Let $R$ be a finite commutative ring with unit $1$. Assume that $n > 2$.
Is it true that the Lie algebra $\mathfrak{sl}_n(R)$ is not nilpotent or solvable?
It is well known that if char$F = 2$, then $\mathfrak{sl}_2(F)$ is nilpotent. But I am interested in the case $n > 2$.
The answer is positive at least for finite fields. For a finite field $F$ of characteristic $p>0$, the Lie algebra $\mathfrak{sl}_n(F)$ is simple for $p\nmid n$. For $p\mid n$, its center $Z$ is $1$-dimensional and consists of the scalar multiples of the identity. However, then $\mathfrak{psl}_n(F)=\mathfrak{sl}_n(F)/F$ is simple for $n\ge 3$, so that $\mathfrak{sl}_n(F)$ cannot be solvable for $p\mid n$ and $n\ge 3$.
Perhaps the argument can be extended to $\mathfrak{sl}_n(R)$ over a commutative ring $R$ with unit.