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.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: