Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 123 / 4 / 2

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 $M/K$.