Let $A$ be a selfinjective algebra. A famous conjecture by Tachikawa says that $Ext^{i}(M,M) \neq 0$ for some $i>0$ and any nonprojective module $M$ (all algebras and modules are finite dimensional over a field K). Now I wondered wheter the following plan might work to prove this conjecture in the commutative case (where i=1 might always do the job): (we can assume that the algebra is local) Let $A^{e}$ be the enveloping algebra, then it should be true that $Ext^{1}_{A^e}(A,A)= \underline{Hom_{A^e}(\Omega^{1}(A),A)}$ is always nonzero. Now the question is: Given any indecomposable nonprojective module $M$, can one find a map in $\underline{Hom_{A^e}(\Omega^{1}(A),A)}$ such that tensoring with $\otimes_A M$ gives a nonzero map in $Ext^{1}_{A}(M,M)= \underline{Hom_{A}(\Omega^{1}(M),M)}$ ? For the simplest case $A=K[x]/(x^n)$ it should work.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ Can you give a Ref? Thanks. $\endgroup$– wonderichCommented Jan 9, 2017 at 1:02
-
$\begingroup$ math.uni-bielefeld.de/~ringel/lectures/tachi/tachikawa/… and the overview article Frobenius algebras by Yamagata. $\endgroup$– MareCommented Jan 9, 2017 at 14:14
-
$\begingroup$ For $A=k[x]/x^n$ and a finitely generated $A$-module $M$, $\mathrm{Ext}^1(M,M)=0$ implies $M$ is free, as you say. $\endgroup$– MohanCommented Jan 14, 2017 at 4:19
Add a comment
|