# Cohomological dimension-doubling

I'm sure this is a question which has been asked many times, if not necessarily on this site:

Why does a (smooth, projective) scheme over a field, with dimension d, behave as though it were a complex manifold of real dimension 2d?

I am pretty ignorant of most details, so this can be taken to represent the underlying question: what is the "reason" that etale cohomology has dimension 2d? I am aware of the comparison theorem with singular cohomology but (being ignorant) I do not know if the proof contains something of an answer to my question or if it just "is". One thing which seems unlikely is that the comparison theorem holds (or can even be stated) for schemes in arbitrary characteristic, where the notion of the associated complex variety is undefined. So this is not really a reason.

I have also been told that the dimensional properties of etale cohomology are consequences of the same facts for motivic cohomology. Of course, this is not an explanation either. The real mystery seems to be: how does cohomology know that $\mathbb{C}/\mathbb{R}$ has degree 2, and how does it know that we had that in mind when we were working with schemes over finite fields?

Possibly, this is related to the fact (?) that any finite extension of fields of which the larger is algebraically closed must have degree 2. Since I only mention this because it is the single natural occurrence of the number 2 that leaps to mind, it is in fact grasping at straws.

I would appreciate any suggestions towards improving this question.

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You want sep. closed base field, can use any sep'td scheme of finite type. It's a game in curve fibrations (via Bertini) and reducing to case of curves. So answer is Leray + Tsen's theorem + cohom. dim. of fn flds of curves over alg. closed flds. The real closed fld stuff has no relevance. There are many good refs; look at the pfs. And there is a comparison thm in any char.: Berkovich analytification over any non-arch. fld. That has a "2" also, for a different and fantastic reason (related to Galois cohomology and Leray in another way); must be many in Boston who can explain that to you. –  BCnrd Jul 5 '10 at 6:20
A heuristic version of BCnrd's answer is that etale = Zariski + Galois. The Zariski cohomological dimension is equal to the geometric dimension for standard reasons. Tsen's theorem says that the Galois cohomological dimension of a geometric field is its transcendence degree. I'm not sure how much you care about etale, except as a tool to motives. But you should expect motivic stuff to reduce to curves, as in etale proofs, so a key step to understand the motivic dimension of curves. –  Ben Wieland Jul 5 '10 at 6:57
Your "fact (?)" is in fact a "fact (!)"; it is a theorem due to Emil Artin and Schreier. Here is a note on it by Keith Conrad: math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf –  Dan Petersen Jul 5 '10 at 13:31

I'm not sure I can give you a morally satisfying answer. To my mind, this sort of theorem should be true because it works over $\mathbb{C}$, for topological reasons. Of course, that isn't a proof over other fields. But, to my limited understanding, the intuition comes from $\mathbb{C}$ and the proofs are motivated by taking proofs which work over $\mathbb{C}$ and seeing whether we can generalize them to an arbitrary field.

That said, I can sketch two proofs of this theorem. The first proof is by induction on dimension. Write $X$ as a family over $Y$, with fibers of dimension $\dim X - \dim Y$. A sectral sequence shows that, if the cohomology of $Y$ vanishes is in degree $>i$ and the cohomology of the fibers vanishes in dimension $>j$, then the cohomology of $X$ vanishes in degree $>i+j$. This reduces us to showing that curves have no cohomology above $H^2$, which can be done by hand; I think this is in Chapter 14 of Milne's lectures.

The other method is for deRham-like cohomology theories (rigid, crystalline, etc.). Roughly speaking, those methods compute cohomology as the hypercohomology of a complex $\Omega^0 \to \Omega^1 \to \Omega^2 \to \cdots$. If your ground field is characteristic zero, and your variety is smooth, these are actually the familiar sheaves of differentials. If one of these two conditions fails, you have to adjust in some manner; the details of this adjustment describe which of the theories you are working in. In any case (very roughly speaking) the $2n$ appears here as $n+n$: the complex has length $n$ and each of the terms in the complex has no cohomology above degree $n$. Again, a spectral sequence finishes the proof from here.

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Amusingly, for Berkovich space etale cohomology (with no deRham aspect!) "$2n = n+n$" occurs, as follows. The morphism from etale topos to underlying "topological" topos of the Berkovich space leads to a spectral sequence for the etale cohomology involving topological cohomology of sheaves whose stalks are Galois cohomology of the fields at the various points of the space. (This uses the henselian property of the local rings, so doesn't work for schemes.) Berkovich shows analytic dimension $n$ equals topological dimension and is an upper bound on Galois cohomological dimension. Voila. "QED" –  BCnrd Jul 5 '10 at 13:52
Interesting! Thanks, BCnrd. –  David Speyer Jul 5 '10 at 13:54
Two proto-cohomological ways in which a smooth curve X/k resembles a topological surface: if you remove finitely many k-points from X, X remains connected. And there exists a finite map Y --> X from a smooth Y which is ramified over and unramified away from some points of X. For a manifold, the first property implies that a point has codimension at least 2 and the second property implies that a point has codimension at most 2. –  David Treumann Jul 5 '10 at 15:38

Re the fact (?) that any finite extension of fields of which the larger is algebraically closed must have degree 2: think about the algebraic closure of a finite field. We know its automorphism group, which has the structure of the product of the additive groups of the $p$-adic integers for all primes $p$. We don't know the factoid that all its open subgroups have index 2, since that is far from being true. There is an obvious subgroup of index 3, and its fixed field would be a counterexample?

Edit: not a counterexample through misapprehension, see comments.

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"the fact (?) that any finite extension of fields of which the larger is algebraically closed must have degree 2." This fact is true; this is a theorem of Artin and Schrier mathoverflow.net/questions/8756/… . A counterexample would require an element of order $3$ (of which there are none), not a subgroup of index $3$. –  David Speyer Jul 5 '10 at 13:24