Assume that $R$ is a ring, not necessarily commutative nor unital. Let $e$ be an idempotent in $R$. Is there a correspondence between ideals of $R$ and ideals of the corner ring $eRe = \{ere : r\in R\}$? What if $e$ is full, i.e. $ReR=R$?
I'm hoping there is some kind of natural order-preserving bijection, at least one that would imply that if $R$ is simple then $eRe$ is simple.
I tried to use a matrix example, e.g. $R=M_2(\mathbf{C})$ and $eRe = \begin{bmatrix} 0 & \mathbf{C} \\ 0 & 0 \end{bmatrix}$. But these are both simple of course, and I don't know how to get (say) a $2$-dimensional corner in $M_n(\mathbf{C})$ which is not simple. The upper triangle $\begin{bmatrix} 0 & \mathbf{C} & \mathbf{C} \\ 0 & 0 & \mathbf{C} \\ 0 & 0 & 0 \end{bmatrix}$ is a nonsimple subalgebra, but it's not a corner since its Jacobson radical is not zero (which would be true for a matrix corner).
There are maps $\alpha:I\mapsto eIe$ from the set of ideals of $R$ to the set of ideals of $eRe$, and $\beta:J\mapsto RJR$ from the set of ideals of $eRe$ to the set of ideals of $R$. They aren't bijections, but $\alpha\circ\beta$ is the identity, so $\beta$ is injective, which is enough to deduce that $R$ simple implies $eRe$ simple.