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 $n\mathrm{=}{r}_{\mathrm{1}}\mathrm{+}\mathrm{2}\mathit{}{r}_{\mathrm{2}}$ be the degree of $K$ over $\mathrm{Q}$, where $r_{\mathrm{1}}$ and $r_{\mathrm{2}}$ are the number of real and complex embeddings, respectively. The class group of $K$ is denoted by $\mathrm{Cl}\mathit{}\mathrm{(}K\mathrm{)}$. In any ideal class $C\mathrm{\in }\mathrm{Cl}\mathit{}\mathrm{(}K\mathrm{)}$, there exists an ideal $\mathrm{A}\mathrm{\in }C$ such that:

$$|𝐍(𝔄)|\le M_K\sqrt{|D_K|}$$

where $\mathrm{N}\mathit{}\mathrm{(}\mathrm{A}\mathrm{)}$ denotes the absolute norm of $\mathrm{A}$ 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:

$$M_{K_i}=\frac{1}{2}\sqrt{|D_{K_i}|},i=2,3,13$$

so in the three cases:

$$M_{K_i}\le \frac{1}{2}\sqrt{13}=1.802\mathrm{\dots }$$

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

and therefore $𝐍(𝔄)=1$. Since the only ideal of norm one is the trivial ideal ${𝒪}_{K_i}$, which is principal, the class $C$ is also the trivial class in $\mathrm{Cl}(K_i)$. 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:

$$M_K=\frac{1}{2}\sqrt{17}=2.06\mathrm{\dots }$$

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

and therefore $𝐍(𝔄)=1$ or $2$. However,

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

so the ideal $2{𝒪}_K$ is split in $K$ and the prime ideals

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

are the only ones of norm $2$. Since they are principal, the class $C$ is the trivial class, and the class group $\mathrm{Cl}(K)$ 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