I am looking to find the left, right and two sided ideals of the ring R = $\begin{bmatrix} \mathbb{Q} & \mathbb{Q}\\ 0 & 0 \end{bmatrix}$. It is in finding the left ideals that I am stuck.
Finding the Right Ideals
To find the right ideals I considered a nonzero element $\begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix}$ of an arbitrary right ideal $I_{r}$. I then considered two cases.
Case 1: $a\neq0$
We have $\begin{bmatrix} a & b\\ 0 & 0 \end{bmatrix}\begin{bmatrix} a^{-1} & 0\\ 0 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}\in I_{r}$. Hence for any $\begin{bmatrix} c & d\\ 0 & 0 \end{bmatrix} \in R$ we have:
$\begin{bmatrix} c & d\\ 0 & 0 \end{bmatrix}=\begin{bmatrix} c & 0\\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}+\begin{bmatrix} 1 & 0\\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & d\\ 0 & 0 \end{bmatrix} \in R$. That is $I_{r}=R$.
Case 2: $a=0$
We have $K=\{b\in\mathbb{Q}\mid\begin{bmatrix} 0 & b\\ 0 & 0 \end{bmatrix}\in I_{r}\}$ is an additive subgroup of $\mathbb{Q}$ and $I_{r}=\begin{bmatrix} 0 & K\\ 0 & 0 \end{bmatrix}$. Moreover one can verify that $\begin{bmatrix} 0 & K\\ 0 & 0 \end{bmatrix}$ is a right ideal of $R$. Hence in this case any ideal is of the form $\begin{bmatrix} 0 & K\\ 0 & 0 \end{bmatrix}$ where $K$ is an additive subgroup of $\mathbb{Q}$.
Finding the Left Ideals
I have tried a similar strategy as above. I considered a nonzero element $\begin{bmatrix} a' & b'\\ 0 & 0 \end{bmatrix}$ of an arbitrary left ideal $I_{l}$. I then attempted to consider two cases.
Case 1: $b'=0$
We have $K'=\{a'\in\mathbb{Q}\mid\begin{bmatrix} a' & 0\\ 0 & 0 \end{bmatrix}\in I_{r}\}$ is an additive subgroup of $\mathbb{Q}$ and $I_{l}=\begin{bmatrix} K' & 0\\ 0 & 0 \end{bmatrix}$. Moreover one can verify that $\begin{bmatrix} K' & 0\\ 0 & 0 \end{bmatrix}$ is a leftideal of $R$. Hence in this case any left ideal is of the form $\begin{bmatrix} K' & 0\\ 0 & 0 \end{bmatrix}$ where $K'$ is an additive subgroup of $\mathbb{Q}$.
It is for the second case that I am stuck, I have no idea how to proceed. Moreover when finding the ideals, I will simply check whether each left/right ideal I find is a right/left ideal. Overall is this the best way to have approached this question, it feels awfully long.
When you multiply by $\begin{bmatrix} a & b\\0 & 0 \end{bmatrix}$ on the right, you get just a scalar multiple by $a$. So the left ideal generated by $\begin{bmatrix} a^{\prime} &b^{\prime}\\0&0 \end{bmatrix}$ is just the collection of all scalar multiples. This is independent of whether $b^{\prime}$ is $0$ or not.
For example: The left ideal generated by $\begin{bmatrix} 1 & 3\\0&0\end{bmatrix}$ is the set $\left\lbrace\begin{bmatrix} a & 3a\\0&0\end{bmatrix} \ : \ a \in \mathbb{Q} \right\rbrace$
Another way to think of the left ideals, your ring is essentially the vector space $\mathbb{Q}^{2}$ (multiplying an element on the left just amounts to scalar multiplication). So any subspace will be a left ideal.