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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?

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    $\begingroup$ If $K$ is $\mathbb Q$ and $A$ is the quaternion division algebra over $\mathbb Q$, then $A$ is already basic and has trivial Jacobson radical. No algebra Morita equivalent to $A$ has quotient by is radical commutative because all such are semisimple and $A$ is not Morita equivalent to a commutative algebra.. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 10, 2015 at 17:19
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    $\begingroup$ If it exists, then it is unique since such an algebra must be basic and there is a unique basic algebra in each Morita class. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 10, 2015 at 17:20

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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.

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  • $\begingroup$ Thank you for that. By $J(B)$ I meant the Jacobson radical of $B$. I will edit it. $\endgroup$
    – Ehud Meir
    Commented Dec 10, 2015 at 17:08
  • $\begingroup$ what about the second question? is this basic algebra canonically defined? that is: can we have two nonisomorphic algebras $A$ and $B$ such that $A/J(A)\cong B/J(B)$ is commutative, and $A$ and $B$ are Morita equivalent? $\endgroup$
    – Ehud Meir
    Commented Dec 10, 2015 at 17:16
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    $\begingroup$ The basic algebra is the endomorphism algebra of the direct sum of one copy of each projective indecomposable module and hence is uniquely determined. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 10, 2015 at 17:22
  • $\begingroup$ @EhudMeir That the basic algebra is well-defined is proved in Nesbitt+Scott. $\endgroup$
    – Theo Johnson-Freyd
    Commented Dec 10, 2015 at 18:46

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