Let $R$ be a complete ( not sure if it's neccessary) DVR together with a ring $h: \mathbb{Z} \to R$ and $p \in \mathbb{P}$ a prime number.
Let $f: R \to S$ a finite ring homomorphism. We assume two things:
- we identify $p$ in $R$ with it's image $h(p)$. Assume $p \in R^{\times}$, so $p$ is invertible in $R$
If we forget the multiplicative structure on $S$ then $f$ endows $S$ with $R$-module structure.
- we assume that $S$ considered as $R$-module is free of rank $p^r$, with $r \ge 1$.
Question: Do (and if yes why) these two assumption imply that $S$ is etale over $R$ (= $f$ is etale map)?
Clearly $S$ is flat over $R$. Why is it also unramified? (recall: etale= flat + ramified). By definition it suffice to show that the discriminant ideal is the unit ideal but I don't see how to show it using the two assumptions. I conjecture that it can be proved (how) that the discriminant ideal contains $p$, but it's only my conjecture. Possibly there is a differently other way to show the claim.
$$S=\Bbb{Z}_3[3^{1/2}],\qquad R=\Bbb{Z}_3$$$