Primeness in semi-simple Lie algebra Definitions : 
A) An ideal $P$ of $L$ is called prime if $[H, K] \subseteq P$ with $H, K$ ideals of $L$ implies $H \subseteq P$ or $K \subseteq P$
B) Let $H$ be an ideal of $L$. The radical of $H$ is $r(H)$=The intersection of all the
prime ideals of $L$ containing $H$.  

Example : Let $S_{1}, S_{2}$ and $S_{3}$ be finite-dimensional simple Lie algebras.
  Let $L=$ $S_{1} \oplus S_{2} \oplus S_{3} .$ 
Then $S_{1},S_{2}$ and $S_{3}$ are not prime, since
  $$[S_1 \oplus S_2, S_1 \oplus S_3] \stackrel{\text{bilinearity}}= [S_1, S_1] \oplus \underbrace{[S_1, S_3]}_{0} \oplus \underbrace{[S_2, S_1]}_{0} \oplus \underbrace{[S_2, S_3]}_{0} = S_1 \subseteq S_{1}$$ but neither $(S_1 \oplus S_2) \nsubseteq  S_1$ nor $(S_1 \oplus S_3) \nsubseteq  S_1$.
  Similarly for $S_{2}$ and $S_{3}$.

Now
$$[S_1 \oplus S_2, S_1 \oplus S_3] \stackrel{\text{bilinearity}}= [S_1, S_1] \oplus \underbrace{[S_1, S_3]}_{0} \oplus \underbrace{[S_2, S_1]}_{0} \oplus \underbrace{[S_2, S_3]}_{0} = S_1 \subseteq S_1 \oplus S_2 $$
$$\implies S_1 \oplus S_2 \subseteq S_1 \oplus S_2 $$ 
$$[S_1 \oplus S_2, S_1 \oplus S_2] \stackrel{\text{bilinearity}}= [S_1, S_1] \oplus \underbrace{[S_1, S_2]}_{0} \oplus \underbrace{[S_2, S_1]}_{0} \oplus [S_2, S_2] = S_1 \oplus S_2 \subseteq S_1 \oplus S_2$$ $$ \implies S_1 \oplus S_2 \subseteq S_1 \oplus S_2$$ 
Therefore $S_1 \oplus S_2$ is prime ideal. Similarly for $ S_{1} \oplus S_{3}$ and $S_{2} \oplus S_{3}$.
My Questions:-

1.Does $S_{1} \oplus S_{2}, S_{1} \oplus S_{3}$ and $S_{2} \oplus S_{3}$ are the prime ideals in $L$?
2.What is the radical of $S_1$ for example?

The source of this question: On prime ideals in Lie algebra
I would really appreciate your help .

Q1. $S1+S2, S1+S3, S2+S3$ and $S1+S2+S3$ are all the prime ideals in $L$
Q2. $$r(S_1)= (S_1 \oplus S_2) \cap (S_1 \oplus S_3)  \cap (S_1 \oplus S_2 \oplus S_3)=S_1$$
