Since $5$ is inert, $\mathfrak m=5\mathcal O_K$ is prime and
$$
N\mathfrak m=25,\qquad
|(\mathcal O_K/\mathfrak m)^\times|=24.
$$
The unit group is $\{\pm1\}$, and only $1$ is congruent to $1$ modulo $\mathfrak m$, so
$$
[\mathcal O_K^\times:U_{\mathfrak m,1}]=2.
$$
There are no real places and $h_K=1$. The <ray class number formula> gives
$$
|\operatorname{Cl}_{\mathfrak m}(K)|=\frac{24}{2}=12.
$$
Hence the <ray class field> modulo $5\mathcal O_K$ has degree $\boxed{12}$ over $K$.

Solution

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

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The ray class field modulo $\mathfrak m$ is the maximal abelian extension whose Artin map factors through the ray class group; its degree is the ray class number.


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