## The set of ideal classes over an algebraic intger ring is an abelian group

Let $\mathcal{O}$ be the set of integers in an algebraic number field $K$, and let $G = \mathcal{O}/\sim$ denote the set of equivalence classes of ideals in $\mathcal{O}$ under the relation $A \sim B$ if and only if $(\alpha)A = (\beta)B$ for some nonzero $\alpha,\beta \in \mathcal{O}$. Prove that $G$ is an abelian group under the usual ideal product of class representatives.

We will denote by $[A]$ the equivalence class containing the ideal $A$. We showed in this previous exercise that $[A][B] = [AB]$ is a well-defined binary operator on $G$.

For all $[A], [B], [C] \in G$, we have $[A]([B][C]) = [A][BC] = [A(BC)]$ $= [(AB)C]$ $= [AB][C]$ $([A][B])[C]$, so that our product is associative.

For all $[A]$, we have $[A][(1)] = [A(1)] = [A]$, and similarly $[(1)][A] = [A]$.

For all $[A]$, there exists (by Theorem 8.13) an ideal $B$ such that $AB$ is principal. Thus $[A][B] = [(1)]$, and so every element of $G$ has an inverse.

Hence $G$ is a group.

Moreover, we have $[A][B] = [AB] = [BA] = [B][A]$, so that $G$ is abelian.