# Nontrivial p-divisible groups over $\mathbb Z$ for general prime $p$

In Tate's famous paper about $$p$$-divisible groups, for a prime number $$p$$ he asks whether there exists a $$p$$-divisible group $$G$$ over $$\mathbb Z$$ such that $$G$$ is not a direct sum of $$\mu_{p^\infty}$$ and $$\mathbb Q_p/ \mathbb Z_p$$, i.e a direct sum of a constant group and a diagonalizable group. He calls such $$p$$-divisible groups nontrival.

This question has some good applications. For instance we know there is no abelian scheme over $$\mathbb Z$$, whose key idea involves analyzing $$p$$-divisible groups for small $$p$$ using discriminant bound. In particular Tate's question is true for $$p=3,5,7,11,13,17$$ (see theorem $$4$$ in Fontaine's paper Il n'y a pas de variété abélienne sur $$\mathbb Z$$ for a proof).

For the negative side, the case $$p=2$$ is bad but also good up to isogeny. In an old paper $$p$$-divisible group over $$\mathbb Z$$ by V. A. Abrashkin, he claims that there also exist nontrivial $$p$$-divisible groups for some small irregular primes, but this seems to be wrong (see one answer below).

Odlyzko's discriminant bound is not good for large $$p$$. But time has passed for more than $$30$$ years, what about the general case now? Is there any progress towards Tate's question? With the technique of mordern $$p$$-adic Hodge theory, maybe such question could be solved. Indeed, some generalizations of Fontaine's result for abelian schemes rely on the Fontaine-Laffaille theory.

• It's a question about abstract groups? if so the current tags are irrelevant. The groups $\mathbf{Q}_p/\mathbf{Z}_p$ and $\mu_{p^\infty}$ are isomorphic, if $\mu_{p^\infty}$ means the group of roots of unity with some power of $p$. Obviously there are other $p$-divisible groups. Is it meant in which every element has order some power of $p$? in this case every such $p$-divisible group is indeed isomorphic to a (restricted) power of $\mu_\infty$. Introducing terminology would help clarify what the question is. – YCor Nov 21 '18 at 8:30
• @YCor. This is about $p$-divisible groups as in Tate's paper citeseerx.ist.psu.edu/viewdoc/… mentioned in the questions. In particular Galois acts on them. – Chris Wuthrich Nov 21 '18 at 9:44
• @ChrisWuthrich thanks: indeed it's an elaborated structure, not reflected by this lame (but now standard) choice of terminology. – YCor Nov 21 '18 at 10:04
• What is the bounty for? In particular, what do you want to know that has not been addressed in clever_answer_bot's answer? – S. Carnahan Jul 24 '19 at 16:04
• @S.Carnahan Because I am not sure whether there are some mistakes in that paper, and a recent question (torsion points and Mazur theorem) reminds me of this question. – sawdada Jul 25 '19 at 1:15

From works of V. A. Abrashkin, we know there exist nontrivial $$p$$-divisible groups for $$p=2$$ and some irregular primes.

This is not true, and Tate's conjecture is very much still expected to hold. That said, the only known general cases are those which proceed via discriminant bounds, which might just include $$p = 2$$ and $$p = 3$$ (and maybe $$p = 5$$ or $$7$$ i'm not sure).

Suppose that $$V$$ is a $$p$$-divisible group over $$\mathbf{Z}$$, and assume that the corresponding Galois representation is absolutely irreducible. If you could prove that $$V = \mathbf{Q}_p/\mathbf{Z}_p$$ or $$\mu_{p^{\infty}}$$ then you would know all $$p$$-divisible groups over $$\mathbf{Z}$$, because you know how to compute extensions between such objects (there are none if $$p > 2$$ and easy to understand when $$p = 2$$ (there is an extension killed by $$2$$ and split over $$\mathbf{Q}(\sqrt{-1})$$.)

