Does Corollary 3.4.4 in Raynaud's paper ``Schemas en Groupes de Type (p, ..., p)'' apply also to the case where G is quasi-finite? If not, what is the more general statement?

The corollary states:

``Soit G un K-schéma en groupes fini, commutatif, annulé par une puissance de p et qui se prolonge en un R-schéma en groupes fini et plat. Soit H un quotient de Jordan-Hôlder de G. Alors H est un schéma en F-vectoriels, pour un corps fini convenable F. Si F à $p^r$ éléments, le caractère $\psi : I_t \to F^*$, qui décrit l'action du groupe de Galois $Gal(\bar{K}/K)$ sur le F-vectoriel $G_i(K)$ est alors de la forme

$\psi = \psi_{i + 1}^{n_1} \cdots \psi_{i + r}^{n_r}$ avec $0 \leq n_j \leq e$ pour tout j.''

(Here, R is a strictly Henselian discrete valuation ring, K is its fraction field, char K = 0, and p is the residue characteristic of R.)

  • $\begingroup$ Can you clarify your question? Since $G$ is over $K$, which is a field, it must be finite if it's quasi-finite. $\endgroup$
    – JBorger
    Jan 7, 2011 at 21:23
  • $\begingroup$ I think Eric means that G extends to a quasi-finite flat group scheme over R. $\endgroup$ Jan 7, 2011 at 22:14
  • 2
    $\begingroup$ But if that's the case, then every $G$ has such a model. Just take $j_!G$, the extension by zero of $G$ to $\mathrm{Spec}(R)$ in the etale topology. It's represented by an etale group scheme over $R$, and its restriction to $K$ is $G$. (These are general facts, but in this case they're easy to visualize: you're just prolonging the identity section to $R$.) So surely it's not true, otherwise Raynaud wouldn't have assumed there is a finite flat model. $\endgroup$
    – JBorger
    Jan 8, 2011 at 0:57
  • $\begingroup$ (I was going to find a counterexample, but I couldn't understand the conclusion because of some typos. What is $G_i$? And whatever it is, you probably want $G_i(\bar{K})$ not $G_i(K)$.) $\endgroup$
    – JBorger
    Jan 8, 2011 at 0:58

1 Answer 1


As James Borger writes in his comments above, the answer is surely no. The restrictions on the characters appearing in the Galois representation attached to $G$ are being forced by the fact that $G$ has a finite flat model. If you throw that assumption away, you lose any control on the ramification of the Galois action on $G(\overline{K})$, and it can be whatever you like.

(The point is that any representation of $Gal(\overline{K}/K)$ on a $p$-power order abelian group, or on an $F$-vector space for some char. $p$ finite field $F$, is the group of $\overline{K}$-bar points of a finite group scheme, or $F$-vector space scheme, over $K$. But if the ramification conditions in the conclusion of Raynaud's theorem aren't met, then this group scheme won't have a finite flat prolongation over $R$.)

If you want to have a theory generalizing Raynaud's theory of finite flat group schemes which applies to other representations of $Gal(\overline{K}/K)$ (i.e. ones that don't come from finite flat group schemes over $R$) then you will need to learn integral $p$-adic Hodge theory, which is a highly developed theory. The first step is to learn something about Fontaine--Lafaille theory (which deals with the case when $e = 1$). The step after this is to learn about Breuil modules and/or Kisin modules. But all of this is rather technical and involved, and a serious investement to learn. So you might be better off explaining your motivation and goals, so that people can give more specific and appropriate advice.

  • $\begingroup$ Thanks for your response. Fortunately, I was able to find a discussion of the case I was interested in --- where it comes from the p-torsion points of an abelian variety with semistable reduction --- in SGA 7 expose 9 (in which case it is true). $\endgroup$ Jan 8, 2011 at 21:13

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