$newcommand{ZZ}{mathbb{Z}} newcommand{cO}{mathcal{O}} newcommand{QQ}{mathbb{Q}} newcommand{BB}{mathbb{B}} DeclareMathOperator{disc}{disc} DeclareMathOperator{trd}{trd}$

Let $cO_{d_i} := ZZ(frac{d_i+sqrt{d_i}}{2}) subseteq QQ(sqrt{d_i})$ be the unique quadratic order of discriminant $d_i<0$ for $i = 1,2$. We let $tau_i = frac{d_i+sqrt{d_i}}{2}$, which notably satisfies the quadratic polynomial

begin{equation}

tau^2 – d_i tau + frac{d_i^2-d_i}{4}.

end{equation}

Let $BB_{ell,infty}$ be the quaternion algebra ramified at the finite, integral prime $ell$ and the non-archimedean prime $infty$.

Assume that we have embeddings $cO_{d_i}hookrightarrow BB_{ell,infty}$. Then, we can consider the order generated by the images; namely, $R:= left<1,phi_1,phi_2,phi_1phi_2right>subseteq BB_{ell,infty}$ where the $phi_i$ are the images of the $tau_i$ under the respective embeddings.

One can then compute the discriminant

begin{equation}

disc(R):=det(trd(alpha_ialpha_j)) = left(frac{d_1d_2-(d_1d_2-2trd(phi_1phi_2))^2}{4}right)^2.

end{equation}

($trd$ is the reduced trace in the quaternion algebra– notation of Voight.)

The unique maximal order of $BB_{ell,infty}$ has discriminant $ell^2$, and since containment of orders implies division of their sizes, we must have $disc(cO_{max}) = ell^2mid disc(R)$. My desired conclusion is that

begin{equation}

ellmid frac{d_1d_2-(d_1d_2-2trd(phi_1phi_2))^2}{4}in ZZRightarrow ellleq d_1d_2/4;

end{equation}

however, this requires that the integer is actually positive. I have tried a number of different methods to show this is positive, but to no avail.

My familiarity with quaternion algebras is minimal; so, perhaps there’s some well-known method of showing positivity, but I have had trouble.

Any help is appreciated. Thanks in advance!