Lemma 57.11.1. In the situation above the following are equivalent

1. $X$ is unramified in $L$,

2. $Y \to X$ is étale, and

3. $Y \to X$ is finite étale.

Proof. Observe that $Y \to X$ is an integral morphism. In each case the morphism $Y \to X$ is locally of finite type by definition. Hence we find that in each case $Y \to X$ is finite by Morphisms, Lemma 29.44.4. In particular we see that (2) is equivalent to (3). An étale morphism is unramified, hence (2) implies (1).

Conversely, assume $Y \to X$ is unramified. Let $x \in X$. We can choose an étale neighbourhood $(U, u) \to (X, x)$ such that

$Y \times _ X U = \coprod V_ j \longrightarrow U$

is a disjoint union of closed immersions, see Étale Morphisms, Lemma 41.17.3. Shrinking we may assume $U$ is quasi-compact. Then $U$ has finitely many irreducible components (Descent, Lemma 35.13.3). Since $U$ is normal (Descent, Lemma 35.15.2) the irreducible components of $U$ are open and closed (Properties, Lemma 28.7.5) and we may assume $U$ is irreducible. Then $U$ is an integral scheme whose generic point $\xi$ maps to the generic point of $X$. On the other hand, we know that $Y \times _ X U$ is the normalization of $U$ in $\mathop{\mathrm{Spec}}(L) \times _ X U$ by More on Morphisms, Lemma 37.17.2. Every point of $\mathop{\mathrm{Spec}}(L) \times _ X U$ maps to $\xi$. Thus every $V_ j$ contains a point mapping to $\xi$ by Morphisms, Lemma 29.53.9. Thus $V_ j \to U$ is an isomorphism as $U = \overline{\{ \xi \} }$. Thus $Y \times _ X U \to U$ is étale. By Descent, Lemma 35.20.29 we conclude that $Y \to X$ is étale over the image of $U \to X$ (an open neighbourhood of $x$). $\square$

Comment #6534 by Tim Holzschuh on

Possible typo: "Hence we find that in each case the lemma is finite by Morphisms ..." I think "the lemma" should read sth. along the lines of "the morphism".

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