Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 123 / 1

A modulus of a number field is a formal productwhere ${\mathfrak{m}}_0={\prod }_{\mathfrak{p}}{\mathfrak{p}}^{n_{\mathfrak{p}}}$ is a nonzero integral ideal and ${\mathfrak{m}}_{\infty }$ is a product of distinct real embeddings of $K$. Only finitely many $n_{\mathfrak{p}}$ are nonzero; complex places do not occur.

Let $I_K(\mathfrak{m})$ be the group of fractional ideals prime to ${\mathfrak{m}}_0$. Let $P_K(\mathfrak{m})$ consist of principal ideals $(\alpha )$ withfor every $\mathfrak{p}\mid {\mathfrak{m}}_0$ and $\sigma (\alpha )>0$ for every real place $\sigma \mid {\mathfrak{m}}_{\infty }$. The ray class group is

LetThe ray class number formula is

The quadratic field $K=\mathbb{Q}(\sqrt{-7})$ is the unique quadratic subfield of $\mathbb{Q}({\zeta }_7)$. Its nontrivial character is the quadratic characterUnder the Artin reciprocity map, Frobenius at an unramified prime $p$ restricts trivially to $K$ exactly when ${\chi }_7(p)=1$. The quadratic residues modulo $7$ are $1,2,4$, while $5$ is a nonresidue. Thusso the Artin symbol at $5$ is the nonidentity element of $\mathrm{Gal}(K/\mathbb{Q})$. Therefore $5$ has residue degree two and is inert in $K$.

Since $5$ is inert, $\mathfrak{m}=5{\mathcal{O}}_K$ is prime andThe unit group is $\{\pm 1\}$, and only $1$ is congruent to $1$ modulo $\mathfrak{m}$, soThere are no real places and $h_K=1$. The ray class number formula givesHence the ray class field modulo $5{\mathcal{O}}_K$ has degree $\boxed{12}$ over $K$.