Let $A$ be a finite dimensional (unital) $K$-Algebra. By $A^{\circ}$ we denote the associated $K$-Lie-algebra of $A$ with respect to the product $a\circ b:=ab-ba$. In addition, we denote by $rad(A^{\circ})$ the largest nilpotent ideal of $A^{\circ}$ and by $J(A)$ the largest nilpotent ideal of $A$ (the Jacobson-radical).


If $A/J(A)$ is separable and commutative there exist a complement $H$ of $J(A)$ in $A$ by the Wedderburn-Malcev-Theorem. I could proove that in this case $J(A)+Z(A)=rad(A^{\circ})$ holds ($Z(A)$ is the center of $A$.).


What is the nilradical (in terms of structure properties of the associative algebra) of $A^{\circ}$ for an arbitrary finite dimensional (unital) associative algebras $A$? 
What is the solution for certain associative classes like division algebras, simple, semisimple or basic algebras? 



Nilradical of a Lie algebra associated to a associative algebra
If $A/rad(A)$ is separable and $char(K)\ne 2$ then the nilradical of $A^{\circ}$ is exactly $rad(A)+Z(A)$.The steps for the proff are as follows (which I will refine later):
1.) Proof the theorem in the solvable case.
2.) Use Hersteins results to extend them to simple $K$-Algebras.
3.) Use 2.) and extend them to semisimple algebras by using diagonals.
4.) The general case will be reduced to 1.) and 3.).


