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}).

The exact sequence now reads:

\\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 $\\mathbb{Q}(\\zeta_{p})$ .