Let $F$ be a number field, $p\in\mathbb{Z}$ a prime which is unramified in $F$ and $G$ a connected reductive group over $F$. Moreover $G$ is supposed to be quasi-split over $p$. 


Does there exist a finite set of primes $S\subset\mathbb{Z}$ which does not contain $p$ and a split reductive group $\mathcal{G}$ defined over $\mathbb{Z}[1/S]$ such that for some finite extension $K/F$ which is not ramified over $p$ we have the following isomorphism:


$$ \mathcal{G}\times\_{\mathbb{Z}[1/S]}K\cong G\times\_FK.$$


More generally, given a reductive group $G/F$, does there exist some sort of split integral model 
$\mathcal{G}$ defined over rings like $\mathbb{Z}$, $\mathbb{Z}[1/S]$ or $\mathcal{O}\_F$?



split integral model of a reductive group
If I understand the question correctly, the answer is no, but I think it can be changed to yes if the word quasi-split is replaced by unramified.


Let E be a quadratic extension of the rational numbers ramified at p. Let G be the special unitary group associated to the quadratic extension E/Q. This is a quasi-split group over Q. It does not split over any extension of Q unramified at p.


In general, a reductive group over a number field always has a reductive integral model over a ring of S-integers by a spreading out argument, since reductivity is an open condition. One cannot however ensure that this integral model is split. Groups in general are split only etale-locally. So you can make your group split over a ring of S-integers in some field extension, but in order to control the ramification, you need a stronger condition than quasi-split. 


