Let $K$ be a number field with ring of integers $\mathcal{O}_K$. Let $\mathcal{O}$ be an order of $K$ and $A$ a proper ideal of $\mathcal{O}$, by which I mean that $\mathcal{O} = \{\alpha \in K : \alpha A \subseteq A\}$.
Question:
Does there exist an invertible ideal $I$ of $\mathcal{O}$, such that $I\subseteq A$ and $I\mathcal{O}_K = A\mathcal{O}_K$?
The answer is yes for quadratic fields $K$, since then every proper ideal of every order is invertible. Here is an example in degree 3 that motivated this question for me.
Example:
Let $K=\mathbb{Q}(\sqrt[3]{2})$ with $\mathcal{O}_K=\mathbb{Z}[\sqrt[3]{2}]$ and consider the order
$\mathcal{O}=\mathbb{Z}+2\mathcal{O}_K=\langle 1,2\sqrt[3]{2},2\sqrt[3]{2}^2\rangle_\mathbb{Z}$.
By Example 3.5 of the document
http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/conductor.pdf
on Keith Conrad's webpage, the $\mathbb{Z}$-module
$A = \langle 8, 2\sqrt[3]{2}, 2\sqrt[3]{2}^2 \rangle_\mathbb{Z}$
is an ideal of $\mathcal{O}$ that is proper and not invertible. Moreover,
$A\mathcal{O}_K = \langle 8, 2\sqrt[3]{2}, 2\sqrt[3]{2}^2 \rangle_\mathbb{Z} \langle 1,\sqrt[3]{2}, \sqrt[3]{2}^2 \rangle_\mathbb{Z} = \langle 4, 2\sqrt[3]{2}, 2\sqrt[3]{2}^2 \rangle_\mathbb{Z}$.
The ideal $I=2\sqrt[3]{2}\mathcal{O}\subseteq A$ is invertible, since it is principal, and moreover
$I\mathcal{O}_K = 2\sqrt[3]{2}\mathcal{O}_K = \langle 4, 2\sqrt[3]{2}, 2\sqrt[3]{2}^2 \rangle_\mathbb{Z} = A\mathcal{O}_K$.
I wouldn't expect a principal ideal $I$ with the required properties to exist in general, but maybe invertible is good enough?
