$\overline{\mathbb{Z}}$ is not a Dedekind domain. I have to prove the following statement:
Let $\overline{\mathbb{Z}}$ be the ring of all algebraic integers in (a fixed choice of) $\overline{\mathbb{Q}}$. Then $\overline{\mathbb{Z}}$ is not a Dedekind domain. 

Which one of the three properties of a Dedekind domain does the ring $\overline{\mathbb{Z}}$ not satisfy ?

We know that $\overline{\mathbb{Z}}$ is integrally closed in its field of fractions $\overline{\mathbb{Q}}$ (for my part the proof was given in class but if someone wants it, I can give it).
I found that it's not noetherian because there is a strictly ascending chain of ideals: $\lbrace (2^{\frac{1}{2^{n}}}) \rbrace_{n = 1}^{\infty}$.

And I have to prove that every nonzero prime ideal is maximal. 

Yes, I have to do it to be sure the noetherian property is the only one which is not true. But I have no idea how to do it. I thought about doing it by contradiction... can't seem to get it very far. Is it a good direction ? If it is, how do I go on ? If it is not, what do I do ?
Jérôme
 A: Let $P$ be a prime ideal of $\overline{\mathbb{Z}}$. Let $O_{K}$ denote the ring of integers of a number field $K$. Then $O_{K}\subseteq \overline{\mathbb{Z}}$ for every number field $O_{K}$. Notice that $ O_{K}/P\cap O_{K} \rightarrow \overline{\mathbb{Z}}/P $ is a monomorphism. Observe also that $P\cap O_{K} \neq 0$ for every prime ideal $P$ since $P \cap \mathbb{Z} \neq 0$.
Finally, notice that $\overline{\mathbb{Z}}/P $ is integral over $O_{K}/P\cap O_{K}$ implying that $\overline{\mathbb{Z}}/P$ is a field. This is because $P\cap O_{K}$ is a non-trivial prime ideal and hence a maximal ideal in $O_{K}$ and any domain which is integral over a field must itself be a field. As $\overline{\mathbb{Z}}/P$ is a field, $P$ is maximal. 
A: $\mathbb Z\subset\overline{\mathbb Z}$ is an integral ring extension. By Cohen-Seidenberg Theorem we have $\dim\mathbb Z=\dim\overline{\mathbb Z}$. Since $\dim\mathbb Z=1$ you get $\dim\overline{\mathbb Z}=1$, and therefore every nonzero prime ideal is maximal in $\overline{\mathbb Z}$.
