using Minkowski’s constant to find a class number

We will use the theorem of Minkowski (see the parent entry (http://planetmath.org/MinkowskisConstant)).

Theorem (Minkowski’s Theorem).

Let $K$ be a number field and let $D_{K}$ be its discriminant. Let $n=r_{1}+2r_{2}$ be the degree of $K$ over $\mathbb{Q}$ , where $r_{1}$ and $r_{2}$ are the number of real and complex embeddings, respectively. The class group of $K$ is denoted by $\operatorname{Cl}(K)$ . In any ideal class $C\in\operatorname{Cl}(K)$ , there exists an ideal $\mathfrak{A}\in C$ such that:

|{\bf N}(\mathfrak{A})|\leq M_{K}\sqrt{|D_{K}|}

where ${\bf N}(\mathfrak{A})$ denotes the absolute norm of $\mathfrak{A}$ and

M_{K}=\frac{n!}{n^{n}}\left(\frac{4}{\pi}\right)^{r_{2}}.

Example 1.

The discriminants of the quadratic fields $K_{2}=\mathbb{Q}(\sqrt{2}),\ K_{3}=\mathbb{Q}(\sqrt{3})$ and $K_{13}=\mathbb{Q}(\sqrt{13})$ are $D_{K_{2}}=8,\ D_{K_{3}}=12$ and $D_{K_{13}}=13$ respectively. For all three $n=2=r_{1}$ and $r_{2}=0$ . Therefore, the Minkowski’s constants are:

M_{K_{i}}=\frac{1}{2}\sqrt{|D_{K_{i}}|},\quad i=2,3,13

so in the three cases:

M_{K_{i}}\leq\frac{1}{2}\sqrt{13}=1.802\ldots

Now, suppose that $C$ is an arbitrary class in $\operatorname{Cl}(K_{i})$ . By the theorem, there exists an ideal $\mathfrak{A}$ , representative of $C$ , such that:

|{\bf N}(\mathfrak{A})|<1.802\ldots<2

and therefore ${\bf N}(\mathfrak{A})=1$ . Since the only ideal of norm one is the trivial ideal $\mathcal{O}_{K_{i}}$ , which is principal, the class $C$ is also the trivial class in $\operatorname{Cl}(K_{i})$ . Hence there is only one class in the class group, and the class number is one for the three fields $K_{2},\ K_{3}$ and $K_{13}$ .

Example 2.

Let $K=\mathbb{Q}(\sqrt{17})$ . The discriminant is $D_{K}=17$ and the Minkowski’s bound reads:

M_{K}=\frac{1}{2}\sqrt{17}=2.06\ldots

Suppose that $C$ is an arbitrary class in $\operatorname{Cl}(K)$ . By the theorem, there exists an ideal $\mathfrak{A}$ , representative of $C$ , such that:

|{\bf N}(\mathfrak{A})|<2.06\ldots

and therefore ${\bf N}(\mathfrak{A})=1$ or $2$ . However,

2=\frac{-3+\sqrt{17}}{2}\cdot\frac{3+\sqrt{17}}{2}

so the ideal $2\mathcal{O}_{K}$ is split in $K$ and the prime ideals

\left(\frac{-3+\sqrt{17}}{2}\right),\quad\left(\frac{3+\sqrt{17}}{2}\right)

are the only ones of norm $2$ . Since they are principal, the class $C$ is the trivial class, and the class group $\operatorname{Cl}(K)$ is trivial. Hence, the class number of $\mathbb{Q}(\sqrt{17})$ is one.

Title	using Minkowski’s constant to find a class number
Canonical name	UsingMinkowskisConstantToFindAClassNumber
Date of creation	2013-03-22 15:05:38
Last modified on	2013-03-22 15:05:38
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	4
Author	alozano (2414)
Entry type	Example
Classification	msc 11H06
Classification	msc 11R29
Related topic	ClassNumbersAndDiscriminantsTopicsOnClassGroups