The ring of integers has rank the degree

Put $n=[K:\mathbb{Q}]$ and choose a $\mathbb{Q}$-basis $e_1,\dots ,e_n$ of $K$. By [F1], $K$ is an $n$-dimensional $\mathbb{Q}$-vector space.

We prove by induction on $m\ge 0$ that every additive subgroup $H\le {\mathbb{Z}}^m$ has a finite $\mathbb{Z}$-basis. For $m=0$, the zero subgroup has the empty basis. Now let $m>0$ and assume the assertion for $m-1$. Project $H$ to its first coordinate. By [F6], the image is either $\{0\}$ or $d\mathbb{Z}$ for some positive integer $d$.

Define $B(x,y)={\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(xy)$. Multiplication in $K$ gives $m_{(ax+by)z}=a{m}_{xz}+b{m}_{yz}$ for $a,b\in \mathbb{Q}$, and matrix trace is linear; the same holds in the second variable since $K$ is commutative. Thus $B$ is a bilinear form.

For each $i$, [F3] gives a positive integer $m_i$ such that $u_i:=m_i{e}_i\in {\mathcal{O}}_K$. These elements remain a $\mathbb{Q}$-basis. Set $M={\sum }_{i=1}^n\mathbb{Z}{u}_i$. By [F5], $M\subseteq {\mathcal{O}}_K$.

If the image is zero, $H$ identifies with a subgroup of ${\mathbb{Z}}^{m-1}$ and has a finite basis by induction. If the image is $d\mathbb{Z}$, its kernel $H_0$ identifies with a subgroup of ${\mathbb{Z}}^{m-1}$ and has a finite basis $k_1,\dots ,k_r$ by induction. Choose one $h\in H$ whose first coordinate is $d$.

If $x\mathrm{\ne }0$, take $y=x^{-1}$. Then $B(x,y)={\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(1)=n$, because multiplication by $1$ is the identity on the $n$-dimensional space $K$. As $n>0$ in $\mathbb{Q}$, this proves that the radical of $B$ is zero.

In the nonzero-image case of step 2.3, every $g\in H$ has first coordinate $qd$ for some $q\in \mathbb{Z}$, so $g-qh\in H_0$; hence $h,k_1,\dots ,k_r$ span $H$. If $ah+{\sum }_i{b}_i{k}_i=0$, its first coordinate gives $ad=0$, hence $a=0$, and the independence of the basis of $H_0$ then gives every $b_i=0$. Thus this spanning list is a basis. The induction uses only one lift for each fixed subgroup at this finite stage, not a choice of lifts for a family.

Let $A=({\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(u_i{u}_j){)}_{i,j}$. If $Az=0$ for $z\in {\mathbb{Q}}^n$, put $x={\sum }_j{z}_j{u}_j$. Then ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(u_{i}x)=0$ for every $i$, and bilinearity makes ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(yx)=0$ for every $y\in K$, since the $u_i$ form a $\mathbb{Q}$-basis. By step 3.1, $x=0$, so $z=0$. Thus $A$ has zero kernel and [F7] makes it invertible.

For each $j$, solve $A{v}_j={\epsilon }_j$ over $\mathbb{Q}$, where ${\epsilon }_j$ is the $j$th standard coordinate vector, and put $u_j^{\ast }={\sum }_k(v_j{)}_k{u}_k$. By [F7], these solutions exist uniquely and satisfy ${\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(u_i{u}_j^{\ast })={\delta }_{ij}$ for every $i,j$.

The coordinates of the finitely many $u_j^{\ast }$ in the basis $u_1,\dots ,u_n$ are rational. Choose a positive integer $c$ clearing all their denominators. Then $c{u}_j^{\ast }\in M\subseteq {\mathcal{O}}_K$ for every $j$.

If $x={\sum }_i{a}_i{u}_i\in {\mathcal{O}}_K$, the trace-dual equations give $c{a}_j={\mathrm{Tr}}_{K\mathrm{/}\mathbb{Q}}(xc{u}_j^{\ast })$. The product is integral by [F5], so [F4] makes this trace an integer. Thus every $a_j\in c^{-1}\mathbb{Z}$, proving $M\subseteq {\mathcal{O}}_K\subseteq c^{-1}M$.

Since $u_1\mathrm{/}c,\dots ,u_n\mathrm{/}c$ is a $\mathbb{Z}$-basis of $c^{-1}M\cong {\mathbb{Z}}^n$, the inclusion in step 7.1 identifies ${\mathcal{O}}_K$ with an additive subgroup of ${\mathbb{Z}}^n$. By the induction, it has a finite $\mathbb{Z}$-basis $v_1,\dots ,v_r$.

The $v_i$ span $K$ over $\mathbb{Q}$: their integral span contains $M$, which spans $K$. Their $\mathbb{Z}$-independence implies $\mathbb{Q}$-independence after clearing denominators. Thus they are a $\mathbb{Q}$-basis and $r=n=[K:\mathbb{Q}]$. This proves the claim.