ag.algebraic geometry - sheafifying a projective limit of presheaves

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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