Geometric realization of $B{\mathbb G}\_{\mathfrak m}({\mathbb C})$ is ${\mathbb C}{\mathbb P}^\infty=\varinjlim\_n~ {\mathbb C}{\mathbb P}^n\_k$; what if one considers a separable field $k\neq {\overline k}$? Sheaf-theoretically, $B{\mathbb G}\_{\mathfrak m}$ represents the simplicial sheaf $B{\mbox{hom}}\_{k}(-,{\mathbb G}\_{\mathfrak m})$ on the big etale site $Sch\_k$; should I think of the analogue for ${\mathbb C}{\mathbb P}^\infty$ as $\varinjlim\_n~ {\mbox{hom}}\_{k}(-,{\mathbb P}^n\_k)$?


Context: calculating etale cohomology of $B{\mathbb G}\_{\mathfrak m}(k)$.


Note: For separably closed field $k$, theorem of Friedlander-Parshall says that $H^\ast\_{et}(BG(k),{\mathbb Z}/m)\cong H^\ast\_{et}(BG({\mathbb C}),{\mathbb Z}/m)$ for complex reductive group scheme $G({\mathbb C})$.



infinite grassmannian in algebraic geometry
As far as I understand, yes, you can look at it this way. However, at least to me, this seems like a strange way of doing it, and there are a few problems with it:


* Basically, you are saying
that the cohomology of ${\mathbb P}^\infty$ and $B(\mathbb{Gm})$ coincide because
the fiber (which is ${\mathbb A}^\infty-\{0\}$) is contractible. If you use something other than the `usual' cohomology, you need to justify the cohomological triviality of ${\mathbb A}^\infty-\{0\}$.
* Before you even begin doing this, you must make sure that the cohomology theory you work with even allows ind-objects like ${\mathbb P}^\infty$ and ${\mathbb A}^\infty$.
* When you are done, the argument will be tied to the specific group $\mathbb{Gm}$. What if you put a different reductive group there? Say, an exceptional one?


So it may be better to compute the cohomology of $B(\mathbb{Gm})$ directly, and view the claim about comparison with $\mathbb{P}^\infty$ as a bonus.


