Let $A$ be a finite dimensional algebra over an algebraically closed field $K$.
If $A$ is semisimple, then $A$ is Morita equivalence with a commutative algebra, that is $A \backsim K^n$ where $n$ is the number of isomorphism classes of simple representations of $A$.
For a general, non-semisimple algebra, we cannot expect anymore that $A$ will be Morita equivalent with a commutative algebra: indeed, if $A$ is Morita equivalent with a commutative algebra $B$ then $Z(A)\cong Z(B) = B$, and there are examples of finite dimensional algebras which are not Morita equivalent with their centers. My question is the following:


For a finite dimensional $K$-algebra $A$, can we always find a Morita equivalent finite dimensional $K$-algebra $B$ such that $B/J(B)$ is commutative? ($J(B)$ stands here for the Jacobson radical of $B$). Moreover, in case such an algebra exists, will it necessarily be unique?



A canonical representative in Morita equivalence class
I don't know what $J(B)$ means, but perhaps the result you are looking for is one from Morita's original paper (Kiiti Morita. Duality for modules and its applications to the theory of rings with minimum condition. Sci. Rep. Tokyo Kyoiku Daigaku Sect. A, 6:83–142, 1958. Google has a link to a PDF, but I'm having trouble reproducing it). In it, he shows that two finite dimensional algebras are Morita equivalent (he calls it "similar") iff their *basic algebras* are isomorphic, where "basic algebra" is in the sense of Nesbitt and Scott (C. Nesbitt and W. M. Scott. Some Remarks on Algebras Over an Algebraically Closed Field. Annals of Mathematics
Second Series, Vol. 44, No. 3 (Jul., 1943), pp. 534-553. <http://www.jstor.org/stable/1968979>). Nesbitt and Scott had already shown that every algebra is Morita equivalent to its basic algebra.


