Lemma 81.14.8. Let $S$ be a scheme. Let $X$ be a Noetherian algebraic space over $S$. Let $U \subset X$ be a proper dense open subspace. Then there exists an affine scheme $V$ and an étale morphism $V \to X$ such that

1. the open subspace $W = U \cup \mathop{\mathrm{Im}}(V \to X)$ is strictly larger than $U$,

2. $(U \subset W, V \to W)$ is a distinguished square, and

3. $U \times _ W V \to U$ has dense image.

Proof. Choose a stratification

$\emptyset = U_{n + 1} \subset U_ n \subset U_{n - 1} \subset \ldots \subset U_1 = X$

and morphisms $f_ p : V_ p \to U_ p$ as in Decent Spaces, Lemma 68.8.6. Let $p$ be the smallest integer such that $U_ p \not\subset U$ (this is possible as $U \not= X$). Choose an affine open $V \subset V_ p$ such that the étale morphism $f_ p|_ V : V \to X$ does not factor through $U$. Consider the open $W = U \cup \mathop{\mathrm{Im}}(V \to X)$ and the reduced closed subspace $Z \subset W$ with $|Z| = |W| \setminus |U|$. Then $f^{-1}Z \to Z$ is an isomorphism because we have the corresponding property for the morphism $f_ p$, see the lemma cited above. Thus $(U \subset W, f : V \to W)$ is a distinguished square. It may not be true that the open $I = \mathop{\mathrm{Im}}(U \times _ W V \to U)$ is dense in $U$. The algebraic space $U' \subset U$ whose underlying set is $|U| \setminus \overline{|I|}$ is Noetherian and hence we can find a dense open subscheme $U'' \subset U'$, see for example Properties of Spaces, Proposition 66.13.3. Then we can find a dense open affine $U''' \subset U''$, see Properties, Lemmas 28.5.7 and 28.29.1. After we replace $f$ by $V \amalg U''' \to X$ everything is clear. $\square$

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).