Number-field integer rings and ideals are full lattices

For every ordered $\mathbb{Q}$-basis ${\alpha }_1,\dots ,{\alpha }_n$ of $K$, let $A$ have columns $\sigma ({\alpha }_j)$. By [F1], $A$ is invertible; hence these images form a real basis of ${\mathbb{R}}^n$.

Choose a $\mathbb{Z}$-basis ${\alpha }_1,\dots ,{\alpha }_n$ of ${\mathcal{O}}_K$, which exists by [F2]. A rational relation among the ${\alpha }_i$, multiplied by a positive common denominator, would be an integer relation, so $\mathbb{Z}$-independence makes them $\mathbb{Q}$-independent. There are $n=[K:\mathbb{Q}]$ of them, so they form a $\mathbb{Q}$-basis of $K$.

We prove by induction on $m\ge 0$ that every additive subgroup of ${\mathbb{Z}}^m$ has a finite $\mathbb{Z}$-basis with at most $m$ members. The claim holds for $m=0$, since its only subgroup is $\{0\}$, with empty basis.

Let $m>0$, assume the claim for $m-1$, and project $H\le {\mathbb{Z}}^m$ onto its first coordinate. By [F3], the image is $d\mathbb{Z}$ for a nonnegative integer $d$. If $d=0$, $H$ lies in the last $m-1$ coordinates, so induction gives a basis with at most $m-1$ members.

By steps 1.1 and 1.2, the vectors $\sigma ({\alpha }_1),\dots ,\sigma ({\alpha }_n)$ form a real basis. Additivity of $\sigma$ gives $\sigma ({\mathcal{O}}_K)=\mathbb{Z}\sigma ({\alpha }_1)\oplus \cdots \oplus \mathbb{Z}\sigma ({\alpha }_n)$, so this is a full lattice by [F6]. This proves clause 1.

If $d>0$, choose $h\in H$ whose first coordinate is $d$. The kernel $H_0$ of the projection, viewed in ${\mathbb{Z}}^{m-1}$, has a basis $k_1,\dots ,k_s$ by induction, with $s\le m-1$. Every $g\in H$ has first coordinate $qd$ for a unique $q\in \mathbb{Z}$, so $g-qh\in H_0$; hence $h,k_1,\dots ,k_s$ span $H$. If $ah+{\sum }_i{b}_i{k}_i=0$ with integer coefficients, the first coordinate gives $ad=0$, hence $a=0$, and independence of the basis of $H_0$ gives every $b_i=0$. Together with the $d=0$ case, this proves the induction claim.

Let $\mathfrak{a}\subseteq {\mathcal{O}}_K$ be a nonzero integral ideal. It is an additive subgroup by [F4]. Using the basis of ${\mathcal{O}}_K$ from step 1.2 to identify it with ${\mathbb{Z}}^n$, steps 1.3, 2.1, and 3.1 give a $\mathbb{Z}$-basis ${\beta }_1,\dots ,{\beta }_r$ of $\mathfrak{a}$ with $r\le n$. Choose $0\mathrm{\ne }u\in \mathfrak{a}$; then $u{\mathcal{O}}_K\subseteq \mathfrak{a}$ by [F4].

For any nonzero $v\in K$, coordinate multiplication by the values of the embeddings at $v$ defines a block-diagonal real map $L_v$. Its real blocks are the nonzero scalars ${\sigma }_i(v)$; a complex block ${\tau }_j(v)=a+ib$ is represented by $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$, whose determinant is $a^2+b^2=\mathrm{\mid }{\tau }_j(v){\mathrm{\mid }}^2>0$ because each embedding is injective. Thus $L_v$ is invertible. In particular, for the element $u$ chosen in step 4.1, $\sigma (ux)=L_u\sigma (x)$ and $L_u(\sigma ({\mathcal{O}}_K))=\sigma (u{\mathcal{O}}_K)\subseteq \sigma (\mathfrak{a})$. Applying $L_u$ to the basis in step 2.2 gives a real basis, whose integer span is a full lattice by [F6]; hence $\sigma (\mathfrak{a})$ spans ${\mathbb{R}}^n$.

Since ${\beta }_1,\dots ,{\beta }_r$ are $\mathbb{Z}$-independent, they are $\mathbb{Q}$-independent: a rational relation, after multiplication by a positive common denominator, is an integer relation and therefore has all coefficients zero. Additivity gives $\sigma (\mathfrak{a})=\mathbb{Z}\sigma ({\beta }_1)+\cdots +\mathbb{Z}\sigma ({\beta }_r)$, so these images span it over $\mathbb{R}$. Step 5.1 forces $r\ge n$, while step 4.1 gives $r\le n$. Thus $r=n$, the ${\beta }_i$ form a $\mathbb{Q}$-basis of $K$, and [F1] makes their images a real basis. Therefore $\sigma (\mathfrak{a})=\mathbb{Z}\sigma ({\beta }_1)\oplus \cdots \oplus \mathbb{Z}\sigma ({\beta }_n)$ is a full lattice.

Let $I$ be a nonzero fractional ${\mathcal{O}}_K$-ideal. By [F5], choose $0\mathrm{\ne }d\in {\mathcal{O}}_K$ with $\mathfrak{b}:=dI\subseteq {\mathcal{O}}_K$. The set $\mathfrak{b}$ is an ideal because $I$ is an ${\mathcal{O}}_K$-submodule, and it is nonzero because multiplication by $d\mathrm{\ne }0$ in the field $K$ is injective. Thus step 6.1 shows that $\sigma (\mathfrak{b})$ is a full lattice.

The real-coordinate multiplication $L_d$ is invertible by the block calculation of step 5.1. From $\mathfrak{b}=dI$ and $\sigma (dx)=L_d\sigma (x)$ we obtain $L_d(\sigma (I))=\sigma (\mathfrak{b})$. If ${\gamma }_1,\dots ,{\gamma }_n$ is a lattice basis of $\sigma (\mathfrak{b})$ from step 7.1, then $L_d^{-1}{\gamma }_1,\dots ,L_d^{-1}{\gamma }_n$ is a real basis and its integer span is $\sigma (I)$; thus $\sigma (I)$ is a full lattice by [F6]. This proves clause 2.

Clause 1 is step 2.2 and clause 2 is step 8.1, so both assertions of the statement hold.