Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let A be a finite dimensional algebra over a field k and M,N a finitely generated A-module. Im searching for examples where the module $ Ext^{o} (M,N) $ is a finitely generated $ Ext^{o}(M,M) $ -module(via yonedaproduct) for every finitely generated A-modules M,N.

This is the case for example when A is a cocommutative Hopfalgebra by a result of Friedlander and Suslin and its a conjecture that this also holds for a general Hopfalgebra. see for example: http://www.math.uiuc.edu/K-theory/0085/paper.pdf

share|improve this question
1  
Similar results hold for the small quantum groups (and higher Frobenius-Lusztig kernels), see e.g. [Drupieski: Representations and cohomology for higher Frobenius-Lusztig kernels] and the references therein. –  Julian Kuelshammer Jun 11 '13 at 14:21

1 Answer 1

This is true whenever condition (Fg) from the theory of support varieties holds.

A finite dimensional algebra $A$ with Jacobson radical $J$ is said to satisfy condition (Fg) if the Hochschild cohomology ring ${\rm{HH}}^{\ast}(A)$ of $A$ is Noetherian and ${\rm{Ext}}^{\ast}_A(A/J,A/J)$ is finitely generated as an ${\rm{HH}}^{\ast}(A)$-module. The ring ${\rm{HH}}^{\ast}(A)$ acts on ${\rm{Ext}}^{\ast}_A(M,N)$ through the graded center of ${\rm{Ext}}^{\ast}_A(M,M)$ as explained in section 3 of [Solberg, Support varieties for modules and complexes]. Condition (Fg) implies that ${\rm{Ext}}^{\ast}_A(M,N)$ is finitely generated as an ${\rm{HH}}^{\ast}(A)$-module [Propositions 5.5 and 5.7 of that paper]. This in turn implies that ${\rm{Ext}}^{\ast}_A (M,N)$ is finitely generated as an ${\rm{Ext}}^{\ast}_A(M,M)$-module.

See the introduction to arXiv:1003.2867 for classes of algebras that satisfy condition (Fg). In particular (Fg) holds when $A$ is a representation finite self-injective algebra over an algebraically closed field.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.