Let $K$ be a number field over $\mathbb{Q}$ of degree $n$, and $\mathcal{O} \subset \mathcal{O}_K$ an order.
$\textbf{Questions:}$
$\newcommand{\Spec}{\textrm{Spec }}$
$\newcommand{\cO}{\mathcal{O}}$
1.) Is the natural map $\phi: \Spec \mathcal{O}_K \rightarrow \Spec \mathcal{O}$ flat?
2.) How many distinct primes (can) lie under a given prime? All but finitely many local rings of $\cO$ are canonically identified with local rings of $\cO_K$. Since it is not obvious to me that $\phi$ is surjective, how many more primes are in $\cO$ than in $\cO_K$ (w.r.t. the canonical identification above).
3.) In $\mathcal{O}_K$, and dedekind domains, prime ideals can be generated by two elements. How many elements are required to generate prime ideals of $\mathcal{O}$ ? Is it possible to give an answer depending on the degree $n = [K:\mathbb{Q}]$ and the index $[\mathcal{O}_K: \mathcal{O}]$?
4.) Is every ideal $I$ of $\cO$ also a proper $\cO$-ideal as is the case for the maximal order? That is, has ring of multipliers $R$ exactly $\cO$. (The rings of multipliers $R\subset K$ is the subring of elements $\alpha$ so that $\alpha \cdot I \subset I$). Certainly $\cO \subset R$.
$\textbf{Note:}$ If need be, feel free to assume that $K$ is quadratic imaginary. I'm primarily interested in this case, but I would like to have a clearer picture of the general situation.
Best Answer
2) I'm not sure what you mean by "lie under". Any number of equal-characteristic primes of $\mathcal O_K$ can map to a single prime of $\mathcal O$, but that's lying over, not under. The map on primes is surjective - for any local ring of $\mathcal O$, its integral closure is a local ring of $\mathcal O_K$.
3) They can be at least $n-1$. Take $p$ a totally split prime, and consider the subring of $\mathcal O_K$ of elements that are in $\mathbb Z$, modulo $p$. Then the primes lying over $p$ glue together into a single prime ideal, whose local ring is the inverse image of the diagonal $\mathbb F_p$ in the natural map $\mathbb Z_p^n \to \mathbb F_p^n$. If $m$ is the maximal ideal of this local ring, then $m/m^2 = \mathbb F_p^n = (R/m)^n$, so the ideal requires at least $n$ generators.
4) No. Certainly some ideals have ring of multipliers $\mathcal O_K$. in $\mathbb Z[\sqrt{-3}]$, say, the ideal $(2)$ has this property.