Ideal class group

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.