Show a chain is a composition series for a matrix ring 
Consider the ring
  $$
R = 
 \begin{pmatrix}
  \mathbb{Q} & 0 \\
  \mathbb{R} & \mathbb{R}
 \end{pmatrix}
$$
  as a left $R$-module over itself. I want to show that the following is a composition series
  $$
 0=\begin{pmatrix}
  0 & 0 \\
  0 & 0
 \end{pmatrix}\subset
 \begin{pmatrix}
  0 & 0 \\
  \mathbb{R} & 0
 \end{pmatrix}\subset
 \begin{pmatrix}
  0 & 0 \\
  \mathbb{R} & \mathbb{R}
 \end{pmatrix}\subset
 \begin{pmatrix}
  \mathbb{Q} & 0 \\
  \mathbb{R} & \mathbb{R}
 \end{pmatrix}=R
$$
  for $R$. 

So I consider the quotients of the terms, for example $ \begin{pmatrix}
  \mathbb{Q} & 0 \\
  \mathbb{R} & \mathbb{R}
 \end{pmatrix} /  \begin{pmatrix}
  0 & 0 \\
  \mathbb{R} & \mathbb{R}
 \end{pmatrix}$. I see that this is isomorphic to $\mathbb{Q}$. Can I use this to show that the quotient is simple?
Similarly how can I do the same thing for the quotient  $\begin{pmatrix}
  0 & 0 \\
  \mathbb{R} & 0
 \end{pmatrix} /  \begin{pmatrix}
  0 & 0 \\
  0 & 0
 \end{pmatrix}$ ?


$ \begin{pmatrix}
  \mathbb{Q} & 0 \\
  \mathbb{R} & \mathbb{R}
 \end{pmatrix} /  \begin{pmatrix}
  0 & 0 \\
  \mathbb{R} & \mathbb{R}
 \end{pmatrix}$. I see that this is isomorphic to $\mathbb{Q}$. Can I use this to show that the quotient is simple?

That alone is probably not a transparent enough explanation of why it is simple. You can, if you like, start with the fact it is isomorphic to $\begin{pmatrix}
  \mathbb Q & 0 \\
  0 & 0
 \end{pmatrix}$ as a left $R$ module, and note that for any two nonzero elements $\begin{pmatrix}x & 0\\0&0\end{pmatrix}$ and $\begin{pmatrix}y & 0\\0&0\end{pmatrix}$, you can find $\begin{pmatrix}q & 0\\r&s\end{pmatrix}$ such that $\begin{pmatrix}q & 0\\r&s\end{pmatrix}\begin{pmatrix}x & 0\\0&0\end{pmatrix}=\begin{pmatrix}y & 0\\0&0\end{pmatrix}$. (When a ring acts transitively on the nonzero elements of a module, that means it is simple.)
Similar logic holds for $\begin{pmatrix}
  0 & 0 \\
  \mathbb{R} & \mathbb R
 \end{pmatrix} /  \begin{pmatrix}
  0 & 0 \\
  \mathbb R & 0
 \end{pmatrix}\cong\begin{pmatrix}
  0 & 0 \\
  0&\mathbb R
 \end{pmatrix}$ and $\begin{pmatrix}
  0 & 0 \\
  \mathbb{R} & 0
 \end{pmatrix} /  \begin{pmatrix}
  0 & 0 \\
  0 & 0
 \end{pmatrix}\cong \begin{pmatrix}
  0 & 0 \\
  \mathbb R & 0
 \end{pmatrix}$
