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.


1 Answer 1


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 $K^{\text{gen}}(\zeta_{12})=K(\zeta_{12})=\mathbb{Q}(\zeta_{12})$.

  • $\begingroup$ Dear Jarek, many thanks for the nice answer. $\endgroup$ Jan 29, 2015 at 23:24

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