# Sum of idempotent and nilpotent minimal right ideals

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!

• By the way, can you let us know when the publications these questions generate appear? Surely they're getting close, with all the material in response to your questions... Mar 29, 2018 at 14:36

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.
• The class of $R$ modules is partitioned by the equivalence relation "isomorphic." An isoclass is one equivalence class. Lots of proofs of the Artin-Wedderburn theorem use the lemma that the sum of minimal ideals isomorphic to a fixed minimal ideal is an ideal in the ring. The proof that the component ideals intersect trivially goes much like the one you just did for $S_r$ in your last post. Try to prove it yourself, maybe. Mar 29, 2018 at 17:30