2

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 $$ i^2 \in \mathcal{O}_k, \quad j^2 \in \mathcal{O}_k, \quad ij = -ji = k $$ ?

abenthy
  • 903

1 Answers1

3

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.)

Kimball
  • 3,112
  • Great answer! Might be worth noting for the OP that the order with basis $ 1,i,j,k $ is not maximal and is known as the Lipschitz quaternions, $ L $. The Hurwitz quaternions, $ H $, are in fact the unique maximal order containing the $ L $ ( $ L $ is an index 2 subgroup of $ H $). – Ian Gershon Teixeira Aug 26 '23 at 15:14