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Dec 4, 2017 at 13:50 answer added nfdc23 timeline score: 5
Dec 4, 2017 at 6:47 comment added David Loeffler I think this question is quite sensitive to the exact meaning of the word "model". You want to find some scheme $\mathcal{G} / \mathcal{O}_{F, S}$ such that $\mathcal{G} \times F = G$. But do you require $\mathcal{G}$ to be a reductive group scheme, an arbitrary group scheme, or a totally arbitrary scheme?
Dec 3, 2017 at 21:18 comment added Daniel Loughran @CoffeeBliss: Your changes have not solved the problem; the answer to your question is still no, as the example of user94041 still stands.
Dec 3, 2017 at 21:14 comment added Not a grad student @PeterMcNamara For almost all places $G \times_F F_v$ is quasisplit and is split over an unramified extension. But I am wondering what happens if $S$ is smaller than that set.
Dec 3, 2017 at 21:13 comment added Not a grad student @user94041 I edited my question to include conditions on $S$.
Dec 3, 2017 at 20:00 history edited Not a grad student CC BY-SA 3.0
added 34 characters in body
Dec 3, 2017 at 19:35 comment added Peter McNamara If you allow yourself the possibility of adding a finite number of places to S, then the answer becomes yes.
Dec 3, 2017 at 18:01 comment added user94041 What if I take $\mathrm G$ to be the norm 1 units of a quadratic extension $E / F$? It seems that I can write down the coordinate ring of $\mathrm G$ over the ring of integers of $F$, so I can take $S$ to be empty. But at places $v$ where $E/F$ is ramified, you will need a ramified extension to split the torus.
Dec 3, 2017 at 17:25 comment added YCor "Suppose there is a set $S$ of places for which $G$ has a model over the ring of $S$-integers in $F$" is an empty condition. I reformulated accordingly.
Dec 3, 2017 at 17:24 history edited YCor CC BY-SA 3.0
clarified
Dec 3, 2017 at 17:19 history asked Not a grad student CC BY-SA 3.0