# If an algebra is derived equivalent to a (special) biserial algebra is it biserial?

A (finite dimensional) algebra is called biserial, if the radical of each projective indecomposable left/right module is the sum of two uniserial modules whose intersection is either trivial or simple.

It is known that a certain subclass of algebras (called gentle algebras) is closed under derived equivalence.

What about this larger class of biserial algebras? Is there an example of an algebra derived equivalent to a biserial algebra which is not biserial?

(What about the subclass of special biserial algebras?)

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The answer to your question is no, in general. A simple counterexample is provided by a path algebra of the Dynkin quiver $D_4$ and another algebra tilted from it. That is, let $A$ be the path algebra where the quiver $D_4$ is oriented so that the vertex of degree 3 is a source and let $B$ be the quotient of the path algebra of the Euclidean quiver $\tilde{A}_3$--oriented so that we have two paths of length 2 from a source to a sink--by the relation expressing that these two paths are equal. Then $B$ is biserial (I believe it is even special biserial, if I remember the definition correctly), but $A$ is not biserial. That these algebras are derived equivalent follows from the fact that one is tilted from the other, which is remarked, for instance, in IV.4.3(b) of Happel's book Triangulated categories in the representation theory of finite dimensional algebras.

However, assuming that you mean to include Nakayama algebras (i.e., uniserial algebras) as (special) biserial, it is true for self-injective special biserial algebras: Pogorzaly shows that this class of algebras is closed under stable equivalence [Comm. Algebra 22 (1994), no. 4, 1127–1160], and derived equivalent self-injective algebras are always stably equivalent by a theorem of Rickard.

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In fact it should read: "each indecomposable projective left or right module" and then your counterexample isn't one anymore, sorry. –  Julian Kuelshammer Dec 2 '11 at 8:41
No problem. I'll fix my answer. –  Alex Dugas Dec 3 '11 at 1:39
Thanks, also the reference to Pogorzaly's paper was helpful. –  Julian Kuelshammer Dec 3 '11 at 14:38
And yes, it is even special biserial. –  Julian Kuelshammer Dec 3 '11 at 14:41

Gentle algebras can all be described as path algebras with relations of length $2$, so let's try something with a length $3$ relation.

If you take $k A_4$ with a length $3$ relation you get an algebra derived equivalent to $k D_4$. The first algebra is special biserial, while the other is not biserial.

There is a nice table on page 694 of [Happel, Dieter; Seidel, Uwe. Piecewise hereditary Nakayama algebras. Algebr. Represent. Theory 13 (2010), no. 6, 693--704] which shows the derived equivalence classes you get from $A_n$ with linear orientation and relations of uniform length.

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