The following question looks like a basic question on resolutions over commutative rings, unfortunately I was not able to find a reference.

Let $A$ be a regular commutative noetherian ring (and satisfy all other nice properties that could be needed for construction) , $I \subset A$ is an ideal. Suppose $D$ is a differential graded algebra resolution of $A/I$ over $A$ i.e. $D$ is a free resolution of $A/I$ that admits a structure of dg-algebra over $A$. Basic example of such resolutions is a Koszul complex for ideals generated by a regular sequence. Now let $M$ be a finitely generated module over $A/I$, and $F$ is a free resolution of $M$ over $A$. Is it true that that $F$ is an $A_\infty$-module over $D$?

It is well known that there is a resolution with a dg-module structure over $D$, my question is that any resolution $F$ admits an $A_\infty$-module structure over $D$.


1 Answer 1


I'll try to endow $F$ with a strict $D$-algebra structure. The endomorphism DG-$A$-algebra $\operatorname{End}_A(F)$ satisfies

$$H_n\operatorname{End}_A(P)=\operatorname{Ext}^{-n}_A(M,M).$$ In particular the homology vanishes in positive dimensions. A $D$-algebra structure on $F$ is an map of DG-$A$-algebras $D\rightarrow \operatorname{End}_A(F)$. Since $D$ is concentrated in non-negative degrees, this is the same as a map to the truncation $D\rightarrow t_{\geq 0}\operatorname{End}_A(F)$, whose homology is $\operatorname{Hom}_A(M,M)$ concentrated in degree $0$.Therefore, the natural projection $t_{\geq 0}\operatorname{End}_A(F)\twoheadrightarrow\operatorname{Hom}_A(M,M)$ is a quasi-isomorphism, indeed a trivial fibration in the projective model structure on the category of DG-$A$-algebras.

Since $M$ is actually an $A/I$-module, the 'enriched identity in $M$', which is given by an $A$-module morphism $A\rightarrow\operatorname{Hom}_A(M,M)$, factors through the natural projection $A\twoheadrightarrow A/I$.

Now, consider the following commutative square in the category of DG-$A$-algebras, where $A$ is the initial object, $$\begin{array}{ccccc} A&-&-&\rightarrow&t_{\geq 0}\operatorname{End}_A(F)\\ \downarrow&&&&\downarrow\\ D&\stackrel{\sim}\longrightarrow&A/I&\rightarrow&\operatorname{Hom}_A(M,M) \end{array}$$ The upper arrow should be continuous, but I didn't manage to get it. The right vertical arrow is a trivial fibration, as I've already noticed. The map $A\rightarrow D$ should be a cofibration (this is how I understand that $D$ is a resolution of $A/I$) and $D\rightarrow A/I$ is the resolution map (a trivial fibration, or just an equivalence, it doesn't really matter).

Then, the model category axioms give you a map $D\rightarrow t_{\geq 0}\operatorname{End}_A(F)$ such that, when fitted in the previous square, the two triangles commute.

After your clarification, I can complete my argument to give a positive answer to your question. Before $D$ was a DGA resolution of $A/I$. Now I simply take a DGA $D'$, levelwise projective, concentrated in non-negative degrees, and with $H_*(D')=A/I$ concentrated in degree $0$. The natural projection $D'\twoheadrightarrow A/I$ is therefore a trivial fibration. Hence, the following commutative diagram has a lift by the same argument as before $$\begin{array}{ccc} A&\rightarrow &D'\\ \downarrow&&\downarrow\\ D&\rightarrow &A/I \end{array}$$ The lift $D\rightarrow D'$ is a quasi-isomorphism by the 2-out-of-3 property, since $D\rightarrow A/I$ and $D'\rightarrow A/I$ are quasi-isos. Hence, it has an up-to-homotopy inverse $D'\rightarrow D$ which is an A-infinity morphism. It is exactly here where the fact that the underlying complex of $D'$ is a projective resolution is used.

The composite $D'\rightarrow D\rightarrow t_{\geq 0}\operatorname{End}_A(F)\subset \operatorname{End}_A(F)$ is an A-infinity morphism which endows $F$ with the structure of an A-infinity $D'$-module such that the induced $A/I$-module structure on $M$ (obtained by taking $H_0$) is the given one.

  • $\begingroup$ I agree, such constructions works, but it give in some sense trivial module structure. What I want is to describe some operations on complex $F$ that I get from $D_j \otimes F_i \to F_{i+j}$, for example elements of $D_1$ should give non-trivial maps of degree one going in the direction opposite to the direction of differential on $F$. $\endgroup$ Nov 11, 2013 at 13:49
  • $\begingroup$ It doesn't give the trivial module structure at all, since by pulling back along the unit $A\rightarrow D$ recovers the $A$-module structure. It reflects very well the fact that $D$ is a homotopy quotient of $A$. The maps $D_i\otimes F_j\rightarrow F_{i+1}$ are the adjoints of $D\rightarrow t_{\geq 0}\operatorname{End}_A(F)$, and these are not an A-infinity module structure, but an honest $D$-module structure on $F$. $\endgroup$ Nov 11, 2013 at 16:14
  • $\begingroup$ Could you please explain to me then simplest possible example: $I$ is generated by regular sequence, $D$ is koszul complex, what is structute of dg module in this case? For me it looks like only $D_0$ acts non-trivially. $\endgroup$ Nov 11, 2013 at 18:48
  • 2
    $\begingroup$ Koszul complex is a free resolution over $A$, that has structure of dga over $A$, that is what I mean by $D$ in the question. I'm sorry if that caused some confusion. $\endgroup$ Nov 11, 2013 at 20:38
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    $\begingroup$ Dear @Sasha Pavlov: You may want to edit the question and clarify the nature of $D$ there. $\endgroup$ Nov 11, 2013 at 21:51

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