When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$? - MathOverflow

When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$?

2012-03-08T16:30:59Z

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.)

Answer by Eric Wofsey for When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$?

2012-03-08T17:39:47Z

In fact, this holds for the derived category of any connective $E_\infty$ ring spectrum $A$ such that $\pi_i(A)=0$ for $i$ sufficiently large: If $M$ is an $A$-module and $M\otimes_A H\pi_0(A)$ is contractible, then $M$ is contractible.

For $M$ an $A$-module, let $\langle M \rangle$ be the smallest full subcategory of the category of $A$-modules which contains $M$ and is closed under homotopy colimits and desuspension. It suffices to prove that for any $M$, $M\in\langle M \otimes H\pi_0(A)\rangle$. Since the tensor product preserves homotopy colimits and desuspensions in each variable, $A\in \langle H\pi_0(A)\rangle$ implies $M=M\otimes A \in \langle M \otimes H\pi_0(A)\rangle$ for all $M$. Thus it suffices to show we can ``build'' $A$ out of $H\pi_0(A)$ using colimits and desuspensions.

Now if $M$ is an $A$-module such that $\pi_i(M)=0$ for $i\neq 0$, then the $A$-module structure on $M$ factors through $H\pi_0(A)$. Hence in particular, $M=H\pi_0(M)$ is a homotopy colimit of (shifted) copies of $H\pi_0(A)$ and is thus in $\langle H\pi_0(A)\rangle$. Since $A$ has only finitely many homotopy groups, the Postnikov filtration on $A$ builds $A$ out of finitely many pieces, each of which has only one homotopy group. Each one of these pieces is a suspension of a module such that $\pi_i(M)$ is concentrated in degree $0$, and thus is in $\langle H\pi_0(A)\rangle$. Gluing together the Postnikov sections of $A$ by cofiber sequences, we find that $A\in\langle H\pi_0(A)\rangle$.

Answer by Tom Goodwillie for When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$?

2012-03-08T17:55:29Z

(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 $n=0$.