## 76.33 Étale localization of morphisms

The section is the analogue of More on Morphisms, Section 37.41.

Lemma 76.33.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $y \in |Y|$. Let $x_1, \ldots , x_ n \in |X|$ mapping to $y$. Assume that

1. $f$ is locally of finite type,

2. $f$ is separated,

3. $f$ is quasi-finite at $x_1, \ldots , x_ n$, and

4. $f$ is quasi-compact or $Y$ is decent.

Then there exists an étale morphism $(U, u) \to (Y, y)$ of pointed algebraic spaces and a decomposition

$U \times _ Y X = W \amalg V$

into open and closed subspaces such that the morphism $V \to U$ is finite, every point of the fibre of $|V| \to |U|$ over $u$ maps to an $x_ i$, and the fibre of $|W| \to |U|$ over $u$ contains no point mapping to an $x_ i$.

Proof. Let $(U, u) \to (Y, y)$ be an étale morphism of algebraic spaces and consider the set of $w \in |U \times _ Y X|$ mapping to $u \in |U|$ and one of the $x_ i \in |X|$. By Decent Spaces, Lemma 68.18.4 (if $f$ is of finite type) or Decent Spaces, Lemma 68.18.5 (if $Y$ is decent) this set is finite. It follows that we may replace $f$ by the base change $U \times _ Y X \to U$ and $x_1, \ldots , x_ n$ by the set of these $w$. In particular we may and do assume that $Y$ is an affine scheme, whence $X$ is a separated algebraic space.

Choose an affine scheme $Z$ and an étale morphism $Z \to X$ such that $x_1, \ldots , x_ n$ are in the image of $|Z| \to |X|$. The fibres of $|Z| \to |X|$ are finite, see Properties of Spaces, Lemma 66.6.7 (or the more general discussion in Decent Spaces, Section 68.6). Let $\{ z_1, \ldots , z_ m\} \subset |Z|$ be the preimage of $\{ x_1, \ldots , x_ n\}$. By More on Morphisms, Lemma 37.41.4 there exists an étale morphism $(U, u) \to (Y, y)$ such that $U \times _ Y Z = Z_1 \amalg Z_2$ with $Z_1 \to U$ finite and $(Z_1)_ y = \{ z_1, \ldots , z_ m\}$. We may assume that $U$ is affine and hence $Z_1$ is affine too.

Since $f$ is separated, the image $V$ of $Z_1 \to X$ is both open and closed (Morphisms of Spaces, Lemma 67.40.6). Set $W = X \setminus V$ to get a decomposition as in the lemma. To finish the proof we have to show that $V \to U$ is finite. As $Z_1 \to V$ is surjective and étale, $V$ is the quotient of $Z_1$ by the étale equivalence relation $R = Z_1 \times _ V Z_1$, see Spaces, Lemma 65.9.1. Since $f$ is separated, $V \to U$ is separated and $R$ is closed in $Z_1 \times _ U Z_1$. Since $Z_1 \to U$ is finite, the projections $s, t : R \to Z_1$ are finite. Thus $V$ is an affine scheme by Groupoids, Proposition 39.23.9. By Morphisms, Lemma 29.41.9 we conclude that $V \to U$ is proper and by Morphisms, Lemma 29.44.11 we conclude that $V \to U$ is finite, thereby finishing the proof. $\square$

Lemma 76.33.2. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$ with image $y \in |Y|$. Assume that

1. $f$ is locally of finite type,

2. $f$ is separated, and

3. $f$ is quasi-finite at $x$.

Then there exists an étale morphism $(U, u) \to (Y, y)$ of pointed algebraic spaces and a decomposition

$U \times _ Y X = W \amalg V$

into open and closed subspaces such that the morphism $V \to U$ is finite and there exists a point $v \in |V|$ which maps to $x$ in $|X|$ and $u$ in $|U|$.

Proof. Pick a scheme $U$, a point $u \in U$, and an étale morphism $U \to Y$ mapping $u$ to $y$. There exists a point $x' \in |U \times _ Y X|$ mapping to $x$ in $|X|$ and $u$ in $|U|$ (Properties of Spaces, Lemma 66.4.3). To finish, apply Lemma 76.33.1 to the morphism $U \times _ Y X \to U$ and the point $x'$. It applies because $U$ is a scheme and hence $u$ comes from the monomorphism $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$. $\square$

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).