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Hello!

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!

Hanno

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# How does the annihilator of a module act on its free resolution?

Hello!

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?

Thank you!

Hanno