I am trying to determine all the right submodules of the R-module
$\begin{pmatrix}0&\mathbb{Q}\\ 0&\mathbb{Q}\end{pmatrix}$
where R is the ring
$\begin{pmatrix}0&0\\ 0&\mathbb{Q}\end{pmatrix}$
Initially, I thought that the answer might be just the direct sums in the form $I_1 \oplus I_2$ where $I_1$ is an ideal in the ring
$\begin{pmatrix}0&\mathbb{Q}\\ 0&0\end{pmatrix}$
and $I_2$ is an ideal in the ring
$\begin{pmatrix}0&0\\ 0&\mathbb{Q}\end{pmatrix}$
However, now I think it’s not the case.
Note $\mathbb Q\cong \begin{bmatrix}0&0 \\ 0&\mathbb Q\end{bmatrix}$ as rings. The matrix notation suggests the intended action should be via matrix multiplication. Then
$\begin{bmatrix}0 & a \\ 0 & b\end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & \lambda\end{bmatrix}=\begin{bmatrix}0&\lambda a \\ 0&\lambda b\end{bmatrix}=\begin{bmatrix}0& a \\ 0& b\end{bmatrix}\lambda$
shows that right multiplication by elements of the ring is nothing more than matrix "scalar multiplication," and therefore we have that your module is just a $2$-dimensional $\mathbb Q$ vector space which we are all familiar with.
Other than the two trivial subspaces, it has all the $1$-dimensional subspaces.
I did also think of triangular matrix rings as Dietrich Burde linked; however, I'm not certain if all arguments still work when the $M$ in the $\begin{bmatrix}R&M\\0&S\end{bmatrix}$ is not a unital left $R$ module (in this case with $R=\{0\}$ and $M=\mathbb Q$, $M$ would not be a unital left $R$ module.) Perhaps they do, because the answer is still ultimately the same.
