Let $k$ be a field of characteristic zero, $A$ a simplicial commutative kalgebra, and $M$ a simplicial $A$module. Consider the trivial squarezero 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.

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 supercoalgebra 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 cobracket. (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) 

