Proper fractional ideal of an order

Let $\mathcal O$ be an order in a number field. A fractional ideal $I$ of $\mathcal O$ is proper for $\mathcal O$ if its multiplier ring satisfies

(I:I)=\mathcal O.

Here “proper” records the exact ring of scalars preserving $I$ . It does not mean $I\subsetneq\mathcal O$ ; in particular $\mathcal O$ itself is proper in this sense.

Invertibility

Every invertible fractional ideal is proper, by multiplying $xI\subseteq I$ by its inverse. For orders in quadratic number fields, the converse holds as well. This equivalence is special to the quadratic-order setting and should not be used for arbitrary higher-degree orders.

Example

For $\mathcal O=\mathbb Z[\sqrt{-3}]$ , the ideal $I=(2,1+\sqrt{-3})$ has multiplier ring $\mathbb Z[(1+\sqrt{-3})/2]$ , so it is not proper for $\mathcal O$ . As an ideal of the larger ring, it is principal and proper.

References

Andrew Sutherland, MIT 18.783 Lecture 18, §18.3 and Theorem 18.10, imaginary quadratic orders.
John Voight, Quaternion Algebras, author's text, Exercise 16.16(b), properness and invertibility for quadratic orders.