 Unramified biquadratic extension of a real quadratic field

For distinct primes $p,q\equiv 1(\mathrm{mod}4)$, put $E=\mathbb{Q}(\sqrt{p},\sqrt{q})$ and $K=\mathbb{Q}(\sqrt{pq})$. The algebraic integers $a=(1+\sqrt{p})/2$ and $b=(1+\sqrt{q})/2$ generate an order whose basis has discriminant of elements of a number field $p^2{q}^2$: the trace pairing matrix is, up to basis order, the tensor product of the quadratic trace matrices of determinants $p$ and $q$. If its index in ${\mathcal{O}}_E$ is $j$, the discriminant-index formula for an integral lattice and the relative discriminant tower formula giveBoth $j$ and this ideal norm are positive integers, forcing $j=1$ and ${\mathfrak{d}}_{E/K}={\mathcal{O}}_K$. Thus $E/K$ is unramified at all finite primes. It is also totally real, so no real place ramifies. This quadratic abelian extension is contained in the ordinary Hilbert class field of $K$, and Artin reciprocity gives a quotient of its ideal class group of order two. The extensions $K(\sqrt{p})$ and $K(\sqrt{q})$ are equal to $E$, not two different extensions of $K$.