2
$\begingroup$

Let $X_n$ be the set of Nakayama algebras with a linear connected quiver and n simple modules. Define a relation on $X_n$ by saying that two such algebras are equivalent if one is isomorphic to the endomorphism ring of a tilting module of the other algebra. Recall that here a tilting module is a module $T$ with $n$ indecomposable summands and $Ext^i(T,T)=0$ for all $i >0$.

  1. Are they equivalent iff they are derived equivalent? (if not, i hope this is still an equivalence relation) edit: It is not an equivalence relation, see the comment by Jeremy Rickard. Thus the answer to 1. is no.
  2. How many equivalence classes are there under derived equivalence?
  3. Is 1. true if we replace in the relation the tilted condition into "iterated tilted"?
$\endgroup$
2
  • 1
    $\begingroup$ It's not a transitive relation. Take $n=4$, let $A$ be the hereditary Nakayama algebra, and $B$ the one with radical square zero. Neither is the endomorphism ring of a tilting module for the other, but they are derived equivalent, and you can get from $A$ to $B$ with a sequence of tilting modules. $\endgroup$ Aug 26, 2017 at 8:14
  • $\begingroup$ Ok, that answers 1. Thanks. I modified question two. $\endgroup$
    – Mare
    Aug 26, 2017 at 8:21

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.