Lemma 69.9.2. Let $S$ be a scheme. Let $X$ be a quasi-separated algebraic space over $S$. Let $x \in |X|$. The following are equivalent

$x$ is a point of codimension $0$ on $X$,

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

$x$ 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 65.20.1 and the fact that a quasi-separated algebraic space is decent (Decent Spaces, Section 65.6). 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 63.10.2). 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 65.8.6. 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 63.11.1). 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 10.25.4). 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 63.13.3.
$\square$

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