I am forced to know the etale fundamental group of the grassmannian over the rational field. I searched it but couldn't find any hint. I am wondering whether there are some positive results or recipe to compute it.
Thank you!



Etale $\pi_1$ of Grassmannian
The answer is that any Grassmannian is geometrically simply connected, so the etale fundamental group over $\mathbb{Q}$ is simply [**edit**: !!] the absolute Galois group $\operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ of $\mathbb{Q}$.


In more detail: let $X$ be a geometrically integral variety defined over $\mathbb{Q}$, let 
$\overline{X}$ be its basechange to an algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$, and let $\mathfrak{g}\_{\mathbb{Q}} = \operatorname{Aut}(\overline{\mathbb{Q}}/\mathbb{Q})$ be the absolute Galois group of $\mathbb{Q}$. 


1) Via a choice of geometric points (which we suppress), we get a short exact sequence of profinite groups


$1 \rightarrow \pi\_1(\overline{X}) \rightarrow \pi\_1(X) \rightarrow \mathfrak{g}\_{\mathbb{Q}} \rightarrow 1$. 


2) Let $K$ be an algebraically closed field of characteristic $0$. Suppose that *either* $X$ is complete or that $X$ is nonsingular [in our application, both hold]. Then the natural map
$\pi\_1(\overline{X}) \rightarrow \pi\_1(\overline{X} \otimes K)$ is an isomorphism. 


[Comment: if instead of $\mathbb{Q}$, our ground field was a field $k$ of positive characteristic and $K$ is an algebraically closed field containing $\overline{k}$, this map is still an isomorphism for complete varieties but not necessarily for all smooth varieties and indeed, not even for the affine line!]


3) Take $K = \mathbb{C}$. Then the analytification functor induces an isomorphism from the 
profinite completion of the topological fundamental group of $X(\mathbb{C})$ (with the $\mathbb{C}$-analytic topology) to the etale fundamental group $\pi\_1(X \otimes \mathbb{C})$. 


4) If $X/\mathbb{Q}$ is any Grassmannian, then it is nonsingular, complete and its analytification is the usual complex Grassmannian, which is simply connected: see e.g.


<http://books.google.com/books?id=WHjO9K6xEm4C&pg=PA748&lpg=PA748&dq=simply+connected+Grassmannian&source=bl&ots=waYTv_whVx&sig=ErlPHYKL5FdQPUdBlIrIgYQGhbE&hl=en&ei=9FiITJuNNYjW9ASP_p3fDg&sa=X&oi=book_result&ct=result&resnum=6&ved=0CC8Q6AEwBTgK#v=onepage&q&f=false>


Therefore putting the previous parts together the geometric fundamental group $\pi\_1(\overline{X})$ is trivial, so we naturally have $\pi\_1(X) \cong \mathfrak{g}\_{\mathbb{Q}}$.


References for these facts may be found, for instance, in Chapter 5 of Szamuely's *Galois Groups and Fundamental Groups*. 


