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Let $K$ be a finite extension of $\mathbf{Q}_p$, with integer ring $R$ and residue field $k$. Say $G/R$ is a finite flat (commutative) group scheme of order $p^2$, killed by $p$. Say the special fibre of $G$ is isomorphic to the $p$-torsion in a supersingular elliptic curve over $k$. Is there some finite extension $L/K$ and an elliptic curve $E$ over $R_L$, the integers of $L$, such that $G$ becomes isomorphic to $E[p]$ over $R_L$?

The reason I ask is that in a lemma in Vincent Pilloni's thesis (which I unfortunately cannot find online) he proves that if the special fibre of $G$ is as above then the "degree" of such a $G$ is 1 (see Fargues' work on Harder-Narasimhan filtrations, for example, for the definition of "degree") via a brute force calculation. Pilloni's assertion would however also be a consequence of an affirmative answer to my question (and provided the motivation for my question). People are so good at deformation theory of finite flat group schemes nowadays that I thought this might be well-known to some people nowadays.

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Wouldn't that force G to be self-dual over L? Is that automatic? If I had to guess, I'd say the answer is "no", but I don't have a counterexample. –  Felipe Voloch Jan 6 '10 at 18:07
    
Yes you're right and so now I'm rather suspicious of my question. I realised I could reformulate the "degree" assertion in an elementary way: I think it's just the assertion that if G=Spec(A) then Disc(A/R) is (p^{p^2}). –  Kevin Buzzard Jan 6 '10 at 21:39
    
Does the argument you have in mind generalize at all? There are theorems of Raynaud about putting finite flat group schemes inside abelian varieties (as you no doubt know!). Also, one could imagine trying to apply Serre--Tate theory in some context; then the question becomes something like: (a) if we deform the p^2 torsion of a p-divisible group, when can we correspondingly deform the whole p-divisible group. Does Tong Liu's thesis shed any light on this at all? (This last question may be totally of base; I'm just throwing it out there!) –  Emerton Jan 7 '10 at 2:55
    
I'm deliberately being imprecise about the context in which my suggestions are to be placed, so as to get around (in the most vacuous sense of this) Felipe's objection! Whatever the answer, I am curious about the nature of the map from the deformation space of a p-divisible group to the deformation space of its p^n-torsion subschemes. (This is a way to rephrase your question without guessing as to the truth value of the answer.) –  Emerton Jan 7 '10 at 2:57
    
@Matt: Brian Conrad reminded me by email that the theorem is that a BT_n embeds in a p-divisible group, but the definition of a BT_1 is a bit subtle (see Illusie's article in that Asterisque on the Mordell conjecture) and he implied that he could see no reason why a condition on the special fibre would imply that G was a BT_1. In particular when you talk about p^n above, Brian's comment almost answers your question (for n>=2) whilst disproving mine (n=1)! –  Kevin Buzzard Jan 7 '10 at 10:48
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2 Answers 2

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La réponse est oui. La raison est la suivante. Si $S$ est un schéma sur p est localement nilpotent, par définition (cf. thèse de Messing, chapitre I), un groupe de Barsotti-Tate tronqué d'échelon $1$ sur $S$ est un schéma en groupes fini localement libre sur $S$ annulé par $p$ tel que si $G_0$ désigne la réduction de $G$ modulo $p$ on ait $$ Im (F_{G_0}) = \ker (V_{G_0}) $$ comme faisceaux fppf sur $S_0$, $S_0$ désignant la réduction modulo $p$ de $S$. Cependant, il résulte du critère de platitude fibre à fibre de EGA IV que le morphisme de $S_0$-schémas en groupes de présentation finie $$ F:G_0 \rightarrow \ker (V_{G_0}) $$ est fidèlement plat si et seulement si il en est de même fibre à fibre sur $S_0$. De cela on déduit que dans la définition d'un $BT_1$ on peut remplacer $S_0$ par $S_{red}$ !

