## 37.37 Unramified and étale morphisms

Sometimes unramified morphisms are automatically étale.

Lemma 37.37.1. Let $f : X \to Y$ be a morphism of schemes. Let $x \in X$ with image $y \in Y$. Assume

1. $Y$ is integral and geometrically unibranch at $y$,

2. $f$ is locally of finite type,

3. there is a specialization $x' \leadsto x$ such that $f(x')$ is the generic point of $Y$,

4. $f$ is unramified at $x$.

Then $f$ is étale at $x$.

Proof. We may replace $X$ and $Y$ by suitable affine open neighbourhoods of $x$ and $y$. Then $Y$ is the spectrum of a domain $A$ and $X$ is the spectrum of a finite type $A$-algebra $B$. Let $\mathfrak q \subset B$ be the prime ideal corresponding to $x$ and $\mathfrak p \subset A$ the prime ideal corresponding to $y$. The local ring $A_\mathfrak p = \mathcal{O}_{Y, y}$ is geometrically unibranch. The ring map $A \to B$ is unramified at $\mathfrak q$. Also, the point $x'$ in (3) corresponds to a prime ideal $\mathfrak q' \subset \mathfrak q$ such that $A \cap \mathfrak q' = (0)$. It follows that $A_\mathfrak p \to B_\mathfrak q$ is injective. We conclude by More on Algebra, Lemma 15.107.2. $\square$

Lemma 37.37.2. Let $f : X \to Y$ be a morphism of schemes. Assume

1. $Y$ is integral and geometrically unibranch,

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

3. $f$ is unramified, and

4. $X$ 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 37.37.1 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 35.16.3. 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. $\square$

Lemma 37.37.3. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes. Let $x \in X$ with image $y \in Y$. Assume

1. $Y$ is integral and geometrically unibranch at $y$,

2. $f$ is locally of finite type,

3. $g \circ f$ is étale at $x$,

4. there is a specialization $x' \leadsto x$ such that $f(x')$ is the generic point of $Y$.

Then $f$ is étale at $x$ and $g$ is étale at $y$.

Proof. The morphism $f$ is unramified at $x$ by Morphisms, Lemmas 29.35.16 and 29.36.5. Hence $f$ is étale at $x$ by Lemma 37.37.1. Then by étale descent we see that $g$ is étale at $y$, see for example Descent, Lemma 35.14.4. $\square$

Lemma 37.37.4. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of schemes. Assume

1. $Y$ is integral and geometrically unibranch,

2. $f$ is locally of finite type,

3. $g \circ f$ is étale,

4. every irreducible component of $X$ dominates $Y$.

Then $f$ is étale and $g$ is étale at every point in the image of $f$.

Proof. Immediate from the pointwise version Lemma 37.37.3. $\square$

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