PlanetMath: using Minkowski's constant to find a class number

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using Minkowski's constant to find a class number

(Example)

We will use the theorem of Minkowski (see the parent entry).

Theorem 1 (Minkowski's Theorem) Let be a number field and let be its discriminant. Let be the degree of over $\mathbb{Q}$ , where and are the number of real and complex embeddings, respectively. The class group of 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:

$\displaystyle \vert{\bf N}(\mathfrak{A})\vert \leq M_K \sqrt{\vert D_K\vert}$

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

$\displaystyle M_K=\frac{n!}{n^n} \left(\frac{4}{\pi}\right)^{r_2}.$

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

and

. Therefore, the Minkowski's constants are:

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

so in the three cases:

$\displaystyle M_{K_i}\leq \frac{1}{2}\sqrt{13}=1.802\ldots$

Now, suppose that

is an arbitrary class in $\operatorname{Cl}(K_i)$ . By the theorem, there exists an ideal $\mathfrak{A}$ , representative of

, such that:

$\displaystyle \vert{\bf N}(\mathfrak{A})\vert<1.802\ldots <2$

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

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

and the Minkowski's bound reads:

$\displaystyle M_K=\frac{1}{2}\sqrt{17}=2.06\ldots$

Suppose that

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

, such that:

$\displaystyle \vert{\bf N}(\mathfrak{A})\vert<2.06\ldots$

and therefore ${\bf N}(\mathfrak{A})=1$ or

. However,

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

so the ideal $2\mathcal{O}_K$ is split in

and the prime ideals

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

are the only ones of norm

. Since they are principal, the class

is the trivial class, and the class group $\operatorname{Cl}(K)$ is trivial. Hence, the class number of $\mathbb{Q}(\sqrt{17})$ is one.

"using Minkowski's constant to find a class number" is owned by alozano.

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Keywords:

computing class numbers

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Cross-references: prime ideals, bound, fields, class number, norm, class, Minkowski's constants, quadratic fields, absolute norm, ideal, ideal class, class group, real and complex embeddings, number, degree, discriminant, number field
There are 2 references to this entry.

This is version 1 of using Minkowski's constant to find a class number, born on 2005-02-24.
Object id is 6822, canonical name is UsingMinkowskisConstantToFindAClassNumber.
Accessed 1644 times total.

Classification:

AMS MSC:	11H06 (Number theory :: Geometry of numbers :: Lattices and convex bodies)
	11R29 (Number theory :: Algebraic number theory: global fields :: Class numbers, class groups, discriminants)

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