If $L$ is a Lie algebra over an algebraic closed field $K$ of characteristic zero, then all Cartan subalgebras are conjugated. Hence, they have all the same dimension. If $K$ is not algebraic closed but still of characteristic zero, then let $F$ the algebraic closure of $K$ and $H$ a Cartan subalgebra of $L$. Then $H\otimes F$ is a Cartan subalgebra of $L\otimes F$ with $dim_K(H)=dim_F(H\otimes F)$. Thus, the dimension of all Cartan subalgebras in characteristic zero is identical.
In the modular case this statement is wrong. There are several examples in modular Lie algebras such that the dimension of the Cartan subalgebras is not identical.
In the literature I found no examples of a modular Lie algebra based on an associative finite-dimensional unital algebra such that there exist Cartan subalgebras of different dimension.
My question is: Let $A$ be an associative finite-dimensional unital algebra over a field $K$ of characteristic $p$. Is the dimension of the Cartan subalgebras of the associated (restricted) Lie algebra based on $A$ identical?
Remark: I could prove that for $p\ne 2$ and $A$ being solvable this statement is true. In addition, for separable $A$ its true, too.
By a theorem of Premet (which was later on again proven by Farnsteiner) the following theroem holds for a restructed Lie algebra over an algebriach closed field of characteristic p: if the a torus has maximal dimension then the CSA based on this torus is of dimension of the rank of the Lie algebra. In particluar, if all tori are of the same dimension, then the CSAs are of the same dimension, too.
This tehroem can be used for this situation to prove the uniquness even for modular Lie algebras based on associative algebras.
see also the answer of the cross-posted question!