Let $A$ be a simplicial commutative ring over a field $k$ of characteristic zero (or a cdga
in non-positive degrees with differential of degree -1). Let $M$ be a perfect $A$ module. If necessary, assume that $\pi\_{i}(A)=0$ for $i$ sufficiently large (or $H^{-i}(A)=0$ for $i$ sufficiently large).


In a few arguments in the literature, it seems to be assumed that $M \otimes\_{A} \pi\_{0}(A) \simeq 0$ implies $M \simeq 0$. (Here, the pull-back to $\pi\_{0}(A)$ is of course derived.)


Is this true? And if so, why? 


It seems to me somewhat reasonable, since geometrically this seems to be saying that pull-back along $i: Spec (\pi\_{0}(A)) \rightarrow Spec (A)$ is conservative on perfect complexes. But if $A$ were a nilpotent rather than derived thickening of $\pi\_{0}(A)$, this is almost obvious, since I can check if something is zero by seeing that it has non-empty support, and support doesn't seem to see nilpotents.


(Note that probably characteristic zero is completely unnecessary and is just included so that we can freely pass between simplicial commutative algebras and cdgas when convenient.)



When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$?
(1) If $\pi\_j(M)=0$ for $j\le m-1$ and $\pi\_j(N)=0$ for $j\le n-1$, then $\pi\_j(M\otimes\_A N)=0$ for $j\le m+n-1$.


(2) In this case $ \pi\_{m+n}(M\otimes\_A N)=\pi\_m(M)\otimes\_{\pi\_0(A)}\pi\_n(N)$ (where the right hand side is a plain underived tensor product).


(3) Apply this equation with $N=\pi\_0A$ and $n=0$.


