Remark 70.9.4. Here is a sketch of a proof of Lemma 70.9.3 which avoids using More on Groupoids, Lemma 40.14.10.

Step 1. We may assume $X$ is a reduced Noetherian separated algebraic space (for example by Cohomology of Spaces, Lemma 67.17.1 or by Limits of Spaces, Lemma 68.15.3) and we may choose a finite surjective morphism $Y \to X$ where $Y$ is a Noetherian scheme (by Limits of Spaces, Proposition 68.16.1).

Step 2. After replacing $X$ by an open neighbourhood of $x$, there exists a birational finite morphism $X' \to X$ and a closed subscheme $Y' \subset X' \times _ X Y$ such that $Y' \to X'$ is surjective finite locally free. Namely, because $X$ is reduced there is a dense open subspace $U \subset X$ over which $Y$ is flat (Morphisms of Spaces, Proposition 65.32.1). Then we can choose a $U$-admissible blowup $b : \tilde X \to X$ such that the strict transform $\tilde Y$ of $Y$ is flat over $\tilde X$, see More on Morphisms of Spaces, Lemma 74.39.1. (An alternative is to use Hilbert schemes if one wants to avoid using the result on blowups). Then we let $X' \subset \tilde X$ be the scheme theoretic closure of $b^{-1}(U)$ and $Y' = X' \times _{\tilde X} \tilde Y$. Since $x$ is a codimension $1$ point, we see that $X' \to X$ is finite over a neighbourhood of $x$ (Lemma 70.3.2).

Step 3. After shrinking $X$ to a smaller neighbourhood of $x$ we get that $X'$ is a scheme. This holds because $Y'$ is a scheme and $Y' \to X'$ being finite locally free and because every finite set of codimension $1$ points of $Y'$ is contained in an affine open. Use Properties of Spaces, Proposition 64.14.1 and Varieties, Proposition 33.41.7.

Step 4. There exists an affine open $W' \subset X'$ containing all points lying over $x$ which is the inverse image of an open subspace of $X$. To prove this let $Z \subset X$ be the closure of the set of points where $X' \to X$ is not an isomorphism. We may assume $x \in Z$ otherwise we are already done. Then $x$ is a generic point of an irreducible component of $Z$ and after shrinking $X$ we may assume $Z$ is an affine scheme (Lemma 70.9.2). Then the inverse image $Z' \subset X'$ is an affine scheme as well. Say $x_1, \ldots , x_ n \in Z'$ are the points mapping to $x$. Then we can find an affine open $W'$ in $X'$ whose intersection with $Z'$ is the inverse image of a principal open of $Z$ containing $x$. Namely, we first pick an affine open $W' \subset X'$ containing $x_1, \ldots , x_ n$ using Varieties, Proposition 33.41.7. Then we pick a principal open $D(f) \subset Z$ containing $x$ whose inverse image $D(f|_{Z'})$ is contained in $W' \cap Z'$. Then we pick $f' \in \Gamma (W', \mathcal{O}_{W'})$ restricting to $f|_{Z'}$ and we replace $W'$ by $D(f') \subset W'$. Since $X' \to X$ is an isomorphism away from $Z' \to Z$ the choice of $W'$ guarantees that the image $W \subset X$ of $W'$ is open with inverse image $W'$ in $X'$.

Step 5. Then $W' \to W$ is a finite surjective morphism and $W$ is a scheme by Cohomology of Spaces, Lemma 67.17.1 and the proof is complete.

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