[Expose I, Corollary 9.11, SGA1]

Lemma 37.36.2. Let $f : X \to Y$ be a morphism of schemes. Assume

$Y$ is integral and geometrically unibranch,

at least one irreducible component of $X$ dominates $Y$,

$f$ is unramified, and

$X$ is connected.

Then $f$ is étale and $X$ is irreducible.

**Proof.**
Let $X' \subset X$ be the irreducible component which dominates $Y$. This means that the generic point of $X'$ maps to the generic point of $Y$ (see for example Morphisms, Lemma 29.8.5). By Lemma 37.36.1 we see that $f$ is étale at every point of $X'$. In particular, the open subscheme $U \subset X$ where $f$ is étale contains $X'$. Note that every quasi-compact open of $U$ has finitely many irreducible components, see Descent, Lemma 35.15.3. On the other hand since $Y$ is geometrically unibranch and $U$ is étale over $Y$, the scheme $U$ is geometrically unibranch. In particular, through every point of $U$ there passes at most one irreducible component. A simple topological argument now shows that $X' \subset U$ is both open and closed. Then of course $X'$ is open and closed in $X$ and by connectedness we find $X = U = X'$ as desired.
$\square$

## Comments (2)

Comment #7216 by Alex Ivanov on

Comment #7329 by Johan on