About right ideals of a $\mathbb{Q}$-subalgebra of $M_2(\mathbb{R})$, $A = \begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$ I am looking to find the  right ideals of the $\mathbb{Q}$-subalgebra of $M_2(\mathbb{R})$, A = $\begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$, how asked by R.S.Pierce here. The question is to prove that every right ideal of A has one of the following forms:
$$0,\,A,\,\begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & 0 \end{bmatrix},\,\begin{bmatrix} 0 & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}, \text{or} \,\,\,\,\rho_{x,y}=\begin{bmatrix} 0 & x\\ 0 & y \end{bmatrix}\cdot\mathbb{R}$$for some fixed $(x,y)\in \mathbb{R}^2-\{(0,0)\}$.
To show it I considered a nonzero element $\begin{bmatrix} q & r\\ 0 & s \end{bmatrix}$ of an arbitrary right ideal $\rho$ of A. I  then considered many cases. 
Case 1: $q\neq0, r\neq0\, \text{and}\, s\neq0$
In this case it is trivial to show that $\rho=A$.
Case 2: $q\neq0, s\neq0\,  \text{and}\, r=0$
We have $\begin{bmatrix} q & 0\\ 0 & s \end{bmatrix}\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} q & q\\ 0 & s \end{bmatrix}\in \rho$. Hence for the Case 1 we have $\rho=A$. 
Case 3: $q\neq0, r\neq0\,  \text{and}\, s=0$
Because every ideal of A is a $\mathbb{Q}$-vectorial subspace of A then we can suppose that  $\begin{bmatrix} 1 & r'\\ 0 & 0 \end{bmatrix}\in \rho$, for some $r'\neq 0$. But now for every $\begin{bmatrix} \overline{q} & \overline{r}\\ 0 & 0 \end{bmatrix}\in \begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & 0 \end{bmatrix}$
we have $\begin{bmatrix} \overline{q} & \overline{r}\\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 & r'\\ 0 & 0 \end{bmatrix}\cdot \begin{bmatrix} \overline{q} & \overline{r}\\ 0 & 0 \end{bmatrix}\in \rho$. Then $\rho$ must contain $\begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & 0 \end{bmatrix}$. It is not difficult to show that $\begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & 0 \end{bmatrix}$ is a maximal right ideal of A, so we have $\rho=\begin{bmatrix} \mathbb{Q} & \mathbb{R}\\ 0 & 0 \end{bmatrix}$ or $\rho=A$.
Case 4 $s\neq0, r\neq0\,  \text{and}\, q=0$
Because $\begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}\in \rho$ then for every $t \in \mathbb{R}$ we have $$\begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}\cdot \begin{bmatrix} 0 & 0\\ 0 & t \end{bmatrix}= \begin{bmatrix} 0 & tr\\ 0 & ts \end{bmatrix}\in \rho,$$
and so $\rho$ must contain $\rho_{r,s}$.
Question
It is for the Case 4 that I am stuck, because I have no idea how to prove that, in this case, we have $\rho=\rho_{r,s}$ or $\rho=\begin{bmatrix} 0 & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$. 
Any suggestion?
 A: To your line of arguments in case $4$: 
Let $x = \begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}$ and $x^{\prime}=\begin{bmatrix} 0 & r^{\prime}\\ 0 & s^{\prime} \end{bmatrix}$ for $(r,s) \neq (0,0)$ and $(r^{\prime},s^{\prime}) \neq (0,0)$. $x,x^{\prime}\in \rho$.
Since 
$\begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}\cdot \begin{bmatrix} a & b\\ 0 & c \end{bmatrix}= \begin{bmatrix} 0 & cr\\ 0 & cs \end{bmatrix} = \begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}\cdot c$
we have that $ \langle\begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}\rangle$ is an ideal for a fix $(r,s)\neq (0,0)$ and $ \langle\begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}\rangle \subseteq \rho$.
If $x^{\prime} \notin \langle\begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}\rangle$, i.e. $\begin{bmatrix}r\\s\end{bmatrix} \neq c\cdot \begin{bmatrix}r^{\prime}\\s^{\prime}\end{bmatrix}$ for any $c\in \mathbb{R}$, then $x,x^{\prime}\in \langle\begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix},\begin{bmatrix}0 & r^{\prime}\\0 & s^{\prime}\end{bmatrix}\rangle = \begin{bmatrix}0&\mathbb{R}\\0& \mathbb{R}\end{bmatrix}.$ Hence $\rho = \rho_{r,s}$ or $\rho =  \begin{bmatrix}0&\mathbb{R}\\0&\mathbb{R}\end{bmatrix}.$
Note that $\rho_{r,s} = \rho_{cr,cs}$, $c\in \mathbb{R}$.
A: Suppose that $\rho$ properly contains $\rho_{r,s}$. Hence there exist $(x,y)\in \mathbb{R}^2-{(0,0)}$ such that $\begin{bmatrix} 0 & x\\ 0 & y \end{bmatrix}\in \rho$ but $\begin{bmatrix} 0 & x\\ 0 & y \end{bmatrix}\not \in \rho_{r,s}$. So  $\begin{bmatrix} 0 & x\\ 0 & y \end{bmatrix}\neq \begin{bmatrix} 0 & \alpha r\\ 0 & \alpha s \end{bmatrix}$ for all $\alpha \in \mathbb{R}$. Because we can always write $x=(xr^{-1})\cdot r$, we can suppose that $y\neq (xr^{-1})\cdot s$, so $ry-xs\neq0$. 
Now for every $\begin{bmatrix} 0 & u\\ 0 & v \end{bmatrix}\in \begin{bmatrix} 0 & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$, we have that the system
$$\begin{bmatrix} 0 & u\\ 0 & v \end{bmatrix}=\begin{bmatrix} 0 & \alpha r\\ 0 & \alpha s \end{bmatrix}+\begin{bmatrix} 0 & \beta x\\ 0 & \beta y \end{bmatrix} $$ has a solution because $det\begin{bmatrix} r & x\\ s & y \end{bmatrix}\neq 0$. Hence $\rho$ contains $\begin{bmatrix} 0 & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$, and so $\rho=\begin{bmatrix} 0 & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$ or $\rho=A$ because $\begin{bmatrix} 0 & \mathbb{R}\\ 0 & \mathbb{R} \end{bmatrix}$ is a maximal right ideal of A.
A: Hint: Show that the set of pairs $(r,s)$ such that $\begin{bmatrix} 0 & r\\ 0 & s \end{bmatrix}\in \rho$ is a vector subspace of $\mathbb{R}^2$, and then think about its dimension.
