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!

One way to think about this is that both sides equal the absolute value of the determinant of the map $m_\alpha$ -- the multiplication by $\alpha$ map on $A$. Indeed, the left hand side is defined to be the said quantity while the right hand side is the order of the cokernel of $m_\alpha$.

Translate here Part 3.5 of the book of Pierre Samuel, “Théorie algébrique de nombres” (sorry by translation surely deficient).

3.5 Norm of an ideal

Let $K$ be a number field, $n$ its degree, $A$ its ring of integers. We note $N(x)$ instead of $ N_{K / \mathbf{Q}}(x)$.

Proposition 1.- If $x\ne 0$ is an element of $A$ then $|N(x)|=$#$(A/Ax)$

(note that since $x\in A$, one has $N(x)\in \mathbb Z$ so the formula above makes sense).

Proof.-We know that $A$ is a free $\mathbb Z$-module of rank $n$; the $\mathbb Z$-module $Ax$ is also of rank $n$ because the multiplication by $x$, $A\to Ax$ is a $\mathbb{Z}$-module isomorphism. We know there is a base $(e_1,…..,e_n)$ of the $\mathbb Z$-module $A$ and elements $c_i$ of $\mathbb N$ such that $(c_1e_1,…..,c_ne_n)$ is a base of $Ax$. Hence $A/Ax$ is isomorphic to $\prod_{i=1}^n\mathbb Z/c_i\mathbb Z $ and its order is equal to $c_1c_2…..c_n$. Consider the $\mathbb Z$-linear function $u$ of $A$ on $Ax$ defined by $u(e_i)=c_ie_i$; $i=1,…..,n$; one has det$(u)=c_1c_2…..c_n$. Furthermore, $(xe_1,…..,xe_n)$ is also a base of $Ax$ hence we have an automorphism $v$ of the $\mathbb Z$-module $Ax$ such that $v(c_ie_i)=xe_i$; it follows that det$(v)$ is invertible in $\mathbb Z$ so det $(v)=\pm 1$ and consequently $v\circ u$ is the multiplication by $x$ and its determinant is, by definition, $N(x)$. Since det $(v\circ u)=$ det $(v)$ det $(u)$ it follows that $N(x)=\pm c_1…..c_n=\pm$ #$(A/Ax)$.

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.