# the ring of integers of a number field is finitely generated over $\mathbb{Z}$

###### Theorem.

Let $K$ be a number field of degree $n$ over $\mathbb{Q}$ and let $\mathcal{O}_{K}$ be the ring of integers of $K$. The ring $\mathcal{O}_{K}$ is a free abelian group of rank $n$. In other words, there exists a finite integral basis (with $n$ elements) for $K$, i.e. there exist algebraic integers $\alpha_{1},\ldots,\ \alpha_{n}$ such that every element of $\mathcal{O}_{K}$ can be expressed uniquely as a $\mathbb{Z}$-linear combination of the $\alpha_{i}$.

###### Corollary.

Every ideal of $\mathcal{O}_{K}$ is finitely generated.

###### Proof of the corollary.

By the theorem, $\mathcal{O}_{K}$ is a free abelian group of rank $n$, and therefore it is finitely generated. Notice that an ideal is an additive subgroup. Finally a subgroup of a finitely generated free abelian group is also finitely generated. ∎

This is the first step to prove that $\mathcal{O}_{K}$ is a Dedekind domain. Notice that the field of fractions of $\mathcal{O}_{K}$ is the field $K$ itself. Therefore, by definition, $\mathcal{O}_{K}$ is integrally closed in $K$.

Title the ring of integers of a number field is finitely generated over $\mathbb{Z}$ TheRingOfIntegersOfANumberFieldIsFinitelyGeneratedOvermathbbZ 2013-03-22 15:08:22 2013-03-22 15:08:22 alozano (2414) alozano (2414) 7 alozano (2414) Theorem msc 13B22