ideals in rings of algebraic integers are finitely generated I am trying to write about rings of algebraic integers $\mathcal{O}_K$ in a number field $K$ without introducing to much field theory. I want to show that these rings are Dedekind. First of all I want to show that these rings are Noetherian. Is there any simple way to show that every ideal is finitely generated?

The argument I found so far is to consider $\mathcal{O}_K$ as an $\mathbb{Z}$-module. After deducing that $\mathcal{O}_K$ has a finite basis as such, one deduces that every subgroup of this free abelian group is finitely generated and hence that this implies that every ideal of $\mathcal{O}_K$ has to be finitely generated since the underlying group is. Or this is how I understand the argument anyway. The conclusion that $\mathcal{O}_K$ has a finite basis as a $\mathbb{Z}$-module involves a lot of work in the book I am using, I would appreciate if someone wants to make that argument clear here. Or correct me if I misunderstood any other part of the argument. Or give another argument leading to the desired conclusion.
$K$ is a number field of a certain degree, say $n$, hence is the $\mathbb Q$-span of a set which has $n$ elements. Then $K$ has an integral basis made up of $n$ algebraic integers of $K$. The $\mathbb Z$-span of that integral basis is $\mathcal O_k$ hence $\mathcal O_k$ is finitely generated as a $\mathbb Z$-module. Any ideal of $\mathcal O_k$ is generated by a subset of that integral basis hence is finitely generated. Therefore $\mathcal O_k$ is noetherian.
$K$ is a number field of a certain degree, say $n$, hence is the $\mathbb Q$-span of a set which has $n$ elements. Then $K$ has an integral basis made up of $n$ algebraic integers of $K$. The $\mathbb Z$-span of that integral basis is $\mathcal O_k$ hence $\mathcal O_k$ is finitely generated as a $\mathbb Z$-module. Any ideal of $\mathcal O_k$ is generated by a subset of that integral basis hence is finitely generated. Therefore $\mathcal O_k$ is noetherian.