# Locally free group scheme étale

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 $$p \in R^*$$.

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.

• 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 – Tim Grosskreutz Feb 11 at 21:08
• 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... – Tim Grosskreutz Feb 11 at 21:19
• @TimGrosskreutz: right, I forgot to mention the characteristic 0 situation. There everything is reduced, so a finite group scheme is always étale. – R. van Dobben de Bruyn Feb 11 at 21:21
• 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? – Tim Grosskreutz Feb 11 at 21:25
• See for example Tags 00U3 and 030W (but the latter deals with separable transcendental extensions as well). – R. van Dobben de Bruyn Feb 11 at 22:54