The Stacks project

Lemma 68.9.2. 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 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

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

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 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

\[ Z' \amalg T \longrightarrow X \]

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 $U$. $\square$

Comments (5)

Comment #4213 by 羽山籍真 on

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

Comment #5109 by Shiji Lyu on

In the last paragraph we applied Lemma 0ABS to the morphism . However, it seems that this morphism is not necessarily quasi-finite since the morphism 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 of a quasi-separated and quasi-compact algebraic spaces and an integral morphism we can extend to an integral morphism in the sense that is isomorphic to the inverse image of in . To do this, let be the quasi-coherent sheaf of -algebras on corresponding to , in other words, . Pushforward along preserves quasi-coherence. Hence is a quasi-coherent -algebra. Now we let be the integral closure of . By Lemma 29.53.1 (translated over to the category of algebraic spaces; details omitted) we see that is a quasi-coherent -algebra and we see that because the integral closure of in is as is integral! Thus 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.

Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09YB. Beware of the difference between the letter 'O' and the digit '0'.