Is it true that if $D$ is a division ring and $n\in\mathbb{Z}_{\geq1}$, then the only left and right ideals of the ring $M_n(D)$ are the trivial ones?
I know that $M_n(D)$ is simple, and the proof is basically that if $0\neq I\triangleleft M_n(D)$, then there exists $A\in I$ such that an entry of $A$, say $a_{i,j}$, is invertible, so that $$ I_n=\sum_{k=1}^n e_{k,i}[a_{i,j}^{-1}I_n(e_{i,i}Ae_{j,j})]e_{j,k} \in I.$$ But, I don't see how one could multiply $A$ only by the left or only by the right to get $I_n$.
Consider $D=\Bbb R$, $n=2$ and the set of matrices $$\begin{pmatrix}0&a\\0&a'\end{pmatrix}$$
This is a left ideal, and it is proper. You can probably generalize this yourself.