Lemma 68.9.2. 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 64.20.1 and the fact that a quasi-separated algebraic space is decent (Decent Spaces, Section 64.6). 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 62.10.2). 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 64.8.6. 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 62.11.1). 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 27.2.2 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 62.13.3.
$\square$

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