Dedekind domain and converse of Krull's principal ideal theorem It is clear that Dedekind domain's all primes other than $(0)$ are maximal. Its dimension must be $\leq 1$. Suppose this domain is not a field. So $\dim=1$. 
So all maximal must be minimal over a principal element. 
Here is Krull's principal ideal theorem converse. 
Thm: Any prime of height $c$ is minimal over an ideal generated by $c$ elements. 
Consider $\mathbb Z[\sqrt{-5}]$. Then $m=(3,2+\sqrt{-5})$ is a maximal ideal. One can check this easily by lifting to $\mathbb Z[x]/(x^2+5)$.
Q1. This maximal ideal $m$ is generated by 2 elements. Why this maximal ideal is not minimal over itself? $m$'s primary decomposition is itself. So it is minimal over itself. 
The reason to ask this question is that $m$ should be minimal over principal ideal rather than $(3,2+\sqrt{-5})$. If $m$ minimal over itself, this says $ht(m)\leq 2$ from Krull PIT. Now converse says $m$ minimal over principal by $m$ having $ht(m)=1$. 
Q2. $m$ cannot be minimal over itself and a principal ideal simultaneously. 

The ideal $m$ is minimal over itself, but this does not contradict the theorem: it simply says that $m$ is minimal over a principal ideal.  And indeed, since $(3) = (3, 1 + \sqrt{-5})(3, 2 + \sqrt{-5})$, $m$ is minimal over the principal ideal $(3)$.
