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Let $f:X\to Y$ be a representable map of finite type (or is finite dimensional enough?) Artin stacks, whose fibres (which are schemes) have dimension at most $n$. Then is it true that $R^qf_*\mathbf{Q}_\ell=0$ for all $q\gg 0$?

Note: by taking atlases, I think it is sufficient to let $X,Y$ be schemes.


Edit: Will Sawin pointed out that the question as stated was obviously false, I've edited it to remove that false statement.

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$Y$ admits a smooth surjective morphism from a scheme $Z$. Because smooth morphisms are locally of finite type, $Z \to Y$ is locally of finite type, and you can choose an open cover that covers $Y$ and then pass to a finite subcover to make $Z$ of finite type.

Because this morphism is smooth, by smooth base change the pullback of $R^q f_* \mathbf Q_\ell$ to $Z$ is the pushforward of $\mathbb Q_\ell$ from $Z \times_Y X$ to $Z$. Because this morphism is smooth, it suffices to prove a bound for this pushforward.

If $Z \to Y$ is a schematic morphism (this might be a little stronger than the fibers being schemes) then $Z \times_Y X$ is a scheme, also of finite type. Boundedness then follows from classical results - mod $\ell$ cohomology is a limit over the cohomology of neighborhoods, and these are finite type of bounded dimension.

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  • $\begingroup$ Whoops, I must have not been thinking when I wrote the question as it was. I've edited it now, hopefully it is less trivial (or perhaps even correct). $\endgroup$
    – Meow
    Jun 18 at 23:42
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    $\begingroup$ @Meow Isn't the disjoint union of $\mathbb A^m$ for all $m$ an Artin stack? You need some kind of finiteness assumption on $X$. $\endgroup$
    – Will Sawin
    Jun 19 at 1:19
  • $\begingroup$ Right, I should have included a finite type assumption, which has been edited in now. $\endgroup$
    – Meow
    Jun 19 at 9:07
  • $\begingroup$ @Meow See edits. $\endgroup$
    – Will Sawin
    Jun 19 at 14:41
  • $\begingroup$ Thanks for being accomodating of my shifting the goalposts! $\endgroup$
    – Meow
    Jun 19 at 23:52

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