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 (resricted) 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. 



Unique dimension of Cartan subalgebras in modular Lie algebras
By a theorem of Premet ([MR](http://www.ams.org/mathscinet-getitem?mr=864177)) (which was later on again proven by Farnsteiner ([article](http://www.ams.org/journals/tran/2004-356-10/S0002-9947-04-03476-2) [MR](http://www.ams.org/mathscinet-getitem?mr=2058843))) the following theorem holds for a restricted Lie algebra over an algebraically closed field of characteristic p: if a torus has maximal dimension, then the CSA based on this torus has dimension the rank of the Lie algebra.


In particular, if all tori are of the same dimension, then the CSAs are of the same dimension, too.


This theorem can be used for this situation to prove uniquenessalso for modular Lie algebras based on associative algebras:


Let $A$ be an associative finite-dimensional $K$-algebra over a field $K$ of positive characteristics and $L$ be an algebraic closed superfield of $K$. If $H$ is a CSA of $A\_{Lie}$, then $H\otimes L$ is a CSA of $(A\otimes L)\_{Lie}$. In addition, $dim\_K(H)=dim\_L(H\otimes L)$ is valid. The $L$-algebra $A\otimes L$ is associative and finite-dimensional and possesses a separable radical factor algebra (because $L$ is algebraically closed and the factor algebra is isomorphic to diect product of full matrix algebras over $L$). In view of the theorem of Premet/Farnsteiner we have to prove that the dimension of the maximal tori of such algebras is unique. For this we start with the observation that the maximal tori of $A$ are mapped to maximal tori of $A/J(A)$ ($J(A)$ the Jacobson radical of $A$) by $T -> (T+J(A))/J(A)$. The dimension of the maximal tori does not change because the intersection of $T$ with $J(A)$ is zero. Thus, we have to calculate the dimension of maximal tori for separable associative algebras. These are direct products of separable simple associative algebras. Maximal tori of direct products are exactly the direct products of maximal tori of the components. We have reduced the question to simple separable associative algebras. Here we can prove that maximal tori and CSA are identical and the dimension is the product of $dim\_K(Z(A))$ with the index of the central-simple algebra $A$ over its center.


