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

Title	an exact sequence for ray class groups
Canonical name	AnExactSequenceForRayClassGroups
Date of creation	2013-03-22 15:42:46
Last modified on	2013-03-22 15:42:46
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	6
Author	alozano (2414)
Entry type	Result
Classification	msc 11R29
Related topic	Modulus
Related topic	RayClassField
Related topic	ClassNumbersAndDiscriminantsTopicsOnClassGroups