1
$\begingroup$

Let $K$ be an algebraically closed field. $Q=(Q_0,Q_1)$ is a quiver($Q_0$ is a finite set of vertices and $Q_1$ is a finite set of arrows). An algebra $KQ/I$ is called monomial if $I$ is a ideal consisting of monomial relations. A finite dimensional algebra $A$ is special multiserial if it is Morita equivalent to an algebra $KQ/I$ such that for all $a \in Q_1$ there exists at most one arrow $b \in Q_1$ such that $ab \not \in I$ and there exists at most one arrow $c \in Q_1$ such that $ca\not \in I$.

Now assume that $A$ is special multiserial. I have seen in a place that "by successively factoring out the socles of indecomposable projective-injective modules, we get a monomial special multiserial algebra $B$. I don't know why the above conclusion hold. Also I want to know whether it is hold for general quiver algebras. Thank you for some reference about it and help.

$\endgroup$
0

1 Answer 1

0
$\begingroup$

This is not true as stated. For example take the path algebra of the quiver with 5 vertices and 5 arrows: $1\to 2\to 4$, $1\to 3\to 4$, and $1\to 5$. Impose the relation $(1\to 2\to 4)=(1\to 3\to 4)$. Then, this algebra is obviously special multiserial, but it does not have indecomposable projective-injective modules.

An easier example in the general case is the path algebra of $D_4$ with non-special-multiserial orientation. It also does not have indecomposable projective-injective modules.

$\endgroup$
1
  • $\begingroup$ Thank you for your examples. I have seen the words here:arxiv.org/pdf/1704.00612v1.pdf, in the last paragraph of the paper. So, is there something wrong? $\endgroup$ Commented Apr 7, 2017 at 9:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .