Uniqueness of finite flat models over bases of low ramification via Breuil-Kisin modules 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).
 A: 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 $\mathfrak{M}=\mathfrak{N}$.
