# 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 .

• What do you mean by "the index of $\alpha$ in $\Bbb{Z}_k$"? Do you mean the index of $(\alpha)$? Also, is $\Bbb{Z}_k$ supposed to denote the ring of integers of the number field $K$? The standard notation is $\mathcal{O}_K$. Jul 8 '19 at 22:15

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 $$(O_k/ \alpha O_k$$) ./.
• It is non trivial that the $a_j$ can be integers, you should consider instead the matrix $A \in M_n(\Bbb{Z})$ representing the multiplication by $\alpha$ in the $e_j$ basis then $O_k/\alpha O_k \cong \Bbb{Z}^n / A \Bbb{Z}^n$ has $|\det(A)|$ elements by the volume definition of $\det$ (identifying $b \in \Bbb{Z}^n$ with the unit cube $b + [0,1]^n$) Jul 9 '19 at 16:31
• I only applied the general theorem concerning a noetherian submodule N $\neq 0$ contained in a free module M of finite rank when the base ring is a PID (see e.g. Lang's "Algebra", chap.15, §2, thm.5). But it is true that I forgot to mention the conditions that $a_i$ divides $a_{i+1}$. I edited that. Jul 9 '19 at 17:21