Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 123 / 4 / ii

Put $s=\sqrt{-6}$, so ${\mathcal{O}}_K=\mathbb{Z}[s]$ and ${\mathcal{O}}_K^{\times }=\{\pm 1\}$. The primes $2$ and $3$ are ramified:Thus the finite modulus in the question is

The ray class exact sequence, with no real component in the modulus, givesThe Chinese remainder theorem for unit groups and the ramification relations yieldThe image of $-1$ kills the $C_2$ coming from ${\mathbb{F}}_3^{\times }$. Consequently the kernel of the map from the ray class group to the ordinary class group isSince the given class number is two, $|{\mathrm{Cl}}_{\mathfrak{m}}(K)|=12$.

It remains to distinguish $C_{12}$ from $C_6\times C_2$. LetThe prime $5$ splits in $K$, and $\mathfrak{q}$ represents the nontrivial ordinary ideal class because no element of $\mathbb{Z}[s]$ has norm $5$. Direct multiplication, or comparison of norms and valuations at the two primes over $5$, givesModulo $2$, the element $1+2s$ is $1$. Modulo $3$, its class $1+2ϵ$ is nontrivial and has order three because ${ϵ}^2=0$. Hence the ray class of $\mathfrak{q}$ has order six: its square is a nontrivial element of order three in the congruence kernel. The remaining order-two factor of that kernel, coming from $({\mathcal{O}}_K/(2){)}^{\times }$, is independent of $⟨[\mathfrak{q}]⟩$. Therefore