Let us say that S is a Lie algebra of dimension $n$, which is also solvable. Is it true that S contains an ideal of each dimension $d$ for $0 \leq d \leq n$? If so, how?
Thanks for all the help.
A Lie algebra $L$ is called completely solvable, or supersolvable if it has a chain of ideals $$ 0=L_0\subset L_1\subset \cdots L_n=L $$ with $\dim L_i=i$. Over the complex numbers every solvable Lie algebra is completely solvable, hence has ideals of every dimension $0,1,\ldots ,n$. This follows from Lie's theorem. (This is no longer true in prime characteristic, or for non-algebraically closed fields of characteristic zero).
So the answer is yes for complex solvable Lie algebras, but no for real solvable Lie algebras. For example, the following $4$-dimensional solvable real Lie algebra $L$ has no $1$-dimensional ideal: $L=\langle e_1,e_2,e_3,e_4\rangle$ and $$ [e_1,e_3]=e_1,\, [e_1,e_4]=-e_2,\, [e_2,e_3]=e_2,\, [e_2,e_4]=e_1. $$ Over the complex numbers, there is a $1$-dimensional ideal, e.g., $I=\langle e_1+ie_2\rangle$, with $i^2=-1$, but not over the real numbers.