Let $k$ be a field of characteristic zero, $A$ a simplicial commutative k-algebra, and $M$ a simplicial $A$-module. Consider the trivial square-zero extension $A\oplus M$ as an $A$-algebra. Is it true that the relative cotangent complex of $A\oplus M$ over $A$ (i.e the cotangent complex of the map $A \rightarrow A\oplus M$) is isomorphic to $M$ (say in the derived category of $A$-modules) ? This might be easy but thanks anyway for any suggestion or reference.
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3$\begingroup$ Unless I'm mistaken, this already fails in the discrete case. Take A = k, and M = k. Then B := A + M is k[e]/(e^2), and the cotangent complex of B/A (computed using the transitivity triangle for k -> k[e] ->> B) is quasi-isomorphic to the 2-term complex given by multiplication by d(e^2) = 2e on B. In particular, it has two non-zero homology groups as a complex of k-vector spaces. $\endgroup$– BhargavCommented Mar 18, 2011 at 21:54
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$\begingroup$ Thanks a lot Bhargav. I am such an idiot! I guess I was computing the case $A=k$ and any $M$, but I was computing the $H^0$ i.e. the Kahler differentials. Sorry about being so sloppy. By the way, do you have a guess for a general answer? $\endgroup$– Martin LagenbachCommented Mar 19, 2011 at 12:43
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$\begingroup$ I think one possible general answer is that L_{B/A} (with B = A + M) has good connectivity properties if M does so. These issues were discussed in what used to be Lurie's DAG IV, and I suspect can also be found in his "Higher Algebra" book (but I did not look). $\endgroup$– BhargavCommented Mar 19, 2011 at 18:59
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I think if we are in characteristic $0$ (i.e., $A\supset\mathbb Q$), and $M$ is flat, there is an explicit formula for the cotangent complex: $$L_{A\to A\oplus M} = (\mathcal C_A(M[1])[-1] \otimes_A (A\oplus M), d) , $$ where $\mathcal C_A$ is the free (graded) Lie super-coalgebra over $A$ and $d$ is obtained by the natural map $\mathcal C_A(M[1])\to\mathcal C_A(M[1]) \otimes_A M[1]$ given by the Lie co-bracket. (You can of course get rid of the "co-" by dualizing everything.) If you restrict to $\operatorname{Spec}A\subset\operatorname{Spec}(A\oplus M)$, you get $\mathcal C_A(M[1])[-1]$ with zero differential.
This can be deduced from what's called "Koszul duality for operads" between commutative and Lie algebras.
(and if $M$ is not flat, you have to replace it with projective resoljtion and apply this formula, with a term added to $d$ coming from the differential in the resolution)
