I have a reference or counterexample request. Suppose $k$ is a field and $\ell\neq char(k)$. There are several common references that show that $H^i_{et}(-, \mathbb{Q}_\ell )$ is a Weil cohomology theory when $k$ is separably closed.

I've spent several hours skimming through Milne's Etale Cohomology, the 1994 Motives volume, SGA articles, online searches, etc and I can't seem to determine whether or not $\ell$-adic cohomology forms a Weil cohomology theory when you don't assume you are in some "geometric" situation by making assumptions on the field.

Is there a reference that proves this is still a Weil cohomology theory or is it just false in this case? Thanks.

(This might be in SGA somewhere, but my skimming of French is rather slow and any specific related statement I find tends to throw in being over an algebraically closed field.)

Edit/Update: Everyone is commenting on the non-finitely generatedness, so I'll be more specific. That isn't really the interesting thing to me. Do you still have some sort of cycle class map that behaves nicely (functorially)? For instance, that paper Timo listed seems to imply that as long as finiteness is satisfied when you plug in a particular variety, everything else seems to be fine, but I haven't had time to seriously look at it yet.

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    $\begingroup$ The étale cohomology of a field with coefficients in ${\bf Z}/l$ coincides with the Galois cohomology of ${\bf Z}/l$ (where ${\bf Z}/l$ is endowed with the trivial action). This is not finitely generated in general. $\endgroup$ Jan 6, 2012 at 21:45

1 Answer 1


See e.g. the first page of Uwe Jannsen, Continuous étale cohomology: http://www.springerlink.com/content/u526616k42051570/ (For varieties over number fields, the cohomology groups in question will in general not be finite dimensional.)

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    $\begingroup$ Thanks. This is fantastic. For my purposes it seems like I may be able to substitute this for my one step that occurs over Q and then since I base change afterwards and this matches up with $\ell$-adic over $\overline{\mathbb{Q}}$ I shouldn't have any problems. I'll look more closely this weekend to see if what I just said is realistic, though. $\endgroup$
    – Matt
    Jan 6, 2012 at 23:08

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