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# using Minkowski’s constant to find a class number

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

###### 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.

## Mathematics Subject Classification

11H06*no label found*11R29

*no label found*

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