How to prove that this structure only has one nontrivial submodule? Let us have two structures: $M= \left\{ \begin{bmatrix}
    a       & b \\
    0       & -a \\ 
\end{bmatrix}
\mid a,b \in \mathbb{R} \right\}$ and $R= \left\{ \begin{bmatrix}
    c       & d \\
    0       & c \\ 
\end{bmatrix}
\mid c,d \in \mathbb{R} \right\}$.
Prove, that $M$ is a right module over $R$(with regards to regular matrix multiplication) and that $M$ has only one nontrivial submodule.
I was able to solve the first part of the problem (testing for the attributes of a submodule and proving that they are indeed correct for this example).
However, I have no idea how to approach the second part of the task (I know that trivial submodules mean either $M$ or {0}).
Any help appreciated :)
 A: The elements of the form $\begin{bmatrix}a&0\\0&a\end{bmatrix}$ give $M$ the structure of a $2$-dimensional $\mathbb R$ vector space.
Therefore, there are only three possible dimensions for subspaces, and nontrivial submodules would have to have dimension $1$.
The matrices of the form $\begin{bmatrix}0&b\\0&0\end{bmatrix}$ obviously form one such submodule, and the idea would be to show that it's the only possibility.
Suppose that $\begin{bmatrix}a&b\\0&-a\end{bmatrix}$ is in a right submodule with $a\neq 0$. Then $\begin{bmatrix}a&b\\0&-a\end{bmatrix}\begin{bmatrix}0&a^{-1}c\\0&0\end{bmatrix}=\begin{bmatrix}0&c\\0&0\end{bmatrix}$ is in the submodule too, for any choice of $c$. Clearly $\begin{bmatrix}a&b\\0&-a\end{bmatrix}$ and $\begin{bmatrix}0&c\\0&0\end{bmatrix}$ are linearly independent, so the submodule cannot be nontrivial.
Therefore the only elements that can be in any nontrivial submodule are of the form $\begin{bmatrix}0&b\\0&0\end{bmatrix}$, and it's therefore the unique nontrivial right submodule.
