$\mathbb{Q}(\sqrt{p^*})$ is contained in the ring class field of conductor $p$ Let $K$ be an imaginary quadratic field, $p$ a prime of $\mathbb{Q}$ and $H_p$ the ring class field of $K$ of conductor $p$, i.e. the abelian extension of $K$ with Galois group isomorphic to the class group of the order of $K$ of conductor $p$. Let $p^*=(-1)^{\frac{p-1}{2}}p$, so that $p^*\equiv 1\pmod 4$.
By class field theory, it seems known that $\mathbb{Q}(\sqrt{p^*})\subseteq H_p$.
Can someone give a hint about why this inclusion should hold?
 A: Sorry for the mess - the problem was that the result is "obvious" (see below) but that I wanted to avoid using the Artin isomorphisms.
A normal extension $L/K$ of a quadratic number field is a ring class field if and only if $G = $Gal$(L/{\mathbb Q})$ is a group extension of an abelian group by a group of order $2$ and if the nontrivial automorphism of $K/{\mathbb Q}$ acts as $-1$ on Gal$(L/K)$ (see Bruckner [Charakterisierung der galoisschen Zahlkörper, deren zerlegte Primzahlen durch binäre quadratische Formen gegeben sind. Math. Nachr. 32, 317-326 (1966)]; this is essentlially Artin at work since $\sigma$ acts as $-1$ on the class group of a quadratic number field). Since ${\mathbb Q}(\sqrt{p^*})$ is
contained in the ray class field modulo $p\infty$ of ${\mathbb Q}$,
$L = K(\sqrt{p^*})$ is contained in the ray class field modulo $p\infty$ of $K$ ("translation"), hence in the ray class field modulo $p$ of $K$ since $K$ is complex quadratic. And since the nontrivial automorphism $\sigma$ of $K/{\mathbb Q}$ acts trivially on Gal$(L/K)$ ($\sigma \tau \sigma^{-1} = \tau = \tau^{-1}$ since the extension is elementary abelian), it is a ring class field.
I still believe that there's a simple argument that works without Bruckner.
