Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 123 / 4

The idele group is the restricted productover all places, with respect to ${\mathcal{O}}_v^{\times }$ at finite places. Thus an idele has a nonzero component in every completion and belongs to ${\mathcal{O}}_v^{\times }$ at all but finitely many finite places.

For $\alpha =({\alpha }_v)\in {\mathbb{I}}_K$, the congruence $\alpha \equiv 1(\mathrm{mod}\mathfrak{m})$ meansAt finite primes outside the modulus require ${\alpha }_{\mathfrak{p}}\in {\mathcal{O}}_{\mathfrak{p}}^{\times }$. These ideles form ${\mathbb{I}}_K(\mathfrak{m})$. Their imageinside the idèle class group is the idelic congruence subgroup.

Write $m={\prod }_p{p}^{n_p}$. For the modulus $m\infty$,This is understood as the corresponding restricted-product subgroup of ${\mathbb{I}}_{\mathbb{Q}}$.

Multiplying by a positive rational number normalizes the real component and all valuations, leaving the finite unit residue modulo $m$. This identifiesLocally,when $n_p>0$. Multiplying gives

The assumed inclusion and the class-field norm-index theorem giveThe outer subgroup has index $\phi (m)$ by part (d), whileTwo nested subgroups of the same finite index are equal. Thus

The Kronecker–Weber theorem says that every finite abelian extension of $\mathbb{Q}$ lies in some $\mathbb{Q}({\zeta }_m)$.

Indeed, global class field theory assigns to a finite abelian $L/\mathbb{Q}$ the open norm group of an abelian extension $N_{L/\mathbb{Q}}{C}_L$. It contains a congruence subgroup $C_{\mathbb{Q}}(m\infty )$. By part (e), this is the norm group of $\mathbb{Q}({\zeta }_m)$. The inclusion-reversing class-field correspondence therefore gives