I'm currently working through Tate's paper on $p$-divisible groups and have come to an impasse in his discussion of tangent spaces of $p$-divisible groups. I'd like to show that for an abelian variety $A$ over $\mathbb{Z}_p$, the $p$-divisible group associated to its $p^n$ torsion, $A[p^\infty]$, has tangent space at the identity (i.e. derivations $A^{\circ}[p^\infty] \to \mathbb{Q}_p$, where $A^{\circ}[p^\infty]$ is the formal group law associated to the connected part of $A[p^\infty]$) isomorphic to the tangent space of $A$ at the identity. This seems nontrivial and isn't something that Tate proves in his paper. Is there a reference where a result like this is proven? If so, what is it, and what is the idea of the proof? Proving it seems to require some formal scheme theory that I do not know.
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3$\begingroup$ It might help to observe that there is a formal group law $\hat{A}$ associated to $A$, whose underlying space is the formal completion of $A$ at the identity (and hence has the same tangent space as $A$); there is a very concrete and hands-on account of this for elliptic curves in Silverman's books, and the generalisation to abelian varieties is straightforward. Once you have this, the formal group associated to $A[p^\infty]$ is just $\hat{A}$. $\endgroup$David Loeffler– David Loeffler2025-11-12 08:17:25 +00:00Commented Nov 12 at 8:17
1 Answer
This is also fine over any base scheme $S$ on which $p$ is locally nilpotent (for your version, you can work over $\operatorname{Spec} (\mathbb{Z}_p / p^r \mathbb{Z}_p)$ and send $r \rightarrow \infty$).
There is a canonical inclusion $\operatorname{Lie} A[p^{\infty}] \rightarrow \operatorname{Lie}A$ of finite locally free sheaves, and one can check that this is a surjection after pullback to any field-valued point of the base (then apply Nakayama's lemma): over a field of characteristic $p$, any finite order thickening of the identity point on an abelian variety will be killed by $p^N$ for $N \gg 0$, because the finite order thickening will in fact be killed after applying relative Frobenius sufficiently many times, and relative Frobenius is an isogeny of $p$-power degree (and/or consider Verschiebung).
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Edit:
Here's an easier (?) reason. Suppose $A \rightarrow S$ is an abelian scheme where $S$ is a scheme over $\operatorname{Spec} \mathbb{Z} / p^r \mathbb{Z}$. Since multiplication by $N$ on $A$ acts as multiplication by $N$ on $\operatorname{Lie} A$, we conclude that every morphism $S[\epsilon] \rightarrow A$ lifting the identity section must factor through $A[p^r]$. This shows that $\operatorname{Lie} A[p^{\infty}] = \operatorname{Lie} A[p^r] = \operatorname{Lie} A$.
Here $S[\epsilon] := S \times_{\operatorname{Spec} \mathbb{Z}} \operatorname{Spec} \mathbb{Z}[\epsilon]/(\epsilon^2)$.
