Intersection of ideals of $R$ and $eRe$ Let $e=e^2\in R$ such that $ReR=R$. If $\frak {B}$ and $\frak{C}$ are ring direct summands of the ring $eRe$ with $eRe=\frak{B}\oplus \frak{C}$, then necessarily $\frak{B}\cap \frak{C}=0$, because of the direct sum between them. Is it true that $R\mathfrak{B} R\cap R \mathfrak{C}R=0$? I know that any ring direct summand is generated by a central idempotent, and via Theorem 21.11(2) of "A First Course in Noncommutative Rings", 2nd edition (2001), by T. Y. Lam, we have $e(R\mathfrak {A}R)e=\mathfrak{A}$, for any ideal $\frak{A}$ of $eRe$, and also, the map $\mathfrak{A}\rightarrow R\mathfrak{A}R$ defines a 1-1 correspondence between the idals of $eRe$ and those of $R$.

So, you're saying that $A\mapsto RAR$ is a bijection, namely that $I\mapsto eIe$ and $A\mapsto RAR$ are mutually inverse.

Then $e(RBR\cap RCR)e=eRBRe\cap eRCRe=B\cap C=\\{0\\}\lhd eRe$ implies that applying the second map to both sides will yield $RBR\cap RCR=\\{0\\}\lhd R$.