We have a quaternion algebra $(\frac{-1,-11}{\mathbb{Q}})$, which is ramified at $p$ and $\infty$.
I know that $\mathcal{O} = \mathbb{Z} \oplus \mathbb{Z} i \oplus \mathbb{Z} \frac{1+j}{2} \oplus \mathbb{Z} i \frac{1+j}{2}$ is a maximal order and is split at the prime $2$. To simplify we take $\alpha = i, \beta = \frac{1+j}{2}$, which then is a basis for the order. My goal is to calculate the following embedding:
$\begin{align*} \mathcal{i}&: \mathcal{O} \to M_2(\mathbb{Z}_2)\\ \alpha,\beta &\mapsto \dots \end{align*}$
here $\mathbb{Z}_2$ are the 2-adic integers.
This is the same problem as example 17.6.3 in Voights book on Quaternion algebras, except here it is over the quaternion algebra $(\frac{-1,-23}{\mathbb{Q}})$, but it is the same maximal order.
He finds the solution:
$\begin{align*} \mathcal{i}&: \mathcal{O} \to M_2(\mathbb{Z}_2)\\ \alpha,\beta &\mapsto \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0\\ 0 & b \end{pmatrix} \end{align*}$
Where $b = 2 + 8 + 16 + 32 +\dots$ in $\mathbb{Z}_2$ which is a solution to $b^2 -b +6 = 0$, the same equation which $\beta$ satisfies.
In my case the equation would be $\beta - \beta + 3 = 0$ which does not have a solution mod 2 in $\mathbb{Z}$, so i don't know how i would go about getting the embedding in my case. Does anyone have any hints or solutions to my problem, it would be much appreciated.