On the intersection of two ring class fields

Question

Given two distinct imaginary quadratic fields $K_1$ and $K_2$ with coprime discriminants. For a prime $p$, let $H_1(p)$ and $H_2(p)$ be the ring class fields of conductor $p$ of $K_1$ and $K_2$, respectively. Is it true that the intersection $H_1(p) \cap H_2(p)=\mathbb{Q}$? Any help is highly appreciated. Thanks!

Answer 1

As a matter of fact, all ring class fields of conductor $p \equiv 1 \bmod 4$ contain the quadratic number field ${\mathbb Q}(\sqrt{p})$.

The formula for the ring class number (Cox) gives $h({\mathcal O}) = \frac{2h}{w}(p + (\frac{d}{p}))$, where the quadratic base field has discriminant $d$, $h$ is the class number of $K$, and $w$ the number of roots of unity in $K$. It is easy to see that this ring class number is always even. The maximal $2$-extension inside this ring class field can be constructed by a sequence of quadratic central extensions. Let $L/K$ be such a quadratic subextension. Then $L/{\mathbb Q}$ is normal; but the extension cannot be cyclic since every prime ramifying in $K/{\mathbb Q}$ would then also ramify in $L/K$, which is impossible since $L/K$ is only ramified at $p$ and since there always is a prime $q \ne p$ ramified in $K/{\mathbb Q}$ since $K$ is complex and therefore $K \ne {\mathbb Q}(\sqrt{p})$.

This implies that $L = {\mathbb Q}(\sqrt{d},\sqrt{p})$.