# Commutator of finite global dimension algebras

Let $$A=KQ/I$$ be a finite dimensional quiver algebra of finite global dimension. Is it true that the dimension of $$A/[A,A]$$ is equal to the number of simples of $$A$$? Here $$[A,A]$$ is the vector space generated by all elements of the form $$ab-ba$$.

Note that it is known that in general for such $$A$$ that the dimension of $$A/([A,A]+rad(A))$$ is equal to the number of simple $$A$$-modules. Thus the question should be equivalent to asking whether we have $$rad(A) \subseteq [A,A]$$ in case $$A$$ has finite global dimension.

• I am maybe missing something, but what if your quiver has one vertex, and your algebra $A$ is the algebra of polynomials? What does your suggested result want to say then? – Vladimir Dotsenko Mar 19 '20 at 8:31
• @VladimirDotsenko A quiver algebra is finite dimensional for me. I add that assumption, thanks. – Mare Mar 19 '20 at 8:31