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Let G be a smooth commutative $\mathbb{Z}_p$-group scheme of finite type and let $G_0$ be the $\mathbb{Q}_p$-fiber. We have an embedding $G(\mathbb{Z}_p)\subseteq G_0(\mathbb{Q}_p)$. My question is does every torsion point in $G_0(\mathbb{Q}_p)$ come from a torsion point in $G(\mathbb{Z}_p)$? I am mostly interested in the prime-to-p part of the torsion group. I think that the answer to this quesiton is yes, but I can't figure out how to prove it. Any ideas or references would be greatly appreciated.

Thanks in advance!

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If $G$ is quasi-finite but not finite, this can certainly fail: just throw out some closed points from the closed fiber. Do you want to assume that $G$ is separated and the Néron model of $G_0$? In that case, every $\mathb{Q}_p$ point of $G_0$ extends to a $\mathbb{Z}_p$ point of $G$, and torsion points extend to torsion points. – Jason Starr Jul 8 '11 at 13:29
[I don't understand that bit of your comment but I do understand the rest of it -- for example there is a smooth commutative group scheme whose generic fibre is $\mathbf{Z}/n\mathbf{Z}$ and whose special fibre is just the trivial group: it's the spectrum of $\mathbf{Z}_p\oplus\mathbf{Q}_p^{n-1}$.] – Kevin Buzzard Jul 8 '11 at 19:08
@Kevin: for any variety $X$ over (say) $\mathbb{Q}_p$, the notion of Neron model makes sense: it is a model over $\mathbb{Z}_p$ which satisfies the Neron mapping property. Unfortunately the Neron model of a group variety will not exist unless the component of the identity is semi-abelian. But Jason's comment still seems relevant: given that the answer in general is "no", this is one important case in which the answer is "yes". – Pete L. Clark Jul 8 '11 at 21:50
@Tzanko: (1) if you want to change a question (e.g. by adding new hypotheses) then, rather than just changing it, it might be better to also flag the changes (e.g. write "EDIT: and furthermore assume G is of finite type"). Either that or ask a new question. But more importantly (2) $\mathbf{Q}_p$ is of finite type over $\mathbf{Z}p$ (it's just an open immersion) so all the counterexamples still apply. – Kevin Buzzard Jul 10 '11 at 11:30
Tzanko Matev, "finite type" is a much weaker condition than "finite". In particular, the answer to your revised question is certainly "no". – S. Carnahan Jul 10 '11 at 14:55

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