For an l-adic sheaf (F_n), why is the complex F_n of finite Tor dimension?

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Let $X$ be a variety and let `$\mathcal{F}=(\mathcal{F}_n)_{n\geq 0}$` be a (constructible) $\ell$-adic sheaf on $X$. Let $K_n$ be the object in the derived category $D(X,\mathbf{Z}/\ell^{n+1})$ of sheaves of $\mathbf{Z}/\ell^{n+1}$-modules defined by $\mathcal{F}_n$, i.e., the complex with $\mathcal{F}_n$ concentrated in degree zero and zeroes everywhere else.

In the proof of the generalized Trace formula (see Deligne's Rapport sur la formule des traces or de Jong's Stacks Project: Etale cohomology) the following fact is used.

Fact. We have that $K_n$ is of finite Tor dimension.

Equivalently:

Fact. We have that $K_n$ is isomorphic in $D^-(X,\mathbf{Z}/\ell^{n+1})$ to a bounded complex of constructible sheaves of flat $\mathbf{Z}/\ell^{n+1}$-modules.

Unfortunately, I haven't been able to find a proof of this fact in the literature.

Question. How does one prove the above Fact?

[Edit: Explanation of Torsten Ekedahl's answer] By an $\ell$-adic sheaf, one usually means a `$\mathbf{Q}_\ell$`-sheaf. My mistake was that I considered an arbitrary `$\mathbf{Z}_\ell$`-sheaf $\mathcal{F}$. In fact, given a `$\mathbf{Q}_\ell$`-sheaf $\mathcal{G}$, we have that `$\mathcal{G}= \mathcal{F}\otimes \mathbf{Q}_\ell$` for some torsion-free `$\mathbf{Z}_\ell$-sheaf $\mathcal{F}$`. But then it's clear that the complex defined by $\mathcal{F}_n$ is of finite Tor-dimension.

This Fact is very strange to me for the following reason:

Example. Consider the complex of $\mathbf{Z}/\ell^{n+1}$-modules $$\ldots \stackrel{\cdot l^n}{\longrightarrow} \mathbf{Z}/\ell^{n+1} \stackrel{\cdot l}{\longrightarrow} \mathbf{Z}/\ell^{n+1} \stackrel{\cdot l^n}{\longrightarrow} \mathbf{Z}/\ell^{n+1} \stackrel{\cdot l}{\longrightarrow}\mathbf{Z}/\ell^{n+1} \longrightarrow \mathbf{Z}/\ell \rightarrow 0 .$$ From this it follows that Tor_i$(\mathbf{Z}/\ell, \mathbf{Z}/\ell) = \mathbf{Z}/\ell \neq 0$ if $i>0$. In particular, the $\mathbf{Z}/\ell^{n+1}$-module $\mathbf{Z}/\ell$ is NOT of finite Tor dimension. Thus, we see that $K_n = \mathbf{Z}/\ell$ does not give an $\ell$-adic sheaf. (This would contradict the Fact.)

This Example suggests that the proof of the above Fact relies on the compatibility between the K_n.

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1 Answer

I think you have (slightly) misread your sources. If you take Rapport for instance (which is the one I am familiar with) Deligne never make this claim (as far as I can see). As a typical example consider 4.11 (proof of 2.30) where he assumes that $(K_n)$ is a torsion free $\mathbb Z_\ell$-complex in which case $K_n$ is indeed of finite Tor-dimension. Alternatively, he could have considerd (even when $K$ is just a sheaf) the derived reduction modulo $\ell^n$, `$K\bigotimes^{\mathbb L}_{\mathbb Z_\ell}\mathbb Z/\ell^n$` which always will have finite Tor-dimension. Your example is a proof that not taking the derived reduction but just the reduction will not necessarily give something of finite Tor-dimension.

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Thank you very much!  – Ariyan Javanpeykar May 29 2011 at 7:11
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