I'm reading Bergeron's book "The Spectrum of Hyperbolic Surfaces". In chapter 8, he talks about orders of quaternion algebras. I am interested in Proposition 8.3 which states
Let $\mathcal{O}$ and $\mathcal{O}'$ be two orders of $A$ (a quaternion algebra over $\mathbb{Q}$). If $\mathcal{O}\supset\mathcal{O}'$, then $d(\mathcal{O})$ divides $d(\mathcal{O}')$. Moreover, if $\mathcal{O}\supset\mathcal{O}'$ and $d(\mathcal{O})=d(\mathcal{O}')$, then $\mathcal{O}=\mathcal{O}'$.
The proof is omitted but it is remarked that it follows from a previous proposition that says for an order $\mathcal{O}$, we have that $\mathcal{O}\cap\mathbb{Q}=\mathbb{Z}$.
In the above, we have that $d(\mathcal{O})$ is the reduced discriminant. It is shown that if $(e_1,e_2,e_3,e_4)$ is a basis for the order $\mathcal{O}$, and $P$ is the change of basis matrix from $(1,i,j,k)$ to $(e_1,e_2,e_3,e_4)$, then we have that $d(\mathcal{O})=2^2ab(\det P)$ where we have that the quaternion algebra $A=D_{a,b}(\mathbb{Q})$ (i.e. with $i^2=a,j^2=b$).
I am having trouble seeing a couple of things. First, they claim that since $\text{disc}(\mathcal{O})\in\mathbb{Z}$ and $\det P\in\mathbb{Q}$ this implies that $d(\mathcal{O})\in\mathbb{Z}$. It is clear to me why $\text{disc}(\mathcal{O})\in\mathbb{Z}$, but it isn't clear why $\det P\in\mathbb{Q}$ since I could see a world where this lies in some quadratic extension of $\mathbb{Q}$ since we have that $$ \text{disc}(\mathcal{O})=-2^4a^2b^2(\det P)^2 $$ I see that the left-hand side of this is an integer, so I see that $(\det P)^2\in\mathbb{Q}$, but it isn't clear to me why $\det P\in\mathbb{Q}$. If this is true, I then see why $d(\mathcal{O})=2^2ab(\det P)\in\mathbb{Z}$. It seems to me that for some reason $P$ being the change of basis into an order must be what forces $\det P$ to actually be rational, but if there is a more concrete explanation as to why this is true that would help clear that up for me.
The other thing I am having trouble seeing is the first part of the proposition, and why $\mathcal{O}\cap \mathbb{Q}=\mathbb{Z}$ implies this "to divide is to contain" sort of proposition. I can understand heuristically why it should be true in that we have that if $P$ is the change of basis matrix for $\mathcal{O}$ and $Q$ is the change of basis matrix for $\mathcal{O}'$, then the division criterion on the reduced discriminant is equivalent to saying that $\det P\mid \det Q$. I think the idea is that since $\mathcal{O}\supset\mathcal{O}'$ we will have that $\mathcal{O'}$ is the larger lattice, so it has a larger fundamental parallelogram, and the quotient of the determinants should be the index of the lattice.
If someone could help clear these two points up for me, I would greatly appreciate it.