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Set $K=\mathbb{Q}(\sqrt{pq})$ and $E=\mathbb{Q}(\sqrt{p},\sqrt{q})$. The distinct primes give independent square classes, so $[E:\mathbb{Q}]=4$ and $[E:K]=2$. In particular $K(\sqrt{p})=K(\sqrt{q})=E$: these are the same quadratic extension of $K$. We prove it is unramified by an explicit field discriminant calculation.

Since $p,q\equiv 1(\mathrm{mod}4)$, put $a=(1+\sqrt{p})/2$, $b=(1+\sqrt{q})/2$. Both are algebraic integers. The ring $R=\mathbb{Z}[a,b]\subseteq {\mathcal{O}}_E$ has basis $1,a,b,ab$. Up to the ordering of the basis, its trace pairing matrix is the tensor product of the quadratic trace matrices, because the four embeddings of $E$ choose the two radical signs independently. The two quadratic matrices areThusOn the other hand, $pq$ is squarefree and congruent to one modulo four, so $d_K=pq$. Let $j=[{\mathcal{O}}_E:R]$. The discriminant-index formula for an integral lattice and the relative discriminant tower formula giveThe ideal norm on the right is a positive integer, while $j$ is a positive integer. Hence $N({\mathfrak{d}}_{E/K})=1/j^2$ forces $j=1$ and ${\mathfrak{d}}_{E/K}={\mathcal{O}}_K$. Its prime divisors are exactly the ramified finite primes, so $E/K$ is unramified at every finite prime, including primes above $2,p,q$. This also proves $d_E=p^2{q}^2$, agreeing with the discriminant of a biquadratic field formula.

Every embedding of $E$ is real, so the real places of $K$ split as well. The extension is quadratic, hence abelian. The ordinary Hilbert class field characterization therefore impliesFinally, the Artin reciprocity map identifies $\mathrm{Cl}(K)$ with $\mathrm{Gal}(H_K/K)$. Restriction onto $\mathrm{Gal}(E/K)\simeq C_2$ is surjective, so the ideal class group has a quotient of order two. Therefore $\boxed{|\mathrm{Cl}(K)|\text{ is even}}$.

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.