Why define étale cohomological dimension as it is defined? I am learning étale cohomology, and we defined the étale cohomological dimension of a scheme $X$ as the minimum of $n$ where $H_{ét} ^n (X,F)$ vanishes for all the torsion sheaves $F$, yet I don't get why we only restrict our view to torsion sheaves.
This seems to be a very stupid question, but I really have no idea why we only talk about torsion sheaves as while discuss étale cohomological dimensions. Is there some kind of fundamental reason or intuition behind this?
This may be too elementary, so if it is, please feel free to tell me.
 A: I think it's a combination of two things:
First, we do etale cohomology with torsion sheaves only, because of pathologies with non-torsion sheaves, and so we care about the cohomological dimension with torsion sheaves only.
Second, the cohomological dimension is actually different if we include nontorsion sheaves. Thus we lose information about torsion sheaves if we include nontorsion sheaves.
Both of these can be demonstrated already with the multiplicative group $\mathbb G_m$. We expect $H^1$ to be one-dimensional by analogy with classical cohomology, but it is easy to calculate $H^1_{et}(\mathbb G_m, \mathbb Z)=0$ by the relationship with the fundamental group, say. So the nontorsion cohomology is bad for $\mathbb G_m$.
Next consider the exact sequence $0 \to \mathbb Z \to \mathbb Z \to \mathbb Z/n\mathbb Z \to 0$, which by the previously mentioned vanishing induces an exact sequence $0 \to H^1(\mathbb G_m, \mathbb Z/n\mathbb Z) \to H^2(\mathbb G_m, \mathbb Z) \to H^2(\mathbb G_m, \mathbb Z)$ which, because $H^1(\mathbb G_m , \mathbb Z/n)$ is non-trivial, implies $H^2(\mathbb G_m, \mathbb Z)$ is nontrivial, and hence the cohomlogical dimension allowing nontorsion sheaves is at least $2$, while the cohomological dimension with torsion sheaves only is $1$.
Because there is a distinction, we must pick either torsion or nontorsion, and we pick torsion for the better analogies with singular cohomology.
