Derived equivalences of Artin algebras with finitistic dimension zero

Question

Let $A$ be an Artin algebra of finitistic dimension zero and $B$ an algebra derived equivalent to $A$. Does $B$ also have finitistic dimension zero?

In case this is true, this might generalise the result that selfinjective algebras are closed under derived equivalence. (Since selfinjective algebras have finitistic dimension zero and they are exactly the Gorenstein algebras with finitistic dimension zero (and being Gorenstein is preserved under derived equivalence)).

edit: It might be a better question to ask for the two-sided condition. That is: Let $A$ be an Artin algebra of left and right finitistic dimension zero and $B$ an algebra derived equivalent to $A$. Does $B$ also have left and right finitistic dimension zero?

I think in this form it might be true. Non-selfinjective examples with left and right finitistic dimension zero are quiver algebras where at each vertex there is at least one loop.

Answer 1

Let $A$ be the radical square zero algebra whose quiver has two vertices $1$ and $2$, with a loop at each vertex and an arrow from vertex $1$ to vertex $2$. Then the projective (right) modules have structure $$P_1=\matrix{&1&\\1&&2}\mbox{$\quad$ and $\quad$}P_2=\matrix{2\\2},$$ and the left and right finitistic dimensions are both zero.

There is a tilting complex that is the direct sum of complexes $T_a:=\dots\to0\to P_2\to P_1\to0\to\dots$ and $T_b:=\dots\to0\to0\to P_1\to0\to\dots$, with endomorphism algebra having projective right modules (according to my hand calculations) with structure $$Q_a=\matrix{&a&\\a&&b\\&&a\\&&b}\mbox{$\quad$ and $\quad$}Q_b=\matrix{b\\a\\b},$$ and since there is an injective map $Q_b\to Q_a$, the right finitistic dimension is greater than zero.