Let $S=Spec A$ be the spectrum of an integrally closed, excellent and Noetherian ring. In his paper, David Mumford constructs an $S$-group scheme $G$. He shows that the torsion of $G_s$ is $p$-divisible for every prime $p$ and every $s\in S$ and concludes that $G_s$ is thus connected and without unipotent radical. I do not know how he comes to that conclusion. Is there something I don't know?
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asked Mar 13, 2022 at 16:12
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1$\begingroup$ Multiplication by $p$ on an additive group / vector group in characteristic $p$ is a zero map. So the additive group is not $p$-divisible in characteristic $p$. $\endgroup$– Jason StarrMar 13, 2022 at 16:44
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$\begingroup$ @JasonStarr So what? Excuse me if I am stupid. $\endgroup$– The Thin WhistlerMar 13, 2022 at 16:59
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$\begingroup$ I mean he shows that $G_s$ is of char $0$. What do I gain from this? $\endgroup$– The Thin WhistlerMar 13, 2022 at 17:28
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1$\begingroup$ I do not have access to Mumford's paper. What I can say is that $p$-divisibility rules out certain kinds of bad reduction. Without looking at Mumford's paper at all, one possibility is that Mumford uses the technique (also used elsewhere to prove theorems about Abelian varieties) of showing that the family of Abelian varieties is the base change of a family with similar properties defined over a finite type scheme over $\text{Spec}\ \mathbb{Z}$. If this family also satisfies $p$-divisibility then you are set: just reduce modulo maximal ideals having finite residue fields. $\endgroup$– Jason StarrMar 13, 2022 at 20:28
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$\begingroup$ That doesn't help at all. $\endgroup$– The Thin WhistlerMar 13, 2022 at 22:30
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