Fractional ideals of maximal orders in quaternion algebras

Question

Let $D$ be a skew field that is central and finite-dimensional over a number field $F$ (in particular: a quaternion algebra over $F$). Let $\Delta \subseteq D$ be a maximal $\mathcal{O}_{F}$-order. Let $\mathfrak{b}$ be a fractional left $\Delta$-ideal and say we have a $\mathbb{Z}$-basis $\omega_1, \dotsc , \omega_n$ for $\Delta$ with $\omega_1 = 1$.

According to the definition of fractional ideals, we can find $d \in \Delta$ such that $d\mathfrak{b}$ (or $\mathfrak{b}d$?) $\subseteq \Delta$.

Given this setting, what do we know about $d$? A paper I have recently read has led me to believe that $d \in \mathbb{Z}$ but I don't see why that would be the case.

Answer 1

This is true because we can always multiply $d$ from the left with an element of $\Delta$ to obtain a new $d'$ satisfying the property. Since $\Delta$ is a finitely generated order, we can find such a $d'\in\mathbb{Z}$.

Let $\mathfrak{b}$ be a fractional left $\Delta$-ideal and let $d\in \Delta $ be such that $d\mathfrak{b}\subset\Delta$. Since $\Delta$ is finitely generated over $\mathbb{Z}$, the set $\{1,d,d^2,d^3,\dotsc,d^{n+1}\}$ is linearly dependent over $\mathbb{Z}$. Thus, we find a polynomial $f\in\mathbb{Z}[X]$ such that $f(d)=0$. Since $\mathbb{Z}$ is integrally closed, we can in fact assume that $f$ is monic and irreducible.

Let $a_0$ be the constant term of $f$. Division with remainder in $D[X]$ shows that $f=g\cdot(X-d)$. Thus $a_0=\alpha d$ for some $\alpha\in \Delta$ and we have $a_0\mathfrak{b}=\alpha d\mathfrak{b}\subset \alpha\Delta\subset\Delta$.

In the case that $D$ is a quaternion algebra we have $\bar{d}d\in\mathcal{O}_F$, and consequently $\operatorname{Nm}_{F/\mathbb{Q}}(\bar{d}d)\in \mathbb{Z}$.