= Unramified biquadratic extension of a real quadratic field
{title2=$\mathbb Q(\sqrt p,\sqrt q)/\mathbb Q(\sqrt{pq})$}
For distinct primes $p,q\equiv1\pmod4$, 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> $p^2q^2$: 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 give
$$
p^2q^2/j^2=d_E=(pq)^2N_{K/\mathbb Q}(\mathfrak d_{E/K}).
$$
Both $j$ 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 $K$.
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