In the search for a Weil cohomology theory $H$ over a field $K$ (with $\text{char}(K)=0$) for varieties in characteristic $p$, a classical argument by Serre shows that the coefficient field cannot be a subfield of $\mathbb{R}$ or of $\mathbb{Q}_p$; an obvious choice is to take $\mathbb{Q}_\ell$ for a prime $\ell \neq p$.
Now, we can try to make a Weil cohomology theory by taking the sheaf cohomology with constant sheaves with the Zariski topology, but this does not work as all cohomology vanishes.
Grothendieck's insight was that we can find a different topology, for example the étale topology. Then we can build a Weil cohomology theory with coefficients in $\mathbb{Q}_\ell$ by taking cohomology with coefficients in the constant sheaves $\mathbb{Z}/ \ell^n\mathbb{Z}$ and then taking the inverse limit with respect to $n$ and tensor with $\mathbb{Q}_\ell$: this gives $\ell$-adic cohomology.
But it is not so clear to me why the étale topology is best suited at this task. What happens if we repeat the above procedure on other sites? Does the cohomology theory we get fail to be a Weil cohomology theory?
P.S.: Information for fields other than $\mathbb{Q}_\ell$ would also be nice!
Best Answer
As observed, the Zariski topology has too few open subsets to compute cohomology with constant coefficients. It had already been observed by Serre that etale covers were enough to trivialize principal bundles for many algebraic groups. That suggested using etale covers. The etale "topology" is the coarsest for which the inverse function theorem holds. The flat topology also gives Weil cohomologies (the same ones as the etale topology), but why use flat covers when etale covers are enough.