Let L be a Lie algebra of dimension 3.
suppose $ad_z = 0$ with $z\ne 0$. Show that there are $x$ and $y$ in $L$ such that $[x,y] = z$ and realize L in $\mathfrak{b}_3$ the Borel algebra of upper triangular matrices.
What I did:
I proved the existence of x and y. I am stuck in realizing L in $\mathfrak{b}_3$: i.e. finding a morphism $f:L\to \mathfrak{b}_3$ by giving to $x,y$ and $z$ and image in $\mathfrak{b}_3$.
$ad_z = 0$ means z is in the center of L so clearly $f(z)=I_3$. (This is wrong as pointed in comments)
But what about $f(x)$ and $f(y)$. I tried to look for solutions in the form $f(x)=I_3+E_{i,j}$ and $f(y)=I_3+E_{k,l}$ but I can't find a a solution such that $[f(x),f(y)] = f(z)=I_3$
Please note that [,] is the Lie Bracket.
Thanks for any help
The classification of $3$-dimensional Lie algebras with $1$-dimensional derived algebra has been explained at this duplicate.
For completing the classification in general in dimension $3$ is a bit more work necessary, even over the complex numbers. For the classification of $3$-dimensional Lie algebras over an arbitrary field $K$ see for example this thesis. It has $58$ pages and the classification is quite complicated, in particular for characteristic $2$. Over the complex numbers the classification is much easier. This can be found, for example, in Jacobson's book on Lie algebras. Bianchi classified $3$-dimensional real and complex Lie algebras $115$ years ago. There are still new articles on this classification, e.g., to put it in representation theoretic context, see for example here.