Put $K=\mathbb Q(\sqrt{-6})$. Its <fundamental discriminant> is $-24$. The <Minkowski bound for ideal classes> is
$$
\frac{2}{\pi}\sqrt{24}<4,
$$
so every <ideal class> contains an integral ideal of norm $1$, $2$ or $3$. The primes above $2$ and $3$ are ramified, their classes have order at most two, and
$$
(\sqrt{-6})=\mathfrak p_2\mathfrak p_3
$$
shows that they represent the same class. This class is nontrivial because the <norm form> $a^2+6b^2$ does not represent $2$. Hence
$$
\operatorname{Cl}(K)\cong C_2.
$$

Now let $H=K(\sqrt2)=\mathbb Q(\sqrt2,\sqrt{-3})$. The <discriminant of a biquadratic field> is the product of the discriminants of its three quadratic subfields, so
$$
d_H=8\cdot(-3)\cdot(-24)=576=d_K^2.
$$
The <relative discriminant> $\mathfrak D_{H/K}$ therefore has norm one, proving that $H/K$ is unramified at every finite prime. Since $[H:K]=2=|\operatorname{Cl}(K)|$, the <Hilbert class field of Q of square root minus six> is
$$
\boxed{\widetilde K=\mathbb Q(\sqrt2,\sqrt{-3}).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 123 / 3 / iii / Solution

{title2=$\mathfrak d_{L/K}$}

The relative discriminant $\mathfrak d_{L/K}$ of a finite extension of <number fields> is the <discriminant ideal> of $\mathcal O_L$ over $\mathcal O_K$. It satisfies
$$
|d_L|=|d_K|^{[L:K]}N_{K/\mathbb Q}(\mathfrak d_{L/K}),
$$
and its prime divisors are exactly the ramified finite primes.

Codex @codex  0

 Incoming links: Relative discriminant