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](http://en.wikipedia.org/wiki/Genus_field) for an explicit description of $k^{gen}$ when $k$ is quadratic. 



Intersection of a ring class field of a quadratic field K with the cyclotomic extension of K
The answer is no. I'll only focus on imaginary quadratic fields since they are easier to deal with than real ones. Great reference for all this theory is Cox's book *Primes of the form $x^2+ny^2$*.


So let $K$ be imagimary, thus $d<0$, and let $D=f^2d$. Since $L=\bigcup\_{n\in\mathbb{N}}\mathbb{Q}(\zeta\_n)$ is the maximal abelian extension of $\mathbb{Q}$, the field $K\_f^{\text{ring}}\cap L$ is the maximal subfield of $K\_f^{\text{ring}}$ which is abelian over $\mathbb{Q}$. We have $\operatorname{Gal}(K\_f^{\text{ring}}/K)\cong C(D)$, where $C(D)$ is the group of classes of primitive positive definite binary quadratic forms of discriminant $D$, and $G:=\operatorname{Gal}(K\_f^{\text{ring}}/\mathbb{Q})$ is isomorphic to the generalized dihedral group $C(D)\rtimes\{\pm1\}$ (which means that $-1$ acts on $C(D)$ as inverse). 
Under this isomorphism $[G,G]$ corresponds to $C(D)^2\times\{1\}$. Since $\operatorname{Gal}(K\_f^{\text{ring}}/K\_f^{\text{ring}}\cap L)=[G,G]$, we get $[K\_f^{\text{ring}}\cap L:K]=[C(D):C(D)^2]$.


Let $r$ be the number of odd primes dividing $D$. We define number $\mu$ in the following way: if $D$ is odd then $\mu=r$, and if $D=-4n$ is even then
$$\mu=
\begin{cases}
r&\text{if }n\equiv3\pmod4,\\
r+1&\text{if }n\equiv1,2\pmod4,\\
r+1&\text{if }n\equiv4\pmod8,\\
r+2&\text{if }n\equiv0\pmod8.\\
\end{cases}$$
Gauss proved that $[C(D):C(D)^2]=2^{\mu-1}$, so we get $[K\_f^{\text{ring}}\cap L:K]=2^{\mu-1}$ and $[K\_f^{\text{ring}}\cap L:\mathbb{Q}]=2^\mu$. In fact we can describe $K\_f^{\text{ring}}\cap L$ explicitly. Let $p\_1,\ldots,p\_r$ be odd primes dividing $D$ and let $p^\*=(-1)^{(p-1)/2}p$. If $D$ is odd then $K\_f^{\text{ring}}\cap L=\mathbb{Q}(\sqrt{p\_1^\*},\ldots,\sqrt{p\_r^\*})$, and if $D$ is even then 
$$K\_f^{\text{ring}}\cap L=
\begin{cases}
\mathbb{Q}(\sqrt{p\_1^\*},\ldots,\sqrt{p\_r^\*})&\text{if }n\equiv3\pmod4,\\
\mathbb{Q}(\sqrt{p\_1^\*},\ldots,\sqrt{p\_r^\*},i)&\text{if }n\equiv1\pmod4,\\
\mathbb{Q}(\sqrt{p\_1^\*},\ldots,\sqrt{p\_r^\*},\sqrt{-2})&\text{if }n\equiv2\pmod8,\\
\mathbb{Q}(\sqrt{p\_1^\*},\ldots,\sqrt{p\_r^\*},\sqrt{2})&\text{if }n\equiv6\pmod8,\\
\mathbb{Q}(\sqrt{p\_1^\*},\ldots,\sqrt{p\_r^\*},i)&\text{if }n\equiv4\pmod8,\\
\mathbb{Q}(\sqrt{p\_1^\*},\ldots,\sqrt{p\_r^\*},i,\sqrt{2})&\text{if }n\equiv0\pmod8.
\end{cases}$$


Since $K^{\text{gen}}=K\_1^{\text{ring}}\cap L$, by applying the previous formula with $f=1$ we get $[K^{\text{gen}}:\mathbb{Q}]=2^{\mu'}$, where $\mu'$ is defined same as $\mu$ with $d$ in place of $D$. Thus $[K^{\text{gen}}(\zeta\_{12}):\mathbb{Q}]\le[K^{\text{gen}}:\mathbb{Q}]\cdot[\mathbb{Q}(\zeta\_{12}):\mathbb{Q}]= 2^{\mu'}\cdot\varphi(12)=2^{\mu'+2}$. In particular $[K\_f^{\text{ring}}\cap L:\mathbb{Q}]$ can be arbitrarily much larger than $[K^{\text{gen}}(\zeta\_{12}):\mathbb{Q}]$ (if $f$ is divisible by lot of primes that don't divide $d$). However even if $[K\_f^{\text{ring}}\cap L:\mathbb{Q}]\le[K^{\text{gen}}(\zeta\_{12}):\mathbb{Q}]$ that still doesn't imply $K\_f^{\text{ring}}\cap L\subseteq K^{\text{gen}}(\zeta\_{12})$. For example take $K=\mathbb{Q}(i)$ and $f=8$. Then $K\_f^{\text{ring}}=\mathbb{Q}(i,\sqrt[4]{2})$ and $K\_f^{\text{ring}}\cap L=\mathbb{Q}(i,\sqrt{2})=\mathbb{Q}(\zeta\_8)$ but $K^{\text{gen}}(\zeta\_{12})=K(\zeta\_{12})=\mathbb{Q}(\zeta\_{12})$.


