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Let $R$ be a commutative ring, $p >0$ prime and $G$ a finite, locally free group scheme over $R$ of rank $p^n$; $n \in \mathbb{N}_{\ge 1}$. Assume $p \in R^*$ (i.e. is a unit in $R$).

Question: Why this condition on the rank implies that $G$ is étale?

By definition etale is equivalent to flat & unramified. As $G$ is locally free it's obviously flat. Be unramified is also a local condition. Thus we can translate the problem to commutative algebra and asking why the free $R$-module $R^{p^n}$ is unramified at a prime $\mathfrak{q} \subset R$ if $p \in R^*$.

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This is not so easy, but relies on a well-known structure theorem for connected group schemes over a perfect field.

Lemma 1. A finitely presented morphism $Y \to X$ of schemes is unramified if and only if $Y_x \to x$ is unramified for all $x \in X$.

Proof. See [EGA IV$_4$, Cor. 17.4.2]. $\square$

Thus, we may reduce to the case $R = k$ for $k$ a field, and by flat descent to the case where $k$ is algebraically closed. Then unramified for a finite extension means (geometrically) reduced.

Lemma 2. A finite group scheme over an algebraically closed field is a semidirect product of an étale group scheme and a connected group scheme.

Proof. See for example [Wat, §6.8]. The étale part is $\pi_0(G)$ and the connected part is $G^0$. $\square$

Theorem. If $G$ is a geometrically connected finite group scheme over a perfect field $k$ of characteristic $p > 0$, then $\Gamma(G,\mathcal O_G)$ is isomorphic to $k[X_1,\ldots,X_n]/(X_1^{p^{e_1}},\ldots,X_n^{p^{e_n}})$ for some $e_1,\ldots, e_n \in \mathbf Z_{>0}$.

Proof. See for example [Wat, §14.4]. I don't know a quick summary of why this is supposed to be true (I would be interested if someone does), but the proof is not that hard. $\square$

Theorem. If $G$ is group scheme over a field of characteristic $0$, then $G$ is geometrically reduced.

Proof. See for example [Wat, §11.4]. $\square$

In particular, if the rank of $G$ not divisible by $p$ (e.g. $p = 0$), then $G^0$ has to be trivial and $G$ is étale. $\square$

References.

[EGA IV$_4$] A. Grothendieck, Éléments de géométrie algébrique. IV: Étude locale des schémas et des morphismes de schémas (Quatrième partie).. Publ. Math., Inst. Hautes Étud. Sci. 32, p. 1-361 (1967). ZBL0153.22301.

[Wat] W.C. Waterhouse, Introduction to affine group schemes. Graduate Texts in Mathematics 66 (1979). ZBL0442.14017.

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  • $\begingroup$ Your final note I not fully understand. By your reduction steps we apply two base changes: first one we take a prime $\mathfrak{q}_x \subset R$ and do base change along $R \to R_{\mathfrak{q}_x} \to R_{\mathfrak{q}_x}/\mathfrak{q}_x=k(x)$ (can now apply Lemma 1) and then change base to alg closure via $k(x) \to \overline{k(x)}$ by the flat descent argument as you said). Set $k=k(x)$ and we are now in the setting of the Theorem, i.e. set $G=G^0$ (only con c), $\Gamma(G,\mathcal O_G)= k[X_1,\ldots,X_n]/(X_1^{q^{e_1}},\ldots,X_n^{q^{e_n}})$ where $q$ is the characteristic of $k$. Obviously $q$ is $\endgroup$
    – user267839
    Commented Feb 11, 2020 at 21:08
  • $\begingroup$ prime to $p$ as $p \in R^*$ by assumption. If $q>0$: rank of $k[X_1,\ldots,X_n]/(X_1^{q^{e_1}},\ldots,X_n^{q^{e_n}})$ is a power of $q$ but also of $p$, a contradiction. That's fine. But what about the case $q=0$. Then the theorem isn't applicable... $\endgroup$
    – user267839
    Commented Feb 11, 2020 at 21:19
  • $\begingroup$ @TimGrosskreutz: right, I forgot to mention the characteristic 0 situation. There everything is reduced, so a finite group scheme is always étale. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 11, 2020 at 21:21
  • $\begingroup$ one remark: could you give a reference for the statement that in case where $k$ is algebraically closed, then for a finite $k$-module (thus a field) unramified is equivalent to (geometrically) reduced? $\endgroup$
    – user267839
    Commented Feb 11, 2020 at 21:25
  • $\begingroup$ See for example Tags 00U3 and 030W (but the latter deals with separable transcendental extensions as well). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 11, 2020 at 22:54

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