Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.)

share|improve this question
    
Can you clarify your question? Since $G$ is over $K$, which is a field, it must be finite if it's quasi-finite. –  JBorger Jan 7 '11 at 21:23
    
I think Eric means that G extends to a quasi-finite flat group scheme over R. –  David Zureick-Brown Jan 7 '11 at 22:14
2  
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. –  JBorger Jan 8 '11 at 0:57
    
(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 $G_i(K)$.) –  JBorger Jan 8 '11 at 0:58
add comment

1 Answer

up vote 6 down vote accepted

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.

share|improve this answer
    
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). –  Eric Larson Jan 8 '11 at 21:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.