Are there Lie algebra ideals with proper radical?

Question

Let $L$ be a Lie algebra over $\mathrm{k}$

1. Ideal $P$ of $L$ is called prime if $[H, K] \subseteq P$ with $H, K$ ideals of $L$ implies $H \subseteq P$ or $K \subseteq P$

2. The radical of ideal M is definied by $\sqrt{M}=\cap\{P \subseteq L: \mathrm{P} \text { is prime ideal containing } M\}$

My Question: Are there Lie algebra ideals with proper radical?

Update:- For $L=gl_n \oplus S$ where $S$ is any simple Lie algebra.

The ideals are $0, S, Z, Z+S, s l_{n}, s l_{n}+S,$ $g l_{n}$ and $L,$ where $Z$ is the centre of $g l_{n}$.

$L, gl_n, Z+S$ are the prime ideals only, therefore $\sqrt{sl_n}=L \cap gl_n =gl_n$.

$0$ is not prime ideal because $[S,gl_n]=0 \subseteq 0 $ but neither $S \subseteq 0$ nor $gl_n \subseteq 0$

$S$ is not prime ideal because $[Z,gl_n]=0 \subseteq S $ but neither $Z \subseteq S$ nor $gl_n \subseteq S$

$Z$ is not prime ideal because $[S,gl_n]=0 \subseteq Z $ but neither $S \subseteq Z$ nor $gl_n \subseteq Z$

$sl_n$ is not prime ideal because $[S,gl_n]=0 \subseteq sl_n $ but neither $S \subseteq sl_n$ nor $gl_n \subseteq sl_n$

$sl_n + S $ is not prime ideal because $[S+Z,gl_n +S]=S \subseteq sl_n+S $ but neither $S+Z \subseteq sl_n + S$ nor $gl_n +S \subseteq sl_n + S$

Is all this true?? I appreciate your help.

Answer 1

Take a simple Lie algebra $S$ and set $L:=\mathfrak{gl}_n \oplus S$, $M:=\mathfrak{sl}_n\oplus 0$. Then $M \subsetneq \sqrt{M} \subsetneq L$.