Let $K$ be a number field with $[K : \mathbb{Q}] = n$. We prove that $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $n$.

Step 1: $\mathcal{O}_K$ is finitely generated over $\mathbb{Z}$

Choose a $\mathbb{Q}$-basis $\{\alpha_1, \ldots, \alpha_n\}$ for $K$. Since each $\alpha_i$ is algebraic over $\mathbb{Q}$, we can clear denominators: there exists $d \in \mathbb{Z}_{>0}$ such that $d\alpha_i \in \mathcal{O}_K$ for all $i$.

Every element of $K$ can be written as $\beta = \sum_{i=1}^n q_i \alpha_i$ with $q_i \in \mathbb{Q}$. If $\beta \in \mathcal{O}_K$, write $q_i = a_i/d$ with $a_i \in \mathbb{Z}$. Then $d\beta = \sum_{i=1}^n a_i(d\alpha_i) \in \mathcal{O}_K$

This shows $\mathcal{O}_K \subseteq \frac{1}{d}\mathbb{Z}[d\alpha_1, \ldots, d\alpha_n]$, proving $\mathcal{O}_K$ is finitely generated over $\mathbb{Z}$.

Step 2: $\mathcal{O}_K$ is torsion-free

If $m\alpha = 0$ for some $\alpha \in \mathcal{O}_K$ and $m \in \mathbb{Z} \setminus \{0\}$, then $\alpha = 0$ since $K$ is a field. Thus $\mathcal{O}_K$ has no $\mathbb{Z}$-torsion.

Step 3: Structure theorem for finitely generated modules

By the structure theorem for finitely generated modules over PIDs, any finitely generated torsion-free $\mathbb{Z}$-module is free. Therefore $\mathcal{O}_K \cong \mathbb{Z}^r$ for some $r \geq 0$.

Step 4: Computing the rank

As a $\mathbb{Q}$-vector space, $K \cong \mathbb{Q}^n$. Since $\mathcal{O}_K \otimes_{\mathbb{Z}} \mathbb{Q} = K$, we have $\mathbb{Q}^n \cong K = \mathcal{O}_K \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Z}^r \otimes_{\mathbb{Z}} \mathbb{Q} \cong \mathbb{Q}^r$

Comparing dimensions over $\mathbb{Q}$ gives $r = n$.