Norm of ideals in the ring of integers Let $\alpha \in \mathbb{Z}_k=\mathcal{O_k}$ (Ring of integers) be non-zero.

Then $$ N_{K/\mathbb{Q}}((\alpha))=N_{K/\mathbb{Q}}(\alpha) $$

$(\alpha)$ denotes the principal ideal generated by $\alpha$. $N$ is the norm.

Proof. Let $\mathbb{Z}_k=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2+\dots+\mathbb{Z}\omega_n$.
Then $(\alpha)=\mathbb{Z}\alpha\omega_1+\mathbb{Z}\alpha\omega_2+\dots+\mathbb{Z}\alpha\omega_n$. Then $N_{K/\mathbb{Q}}((\alpha))=|\mathbb{Z}_k/(\alpha)| $ ; if we write $\alpha \omega_i=\sum_{j=1}^n a_{ji} \omega_j$ , then the index of $\alpha$ in $\mathbb{Z}_k$ is just $\det(a_{ij})$ . But we know that $ N_{K/\mathbb{Q}}(\alpha)=\det(a_{ij})$  , and so the result follows .

I do not understand why the index of ($\alpha$) in $\mathbb{Z}_k$ is $\det(a_{ij})$ .

I know that the index gives the number of the cosets and each element is of the form $\beta+(\alpha)$ but this does not bring me further .

Thanks for the help .
You have certainly seen before that the ring of integers $O_k$ is a free $\mathbf Z$-module of rank $n$ equal to the degree of $k/\mathbf Q$. The $\mathbf Z$-submodule $\alpha O_k$ has also rank $n$ because the multiplication by $\alpha$  is a bijection of $O_k$ onto $\alpha O_k$. But the classical structure theorem of finitely generated free modules over a PID ($\mathbf Z$ here) shows the existence of a basis $(e_1,..., e_n)$ of $O_k$ and natural non null integers $a_1,..., a_n$ s.t. $a_i$ divides $a_{i+1}$ for $i\le n-1$ and $(a_1e_1,...,a_ne_n)$ is a basis of $\alpha O_k$ (the bases are called $adapted$). Hence $O_k/\alpha O_k\cong \mathbf Z/a_1\mathbf Z \times ...\times \mathbf Z/a_n\mathbf Z$ and has order $a_1...a_n$. If $f$ denote the $\mathbf Z$-linear application of $O_k$ onto $\alpha O_k$ defined by $f(e_i)=a_ie_i$, then det($f)=a_1...a_n$.

Besides, since $(\alpha e_1,..., \alpha e_n)$ is a basis of $\alpha O_k$, there is an automorphism $g$ of the $\mathbf Z$-module $\alpha O_k$ s.t. $g(a_ie_i)=\alpha e_i$. Then det($g)$ is invertible in $\mathbf Z$, i.e. det($g)=\pm 1$. But by definition  $g.f$ is the multiplication by $\alpha$ and its determinant is N$(\alpha)$ (your argument). As det($g.f$)=det($g$).det($f$), it follows that N$(\alpha)=\pm a_1...a_n=\pm$ card $(O_k/ \alpha O_k$) ./.
You have certainly seen before that the ring of integers $O_k$ is a free $\mathbf Z$-module of rank $n$ equal to the degree of $k/\mathbf Q$. The $\mathbf Z$-submodule $\alpha O_k$ has also rank $n$ because the multiplication by $\alpha$  is a bijection of $O_k$ onto $\alpha O_k$. But the classical structure theorem of finitely generated free modules over a PID ($\mathbf Z$ here) shows the existence of a basis $(e_1,..., e_n)$ of $O_k$ and natural non null integers $a_1,..., a_n$ s.t. $a_i$ divides $a_{i+1}$ for $i\le n-1$ and $(a_1e_1,...,a_ne_n)$ is a basis of $\alpha O_k$ (the bases are called $adapted$). Hence $O_k/\alpha O_k\cong \mathbf Z/a_1\mathbf Z \times ...\times \mathbf Z/a_n\mathbf Z$ and has order $a_1...a_n$. If $f$ denote the $\mathbf Z$-linear application of $O_k$ onto $\alpha O_k$ defined by $f(e_i)=a_ie_i$, then det($f)=a_1...a_n$.

Besides, since $(\alpha e_1,..., \alpha e_n)$ is a basis of $\alpha O_k$, there is an automorphism $g$ of the $\mathbf Z$-module $\alpha O_k$ s.t. $g(a_ie_i)=\alpha e_i$. Then det($g)$ is invertible in $\mathbf Z$, i.e. det($g)=\pm 1$. But by definition  $g.f$ is the multiplication by $\alpha$ and its determinant is N$(\alpha)$ (your argument). As det($g.f$)=det($g$).det($f$), it follows that N$(\alpha)=\pm a_1...a_n=\pm$ card $(O_k/ \alpha O_k$) ./.