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Let $S = \operatorname{Spec} R$ an affine scheme (in our case latter a complete dvr) and $p$ a prime. Then Barsotti-Tate group or $p$-divisible group $G$ of height $h$ over $S$ is an inductive system

$$(G_n, i_n:G_n \to G_{n+1})$$

where $G_n$ are finite, locally free group schemes of order $p^{nh}$, such that all $i_n$ are closed embeddings with $G_n=G_{n+1}[p^n]$ (equality of sub-groupschemes).

I have a question about a proposition I found in my seminar notes (more precisely a strange reduction step there):

Assume, that for our BT group the base scheme $S= \operatorname{Spec} R$ is spectrum $R$ is a complete, discrete valuation ring. with alg. closed residue field $k=R/ \mathfrak{m}_R$. That's a meaningful reduction in light of the fpqc-descent. Denote it's fraction field by $F:=Frac(R)$.

Let $G_{n}$ an arbitrary member of the inductive system of the Barsotti-Tate group. Denote by $G_{\nu, F}:= G_{n} \times_R \operatorname{Spec} F$ the generic fiber. The proposition tells:

Proposition: The discriminant ideal $\mathcal{D}_{G_n/R} \subset \mathcal{O}_{G_n}$ of the $G_{n}$ depends only on $G_{n, F}$. (in the sense that all parameters determine the ideal sheaf can be extracted from $G_{n, F}$)

The proof begins with a statement I not understand:

It says that the claim above is trivial if $p \in R^{\times}$, ie $p$ invertible in $R$. So assume $p \in \mathfrak{m}_R$ [...]

That's exacly the question: Why is the statement of the Proposition in the case case $p \in R^{\times}$ trivial? Any idea why?

An alternative (or better "classical") proof can be found in J. Tate's paper $p$-Divisible Groups. But the concern of this question is only to understand the "triviality" above.

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    $\begingroup$ If $p \in R^\times$ then the discriminant ideal is trivial because all $G_n$ are etale. $\endgroup$
    – David Hansen
    Commented May 10, 2020 at 23:02
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    $\begingroup$ @DavidHansen: but why $p \in R^\times$ imply that $G_n$ is etale? $\endgroup$
    – user267839
    Commented May 10, 2020 at 23:10
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    $\begingroup$ In Tate's paper I mentioned above I found indeed that the discriminant ideal is generated by a power of $p$ (page 164, Prop. 2), but it's a non trivial result. SoI not see why the implication $p \in R^\times \Rightarrow G_n$ etale is "elementary". Could you sketch the proof or give a reference for the argument? $\endgroup$
    – user267839
    Commented May 10, 2020 at 23:14
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    $\begingroup$ @MortyPB. The following is a standard fact (found in most introductory references on group schemes): Any finite flat group scheme of $p$-power order is \'etale if $p$ is invertible on the base. To prove this, reduce to checking this over a field, and then use that multiplication by $p^N$ is simultaneously $0$ for $N \gg 0$ (by assumption) and unramified (as the map on tangent spaces is given by $p^N$, which is invertible), so the group scheme is geometrically reduced (and thus \'etale). $\endgroup$
    – Anonymous
    Commented May 10, 2020 at 23:51
  • $\begingroup$ Ok, so over a field the story follows from a decomposition into etale and connected part. Structure theorem tells about the connected part that it's of order coprime to $p$, so it's trivial by Lagrange. So over field it's fine. $\endgroup$
    – user267839
    Commented May 11, 2020 at 1:15

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