Is the integral closure of $\mathbb{Z}$ in the algebraic closure of $\mathbb{Q}$ a Dedekind domain?

Question

Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$ and let $\overline{\mathbb{Z}}$ be the integral closure of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$. So far I've shown that $\overline{\mathbb{Z}}$ is integrally closed (that is, the integral closure of $\overline{\mathbb{Z}}$ with respect to its fraction field is itself) and $\dim\overline{\mathbb{Z}}=1$. However, as I don't really know how the ideals in $\overline{\mathbb{Z}}$ look like, I am at lost when thinking if $\overline{\mathbb{Z}}$ is Noetherian or not.
Also, I would like to know for a non-trivial prime ideal $\mathfrak p$ of $\overline{\mathbb{Z}}$ what would the isomorphism class of $\overline{\mathbb{Z}}/\mathfrak p$ be.

Answer 1

No. A Dedekind domain is necessarily Noetherian, but the ring of all algebraic integers has strictly increasing chains of ideals, for instance, $(2)\subset (2^{1/2})\subset(2^{1/4})\subset\cdots$.