# Fractional ideals of maximal orders in quaternion algebras

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 $\omega_1, . . . , \omega_n$ 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.

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 $$\{1,d,d^2,d^3,...,d^{n+1}\}$$ 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 $$\text{Nm}_{F/\mathbb{Q}}(\bar{d}d)\in \mathbb{Z}$$.