Let $R$ be a 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 etale?

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 $p \in R^*$.

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 $p \in R^*$.

Let $R$ be a 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 etale?

Let $R$ be a 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 etale?

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 $p \in R^*$.