Finding a simple module over $R = M_2(\mathbb{Q})$.
So, I don't think that we can use any ideals of $M_2(\mathbb{Q})$, because I believe $M_2(\mathbb{Q})$ is already simple. It is a common result that ideals of $M_2(\mathbb{Q})$ are of the form $M_2(I)$ where $I \triangleleft \mathbb{Q}$, and since $M_2(\mathbb{Q})$ is a field it has no non-trivial ideals.
Anyway, insight is appreciated, thanks!!
$M := \mathbb{Q}^{2\times1}$ is a simple $R$-module via the usual matrix-vector multiplication. This is because $R$ acts transitively on $M$ (for any two vectors $v, w \in M$ you can always find a linear map $A \in \operatorname{End}_{\mathbb{Q}}(M)$ satisfying $Av = w$), so whenever $0 \ne x \in N \le M$ then $M = Rx \le N$, which means $M=N$.
Furthermore, $M$ is actually the only simple $R$-module up to isomorphism, since $$ R = M \oplus M $$ so $R$ is a semisimple ring, and the direct summands represent all isomorphism classes of simple $R$-modules.