For $\alpha=(\alpha_v)\in\mathbb I_K$, the congruence $\alpha\equiv1\pmod{\mathfrak m}$ means
$$
\alpha_\mathfrak p\in1+\mathfrak p^{n_\mathfrak p}\mathcal O_\mathfrak p
\quad(\mathfrak p\mid\mathfrak m_0),
\qquad
\alpha_v>0\quad(v\mid\mathfrak m_\infty).
$$
At finite primes outside the modulus require $\alpha_\mathfrak p\in\mathcal O_\mathfrak p^\times$. These ideles form $\mathbb I_K(\mathfrak m)$. Their image
$$
C_K(\mathfrak m)
=K^\times\mathbb I_K(\mathfrak m)/K^\times
$$
inside the <idèle class group> is the <idelic congruence subgroup>.

Solution

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

{title2=$C_K(\mathfrak m)$}

The ideles congruent to one modulo $\mathfrak m$ have finite components in $1+\mathfrak p^{n_\mathfrak p}$ at primes in the finite modulus and positive components at selected real places. Their image in $C_K$ is the idelic congruence subgroup.


Codex @codex  0

 Incoming links: Idelic congruence subgroup