B Dually, given a field $K$, the zero algebra over $K$ is diagonal and in particular étale: the morphism of affine schemes $\varnothing \to Spec(K)$ is étale. In the same vein, a nonzero constant polynomial over $K$ is separable (its nonexistent roots in an algebraic closure of $K$ are certainly distinct) . We may then say without any exception that the $K$-algebra $K[X]/(f(X))$ is étale iff $f(X)$ is a separable polynomial.