Let $R$ be a ring such that $p^nR=0$ for some integer $n$, and $G$ be a $p$-divisible group over $R$.

We think of a $p$-divisible groups as an fppf sheaf $G\colon \mathrm{Alg}^{op}_{R}\to \mathbf{Gps}$ such that

$1) \ G=\mathrm{colim} \ G[p^n]$,

$2) \ [p]\colon G \to G$ is surjective,

$3) \ G[p]$ is a finite, locally-free group scheme.

Question: Is $G$ represented by an $R$-formal scheme (with a finitely generated ideal of definition)?

The answer is positive if $G$ is 'etale or connected. More generally, the answer is positive if $G$ is isogenous to an extension of an 'etale $p$-divisible group by a connected $p$-divisible group (Lemma 3.3.1). In particular, this always holds if $R$ is a field.

However, Scholze and Weinstein write Section 3 of their paper as though there are examples of non-representable $p$-divisible groups but never provide an example of such a group. So, I guess the answer to the question above should be negative in general. But it will be nice to see a particular counter-example.

We can try to take $Y=\mathrm{Spec} \ R$ to be the affine modular curve over $\bar{\mathbf{F}}_p$ (with some full level structure) and $G$ the $p$-divisible group of the universal elliptic curve over $Y$. Then, presumably, $G$ should not be representable. But I don't know a rigorous way to prove it.


1 Answer 1


Your supposed example works indeed. More generally, I think whenever the étale part is not of locally constant height one will run into problems.

Here's a proof that the $p$-divisible group $G$ of the universal elliptic curve $E$ in characteristic $p$ (with auxiliary level structure) is not representable by a formal scheme. Assume it was, and look at an open affine subset $U\subset G$ containing a supersingular point (whose preimage in $G$ is still topologically just a point). Then $U$ necessarily contains the whole preimage of the generic point of the base $Y=\mathrm{Spec}\, R$, as this whole preimage specializes to the given supersingular point (by properness of all $G[p^n]$). But this preimage is not quasicompact, because the étale part of $G$ over the ordinary locus gives a decomposition into countably many connected components. On the other hand, $U$ being affine, it had to be quasicompact, giving a contradiction.


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