A morphism of smooth curves is separable iff it is etale almost everywhere

Lemma 52.12.1. Let $k$ be a field. Let $f : X \to Y$ be a morphism of smooth curves over $k$. The following are equivalent

1. $\text{d}f : f^*\Omega _{Y/k} \to \Omega _{X/k}$ is nonzero,

2. $\Omega _{X/Y}$ is supported on a proper closed subset of $X$,

3. there exists a nonempty open $U \subset X$ such that $f|_ U : U \to Y$ is unramified,

4. there exists a nonempty open $U \subset X$ such that $f|_ U : U \to Y$ is étale,

5. the extension $k(Y) \subset k(X)$ of function fields is finite separable.

Proof. Since $X$ and $Y$ are smooth, the sheaves $\Omega _{X/k}$ and $\Omega _{Y/k}$ are invertible modules, see Morphisms, Lemma 28.32.12. Using the exact sequence

$f^*\Omega _{Y/k} \longrightarrow \Omega _{X/k} \longrightarrow \Omega _{X/Y} \longrightarrow 0$

of Morphisms, Lemma 28.31.9 we see that (1) and (2) are equivalent and equivalent to the condition that $f^*\Omega _{Y/k} \to \Omega _{X/k}$ is nonzero in the generic point. The equivalence of (2) and (3) follows from Morphisms, Lemma 28.33.2. The equivalence between (3) and (4) follows from Morphisms, Lemma 28.34.16 and the fact that flatness is automatic (Lemma 52.2.3). To see the equivalence of (5) and (4) use Algebra, Lemma 10.139.9. Some details omitted. $\square$

Comment #2111 by Matthew Emerton on

Suggested slogan: A morphism of smooth curves is separable iff it is etale almost everywhere

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).