When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:10:45Z http://mathoverflow.net/feeds/question/90594 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90594/when-does-m-otimes-a-pi-0a-simeq-0-imply-m-simeq-0 When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$? dhagbert 2012-03-08T16:30:59Z 2012-03-08T17:55:29Z <p>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).</p> <p>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.)</p> <p>Is this true? And if so, why? </p> <p>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.</p> <p>(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.)</p> http://mathoverflow.net/questions/90594/when-does-m-otimes-a-pi-0a-simeq-0-imply-m-simeq-0/90600#90600 Answer by Eric Wofsey for When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$? Eric Wofsey 2012-03-08T17:39:47Z 2012-03-08T17:39:47Z <p>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.</p> <p>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.</p> <p>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 $A\in\langle H\pi_0(A)\rangle$.</p> http://mathoverflow.net/questions/90594/when-does-m-otimes-a-pi-0a-simeq-0-imply-m-simeq-0/90605#90605 Answer by Tom Goodwillie for When does $M \otimes_{A} \pi_{0}(A) \simeq 0$ imply $M \simeq 0$? Tom Goodwillie 2012-03-08T17:55:29Z 2012-03-08T17:55:29Z <p>(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 $j\le m+n-1$.</p> <p>(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).</p> <p>(3) Apply this equation with $N=\pi_0A$ and $n=0$.</p>