Dans les considérations précédentes j'ai pris un schéma $S$ sur lequel $p$ est localement nilpotent, mais il en résulte que l'on a le même type de définition-résultat sur une base qui est un schéma formel $p$-adique.

Revenons maintenant à nos moutons, c'est à dire la question de Kevin. On a donc $G$ un schéma en groupes fini et plat sur l'anneau des entiers de $K$ dont la fibre spéciale sur le corps résiduel de $K$ est un $BT_1$. Le schéme en groupes $G$ est donc un $BT_1$.

Nous allons maintenant utiliser le théorème suivant (cf. l'article d'Illusie aux journées arithmétiques de Rennes: "Déformations de groupes de Barsotti-Tate (d'après A. Grothendieck)", Astérisque 127). Si $\mathcal{BT}_1$, resp. $\mathcal{BT}$, désigne le champ des groupes de Barsotti-Tate tronqués d'échelon $1$, resp. des groupes de Barsotti-Tate, sur des bases qui sont des schémas sur lesquels $p$ est localement nilpotent, alors le morphisme "points de $p$-torsion"

$$ \mathcal{BT} \longrightarrow \mathcal{BT}_1 $$

est formellement lisse. De cela on déduit la chose suivante. Soit $k$ un corps parfait de caractéristique $p$ et $H$ un groupe $p$-divisible sur $k$. Soit $\mathfrak{X}$ l'espace des déformations de $H$, un $spf (W(k))$-schéma formel non-canoniquement isomorphe à $spf \big (W(k)[[x_1,\dots,x_{d(h-d)}]]\big )$ où $h$ désigne la hauteur de $H$ et $d$ sa dimension. Soit $\mathcal{H}$ la déformation universelle sur $\mathfrak{X}$. Alors, $\mathcal{H}[p]$ est une déformation verselle de $H[p]$.

Pour conclure et répondre à la question de Kevin il suffit maintenant d'invoquer le théorème de Serre-Tate qui montre que si $H=E[p^\infty]$ où $E$ est une courbe elliptique supersingulière sur $k$ alors $\mathfrak{X}$ est également l'espace des déformation de $E$. Appliquant le théorème d'algébrisation de Grothendieck (GAGF) on en déduit que si $\mathfrak{X}= spf (R)$, la déformation universelle de la courbe elliptique $E$ sur $\mathfrak{X}$ provient en fait d'une courbe elliptique sur $spec (R)$. Le résultat s'en déduit par spécialisation sur $\mathcal{O}_K$.

P.S.: I read the rules for this forum: it is nowhere written the questions and answers should be written in english !!!!!!!!

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Hey Laurent! Thanks for the answer. I can see the heart of the problem is showing G is a BT_1. But this is a condition on G/(R/pR) rather than on the special fibre. I must confess I'm a bit surprised that you think we can check this on the special fibre. I'll have to think about what you say. –  Kevin Buzzard Jan 7 '10 at 13:23
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Thanks to one-click google translate extensions for browsers, French is not a serious problem! –  Anweshi Jan 7 '10 at 13:25
    
If ker(F:G_0-->G_0^(p)) were flat over S_0then I believe what you say. But I guess ker(F) might not be flat, right? –  Kevin Buzzard Jan 7 '10 at 13:26
    
PS @Anweshi: what does google translate say "corps parfait" means? Bet it's not "perfect field"! –  Kevin Buzzard Jan 7 '10 at 13:28
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Google Translate informs me that "Revenons maintenant à nos moutons, c'est à dire la question de Kevin" is best translated as "We now return to our sheep, ie the question of Kevin." –  Ben Linowitz Jan 7 '10 at 13:44
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This is a rough translation of Laurent F.'s answer (see this meta.MO thread). It's community wiki so you can improve the translation if you have 100 rep.


