Let $R$ be a complete DVR of mixed characteristic $(0, p)$, let $K$ be its fraction field, and assume that the absolute ramification index $e$ of $R$ satisfies $e < p - 1$ and that the residue field of $R$ is perfect. Let $G$ be a commutative finite $K$-group scheme of $p$-power order.

It is a classical result of Raynaud that $G$ admits at most one extension to a finite flat $R$-group scheme $\mathcal{G}$. By now, all finite flat commutative $R$-group schemes of $p$-power order have been classified in terms of Breuil-Kisin modules. Can one give a simple reproof of Raynaud's result using this classification?

Such a reproof must be well known to the right people. Moreover, it must be an illuminating example of the Breuil-Kisin module theory, so I would be grateful if someone could record the argument here (or point to an article that does this).

  • $\begingroup$ Any such reproof would rest on a vastly more difficult foundation than Raynaud's proof, so it wouldn't be "simple" in the end compared to Raynaud's proof (but would give another perspective, to be sure). That isn't to say it isn't an instructive example to think through, but (as you may know) the real significance of the B-K formalism in practice is with the cases $e \ge p$. Have you tried to do it yourself using the description of Raynaud's "$F$-vector schemes of rank 1" via Breuil modules in section 2 of math.jhu.edu/~savitt/papers/pdfs/appendix-final.pdf? $\endgroup$
    – nfdc23
    Feb 27, 2017 at 6:04
  • $\begingroup$ Thank you for your comment. I've looked at the paper you mention, but it wasn't clear to me how to use it to answer the question (especially, since the paper deals with the formalism of Breuil modules, not Breuil-Kisin modules, although, I presume, the distinction is minor for the purposes of my question). $\endgroup$
    – Lisa S.
    Feb 28, 2017 at 0:39
  • $\begingroup$ Have you tried writing to the author of that paper? He should be able to provide a more apt reference (if one exists) or argument. $\endgroup$
    – nfdc23
    Feb 28, 2017 at 22:59

1 Answer 1


Yes, I think you can extract Raynaud's result straightforwardly from the Breuil--Kisin theory. Write $k$ for the residue field of $R$. Let's recall how the Breuil--Kisin theory works. Let $\phi : k[[u]] \to k[[u]]$ be the $p$th power map. Then there's an anti-equivalence between the category of $p$-torsion finite flat group schemes over $R$, and the category of finite rank free $k[[u]]$-modules $\mathfrak{M}$ equipped with a $\phi$-semilinear map $\varphi : \mathfrak{M} \to \mathfrak{M}$ such that the image of the ($k[[u]]$-linear) map $\varphi^* : \varphi^* \mathfrak{M} := k[[u]] \otimes_{\phi, k[[u]]} \mathfrak{M} \to \mathfrak{M}$ contains $u^e \mathfrak{M}$. Finally let's recall that the operation of passing to the generic fibre of the group scheme is recovered on the Breuil--Kisin module side by passing from $\mathfrak{M}$ to the étale $\phi$-module $\mathfrak{M}[1/u]$.

So, suppose you have Breuil--Kisin modules $\mathfrak{M} \supset \mathfrak{N}$ with $\mathfrak{M}[1/u] = \mathfrak{N}[1/u]$ (one can always reduce to this case). Choose the least integer $r \ge 0$ such that $\mathfrak{N} \supset u^r \mathfrak{M}$, and take $m \in \mathfrak{M}$ with $n := u^r m \in \mathfrak{N}, u^{r-1} m \not\in \mathfrak{N}$. Since $\varphi(m) \in \mathfrak{M}$ we in particular have $u^r \varphi(m) \in \mathfrak{N}$, and so we require $u^{e+r}\varphi(m)$ to be in the image of $\varphi^*\mathfrak{N}$ under the map $\varphi^*$.

But $u^{e+r}\varphi(m) = \varphi^*(u^{e+r} \otimes_\phi m) = \varphi^*(u^{e-(p-1)r} \otimes_\phi n)$. Since $\varphi^*$ is an isomorphism after inverting $u$, we deduce that $u^{e-(p-1)r} \otimes_\phi n$ must lie in $\varphi^*\mathfrak{N}$, and therefore $e \ge (p-1)r$. If in particular $e < p-1$ then $r=0$ and $\mathfrak{M}=\mathfrak{N}$.


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