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.
 A: 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 spectral 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.
A: 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.
