Example 1 :
If $L= \bigoplus_{i=1}^\infty S_i$, where $S_i$ is simple Lie algebra, then $L$ is non-Noetherian Lie algebra.
Since every ideal $A$ of $L$ has the property $A^2=[A, A]=A$, then every ideal $S_i$ is semi-prime.
Since $L$ is the only prime ideal of itself, then $r(S_i)=L$.
Is this true??
Notions:-
A Lie algebra $L$ satisfy the maximal condition for ideals, if for each , ascending chain $H_{1} \subseteq H_{2} \subseteq \ldots $ an index $m$ exists such that $H_{i}=H_{k}$ if $m<i$, $m<k.$ We say in short: $ L\in{\rm Max}-\triangleleft$ (Noetherian Lie algebra).
An ideal $Q$ of $L$ is called semi-prime if $H^2 \subseteq Q$ with $H$ an ideal of $L$ implies $H\subseteq Q$.
$r(H)$ denote the intersection of all the prime ideals of $L$ containing $H$.