cotangent complex of a trivial extension

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.

flag	asked Mar 18 2011 at 19:08 Martin Lagenbach 21●1

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. – Bhargav Mar 18 2011 at 21:54

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? – Martin Lagenbach Mar 19 2011 at 12:43

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). – Bhargav Mar 19 2011 at 18:59