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.