Definition::An $R$ module $M$ is said to be discrete if it satisfy following two properties
$(D_1)$ For every submodule $A$ of $M$, there is a decomposition $M=M_1\oplus M_2$ such that $M_1\leq A$ and $A\cap M_2$ is superfluous (small) in $M$, i.e. $A\cap M_2\leq_s M$.
$(D_2)$ If $A\leq M$ such that $M/A$ is isomorphic to a summand of $M$, then $A$ is a summand of $M$.
Question: simple submodule of a discrete module a direct summand of the module?
My attempt: Let $A$ be A simple submodule of a discrete module $M$. Then by $(D_1)$, $M=M_1\oplus M_2$ where $M_1\leq A$ and $A\cap M_2\leq_s M$. If $M_1$ is nonzero then $M_1=A$ and we are done. If possible let $M_1$ is zero then? I stuck in finding proof and also I was unable to find a counter example. Please help me.
I think if you check $R=F_2[x]/(x^2)$, where $F_2$ is the field of two elements, I think you'll find $R_R$ is a discrete module, and it has one simple submodule, but it has no nontrivial summands.