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Let $A_\bullet \to B_\bullet$ be a quasi-isomorphism of simplicial rings in the sense of (P.62, I.3.1.7, Complexe Cotangent et Déformations I, Luc Illusie).

Then, we define the simplicial $B_\bullet$-module of Kähler differentials $\Omega^1_{B_\bullet/A_\bullet}$ by $\left(\Omega^1_{B_\bullet/A_\bullet}\right)_n := \Omega^1_{B_n/A_n}$ (P.119, ibid.).

My question is whether it follows that $\Omega^1_{B_\bullet/A_\bullet}$ is acyclic (i.e. quasi-isomorphic to the zero module) from these assumptions.


The broader context is when I am trying to show that the cotangent complex is independent of the free resolution taken.

Suppose we are given an algebra $R \to S$ which has two free simplicial resolutions $P_\bullet \to Q_\bullet$ (i.e. they are quasi-isomorphic to $S$, and $P_n$ and $Q_n$ are free $R$-algebras).

Then, using these two simplicial resolutions, we define the cotangent complex $L_{S/R}$ to be $\Omega_{P_\bullet/R} \otimes_{P_\bullet} S$ and $\Omega_{Q_\bullet/R} \otimes_{Q_\bullet} S$, the former of which is isomorphic to $\Omega_{P_\bullet/R} \otimes_{P_\bullet} Q_\bullet \otimes_{Q_\bullet} S$. My approach was to break down the question of whether $\Omega_{P_\bullet/R} \otimes_{P_\bullet} S$ and $\Omega_{Q_\bullet/R} \otimes_{Q_\bullet} S$ are quasi-isomorphic into two sub-questions:

  1. Whether the map $\Omega_{P_\bullet/R} \otimes_{P_\bullet} Q_\bullet \to \Omega_{Q_\bullet/R}$ is a quasi-isomorphism.
  2. Whether $- \otimes_{Q_\bullet} S$ sends quasi-isomorphisms of free $Q_\bullet$-modules to quasi-isomorphisms.

For 1, I have fitted them in a short exact sequence (in general $0 \to$ is not present, but for my case it is ok): $$0 \to \Omega_{P_\bullet/R} \otimes_{P_\bullet} Q_\bullet \to \Omega_{Q_\bullet/R} \to \Omega_{Q_\bullet/P_\bullet} \to 0$$

So it suffices to show that $\Omega_{Q_\bullet/P_\bullet}$ is acyclic, which is what I asked above.

For 2, I guess the usual proof with cones would go through ($f$ quasi-isomorphism $\implies$ $\operatorname{cone}(f)$ acyclic $\implies$ $\operatorname{cone}(f) \otimes_{Q_\bullet} S = \operatorname{cone}(f \otimes_{Q_\bullet} S)$ acyclic $\implies$ $f \otimes_{Q_\bullet} S$ quasi-isomorphism), but I have not checked the details yet. I would appreciate if someone would give me a pointer on this, but I could also ask this in a later question.

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    $\begingroup$ Maybe you could learn the subject in slightly less old references. For instance Toen and Vezzosi's Homotopical Algebraic Geometry II: geometric stacks and applications. Mem. Amer. Math. Soc. 193 (2008), no. 902. also available at perso.math.univ-toulouse.fr/btoen/files/2012/04/HAGII.pdf Reading Illusie is still instructive, but there are many homotopy theoretic aspects which are problems for Illusie but not any more for a modern reader. $\endgroup$ May 5, 2021 at 10:53
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    $\begingroup$ @Denis-CharlesCisinski Thanks for the pointer. I am a bit confused by the remark on P.65 that "Note that this $\mathbb{L}_{B/A}$ is not the Quillen-Illusie cotangent complex of $A \to B$." $\endgroup$
    – Kenny Lau
    May 5, 2021 at 11:03
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    $\begingroup$ @Denis-CharlesCisinski does your source talk about my question, i.e. computing $\mathbb{L}_{B/A}$ using any free resolution gives the same result? $\endgroup$
    – Kenny Lau
    May 5, 2021 at 13:12
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    $\begingroup$ Actually, I'd recommend an older reference as you're in the affine setting, by an author who certainly understood the homotopy theory. Daniel Quillen. On the (co-) homology of commutative rings. In Applications of Categorical Algebra (Proc. Sympos. Pure Math., Vol. XVII, New York, 1968), pages 65–87. Amer. Math. Soc. Providence, R.I., 1970. $\endgroup$ May 5, 2021 at 13:17
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    $\begingroup$ and for your conclusion to be true, you need the morphism to be a cofibration or similar (levelwise ind-smooth should do); otherwise the relative module of differentials won't behave. $\endgroup$ May 5, 2021 at 13:31

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