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

Suppose that $L/\mathbb{Q}$ were a Finite Galois extension with Galois group $S_3$ and unramified away from $7$. Its quadratic subfield is the fixed field of the alternating subgroup $A_3$. A quadratic field ramified only at $7$ must be $\mathbb{Q}(\sqrt{-7})$: the classification by fundamental discriminants shows that $-7$ is the only nontrivial quadratic discriminant supported at $7$. In particular, $7$ ramifies in this quadratic subfield.

Because $7$ does not divide $|S_3|=6$, all ramification at $7$ is tame ramification. Its inertia group is therefore cyclic. Its image in $S_3/A_3\simeq C_2$ is nontrivial because the quadratic subfield is ramified, so the inertia group is generated by a transposition and has order two; it cannot have order six because it is cyclic.

The discriminant exponent of a tame Galois extension is thereforeThere is no other finite ramification, so $|d_L|=7^3=343$.

On the other hand, the Minkowski bound for ideal classes implies the Minkowski lower bound for a number-field discriminantSince $0\le r_2\le 3$, the right side is smallest at $r_2=3$ and is greater than $980$. This contradicts $|d_L|=343$, so no such extension exists.

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