A number theory question on Dedekind rings

Question

My friend asks me a question,however I have no idea.

Is the integer ring $O_{\mathbb{C}}$ of complex field $\mathbb{C}$ a Dedekind domain?I am not familiar with it,I search help for my dear friend...If the field is a number field(Finite extension of quotient field $\mathbb{Q}$),then the answer is YES.However $\mathbb{C}$ is not a finite algebraic extension of $\mathbb{Q}$.

I don't know whether it's Noetherian...That's difficult for me because it's hard to imagine the structure of rings or all algebraic integers... — Muses_China, Jun 07 '17 at 07:42

score 2 · Accepted Answer · answered Jun 07 '17 at 07:48

If you mean the ring of all algebraic integers, the answer is no, as it is not even a noetherian domain, as observed in a comment.

However, as the algebraic closure of $\mathbf Q$ in $\mathbf C$ is the direct limit of its finite subextensions, the ring of algebraic integers is the direct limit of Dedekind domains, the rings of integers of these subextensions, and hence it is a Prüfer domain. One of the many characterisations of Prüfer domains is that all finitely generated ideals are invertible – the analog of one of the characterisations of Dedekind domains.

Actually, this ring is even more: it is a Bézout domain, i.e. every finitely generated ideal is principal.

You can find much more information in the Wikipedia notice on Prüfer domains.

Actually, we probably have more info here than does Wikipedia, e.g. see this answer for almost 30 characterizations of Prufer domains. — Bill Dubuque, Jun 07 '17 at 15:10

A number theory question on Dedekind rings

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