The Stacks project

Proof. 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 $U'$, we see that we may assume $U$ is affine. In particular $U \to X$ is separated (Schemes, Lemma 26.21.15). Then there exists an integer $d$ bounding the degree of the geometric fibres of $U \to X$ (see Morphisms, Lemma 29.57.9). 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 Lemma 37.43.2 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 $Y$. Note that

where the first summand is the image of $U \to U \times _ X Y$ (which is closed by Schemes, Lemma 26.21.10 and open because it is étale as a morphism between schemes étale over $Y$) and the second summand is the (open and closed) complement. The image $V \subset Y$ of $W$ is an open subscheme 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 $Y$ for example by Lemma 37.45.3 (the degree of the fibres of a quasi-compact separated 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 37.43.2). 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 reduced closed subscheme 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 a neighbourhood $\Omega \subset Z$ which factors through $W \subset U \times _ X Y$ and hence through $U$. This proves existence.

Assume we have found $Y \to X$ integral and surjective which Zariski locally factors through $U$. Choose a finite affine open covering $Y = \bigcup V_ j$ such that $V_ j \to X$ factors through $U$. We can write $Y = \mathop{\mathrm{lim}}\nolimits Y_ i$ with $Y_ i \to X$ finite and of finite presentation, see Limits, Lemma 32.7.3. For large enough $i$ we can find affine opens $V_{i, j} \subset Y_ i$ whose inverse image in $Y$ recovers $V_ j$, see Limits, Lemma 32.4.11. For even larger $i$ the morphisms $V_ j \to U$ over $X$ come from morphisms $V_{i, j} \to U$ over $X$, see Limits, Proposition 32.6.1. This finishes the proof. $\square$