Let $K$ be a quadratic field. Let $f\in\mathbb{Z}_{\geq 1}$. Let $\mathcal{O}_f=\mathbf{Z}+f\mathcal{O}_K$ be the unique order of $K$ of index $f$ in $\mathcal{O}_K$. Let $H_f^{ring}$ denote the ring class field in the narrow sense (so we allow ramification at infinite real places of $K$, if such places exist) associated to the order $\mathcal{O}_f$. For a number field $L$ let us denote by $\mu(L)$ the group of roots of unity of $L$. For an integer $n\in\mathbf{Z}_{\geq 2}$ we let $\mu_n$ be the group of $n$-th roots of unity and $\zeta_n$ a primitive $n$-th root of unity. An (not so trivial) exercise in class field theory shows that $\mu(K_f^{ring})\subseteq \mu_{12}$. Let us denote by $K^{gen}$ the genus field in the narrow sense of $K$. Recall that the genus field $M^{gen}$ (in the narrow sense) of an abelian extension $M$ over $\mathbf{Q}$ is the maximal unramified extension $M^{gen}$ over $M$ (at all finite places of $M$) which is abelian over $\mathbf{Q}$.

Note that $H_K=K_{1}^{ring}$ corresponds to the Hilbert class field, in the narrow sense, of $K$. We obviously have $K_{f}^{ring}\supseteq H_K\supseteq K^{gen}$. Let $d$ be the discriminant of $K$. We also have $K^{gen}\subseteq K(\zeta_d)$. Let $L=K(\zeta_n:n\in\mathbf{Z}_{\geq 2})$.

So this begs the following question:

**Q** Do we always have that $K_{f}^{ring}\cap L\subseteq K^{gen}(\mu_{12})$ ?

P.S. Note that $K^{gen}(\mu_{12})\subseteq K(\mu_m)$ where $m=lcm(d,12)$.

P.S.S. An explicit description of the genus field: In genreral, if $k$ is an abelian extension over $\mathbf{Q}$, we have $k\subseteq\mathbf{Q}(\zeta_m)$ for some $m$. We may associate to $k$ a subgroup $X_k$ of the Pontryagin dual of $(\mathbf{Z}/m\mathbf{Z})^{\times}$, which we denote by $X_m$. Note that $X_m$ has a natural direct sum decomposition with respect to the inertia subgroups of $\mathbf{Q}(\zeta_m)$; which are in correspondence with the primes $p|m$. This simply corresponds to the usual decomposition of a Dirichlet charcter via the Chinese remainder theorem for the prime powers dividing $m$. It is a simple exercise to show that the "cartesian closure" of the subgroup $X_k$ (with respect to this direct sum decomposition of $X_m$) corresponds to $k^{gen}$. So in the case where $k$ is quadratic over $\mathbf{Q}$, this implies that $k^{gen}$ is a multiquadratic extension over $k$. See the wikepedia page on genus fields for an explicit description of $k^{gen}$ when $k$ is quadratic.