# Idele Class Field Theory 1

Today we’ll start reformulating class field theory in the idelic setting. Let ${K}$ be a global field (a number field, or a finite separable extension of ${\mathbb{F}_q(t)}$). Let ${M_k}$ be all the places and ${|\cdot |_v}$ the normalized absolute value at ${v\in M_k}$. Define ${K_v}$ be to be the completion of ${K}$ at ${v}$. Let ${\mathcal{O}_v\subset K_v}$ the ring of integers if ${v}$ is non-archimedean and just ${K_v}$ itself otherwise.

We’re going to assume that a first algebraic number theory course covers some of the basics of the ideles. Define ${\mathcal{J}_K}$ to be the ideles of ${K}$. Recall this is by definition ${\mathcal{J}_K=\{(j_v)_{v\in M_k} : j_v\in K_v^* \ \text{and} \ j_v\in \mathcal{O}_v^* \ \text{for all but finitely many}\}}$. This is a multiplicative group, but moreover there is a natural topology on it from the restricted direct product. We’ll recall properties of this as needed. We should quickly remind the reader that this is not the same as the product topology. It is the topology generated by the local basis around ${1}$ of the form where you take finitely many components to be open sets and the rest must be ${\mathcal{O}_v^*}$.

The ring of adeles is exactly analogous. We take ${\mathbb{A}_K=\{(a_v)_{v\in M_k} : a_v\in K_v \ \text{and} \ a_v\in \mathcal{O}_v \ \text{for all but finitely many}\}}$. This is a locally compact topological ring. By construction ${\mathbb{A}_K^*=\mathcal{J}_K}$ as a set. But the subspace topology is weaker than the topology we give to ${\mathcal{J}_K}$. You do recover the right topology by taking the subspace of the product ${\mathbb{A}_K\times \mathbb{A}_K}$ by ${\mathcal{J}_K=\{(x,y) : xy=1\}}$.

We define now the idele class group to be ${C_K=\mathcal{J}_K /K^*}$. Fix a conductor ${\frak{m}=\{\frak{m}_f, w_1, \ldots, w_i\}}$. Our first theorem is: there is a surjection ${\lambda_m: C_K\rightarrow Cl_\frak{m}(K)}$ where ${\lambda_m}$ is constructed as follows. Given ${j=(j_v)\in J_K}$ use the approximation theorem to find a non-zero ${\alpha\in K^*}$ with the property that ${ord_v(j_v\cdot \alpha^{-1}-1)\geq ord_v(\frak{m}_f)}$ if ${ord_v(\frak{m}_f)>0}$ and ${w_v(j_v\alpha^{-1})>0}$ where ${w_v}$ ranges through ${w_k}$ for ${k=1, \ldots , i}$. Define ${\lambda_m(j)}$ to be the fractional ideal ${\frak{a}}$ such that ${ord_v(\frak{a})=ord_v(j_v \alpha^{-1})}$ for all non-archimedean ${v}$.

Note that if we do this, at very least we’ve constructed a fractional ideal that is relatively prime to ${\frak{m}}$. Thus we have a candidate map. We should check well-definedness, but that is more tedious than it’s worth. Let’s just take the theorem as true. Out of it we get the nice corollary that there is a surjection ${\displaystyle \lambda: C_K\rightarrow \lim_{\leftarrow_{\frak{m}}}Cl_\frak{m}(K)=Gal(K^{ab}/K)}$ where ${K^{ab}}$ is the maximal abelian extension of ${K}$ unramified outside ${\frak{m}}$. The equality comes from piecing together the Artin maps using class field theory. The kernel is the connected component of the identity of ${C_K}$. This corollary is by no means obvious from the theorem.

The fact that the ${\lambda_m}$ form a compatible system to get a map to the inverse limit is easy because from the construction of the ${\frak{a}}$ that we built, we can just use the same for any compatible system. From a few posts ago we worked out what the kernel of the natural surjection ${Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)}$ was. Call that kernel ${Q}$. To figure out the kernel of ${\lambda}$ let’s consider the diagram:

$\displaystyle \begin{matrix} & & 1 \\ & & \downarrow \\ U_k & \twoheadrightarrow & Q \\ \cap & & \downarrow \\ \mathcal{J}_K & \stackrel{\lambda_m}{\twoheadrightarrow} & Cl_\frak{m}(K)\\ \parallel & & \downarrow \\ \mathcal{J}_K & \stackrel{\lambda_{\mathcal{O}_K}}{\twoheadrightarrow} & Cl(\mathcal{O}_K)\\ & & \downarrow \\ & & 1 \end{matrix}$

Now notice that going around the square ${\mathcal{J}_K\rightarrow Cl_\frak{m}(K)\rightarrow Cl(\mathcal{O}_K)}$ is just taking ${j}$ and sending it to the ideal we got out of this ${j\alpha^{-1}}$, but then pushing forward to ${Cl(\mathcal{O}_K)}$ we see that the ${\alpha^{-1}}$ is changing things by some principal and hence ${\mathcal{J}_K\rightarrow Cl(\mathcal{O}_K)}$ is the map that builds the ideal from ${j}$ alone. So we want to know the kernel and hence when is making the ideal formed from looking at ${ord_v(j)}$ principal.

Suppose ${j}$ maps to ${0}$, then there is some ${\beta\in K^*}$ so that ${ord_v(j)=ord_v(\mathcal{O}_K\beta)}$ for all ${v}$. Thus ${j\beta^{-1}}$ maps to ${\displaystyle U_k=\prod \mathcal{O}_v^*\times \prod K_v^*}$, the so-called unit ideles. Since this map factors through the idele class group ${C_K}$, we haven’t changed where ${j}$ goes by altering it to ${j\beta^{-1}}$. This shows the ${U_k}$ part of the diagram except for the surjection. To finish the proof of this corollary just involves messing with the ord conditions and using the Mittag-Leffler condition, so we’ll omit it. It is the statements in class field theory together with some motivating examples that are of interest for this blog at this time.

Let’s wrap up today by stating the big theorem. Suppose ${L/K}$ is finite abelian with conductor ${\frak{m}}$ and ${G=Gal(L/K)}$. We get maps ${C_K\twoheadrightarrow Cl_\frak{m}(K)\stackrel{\Psi_{L/K}}{\rightarrow} G}$. Call the composite ${\psi_{L/K}}$. Then

1) ${ker(\psi_{L/K})=N_{L/K}(C_L)}$

2) For all places ${v}$ of ${K}$ we get an inclusion ${K_v^*\hookrightarrow \mathcal{J}_K}$ by ${\alpha\mapsto (1, 1, \ldots, 1, \alpha, 1, \ldots )}$ and ${\pi: \mathcal{J}_K\rightarrow C_K}$ and ${\psi_{L/K}\circ \pi\circ i_v(K_v^*)=G_w}$ where ${G_w}$ is the decomposition group and ${w}$ is any place of ${L}$ over ${v}$.

Now this gives us an early glimpse at why the idele formulation is going to give us more. We’ll get some control on inertia and decomposition groups.

### Author: hilbertthm90

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