let $K$ be a number field of degree $d$ over $\mathbb{Q}$), Let $\mathcal{O}\subset K $ be an order (i.e. a $\mathbb{Z}$-lattice of $K$ contained in the integer ring $\mathcal{O}_K$ of $K$). If $ \mathcal{O}= \mathcal{O}_K $, we know that any maximal ideal in $ \mathcal{O}_K $ can generated by 2 element. So what's the case in general?

Maybe I should ask a more general question, let $A$ be $\mathbb{Z}$-algebra, and finite free of rank$=d$ as a $\mathbb{Z}$-module, what can we say about the number of generators of maximal ideals of $A$ (in terms of $d$), for example, we have a trivial upper bound $d.$

PS: I asked this question on MSE,https://math.stackexchange.com/questions/1486422/number-of-generators-of-maximal-ideals-in-an-order-of-a-number-field but got no answer.

A special case of the general question, say $A$ is an order in a finite product of field $\prod_{i=1}^n K_i$, i.e. $A\subset \prod \mathcal{O}_{K_i} $ with rank $d=\Sigma [K_i:\mathbb{Q}]$, what's the upper bound of number of generators of maximal ideals of $A$ (in terms of $d$ and $n$). If $A$ is the product of orders in all fields, then we can say something easily (see the answer by @Matthias Wendt). So maybe the question is the structure of $A$?