classes mod the principal ideals generated by elements congruent to {1\mod \frak{m}}!) if and only if {\beta=u\beta'} for some {u\in \mathcal{O}_K^*}.

Claim: The ideal {\mathcal{O}_K\beta} is in {Prin_{\frak{m}}(K)} if and only if {\overline{\beta}\in \text{im}(\Delta)}. If {\overline{\beta}\in \text{im}(\Delta)}, then let {u} a unit such that {\Delta(u)=\overline{\beta}}. Take {\beta'=u\beta}, and you get that {\beta'\equiv 1\mod \frak{m}}, thus {\mathcal{O}_K\beta=\mathcal{O}_K\beta'} which is in {Prin_{\frak{m}}(K)}.

Conversely, if {\mathcal{O}_K\beta} has the property that {\mathcal{O}_K\beta=\mathcal{O}_K\beta'} where {\beta'\equiv 1\mod \frak{m}}, then {\beta=u\beta'} for some unit. But now by definition {\overline{\beta'}} is just {1}. Thus {\overline{\beta}=\Delta(u)}. This proves the claim.

The proof of this claim actually shows that the map descends to an isomorphism and hence the sequence is exact by identifying the appropriate groups using the Chinese remainder theorem. That seems like enough for today.


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