5
$\begingroup$

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 $D$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $F$. $\endgroup$ – Sasha Pavlov Nov 11 '13 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 $F$. $\endgroup$ – Fernando Muro Nov 11 '13 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. $\endgroup$ – Sasha Pavlov Nov 11 '13 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. $\endgroup$ – Sasha Pavlov Nov 11 '13 at 20:38
  • 2
    $\begingroup$ Dear @Sasha Pavlov: You may want to edit the question and clarify the nature of $D$ there. $\endgroup$ – Ricardo Andrade Nov 11 '13 at 21:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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