The main thing that has changed since Tate's original conjecture is the Langlands program. In this context, $$V$$ conjecturally gives rise to a cuspidal automorphic form $$\pi$$ for $$\mathrm{GL}(n)/\mathbf{Q}$$ of level one. By the Fontaine-Mazur conjecture and the standard conjectures, it should come from a pure motive $$M$$ with Hodge Tate weights $$0$$ and $$1$$. Possibly $$M$$ might have coefficients, but by taking the direct sum of the conjugates of this compatible system we may assume that $$M$$ is a pure motive over $$\mathbf{Z}$$ (though maybe now no longer absolutely irreducible). If $$M$$ is of weight $$0$$, then $$M$$ is etale and then it is easy to see that $$M$$ comes from the trivial motive $$\mathbf{Z}$$, and $$V = \mathbf{Q}_p/\mathbf{Z}_p$$. Similarly, if $$M$$ has weight $$2$$, then it is Cartier dual to something etale, and then $$M$$ has to be $$\mathbf{Z}(1)$$ and $$V = \mu_{p^{\infty}}$$. So it remains to consider the case when $$M$$ has weight one. The infinity type of $$\pi$$ is then determined by the Hodge--Tate weights, since complex conjugation on the Hodge structure sends $$H^{1,0}$$ to $$H^{0,1}$$ and is thus completely determined. In particular, the $$L$$-function $$L(M,s) = L(\pi,s)$$ can then be shown not to exist precisely in the same way that Mestre proves that the $$L$$-function of an abelian variety $$A$$ over $$\mathbf{Z}$$ does not exist (the same argument works).

So the real "modern work" on this problem is the work of modularity in the Langlands program. For example, if $$V$$ is irreducible and is assumed to come from a $$2$$-dimensional representation (possibly with coefficients), then the work of Chandrashekar Khare (Serre's modularity conjecture: the level one case, DUKE) proves that $$V$$ does not exist.

I tried to look at Abrashkin's paper (in cyrilic) to see what was going on. This is just a guess, because I don't speak Russian. On page 1006 or so, there seems to be exact sequences of finite flat group schemes of the form:

$$0 \rightarrow (\mathbf{Z}/p^n \mathbf{Z}) \rightarrow \Gamma \rightarrow \mu_{p^n} \rightarrow 0$$

(or rather the generic fibre of such finite flat group schemes). For this to actually come from a non-split extension of finite flat group schemes, there would have to be a non-trivial extension of finite flat group schemes of $$\mu_p$$ by $$\mathbf{Z}/p \mathbf{Z}$$. Since, (by the connected-etale sequence) such sequence would locally split, this would give rise to an unramified extension of $$\mathbf{Q}(\zeta_p)$$ and imply that $$p$$ is irregular. However, the converse is not true. For the converse to apply, there would have to be an extension of a very specific form, namely a $$p$$-quotient of the class group with a precise action of the Galois group of $$\mathbf{Q}(\zeta_p)$$. Such things do not exist, by Herbrand's theorem. Already the lack of such extensions was proved by Mazur in his Eisenstein ideal paper (See proposition 2.1, page 49). But again I can't be sure if this is exactly what is going on in the cited paper.

tl;dr: This problem "reduces" to the Langlands conjecture. Our best people are currently working on it. If you want to do something useful, read Mazur's Eisenstein Ideal paper and then everything on the Langlands program in the last 50 years.

• May I ask for some references for the positive results? Thank you. – sawdada Nov 21 '18 at 23:04
• For $p = 2$ there is of course an extension of $\mathbf{Z}/2 \mathbf{Z}$ by $\mu_2$ as finite flat group schemes defined over $\mathbf{Q}(\sqrt{-1})$, which leads to a $2$-divisible group isogenous to $\mathbf{Q}_2/\mathbf{Z}_2 \oplus \mu_{2^{\infty}}$, but it is still generally conjectured that the only simple finite flat group schemes of $p$-power order are $\mu_p$ and $\mathbf{Z}/p \mathbf{Z}$, which certainly implies Tate's conjecture (up to the one isogeny above when $p=2$). – clever_answer_bot Nov 22 '18 at 3:37
• As I said in my post, the way to prove this is using discriminant bounds. Fontaine's proof that there are no abelian varieties over Z basically is through showing the only 3-divisible groups are the expected ones. – clever_answer_bot Nov 22 '18 at 3:37
• "p-DIVISIBLE GROUPS OVER Z" is at iopscience.iop.org/article/10.1070/IM1977v011n05ABEH001752, the section $3$ is short and includes the result of irregular primes. I am a beginner at this area, as there are some citations for those two papers and I am unable to find an Erratum... I apologize if there is any offense, if you like I can remove the words about V. A. Abrashkin's works. I just want to know more about this area, in particular some references for more recent works. – sawdada Nov 22 '18 at 3:53
• @ChrisWuthrich No. Vote it down if you want to. As the OP seems to have ignored my answer anyway - which answers the question in a complete a manner as anything available - I have no intention to waste any more effort here looking at this question again. – clever_answer_bot Nov 26 '18 at 4:39