Checking positivity of trace of a certain quaternion in $\mathbb{B}_{\ell,\infty}$ Say $\mathbb{B}_{\ell,\infty}$ is the quaternion algebra over $\mathbb{Q}$ ramified at the finite prime $\ell$ and the infinite prime.
Say that you have two elements $\phi_1,\phi_2\in \mathbb{B}_{\ell,\infty}$ satisfying the following polynomials with integer coefficients:
$$\phi_1^2 - d_1 \phi_1 + \frac{d_1^2 - d_1}{4}=0$$
and
$$\phi_2^2 - d_2 \phi_2 + \frac{d_2^2 - d_2}{4}=0$$
for some integers $d_1,d_2<-1$.
Then, their product, $\phi_1\phi_2$, satisfies some polynomial
$$(\phi_1\phi_2)^2 - t \phi_1\phi_2 + \frac{(d_1^2-d_1)(d_2^2-d_2)}{16}=0$$
of negative discriminant.
Define $m:=\frac{d_1d_2-(d_1d_2-2t)^2}{4}$; then, one can show the suborder of $\mathbb{B}_{\ell,\infty}$ generated by $1,\phi_1,\phi_2$ and $\phi_1\phi_2$ has discriminant $m^2$. I would like to say that $m \geq 0$, but I'm having trouble... (this fact is stated in passing on page 22, line 13 of this paper: https://arxiv.org/pdf/1206.6942.pdf).
Solving for our desired bounds on $t$ we want to show
$$\frac{d_1d_2-\sqrt{d_1d_2}}{2}\leq t\leq \frac{d_1d_2+\sqrt{d_1d_2}}{2}.$$
Notice that this lower bound is positive. The fact that the discriminant of the minimal polynomial of $\phi_1\phi_2$ is negative is only enough to give an upper bound, and so that isn't enough. Is there some useful fact of quaternion algebras that I'm overlooking?
Thanks in advance!
 A: The trick is to choose the representation of $\mathbb{B}_{\ell,\infty}$ so that the embedding of one of the $\mathbb{Z}[\frac{d_i+\sqrt{d_i}}{2}] = \mathbb{Z}[\phi_i]$ is particularly nice.
For example, choose a representation of $\mathbb{B}_{\ell,\infty}\hookrightarrow M_2\left(\mathbb{Q}(\sqrt{d_1})\right)$ so that the composition with $\mathbb{Z}[\phi_i]\hookrightarrow \mathbb{B}_{\ell,\infty}$ sends
$$\alpha = a_0 + a_1\sqrt{d_1}\mapsto \begin{pmatrix}a_0 + a_1\sqrt{d_1} & 0\\ 0 & a_0 - a_1\sqrt{d_1}\end{pmatrix}.$$
Then, your $\phi_1$, which satisfies $\phi_1^2 - d_1\phi_1 + \frac{d_1^2 -d_2}{4}$, will map to
$$\phi_1 \mapsto \begin{pmatrix}\frac{d_1+\sqrt{d_1}}{2} & 0\\ 0 & \frac{d_1-\sqrt{d_1}}{2}\end{pmatrix}.$$
Now, under this representation the element $\phi_2$ maps somewhere; say,
$$\phi_2\mapsto \begin{pmatrix}\gamma & \delta\\ j^2\bar\delta & \bar\gamma\end{pmatrix}.$$
Since you know the trace of $\phi_2$ is $d_2$ you can actually rewrite this as
$$\phi_2\mapsto \begin{pmatrix}\frac{d_2}{2} + c\sqrt{d_1} & \delta\\ j^2\bar\delta & \frac{d_2}{2} - c\sqrt{d_1}\end{pmatrix}.$$
Now, you can compute the trace of the product to be $t = cd_1 + d_1d_2/2$, and similarly, you can recompute $m = (d_1d_2-(d_1d_2-2t)^2)/4 = -4c^2d_1^2 + d_1d_2$.
Now your desired inequality of $m\geq 0$ becomes
$$|c|\leq \sqrt{\frac{d_2}{4d_1}}.$$
Since the norm of $\phi_2$ is just the determininant, which equals
$$\det(\phi_2) = N(\gamma)-j^2N(\delta) = \frac{d_2^2-d_2}{4},$$
and $-j^2N(\delta) \geq 0$ (You can choose $j^2 = -\ell q$ or $-q$ for some rational prime $q$ depending on whether $\ell$ is inert or ramified in $\mathbb{Q}(\sqrt{d_1})$, for instance.), you find
$$N(\gamma) = \frac{d_2^2}{4} - c^2d_1\leq \frac{d_2^2-d_2}{4}.$$
Solving for $c$ gives the desired inequality.
