the ring of integers of a number field is finitely generated over $\mathbb{Z}$
RingOfIntegersOfANumberFieldIsFinitelyGeneratedOverMathbbZ
2005-03-17 15:25:43
2005-03-17 15:57:16
Theorem
integral closure
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\begin{thm}
Let $K$ be a number field of degree $n$ over $\Rats$ 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 $\Ints$-linear combination of the $\alpha_i$.
\end{thm}
\begin{cor}
Every ideal of $\mathcal{O}_K$ is finitely generated.
\end{cor}
\begin{proof}[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.
\end{proof}
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 $K$.