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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.)

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2 Answers 2

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(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 $n=0$.

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Thanks. Here's where I'm ignorant: How do I see (2)? I guess I don't really know how to compute these things, and that's my problem. Are there any basic texts on simplicial rings and their modules? –  dhagbert Mar 8 '12 at 19:15
    
OK.You need to represent $M \otimes_{A} N$ as a coequalizer in the usual way, and then look at the associated long exact sequence of homotopy groups. The homotopy groups of the tensor products over $k$ should be easy to compute. Is it true that $\pi_{k}(M \otimes_{k} N) \simeq \oplus_{i+j=k} \pi_{i}(M)\otimes_{k} \pi_{j}(N)$? This would be true under Dold-Kan. (I'm a bit puzzled how to get modules with negative homotopy groups in the simplicial setting. Do you need spectra?) –  dhagbert Mar 9 '12 at 9:10
    
You spoke of two settings: simplicial and dg. I didn't know whether you meant to somehow allow negative htpy gps in the simplicial setting. If so, then you would need to be doing something like spectra. A sort of hybrid of dg and simplicial would be a relatively low-tech way of making sense of this. –  Tom Goodwillie Mar 9 '12 at 11:23
    
I meant negative homotopy groups for the modules over the simplicial commutative ring. I'm happy for the ring itself to be simplicial. But I'd like to consider whatever corresponds to perfect dg modules. –  dhagbert Mar 9 '12 at 19:13

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 $A\in\langle H\pi_0(A)\rangle$.

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I should add that this argument doesn't require $M$ to be perfect, but does require $A$ to have only finitely many homotopy groups. Tom's argument works for arbitrary (connective) $A$ and perfect $M$ (so that you know $M$ has to have a smallest homotopy group). The statement is not true for arbitrary $M$ and arbitrary connective $A$. For instance, $A$ could be a polynomial ring generated by a class in degree 2 and $M$ could be $A$ with that class inverted. –  Eric Wofsey Mar 8 '12 at 20:00
    
@eric: I am quite the beginner in this area, could you be so kind to clarify the counterexample you gave? What kind of object is A? To me it just seems a graded ring. –  Yosemite Sam Mar 8 '12 at 22:07
    
I'm considering this graded ring as (say) a dga with trivial differential. If you want to think of this as a spectrum, there's an equivalence between $A_\infty$-algebra spectra over $H{\mathbb Z}$ and associative dgas, and rationally an equivalence between $E_\infty$-algebras over $H{\mathbb Q}$ and commutative dgas over $\mathbb Q$. There is correspondingly an equivalence between categories of modules over $H{\mathbb Z}$-algebras and derived categories of dg-modules. –  Eric Wofsey Mar 8 '12 at 22:22

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