# an exact sequence for ray class groups

Let $K$ be a number field  , let $\mathcal{O}_{K}$ be its ring of integers  and let $\mathfrak{m}$ be a modulus    in $K$, i.e.

 $\mathfrak{m}=\mathfrak{m}_{0}\mathfrak{m}_{\infty}$

where $\mathfrak{m}_{0}$ is an integral ideal in $\mathcal{O}_{K}$ and $\mathfrak{m}_{\infty}$ is a product of real infinite places (i.e. real archimedean   primes). Let $\operatorname{Cl}(K)$ be the ideal class group   of $K$ and let $\operatorname{Cl}(K,\mathfrak{m})$ be the ray class group of $K$ of conductor  $\mathfrak{m}$. Also, define

 $(\mathcal{O}_{K}/\mathfrak{m})^{\times}=(\mathcal{O}_{K}/\mathfrak{m}_{0})^{% \times}\times(\mathbb{Z}/2\mathbb{Z})^{|\mathfrak{m}_{\infty}|}$

where $|\mathfrak{m}_{\infty}|$ denotes the number of real places in $\mathfrak{m}$. Finally, let $U=\mathcal{O}_{K}^{\times}$ be the unit group of $K$.

###### Proposition.

The elements above fit in the following exact sequence:

 $U\longrightarrow(\mathcal{O}_{K}/\mathfrak{m})^{\times}\longrightarrow% \operatorname{Cl}(K,\mathfrak{m})\longrightarrow\operatorname{Cl}(K)% \longrightarrow 1.$
###### Example 1.

Let $K=\mathbb{Q}$. Thus, $\operatorname{Cl}(\mathbb{Q})$ is trivial and $U=\{\pm 1\}\cong\mathbb{Z}/2\mathbb{Z}$. Let $\mathfrak{m}=p\infty$ where $p>2$ is any prime. Then:

 $(\mathbb{Z}/\mathfrak{m})^{\times}=(\mathbb{Z}/p\mathbb{Z})^{\times}\times(% \mathbb{Z}/2\mathbb{Z}).$

 $\mathbb{Z}/2\mathbb{Z}\longrightarrow(\mathbb{Z}/p\mathbb{Z})^{\times}\times(% \mathbb{Z}/2\mathbb{Z})\longrightarrow\operatorname{Cl}(\mathbb{Q},p\infty)% \longrightarrow 1.$
Therefore, $\operatorname{Cl}(\mathbb{Q},p\infty)\cong(\mathbb{Z}/p\mathbb{Z})^{\times}$. In fact, as we know, the http://planetmath.org/node/RayClassFieldray class field of $\mathbb{Q}$ of conductor $\mathfrak{m}=p\infty$ is the cyclotomic field  $\mathbb{Q}(\zeta_{p})$ where $\zeta_{p}$ is any primitive $p$th root of unity  . Moreover
 $\operatorname{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})\cong\operatorname{Cl}(% \mathbb{Q},p\infty)\cong(\mathbb{Z}/p\mathbb{Z})^{\times}.$
Finally notice that the ray class group of $\mathbb{Q}$ of conductor $\mathfrak{m}=p$ is simply $(\mathbb{Z}/p\mathbb{Z})^{\times}/\{\pm 1\}$ which corresponds to the ray class field $\mathbb{Q}(\zeta_{p})^{+}=\mathbb{Q}(\zeta_{p}+\zeta_{p}^{-1})$, the maximal real subfield  of $\mathbb{Q}(\zeta_{p})$.