Ideal Contains Its Absolute Norm

Proof

Let $I⊴{\mathcal{O}}_{\mathbb{K}}$. 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 $⟨n⟩\subseteq I$ which implies that $I\mid ⟨n⟩$ given ${\mathcal{O}}_{\mathbb{K}}$ is a Dedekind ring. However $I\mid ⟨n⟩\mid ⟨N(I)⟩$ and therefore $N(I)\in I$.