The answer is yes. The reason is as follows. If $S$ is a scheme where $p$ is locally nilpotent, then by definition (cf. Messing's thesis, Chapter I), a truncated Barsotti-Tate group of level of $1$ over $S$ is a finite locally free group scheme over $S$ annihilated by $p$ such that if $G_0$ denotes the reduction of $G$ modulo $p$ we have $$ Im (F_{G_0}) = \ker (V_{G_0}) $$ as fppf sheaves on $S_0$, $S_0$ denoting the reduction modulo $p$ of $S$. However, it follows from the fiberwise criterion for flatness of EGA IV [Ed: he means EGA IV cor 11.3.11] that the morphism of $S_0$-group schemes of finite presentation $$ F:G_0 \rightarrow \ker (V_{G_0}) $$ is faithfully flat if and only if it is fiberwise over $S_0$. From this we deduce that in the definition of a $BT_1$ one can replace $S_0$ by $S_{red}$!

In the above considerations I took a scheme $S$ where $p$ is locally nilpotent, it follows that we have the same type of definition-result over a base which is a $p$-adic formal scheme.

We now return to the subject, namely the question of Kevin. We have that $G$ is a finite flat group scheme over the ring of integers of $K$ whose special fiber over the residue field of $K$ is a $BT_1$. The group scheme $G$ is hence a $BT_1$. [Ed: in fact this statement already implies Pilloni's result, by other results in the above cited article of Illusie.]

We will now use the following theorem (cf. the article of Illusie in Journees Arithmetiques de Rennes: "Déformations de groupes de Barsotti-Tate (d'après A. Grothendieck)", Astérisque 127). If $\mathcal{BT}_1$, resp. $\mathcal{BT}$ denotes the stack of truncated Barsotti-Tate groups of level $1$, resp. of Barsotti-Tate groups, over bases that are schemes in which $p$ is locally nilpotent, then the morphism "points of $p$-torsion" $$ \mathcal{BT} \longrightarrow \mathcal{BT}_1 $$ is formally smooth. From this we deduce the following. Let $k$ be a perfect field of characteristic $p$ and $H$ a $p$-divisible group over $k$. Let $\mathfrak{X}$ be the space of deformations of $H$, a $spf (W(k))$-formal scheme non-canonically isomorphic to $spf \big(W(k)[[x_1,\dots,x_{d(h-d)}]]\big)$ where $h$ denotes the height of $H$ and $d$ dimension. Let $\mathcal{H}$ be the universal deformation of $\mathfrak{X}$. Then $\mathcal{H}[p]$ is a versal deformation of $H[p]$.

To conclude and answer the question of Kevin it suffices to invoke the theorem of Serre-Tate, which shows that if $H=E[p^\infty]$ where $E$ is a supersingular elliptic curve over $k$ then $\mathfrak{X}$ is the space of deformations of $E$. Applying the algebraization theorem of Grothendieck (GAGF) we deduce that if $\mathfrak{X}=spf(R)$, the universal deformation of the elliptic curve $E$ over $\mathfrak{X}$ actually becomes an elliptic curve over $spec(R)$. The result follows by specialization to $\mathcal{O}_K$.

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Thanks for the translation Matthew, that's perfect. In some sens I am a little bit disappointed, since I though writing a text in French would have lead to protestations from some people on the forum and thus have lead to some kind of polemic...doing this was some kind of provocation from my part...but nothing. –  Laurent F. Jan 8 '10 at 12:11
    
In fact the translation was by Anton; Kevin and I made some minor corrections. By the way, is the observation that you can check BT_1 on the special fibre due to you? No-one else seems to have realized it before. –  Emerton Jan 8 '10 at 19:14
    
@Laurent: you shouldn't be disappointed. The discussion you were aiming to create did happen on the meta discussion forum for this site, though perhaps it was not as heated as you might have hoped. The consensus is that it's fine to post in non-English languages, with the understanding that posts in English will appeal to more of the people who use the site. We also discussed ideas for producing translations to reduce the energy barrier that non-English posts erect for many people. I posted this translation as an experiment to see how well those ideas would work. –  Anton Geraschenko Jan 8 '10 at 20:08
    
@Anton. I imagine, the translation in future instances would be easier for common users by capturing the LaTeX from the original post. How do you do this? Please explain on meta. –  Anweshi Jan 8 '10 at 22:48
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