Norm of element $\alpha$ equal to absolute norm of principal ideal $(\alpha)$ Let $K$ be a number field, $A$ its ring of integers, $N_{K / \mathbf{Q}}$ the usual field norm, and $N$ the absolute norm of the ideals in $A$.
In some textbooks on algebraic number theory I have seen the fact: $\vert N_{K / \mathbf{Q}}(\alpha) \vert = N(\alpha A)$ for any $\alpha \in A$. However, I wasn't able to find a proof (neither in books nor by myself). 
Can someone explain to me, why this is true? 
Thanks!

I have written up the proof in Lemma $3.3.3$ of my lecture notes in algebraic number theory, on page $35$. It uses three different $\mathbb{Q}$-bases of the number field $K$, $\mathbb{Z}$-bases for the the ring of integers $\mathcal{O}_K$, and the determinant for the commutative diagram given.
