Let $\mathbb{k}$ be a field, $\mathfrak{g}$ be a finite-dimensional Lie algebra over $\mathbb{k}$.
In Bourbaki's "Lie Groups and Lie Algebras", Ch I, he defines four radical-like ideals of $\mathfrak{g}$:


1. the radical $\mathfrak{r}$, i.e. the maximal solvable ideal;
2. the radical of Killing form $\mathfrak{k}$, i.e. $\mathfrak{k}=\mathfrak{g}^\perp$;
3. the maximal nilpotent ideal $\mathfrak{n}$;
4. the nilpotent radical $\mathfrak{s}$, i.e. intersection of all kernels of irreducible finite-dimensional representations of $\mathfrak{g}$.


He also shows that, $\mathfrak{r}=[\mathfrak{g},\mathfrak{g}]^\perp$,
$\mathfrak{s}=[\mathfrak{g}, \mathfrak{g}]\cap\mathfrak{r}=[\mathfrak{g},\mathfrak{r}]$, and the following inclusion relations:
$$\mathfrak{r} \supset \mathfrak{k} \supset \mathfrak{n} \supset \mathfrak{s}.$$


On the other hand, in Jacobson's book, the nilpotent radical is defined as $\mathfrak{n}$.


My question is, does $\mathfrak{s}$ coincide with $\mathfrak{n}$? or, are Bourbaki's nilradical and Jacobson's nilradical equivalent?


I guess the answer is NO (otherwise Bourbaki should have proved it), but I cannot find any example. Could any one give me an example? Thank you!



On radicals of a lie algebra
If we let $\mathbb{k}$ be the field itself thought of as a Lie algebra with trivial bracket, then $\mathfrak{n}=\mathbb{k}$ since this is a nilpotent Lie algebra but $\mathbb{s}=\{0\}$, since the obvious representation of $\mathbb{k}$ be scalar multiplication is faithful. In general, these will never coincide for a nilpotent Lie algebra.


