# 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:

$$|\mathbf{N}(\U0001d504)|\le {M}_{K}\sqrt{|{D}_{K}|}$$ |

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 $\U0001d504$, representative of $C$, such that:

$$ |

and therefore $\mathbf{N}(\U0001d504)=1$. 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 $\mathrm{Cl}({K}_{i})$. 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 $\U0001d504$, representative of $C$, such that:

$$ |

and therefore $\mathbf{N}(\U0001d504)=1$ 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^{}

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

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 |