Decomposition of socle of a ring My question is divided by two:
"(1) Is the right socle $S_r$ of an arbitrary unital ring $R$ is decomposed as $S_r=S_1\bigoplus S_2$, where $S_1$ is the sum of all the nilpotent minimal right ideals, and $S_2$ is the sum of all the idempotent minimal right ideals?
(2) Is it true that $S_1^2=0$?"
Thanks for any help!
 A: Partial answer is as follows:
Claim:

Given two nilpotent minimal (nonzero) right ideals $S, T$, we have $ST=0$.

Proof of this goes as follows: $ST$ is a right ideal contained in $SR=S$, which is a minimal right ideal, so either $ST=0$ or $ST=S$. But in the latter case, it follows that $0=S\cdot 0=S(TT)=(ST)T=ST$, contradiction. Thus, we necessarily have $ST=0$.
This shows that $S_1^2=0,$ and that any sub-ideal of $S_1$ is nilpotent (more precisely, that any $S \leq S_1$ satisfies $S^2=0$).
Now, $S_2$ together with $S_1$ clearly generate $S_r$, so one has to only show that $S_1 \cap S_2=0$. So assume $X=S_1 \cap S_2\neq 0$. Then $X$ is a submodule of the semisimple module $S_2,$ hence it contains a simple submodule $0 \neq T$. Then $T \leq S_1$, hence $T^2=0$. If one shows that $(*)$ any minimal right ideal in $S_2$ still has to be idempotent, then we are done, because $T$ would have to be both nilpotent and idempotent. 
However, I am not sure about the claim $(*)$: neither about how to prove it, nor whether it actually holds. 
