In "Homological Algebra on a Complete Intersection", Eisenbud proves the following:

Let $A$ be a commutative ring, $M$ be an $A$-module and $F^{\ast}\to M$ an $A$-free resolution. Further, assume that $M$ is annihilated by $I := (x_1,...,x_n)$, and that $I$ contains a non zero divisor of $A$. Then there exist maps $s_{\alpha}: F^{\ast}\to F^{\ast}$, indexed by multiindices $\alpha$ of length $n$, of degree $2|\alpha|-1$, with the following properties:

  • $s_0$ is the differential of $F^{\ast}$.
  • $s_j := s_{(0,0,...,0,1,0,...,0)}$ is a nullhomotopy for the multiplication by $x_j$.
  • For any $\gamma$ with $|\gamma|\geq 2$ we have $\sum\limits_{\alpha+\beta=\gamma} s_{\alpha} s_{\beta} = 0$.

Now he asks the following: Assuming $(x_1,...,x_n)$ is regular and $F^{\ast}$ is the minimal free resolution of $M$, is is possible to choose the nullhomotopies $s_j$ in such a way that $s_j^2=0$ and $s_i s_j = -s_j s_i$? I.e. can we make $F^{\ast}$ into a differential graded module over the Koszul-Algebra of $(x_1,...,x_n)$?

I'm interested in this question and would like to know if progress has been made to answer it. Is it possible to choose the $s_i$ as above, and, if not, how can the obstruction be described?

Edit: Is it possible to handle all the ways the $s_j$ can be constructed? Does it have something to do with the $A_{\infty}$-stuff?

Thank you!


  • 1
    $\begingroup$ Dear Hanno, I changed the title a bit to make it more precise,and add the homological algebra tag. I hope you don't mind, please feel free to roll back the edits if you disagree. $\endgroup$ Sep 28, 2010 at 12:37
  • $\begingroup$ Dear Hailong: No, it's fine, thank you! $\endgroup$
    – Hanno
    Sep 28, 2010 at 15:30

2 Answers 2


It is true if the projective dimension of $M$ over $A$ is at most $3$, and counter examples exist when the projective dimension is $4$.

The first counter example was given in Lucho Avramov's paper "Obstructions to the existence of a multiplicative structure on minimal resolutions". A simplified example, ($A=k[t_1,t_2,t_3,t_4], M= A/(t_1^2, t_1t_2, t_2t_3, t_3t_4, t_4^2)$, here you can choose the $x_i$s to be any regular sequence in the annihilator of $M$) together with discussion of related and more recent results can be found in Section 2 of this note (available on Avramov's website).


Despite the existence of obstructions, there has also been progress in the other direction. Namely, there are lots of classes of free resolutions where the question is known to have an affirmative answer. (When $F^* $ can be given the structure of a dg module, it is sometimes said that $F^*$ has a "multiplicative structure".)

For instance, I believe that Srinivasan first showed that the Eagon-Northcott complex has a multiplicative structure. This was then generalized by Pellikaan who showed that determinantal ideals also have multiplicative structures. There is a reasonably large literature along these lines, and many of the papers have the phrase "multiplicative structure" in the title.

  • $\begingroup$ Hi Dan, it's been a while! Very interesting answer, I did not know about determinantal ideals. $\endgroup$ Sep 28, 2010 at 14:28
  • $\begingroup$ Hi Daniel, thanks alot your answer, too! This seems to be a very interesting topic :-) $\endgroup$
    – Hanno
    Sep 28, 2010 at 20:00

Your Answer

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

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