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

Lemma 53.12.1. 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 $k(X)/k(Y)$ 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 29.34.12. Using the exact sequence

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

of Morphisms, Lemma 29.32.9 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 29.35.2. The equivalence between (3) and (4) follows from Morphisms, Lemma 29.36.16 and the fact that flatness is automatic (Lemma 53.2.3). To see the equivalence of (5) and (4) use Algebra, Lemma 10.140.9. 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

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