sheafifying a projective limit of presheaves

Let $F=(F\_n)\_n$ be an $\ell$-adic sheaf on $X\_{et}$, for a variety $X$ over an algebraically closed field $k$ of characteristic not equal to $\ell$. Does the presheaf sending $U$ to $H^i(U,F):=\lim\_n H^i(U,F\_n)$ sheafify to zero?

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hi just so you know, you can type latex here, just put math formulas in ! – natura Jan 5 '10 at 2:29

CORRECTED ANSWER: I believe that the answer is no, at least in some contexts.

For example, suppose that $X =$Spec $k$, with $k$ a field, and $F = {\mathbb Z}\_{\ell}(1)$. Then $U =$Spec $l$ for some finite separable extension $l$ of $k$, and $H^1(U,F) = \ell$-adic completion of $l^{\times}$, which I will denote by $\widehat{l^{\times}}$.

Thus the stalk of the presheaf $U \mapsto H^1(U,F)$ (and hence of the associated sheaf) at the (unique) geometric point of $X$ is the direct limit over $l$ of $\widehat{l^{\times}}.$

This direct limit need not vanish. For example, if $k$ is finite, then so is $l$, and $\widehat{l^{\times}}$ is just the $\ell$-Sylow subgroup of $l$. Thus the stalk in this case is just $\bar{k}^{\times}[\ell^{\infty}],$ the group of $\ell$-power roots of unity in $\bar{k}$.

This fits with a certain intuition, namely that one has to go to smaller and small etale neighbourhoods to trivialize $F_n$ as $n$ increases, and hence one can't kill of cohomology classes in $H^i(U,F)$ just by restricting to some $V$.

I think that the answer is yes. Here is a proof (hopefully blunder-free):

It is true for the presheaf $U \mapsto H^i(U,F\_1).$ In other words, if we fix $U$, then for each element $h \in H^i(U,F\_1)$ and each geometric point $x$ of $U$, there is an etale n.h. $V$ of $x$ such that $h\_{| V} = 0.$ Since $H^i(U,F\_1)$ is finite dimensional, there is a $V$ that works for the whole of $H^i(U,F\_1)$ at once.

I claim that then $H^i(U,F\_n)$ restricts to $0$ on $V$ as well.

To see this, consider the exact sequence $0 \to F\_n \to F\_{n+1} \to F\_1 \to 0.$ Applying $H^i(U,\text{--})$ to this yields a middle exact sequence $H^i(U,F\_n) \to H^i(U,F\_{n+1}) \to H^i(U,F\_1).$ Applying $H^i(V,\text{--})$ yields a middle exact sequence $H^i(V,F\_n)\to H^i(V,F\_{n+1}) \to H^i(V,F\_1).$ Restriction gives a map from the first of these sequences to the second. It is zero on the two outer terms, by induction together with the case $n = 1$ proved above, and so is zero on the inner term.

This shows that restricting from $U$ to $V$ kills $H^i(U,F_n)$ for all $n$, and hence $H^i(U,F)$, as required.

EDIT: As was noted in the comment below, this proof assumes that $F$ is ${\mathbb Z}_{\ell}$ -flat. Let me sketch an argument that hopefully handles the general case:

Put $F$ in a short exact sequence $0 \to F\_{tors} \to F \to F\_{fl} \to 0.$ The same kind of argument as above reduces us to checking $F\_{fl}$ and $F\_{tors}$ separately. The above proof handles the case of $F\_{fl}$, while $F\_{tors} = F\_{tors,n}$ for some large enough $n$, and so the projective limit collapses in this case and there is nothing to check.

(Note: I am assuming some basic kind of finiteness assumption on $F$ here, so that the above makes sense. Constructibility should be enough.)

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Thanks! But why do we have that short exact sequence? I mean if F is not flat, it's only right exact. For instance F can be a constant system (F_1). – shenghao Jan 5 '10 at 3:24
Yes, I assumed that $F$ is flat. The case $F = (F_1)$ is of course easier! I have edited the answer to take into account the non-flat case. – Emerton Jan 5 '10 at 3:40
I have one more dumb question: it seems that the outer two maps being zero doesn't imply the inner map is also zero. Here's an "example": The first exact sequence is 0 -> Z -> Z^2 -> Z -> 0, where the first map sends a to (a,0) and the second sends (a,b) to b. The section exact sequence is similar, with Z replaced by Z/2Z. The middle map sends (a,b) to (b mod 2, 0), and the two outer maps are both zero. – shenghao Jan 5 '10 at 4:06
typo: "section" should be "second", as in "section exact sequence". – shenghao Jan 5 '10 at 4:08
I think this counts as a blunder. Let me know if you agree with the correction above. (Hopefully there won't be too many more iterations!) – Emerton Jan 5 '10 at 4:37