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The Stacks project

Lemma 72.9.2. Let S be a scheme. Let X be a quasi-separated algebraic space over S. Let x \in |X|. The following are equivalent

  1. x is a point of codimension 0 on X,

  2. the local ring of X at x has dimension 0, and

  3. x is a generic point of an irreducible component of |X|.

If true, then there exists an open subspace of X containing x which is a scheme.

Proof. The equivalence of (1), (2), and (3) follows from Decent Spaces, Lemma 68.20.1 and the fact that a quasi-separated algebraic space is decent (Decent Spaces, Section 68.6). However in the next paragraph we will give a more elementary proof of the equivalence.

Note that (1) and (2) are equivalent by definition (Properties of Spaces, Definition 66.10.2). To prove the equivalence of (1) and (3) we may assume X is quasi-compact. Choose

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

and f_ i : V_ i \to U_ i as in Decent Spaces, Lemma 68.8.6. Say x \in U_ i, x \not\in U_{i + 1}. Then x = f_ i(y) for a unique y \in V_ i. If (1) holds, then y is a generic point of an irreducible component of V_ i (Properties of Spaces, Lemma 66.11.1). Since f_ i^{-1}(U_{i + 1}) is a quasi-compact open of V_ i not containing y, there is an open neighbourhood W \subset V_ i of y disjoint from f_ i^{-1}(V_ i) (see Properties, Lemma 28.2.2 or more simply Algebra, Lemma 10.26.4). Then f_ i|_ W : W \to X is an isomorphism onto its image and hence x = f_ i(y) is a generic point of |X|. Conversely, assume (3) holds. Then f_ i maps \overline{\{ y\} } onto the irreducible component \overline{\{ x\} } of |U_ i|. Since |f_ i| is bijective over \overline{\{ x\} }, it follows that \overline{\{ y\} } is an irreducible component of U_ i. Thus x is a point of codimension 0.

The final statement of the lemma is Properties of Spaces, Proposition 66.13.3. \square


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