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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.

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  • $\begingroup$ I'm not aware of any really short arguments. This is a nontrivial result. $\endgroup$ – Qiaochu Yuan May 23 '13 at 19:06
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$K$ 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.

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    $\begingroup$ That's not necessarily true, take $K=\Bbb Q(\sqrt{5})$ with integral basis $\{1,\sqrt{5}\}$. But it is known that $\Bbb Z[\sqrt{5}]\ne\mathcal{O}_K$. And not all ideals are generated by subsets of even an integral basis for the integer ring. Take $(2)\subseteq\Bbb Z[i]$ which is not generated by any subset of $\{1,i\}$, an integral basis for the ring. $\endgroup$ – Adam Hughes Nov 25 '14 at 14:58
  • $\begingroup$ @AdamHughes Perhaps I misunderstand, but is an integral basis of a number field $K$ not defined to be a $\mathbb{Z}$-basis of the abelian group $\mathfrak{O}_K$? In this case, $\{1, \sqrt{5}\}$ is not an integral basis of $\mathbb{Q}(\sqrt{5})$. $\endgroup$ – Guy Paterson-Jones Sep 7 '17 at 9:28
  • $\begingroup$ @GuyPaterson-Jones the term basis always begs the question of "for what ring?" because a set is a basis dependent upon the scalars and the ambient set you are trying to generate. If you want a basis for $K$ done things will suffice which won't for $O_K$, so the distinction is important. $\endgroup$ – Adam Hughes Sep 7 '17 at 10:37
  • $\begingroup$ @AdamHughes Sure, but the term integral basis is actually defined the way I put it in Algebraic Number Theory by Stewart and Tall, for instance. If you mean a basis consisting of algebraic integers, then of course you are correct. $\endgroup$ – Guy Paterson-Jones Sep 7 '17 at 10:51

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