Norm of element $\alpha$ equal to absolute norm of principal ideal $(\alpha)$

Question

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!

Answer 1

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 $m_\alpha$.

Answer 2

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.

Answer 3

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 $N(x)=\pm c_1…..c_n=\pm$ #$(A/Ax)$.