Lemma 67.9.2. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$.

1. There exists a surjective integral morphism $Y \to X$ where $Y$ is a scheme,

2. given a surjective étale morphism $U \to X$ we may choose $Y \to X$ such that for every $y \in Y$ there is an open neighbourhood $V \subset Y$ such that $V \to X$ factors through $U$.

Proof. Part (1) is the special case of part (2) where $U = X$. Choose a surjective étale morphism $U' \to U$ where $U'$ is a scheme. It is clear that we may replace $U$ by $U'$ and hence we may assume $U$ is a scheme. Since $X$ is quasi-compact, there exist finitely many affine opens $U_ i \subset U$ such that $U' = \coprod U_ i \to X$ is surjective. After replacing $U$ by $U'$ again, we see that we may assume $U$ is affine. Since $X$ is quasi-separated, hence reasonable, there exists an integer $d$ bounding the degree of the geometric fibres of $U \to X$ (see Lemma 67.5.1). We will prove the lemma by induction on $d$ for all quasi-compact and separated schemes $U$ mapping surjective and étale onto $X$. If $d = 1$, then $U = X$ and the result holds with $Y = U$. Assume $d > 1$.

We apply Morphisms of Spaces, Lemma 66.52.2 and we obtain a factorization

$\xymatrix{ U \ar[rr]_ j \ar[rd] & & Y \ar[ld]^\pi \\ & X }$

with $\pi$ integral and $j$ a quasi-compact open immersion. We may and do assume that $j(U)$ is scheme theoretically dense in $Y$. Then $U \times _ X Y$ is a quasi-compact, separated scheme (being finite over $U$) and we have

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

Here the first summand is the image of $U \to U \times _ X Y$ (which is closed by Morphisms of Spaces, Lemma 66.4.6 and open because it is étale as a morphism between algebraic spaces étale over $Y$) and the second summand is the (open and closed) complement. The image $V \subset Y$ of $W$ is an open subspace containing $Y \setminus U$.

The étale morphism $W \to Y$ has geometric fibres of cardinality $< d$. Namely, this is clear for geometric points of $U \subset Y$ by inspection. Since $|U| \subset |Y|$ is dense, it holds for all geometric points of $Y$ by Lemma 67.8.1 (the degree of the fibres of a quasi-compact étale morphism does not go up under specialization). Thus we may apply the induction hypothesis to $W \to V$ and find a surjective integral morphism $Z \to V$ with $Z$ a scheme, which Zariski locally factors through $W$. Choose a factorization $Z \to Z' \to Y$ with $Z' \to Y$ integral and $Z \to Z'$ open immersion (Lemma 67.9.1). After replacing $Z'$ by the scheme theoretic closure of $Z$ in $Z'$ we may assume that $Z$ is scheme theoretically dense in $Z'$. After doing this we have $Z' \times _ Y V = Z$. Finally, let $T \subset Y$ be the induced closed subspace structure on $Y \setminus V$. Consider the morphism

$Z' \amalg T \longrightarrow X$

This is a surjective integral morphism by construction. Since $T \subset U$ it is clear that the morphism $T \to X$ factors through $U$. On the other hand, let $z \in Z'$ be a point. If $z \not\in Z$, then $z$ maps to a point of $Y \setminus V \subset U$ and we find a neighbourhood of $z$ on which the morphism factors through $U$. If $z \in Z$, then we have an open neighbourhood of $z$ in $Z$ (which is also an open neighbourhood of $z$ in $Z'$) which factors through $W \subset U \times _ X Y$ and hence through $U$. $\square$

Comment #4213 by 羽山籍真 on

For the last sentence of the proof, $V$ has been used to denote an open of $Y$, so it's not a subspace of $Z$ indeed, maybe replace it by $V\prime$?

Comment #5109 by Shiji Lyu on

In the last paragraph we applied Lemma 0ABS to the morphism $Z \to Y$. However, it seems that this morphism is not necessarily quasi-finite since the morphism $Z \to V$ is just integral, not finite. Is there a way to resolve this?

Comment #5110 by on

Dear Shiji, thanks for finding this error. There is a way of fixing this using limit arguments, but that would mean pushing this much later in the project. Another fix is the following.

We have to show: Given a quasi-compact open immersion $j : V \to Y$ of a quasi-separated and quasi-compact algebraic spaces and an integral morphism $\pi : Z \to V$ we can extend $\pi$ to an integral morphism $\pi' : Z' \to Y$ in the sense that $Z$ is isomorphic to the inverse image of $V$ in $Z'$. To do this, let $\mathcal{A}$ be the quasi-coherent sheaf of $\mathcal{O}_V$-algebras on $V$ corresponding to $\pi : Z \to V$, in other words, $\mathcal{A} = \pi_*\mathcal{O}_Z$. Pushforward along $j$ preserves quasi-coherence. Hence $j_*\mathcal{A}$ is a quasi-coherent $\mathcal{O}_Y$-algebra. Now we let $\mathcal{A}' \subset j_*\mathcal{A}'$ be the integral closure of $\mathcal{O}_Y$. By Lemma 29.53.1 (translated over to the category of algebraic spaces; details omitted) we see that $\mathcal{A}'$ is a quasi-coherent $\mathcal{O}_Y$-algebra and we see that $\mathcal{A}'|_V \cong \mathcal{A}$ because the integral closure of $\mathcal{O}_V$ in $mathcal{A}$ is $\mathcal{A}$ as $\pi$ is integral! Thus $Z' = \underline{\text{Spec}}_Y(\mathcal{A}') \to Y$ is the answer to our problem.

Comment #5316 by on

Dear Shiji Lyu, thanks again. I found what I was saying in the chapter on morphisms of algebraic spaces so it actually was less work than I thought. See changes here.

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