Let $K$ be an algebraically closed field of characteristic zero, and let $\mathfrak{g}$ be a finite dimensional, nilpotent Lie algebra over $K$.
My question is, can we find an element $x\in\mathfrak{g}$ such that the image of $ad(x)$ is one dimensional over $K$? In other words can we find $x\in\mathfrak{g}$ such that $x$ is not central and there exists $z\in\mathfrak{g}$ such that for every $y\in\mathfrak{g}$, $[x,y]=\alpha z$ for some $\alpha\in K$.
This is not true if $K$ is not algebraically closed, but I haven't been able to find an example where we cannot find such an element $x$ after passing to a finite extension. Does anyone know whether this is true, or have a counterexample?
$Vect(e_1,e_2,e_3,e_4,e_5,e_6)$. $[e_1,e_2]=e_4, [e_1,e_3]=e_5, [e_2,e_3]=e_6$ the other brackets are zero. The center is $Vect(e_4,e_5,e_6)$, $dim(Im(ad_{e_i})=3, i=1,2,3$, $dim(Imad_{e_i}))=0, i=4,5,6$....