Does every order in a quaternion algebra have a nice integral basis?

Question

Let $A$ be a quaternion algebra over a number field $k$ and let $O \subset A$ be an order. Does there exist always a basis $\{1,i,j,k\}$ of $O$ over the ring of integers $\mathcal{O}_k$ of $k$ such that $$ i^2 \in \mathcal{O}_k, \quad j^2 \in \mathcal{O}_k, \quad ij = -ji = k $$ ?

Answer 1

This is not even true in the simplest situation you could hope for, namely if $A = (\frac{-1, -1}{\mathbb Q})$ is the "rational Hamilton quaternions" and $O$ is the standard maximal order (the Hurwitz order). In this case $O$ has a basis of the form

$$1, i, j, \frac{1+i+j+k}2$$

and in general they can get much more complicated. (Here $i, j, k$ are all square roots of $-1$ satisfying the usual identity. You can check that there is no other choice for $i, j, k$ to give you a basis in the form you requested.)