Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 123 / 4

The adele ring is the restricted productAddition and multiplication are componentwise. If $x_v,y_v\in {\mathcal{O}}_v$ outside finite sets, then $x_v+y_v,x_v{y}_v\in {\mathcal{O}}_v$ outside their union, so these operations are closed on ${\mathbb{A}}_K$. The componentwise ring axioms make it a ring. Its restricted product topology has basic open sets ${\prod }_v{U}_v$ with each $U_v\subseteq K_v$ open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

The idele group isthe restricted product with respect to the local unit groups.

Underthe preimage of a basic product neighbourhood $U\times V$ isAt almost all finite places, $U_v=V_v={\mathcal{O}}_v$, and the condition $x_v,x_v^{-1}\in {\mathcal{O}}_v$ is exactly $x_v\in {\mathcal{O}}_v^{\times }$. Thus these preimages form the usual restricted-product basis for ${\mathbb{I}}_K$. Conversely, every basic idele neighbourhood is obtained by choosing suitable $U_v$ and $V_v$ locally. Hence the restricted product topology on the idele group equals the subspace topology induced by $\varphi$.

The diagonal mapis an injective homomorphism because every completion map $K\to K_v$ is injective. The diagonal copy of $K$ is discrete in the adele ring. Choose an adelic neighbourhood $W$ of $1$ with $W\cap K=\{1\}$. By the subspace description of the idele topology, $W\cap {\mathbb{I}}_K$ is an idele neighbourhood of $1$ meeting diagonal $K^{\times }$ only at $1$. Translation proves that $K^{\times }$ is a discrete subgroup of ${\mathbb{I}}_K$.

For each place $v$ of $K$ and each $w\mid v$ of $L$, the inclusion $K_v\hookrightarrow L_w$ definesAt all but finitely many finite $v$, the component $x_v$ is a unit, and its image is a unit at every $w\mid v$, so the image is an idele. The map is a homomorphism and is injective because every local inclusion is injective.

The idele norm has $v$-componentFor the principal idele associated with $\alpha \in L^{\times }$, compatibility of local and global field norms givesThe right side is the principal idele of the element $N_{L/K}(\alpha )\in K^{\times }$.

Let $G=\mathrm{Gal}(L/K)$ and letbe the fixed field of the commutator subgroup. Then $M/K$ is the maximal abelian subextension of $L/K$, withThe nonabelian norm-residue kernel theorem identifies the kernel of the global reciprocity mapwith $N_{L/K}{C}_L$. Applied to the abelian extension $M/K$, the Artin reciprocity law identifies the kernel of the same map with $N_{M/K}{C}_M$. Thereforeso the norm group of the Galois extension $L/K$ is the norm group of the abelian extension $M/K$.