# Is it possible to compute André-Quillen cohomology by resolving the module variable?

Let $A\to B$ be a morphism of commutative rings. Let $\mathcal C$ be the category of commutative $A$-algebras augmented over $B$. Let $\mathcal M_B$ denote the category of $B$-modules. The cotangent complex can be defined like this (I will brush model-categorical considerations under the carpet). Consider the derived functor of the functor

$$s\mathcal C \to s\mathcal M_B$$

obtained levelwise from the functor $\mathcal C\to \mathcal{M}_B$ given by $R\mapsto B\otimes_R \Omega_{R|A}$. Here $\Omega$ denotes Kähler differentials.

The image of $B\in \mathcal C$ under this derived functor is the cotangent complex $\mathbb L \Omega_{A|B}$.

If $M$ is a $B$-module, one defines the André-Quillen cohomology modules of $A\to B$ with coefficients in $M$ as $$D^i(B|A,M)=H^q(\mathcal M_B(\mathbb L\Omega_{B|A},M)) \cong H^q \mathrm{Der}_A(P,M)$$

where $P\to A$ is a projective resolution of $A$ in $s\mathcal C$.

So I guess this allows one to safely say that André-Quillen cohomology are some kind of "derived functor of the derivations".

I am wondering if it is (sometimes?) possible to resolve the module variable, i.e. to derive the functor $\mathrm{Der}_A(B,-):s\mathcal M_B \to s\mathcal M_B$, and get the same result. This feels somehow like asking whether $\mathrm Der_A(-,-): s\mathcal C \times \mathcal sM_B\to s\mathcal M_B$ is balanced.

• For fixed $A$-algebra $B$, the functor $\text{Der}_A(B,-)$ on $B$-modules has derived functor $\text{RHom}_{B-\text{mod}}(\Omega_{B/A},-).$ Typically the cohomology of this derived functor is quite different from $\text{RHom}_{B-\text{mod}}(L_{B/A},0).$ – Jason Starr Nov 10 '17 at 14:23

The cotangent complex $\Bbb L_{B/A}$ is a (homologically) bounded-below complex of projective $B$-modules, and the bottom homology group is $H_0 (\Bbb L_{B/A}) = \Omega_{B|A}$.
If we apply $RHom_{B-mod}(-,M)$ to this, we get a Grothendieck spectral sequence for computing André-Quillen cohomology: $$Ext^p_B(H_q \Bbb L_{B/A}, M) \Rightarrow D^{p+q} (B|A, M).$$ In particular, as Jason Starr points out, the derived functors of $Der_A(B,-) = Hom_B(\Omega_{B|A}, -)$ are the terms $$Ext^p_B(H_0 \Bbb L_{B/A}, M)$$ on the edge of this spectral sequence, and so we get an equality precisely when the rest of the spectral sequence collapses to zero. The most common way that this could happen is if $H_q \Bbb L_{B/A} = 0$ for all $q > 0$ (smoothness, or certain local complete intersections), or maybe if the higher André-Quillen homology groups and $M$ are supported in different places, but is unlikely otherwise.