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Put $K=\mathbb{Q}(\sqrt{-6})$. Its fundamental discriminant is $-24$. The Minkowski bound for ideal classes isso 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, andshows that they represent the same class. This class is nontrivial because the norm form $a^2+6b^2$ does not represent $2$. Hence

Now let $H=K(\sqrt{2})=\mathbb{Q}(\sqrt{2},\sqrt{-3})$. The discriminant of a biquadratic field is the product of the discriminants of its three quadratic subfields, soThe relative discriminant ${\mathfrak{D}}_{H/K}$ therefore has norm one, proving that $H/K$ is unramified at every finite prime. Since $[H:K]=2=|\mathrm{Cl}(K)|$, the Hilbert class field of Q of square root minus six is

Let $K=\mathbb{Q}(\sqrt{-6})$, whose discriminant is $-24$, and putThis is a biquadratic extension of $\mathbb{Q}$ whose three quadratic subfields have discriminants $-24$, $8$, and $-3$. The discriminant of a biquadratic field is thereforeThe relative discriminant formula givesso $H/K$ is unramified at every finite prime. Both fields are totally imaginary, so no infinite place ramifies. Since $[H:K]=2=h_K$, the extension has the full degree of the Hilbert class field, and hence

The trace/norm integrality conditions give the stated congruences for
$\theta =(a+bi+c\sqrt{m}+di\sqrt{m})/2$. Solving them yields these integral bases:The three quadratic subfields have fundamental discriminants $-4,D_m,D_{-m}$. By discriminant of a biquadratic field,in both residue classes.