Relation between rational Tate module and universal cover of a p-divisible group

Question

We can associate two $\mathbb Q_p$ vector spaces to a $p$-divisible group, and I'm a little confused about the relation between these two groups. First of all, I think part of my problem is that when papers discuss these things, they claim that if $R$ is $p$-complete, sending $x\mapsto p^nx$ is topologically nilpotent on $G(R)$, in particular, it is nilpotent if $p^nR=0$. I don't understand why this is true and I would appreciate an explanation or a reference.

But if we accept the claim it is easy to see that for any connected $R$ such that $p^nR=0$, we have: $$\tilde{G}(R):=\lim_{p:G\to G}(G(R))=\operatorname{Hom}(\mathbb{Q}_p/\mathbb{Z}_p,G(R))[1/p].$$ On the other hand one has $$T_G(R):=\lim(G[n](R))=\operatorname{Hom}(\mathbb{Q}_p/\mathbb{Z}_p,G(R)).$$ So it seems to me that if $p^nR=0$, then $\tilde{G}(R)$ is just the rational Tate module, and they are different only in mixed characteristic. Is my understanding correct?

Answer 1

Let $G$ be a $p$-divisible group over any base $S$. In terms of the functor of points we have for any affine scheme $\mathrm{Spec}(R)$ that $G(R):=\mathrm{colim}_n G[p^n](R)$.

Regarding the universal cover etc, let $C(G)=\lim_p G$ and $T(G)=\lim_{p} G[p^n]$. Then you have a short exact sequence (of fpqc sheaves say): $$0\to T(G)\to C(G)\to G\to 0.$$ Then $\mathrm{colim}_p G=\mathrm{colim}_n \mathrm{colim}_p G[p^n]=0$ hence taking the colimit of this short exact sequence under multiplication by $p$ yields an isomorphism $\mathrm{colim}_p T(G)=T(G)\otimes_{\mathbf{Z}_p} \mathbf{Q}_p \stackrel{\sim}{\to} C(G)$.

Regarding topological nilpotence of $p$ on $G$, if $R=\lim_n R/p^n$ is complete then $G(R)=\lim_n G(R/p^n)=\lim_n \mathrm{colim}_m G[p^m](R/p^n)$ and the topology on $G(R)$ is the inverse limit topology where $G(R/p^n)$ has the discrete topology. But $G(R/p^n)=\mathrm{colim} G[p^m](R/p^n)$ is $p$-power torsion and so $p$ is nilpotent on $G(R/p^n)$ and topologically nilpotent on $G(R)=\lim_n G(R/p^n)$.