4
$\begingroup$

Let $R$ be a discrete valuation ring, $K$ its field of fractions, and $$ 0 \rightarrow G_K' \rightarrow G_K \rightarrow G_K'' \rightarrow 0$$ a short exact sequence of smooth $K$-group schemes of finite type. If $G_K'$ and $G_K''$ admit Neron models over $R$, does $G_K$ admit one, too?

A positive answer is claimed in 7.5/1 (b) of "Neron models" by Bosch, Lutkebohmert, and Raynaud, but the proof given there only works in the commutative case, since 10.2/1, on which the proof crucially relies, only concerns commutative groups. I therefore wonder if the statement is nevertheless also true for noncommutative groups?

I recall that a smooth $K$-group scheme $H_K$ of finite type admits a Neron model if $H_K$ extends to a smooth and separated $R$-group scheme $H$ of finite type for which the pullback map $H(S) \rightarrow H_K(S_K)$ is bijective for every smooth $R$-scheme $S$.

$\endgroup$
3
  • 1
    $\begingroup$ If I'm not mistaken, the necessary "compactification" resuts used in Thm 10.2.1 are now known for not necessarily commutative group schemes by the work of Gabber. Try googling Gabber's compactification theorem to see if you get something useful. $\endgroup$ Dec 21, 2014 at 10:50
  • 1
    $\begingroup$ On the other hand, the implications a implies b and a implies e of Thm 10.2.1 hold in general (as you probably already know). There might be a way to circumvent the compactification results and to prove that a group scheme not containing any copy of G_a or G_m is "bounded" directly (but I doubt it...). $\endgroup$ Dec 21, 2014 at 10:52
  • $\begingroup$ Thank you for your helpful comments. I'll look into this. $\endgroup$ Dec 21, 2014 at 17:51

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.