# What sort of ring is the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$?

I begin losing intuition when I start dealing with infinitely generated fields over $\mathbb{Q}$...

The naive guess is that it is noetherian of krull dimension 1. Is this correct?

A related question is what sort of morphism is $Spec($the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}) \rightarrow Spec(O_K)$ for some number field $K$? For example, is it a flat morphism?

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This is just the ring of algebraic integers, which is a Bézout domain. That means that every finitely generated ideal is principal. As there do exist non-finitely generated ideals which are not principal, it is not a Noetherian domain.

For example, consider the ideal generated by $\{2^{1/n}\colon n\ge1\}$. As no generating set is contained in any finite extension of $\mathbb{Q}$, it is not finitely generated. On the other hand, any quotient by a nonzero prime ideal $\mathfrak{p}$ gives an algebraic extension of $\mathbb{Z}/\mathfrak{p}\cap\mathbb{Z}$, so is a field, and $\mathfrak{p}$ is maximal. So it does have Krull dimension 1.

The other common example of a Bézout domian which is not Noetherian is the ring of entire functions on the complex plane.

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No, it is not noetherian. The ideals generated by $2^{1/2^n}, n = 1, 2, \dots$ form a strictly ascending chain that does not stabilize. It is of Krull dimension one (every nonzero prime ideal is maximal) by lying over and going up.

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To your related question "For example is flat?" the answer is yes. The point is that over a Dedekind domain, more generally over a Prüfer ring, a module is flat if and only if it is torsion free( see here). On the other hand I'm not so sure if I can say more about of the morphism. It is not going to be locally finite, locally of finite type, locally Noetherian etc...

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