The Stacks project

Proof. Since both $j$ and $\pi $ are quasi-compact and separated, so is $j \circ \pi $. Let $\nu : Y' \to Y$ be the normalization of $Y$ in $Z$, see Morphisms of Spaces, Section 67.48. Of course $\nu $ is integral, see Morphisms of Spaces, Lemma 67.48.5. The final statement follows formally from Morphisms of Spaces, Lemmas 67.48.4 and 67.48.10. $\square$

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 68.5.1). 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 67.52.2 and we obtain a factorization

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

Here the first summand is the image of $U \to U \times _ X Y$ (which is closed by Morphisms of Spaces, Lemma 67.4.6 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 68.8.1 (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 68.9.1). 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

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 $U$. $\square$

Proof. The proof is the (roughly) same as the proof of Lemma 68.9.2 with additional technical comments to obtain the dense quasi-compact open $U$ (and unfortunately changes in notation to keep track of $U$).

Part (1) is the special case of part (2) where $V = X$.

Proof of (2). Choose a surjective étale morphism $V' \to V$ where $V'$ is a scheme. It is clear that we may replace $V$ by $V'$ and hence we may assume $V$ is a scheme. Since $X$ is quasi-compact, there exist finitely many affine opens $V_ i \subset V$ such that $V' = \coprod V_ i \to X$ is surjective. After replacing $V$ by $V'$ again, we see that we may assume $V$ is affine. Since $X$ is quasi-separated, hence reasonable, there exists an integer $d$ bounding the degree of the geometric fibres of $V \to X$ (see Lemma 68.5.1).

By induction on $d \geq 1$ we will prove the following induction hypothesis $(H_ d)$:

• for any quasi-compact and quasi-separated algebraic space $X$ with finitely many irreducible components, for any $m \geq 0$, for any quasi-compact and separated schemes $V_ j$, $j = 1, \ldots , m$, for any étale morphisms $\varphi _ j : V_ j \to X$, $j = 1, \ldots , m$ such that $d$ bounds the degree of the geometric fibres of $\varphi _ j : V_ j\to X$ and $\varphi = \coprod \varphi _ j : V = \coprod V_ j \to X$ is surjective, the statement of the lemma holds for $\varphi : V \to X$.

If $d = 1$, then each $\varphi _ j$ is an open immersion. Hence $X$ is a scheme and the result holds with $Y = V$. Assume $d > 1$, assume $(H_{d - 1})$ and let $m$, $\varphi : V_ j \to X$, $j = 1, \ldots , m$ be as in $(H_ d)$.

Let $\eta _1, \ldots , \eta _ n \in |X|$ be the generic points of the irreducible components of $|X|$. By Properties of Spaces, Proposition 66.13.3 there is an open subscheme $U \subset X$ with $\eta _1, \ldots , \eta _ n \in U$. By shrinking $U$ we may assume $U$ affine and by Morphisms, Lemma 29.51.1 we may assume each $\varphi _ j : V_ j \to X$ is finite étale over $U$. Of course, we see that $U$ is quasi-compact and dense in $X$ and that $\varphi _ j^{-1}(U)$ is dense in $V_ j$. In particular each $V_ j$ has finitely many irreducible components.

Fix $j \in \{ 1, \ldots , m\} $. As in Morphisms of Spaces, Lemma 67.52.2 we let $Y_ j$ be the normalization of $X$ in $V_ j$. We obtain a factorization

with $\pi _ j$ integral and $V_ j \to Y_ j$ a quasi-compact open immersion. Since $Y_ j$ is the normalization of $X$ in $V_ j$, we see from Morphisms of Spaces, Lemmas 67.48.4 and 67.48.10 that $\varphi _ j^{-1}(U) \to \pi _ j^{-1}(U)$ is an isomorphism. Thus $\pi _ j$ is finite étale over $U$. Observe that $V_ j$ is scheme theoretically dense in $Y_ j$ because $Y_ j$ is the normalization of $X$ in $V_ j$ (follows from the characterization of relative normalization in Morphisms of Spaces, Lemma 67.48.5). Since $V_ j$ is quasi-compact we see that $|V_ j| \subset |Y_ j|$ is dense, see Morphisms of Spaces, Section 67.17 (and especially Morphisms of Spaces, Lemma 67.17.7). It follows that $|Y_ j|$ has finitely many irreducible components. Then $V_ j \times _ X Y_ j$ is a quasi-compact, separated scheme (being finite over $V_ j$) and

Here the first summand is the image of $V_ j \to V_ j \times _ X Y_ j$ (which is closed by Morphisms of Spaces, Lemma 67.4.6 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 étale morphism $W_ j \to Y_ j$ has geometric fibres of cardinality $< d$. Namely, this is clear for geometric points of $V_ j \subset Y_ j$ by inspection. Since $|V_ j| \subset |Y_ j|$ is dense, it holds for all geometric points of $Y_ j$ by Lemma 68.8.1 (the degree of the fibres of a quasi-compact étale morphism does not go up under specialization). By $(H_{d - 1})$ applied to $V_ j \amalg W_ j \to Y_ j$ we find a surjective integral morphism $Y_ j' \to Y_ j$ with $Y_ j'$ a scheme, which Zariski locally factors through $V_ j \amalg W_ j$, and which is finite étale over a quasi-compact dense open $U_ j \subset Y_ j$. After shrinking $U$ we may and do assume that $\pi _ j^{-1}(U) \subset U_ j$ (we may and do choose the same $U$ for all $j$; some details omitted).

We claim that

is the solution to our problem. First, this morphism is integral as on each summand we have the composition $Y'_ j \to Y \to X$ of integral morphisms (Morphisms of Spaces, Lemma 67.45.4). Second, this morphism Zariski locally factors through $V = \coprod V_ j$ because we saw above that each $Y'_ j \to Y_ j$ factors Zariski locally through $V_ j \amalg W_ j = V_ j \times _ X Y_ j$. Finally, since both $Y'_ j \to Y_ j$ and $Y_ j \to X$ are finite étale over $U$, so is the composition. This finishes the proof. $\square$