---
tag: C7Q2MVD933
kind: definition
name: Ideal class group
msc: [11R29, 13F05]
defines: WQ46BD3ARY
uses: [7NPY2CN5KB, YB4JTB4WJN, 0KYX2N69J0, 082GHNPJ53, MWBDFS2KJ5]
sources:
  - url: https://en.wikipedia.org/wiki/Ideal_class_group
---

## Definition

Let $R$ be a [[Dedekind domain|7NPY2CN5KB]] with field of fractions $K$.
Let $I(R)$ be the set of nonzero [[fractional ideals|YB4JTB4WJN]] of $R$; under the product of fractional ideals it is an [[abelian group|082GHNPJ53]] with identity $R$.
Let $P(R) = \set{xR : x \in K^{\times}}$ be its subgroup of principal fractional ideals.
The *ideal class group* of $R$ is the [[quotient group|0KYX2N69J0]]
$$\operatorname{Cl}(R) = I(R) / P(R).$$
For a [[number field|MWBDFS2KJ5]] $K$ with ring of integers $\mathcal{O}_K$, the class group of $K$ is $\operatorname{Cl}(K) = \operatorname{Cl}(\mathcal{O}_K)$.

## Remarks

Example: $R = \Z$ is a principal ideal domain, so every fractional ideal is $x\Z$ with $x \in \Q^{\times}$, $I(\Z) = P(\Z)$, and $\operatorname{Cl}(\Z)$ is trivial.
Example: for $R = \Z[\sqrt{-5}]$, let $\mathfrak{p} = (2, 1 + \sqrt{-5})$.
Then $\mathfrak{p}^2$ is generated by $4$, $2 + 2\sqrt{-5}$ and $(1 + \sqrt{-5})^2 = -4 + 2\sqrt{-5}$; all are multiples of $2$ and the ideal contains $4$ and $(2 + 2\sqrt{-5}) - (-4 + 2\sqrt{-5}) = 6$, hence $2$, so $\mathfrak{p}^2 = (2)$.
The ideal $\mathfrak{p}$ is not principal, since $a^2 + 5b^2 = 2$ has no integer solution; so its class has order $2$, and in fact $\operatorname{Cl}(\Z[\sqrt{-5}]) \cong \Z/2\Z$.
Non-example: the nonzero integral ideals of $R$ alone do not form a group under multiplication.
A proper ideal $\mathfrak{a}$ has no integral ideal $\mathfrak{b}$ with $\mathfrak{a}\mathfrak{b} = R$, since $\mathfrak{a}\mathfrak{b} \subseteq \mathfrak{a}$.
This is why fractional ideals are used.
Equivalently, $\operatorname{Cl}(R)$ is the set of nonzero ideals of $R$ modulo the relation $\mathfrak{a} \sim \mathfrak{b}$ if $x\mathfrak{a} = y\mathfrak{b}$ for some nonzero $x, y \in R$.
The class group is trivial if and only if $R$ is a principal ideal domain.