Find right ideals, left ideals and ideals of the ring $R=\begin{bmatrix}\Bbb Q&\Bbb Q\\ 0&0\end{bmatrix}$
I would have thought and think that the only ideals are: $\begin{bmatrix} \mathbb Q &0 \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix} 0 & \mathbb Q \\ 0 & 0 \end{bmatrix}$ where these ideals are bilateral. Am I right?
We ought to observe that
$\begin{pmatrix}a&b\\ 0&0\end{pmatrix}\begin{pmatrix}c&d\\ 0&0\end{pmatrix}=\begin{pmatrix}ac&ad\\ 0&0\end{pmatrix}$
$\begin{pmatrix}\Bbb Q&0\\ 0&0\end{pmatrix}$ left-acts on $R$ as the scalar multiplication of $R$ as $\Bbb Q$-vector space
$\begin{pmatrix}0&\Bbb Q\\ 0&0\end{pmatrix}\cdot \begin{pmatrix}\Bbb Q&\Bbb Q\\ 0&0\end{pmatrix}=0$
From (2) we know that every left-ideal of $R$ must be a $\Bbb Q$-vector subspace. In fact, from (1) we know that every $\Bbb Q$-vector subspace $V\subseteq R$ is a left ideal. They are clearly closed under difference, and $1$ says that multiplying on the right by a matrix is tantamout to a multiplication by a scalar.
So, we have that the left ideals are exactly the $\Bbb Q$-vector subspaces.
However, (1) also yields that $\begin{pmatrix}\Bbb Q&0\\ 0&0\end{pmatrix}$ is not a right ideal (set $a=c=d=1,\ b=0$).
From (3), we know that a subset $S\subseteq\begin{pmatrix}0&\Bbb Q\\ 0&0\end{pmatrix}$ is a right ideal if and only if it is an additive subgroup. On the other hand, suppose $M$ is a right ideal of $R$ and $\begin{pmatrix}a&b\\ 0&0\end{pmatrix}\in M$ with $a\ne 0$. Then, $\begin{pmatrix}ac&ad\\ 0&0\end{pmatrix}\in M$ for all $c,d\in\Bbb Q$. But this clearly means that $M=R$. So, right ideals of $R$ are either $R$ or additive subgroups of $\begin{pmatrix}0&\Bbb Q\\ 0&0\end{pmatrix}$.
So, wrapping things up:
Left, non-two-sided, ideals: $\Bbb Q$-vector subspaces of dimension $1$.
Right, non-two-sided, ideals: $\begin{pmatrix}0& G\\ 0&0\end{pmatrix}$, where $G$ is a proper, non-zero additive subgroup of $\Bbb Q$.
Two-sided ideals: $\{0\},\ R,\ \begin{pmatrix}0&\Bbb Q\\ 0&0\end{pmatrix}$.