Ideal Contains Its Absolute Norm

Proof

Let \(I \trianglelefteq \mathcal{O}_\mathbb{K}\). Let \(n \in \mathbb{N}\) be the order of \(1 + I \in \mathcal{O}_\mathbb{K}/I\) when considered as an additive group. Do to the consistency of this operation with ring multiplication in this case, we note that therefore

and hence \(n \in I\). Since the order of any element divides the order of the group and the order of the group is \(|\mathcal{O}_\mathbb{K}| = N(I)\), we have that \(n \mid N(I)\).

Since \(n \in I\) we know that \(\langle n \rangle \subseteq I\) which implies that \(I \mid \langle n \rangle\) given \(\mathcal{O}_\mathbb{K}\) is a Dedekind ring. However \(I \mid \langle n \rangle \mid \langle N(I) \rangle\) and therefore \(N(I) \in I\).