Existence of an integral basis

Every degree-$n$ number field has a ring of integers that is a free abelian group of rank $n$. To prove this without assuming finite generation, scale a rational basis to algebraic integers ${\alpha }_1,\dots ,{\alpha }_n$ and set $M={\sum }_j\mathbb{Z}{\alpha }_j$. Their trace pairing has an invertible integer matrix $T$, since the trace form of a number field is nondegenerate. For $x\in {\mathcal{O}}_K$, all $\mathrm{Tr}(x{\alpha }_j)$ are integers. The adjugate matrix formula gives $M\subseteq {\mathcal{O}}_K\subseteq (\det T{)}^{-1}M$. The submodule theorem for free modules over a principal ideal domain makes the middle group free of rank $n$, and a group basis is an integral basis.