Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

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

1 Answer 1

up vote 3 down vote accepted

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.)

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.