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.

isa 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