2
$\begingroup$

The socle of cokernel of irreducible monomorphisms in the AR quiver of type An/I is simple.

I believe that this result is hidden in a more general result in some articles, I tried to find a lot but failed.

The following lemma maybe useful:

M be a module of length n. TFAE:

1 SocM is simple

2 Any non-zero submodule of M is indecomposable

3 There exists a composition series of M with all terms indecomposable

Any suggestions are welcome!

Improve this question
$\endgroup$
3
  • $\begingroup$ Do you assume your $A_n$-quiver to be linearly oriented? $\endgroup$
    – Julian Kuelshammer
    Commented May 18, 2016 at 9:36
  • $\begingroup$ In fact here the orientation and admissable ideal I are all arbitrary. @Julian Kuelshammer $\endgroup$
    – milanelo
    Commented May 19, 2016 at 5:34
  • 1
    $\begingroup$ Every non-projective indecomposable module is the cokernel of an irreducible monomorphism, namely the left almost split morphism in the AR-sequence. I think you mean irreducible monomorphisms between indecomposable modules. $\endgroup$
    – Dag Oskar Madsen
    Commented May 19, 2016 at 10:15

1 Answer 1

Reset to default
3
$\begingroup$

This is more generally true for any representation-finite string algebra (provided as Dag Madsen pointed out you are talking about irreducible maps between indecomposable modules, otherwise every non-projective module is a cokernel of an irreducible monomorphism). String algebras are a subclass of the class of special biserial algebras, where the admissible ideal of the quiver is spanned by zero relations.

For this class of algebras, all indecomposable modules are given by strings (i.e. words in the arrows of the quiver and their inverses such that there is no cancellation). This is essentially due to Gelfand and Ponomarev, see I. M. Gel'fand and V. A. Ponomarev, MR0229751 Indecomposable representations of the Lorentz group., Russian Mathematical Surveys 23 (1968), no. 2, 3--60. The general case is is stated e.g. in Burkhard Wald and Josef Waschbüsch, MR 801283 Tame biserial algebras, J. Algebra 95 (1985), no. 2, 480--500.

In this case the irreducible monomorphisms are given by appending hooks to a given string. This is proven in M. C. R. Butler and Claus Michael Ringel, MR 876976 Auslander-Reiten sequences with few middle terms and applications to string algebras, Comm. Algebra 15 (1987), no. 1-2, 145--179. Thus, the cokernel of an irreducible monomorphism is just the end of a hook, hence has simple socle.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ You can sometimes append hooks to both ends of the string, can't you? Are you (and the OP) assuming that the target of the irreducible monomorphism is an indecomposable module? $\endgroup$
    – Dag Oskar Madsen
    Commented May 19, 2016 at 10:06
  • $\begingroup$ @DagOskarMadsen Thanks, I was implicitly assuming the OP meant this. $\endgroup$
    – Julian Kuelshammer
    Commented May 19, 2016 at 10:54
  • $\begingroup$ Perhaps when he writes "irreducible monomorphisms in the AR quiver" he is referring to (a subset of) the arrows in the AR quiver. $\endgroup$
    – Dag Oskar Madsen
    Commented May 19, 2016 at 11:02
  • $\begingroup$ Thank you for all the comments! What I mean precisely is all the irreducible morphisms between indecomposable modules in the AR quiver, i.e, all the arrows in the quiver. @Dag Oskar Madsen $\endgroup$
    – milanelo
    Commented May 21, 2016 at 11:03
  • $\begingroup$ For the last sentence, why the end of a hook has a simple socle, or equivalently has a unique simple submodule? Thanks a lot for explanation! $\endgroup$
    – milanelo
    Commented May 21, 2016 at 11:43

You must log in to answer this question.

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