One may decompose the right socle $S_r$ of any unital ring $R$ as $S_r=S_1\oplus S_2$, where $S_1$ is the sum of all nilpotent minimal right ideals, and $S_2$ is the sum of all idempotent minimal right ideals of $R$. I want to prove that $S_1$ and $S_2$ are two-sided ideals.
As for $S_1$ (resp. $S_2$) , let $x\in S_1$ (resp.$S_2$) and write $x=r_1+\cdots +r_n$, where $r_i\in I_i$, for some idempotent (resp. nilpotent) minimal right ideal $I_i$. If $s\in R$, it suffices to show that $sx\in S_1$(resp. $S_2$). Each $sr_i$ belongs to $sI_i$. But I do not know how to link these to expressions $xs$ and $r_is$. Incidentally, I do know that any sum of ifempotent right ideals is again an idempotent right ideal so that $S_2$ is such. Also, it is easily shown that $S_1^2=0$.
Thanks for any help!
This seems more or less obvious that they necessarily lie in different Wedderburn components of the socle.
That is, the sum of the collection of minimial right ideals in a fixed isoclass forms an ideal, and the direct sum of these ideals equals $S_r$. Obviously the ones that are idempotent and nilpotent lie in separate components. The sum of the components for nilpotent minimal ideals and the sum of the component for idempotent minimal ideals clearly decompose the socle into two smaller ideals.