Lemma 67.20.1. Let $S$ be a scheme. Let $X$ be a decent algebraic space over $S$. Let $x \in |X|$. The following are equivalent

1. $x$ is a generic point of an irreducible component of $|X|$,

2. for any étale morphism $(Y, y) \to (X, x)$ of pointed algebraic spaces, $y$ is a generic point of an irreducible component of $|Y|$,

3. for some étale morphism $(Y, y) \to (X, x)$ of pointed algebraic spaces, $y$ is a generic point of an irreducible component of $|Y|$,

4. the dimension of the local ring of $X$ at $x$ is zero, and

5. $x$ is a point of codimension $0$ on $X$

Proof. Conditions (4) and (5) are equivalent for any algebraic space by definition, see Properties of Spaces, Definition 65.10.2. Observe that any $Y$ as in (2) and (3) is decent by Lemma 67.6.6. Thus it suffices to prove the equivalence of (1) and (4) as then the equivalence with (2) and (3) follows since the dimension of the local ring of $Y$ at $y$ is equal to the dimension of the local ring of $X$ at $x$. Let $f : U \to X$ be an étale morphism from an affine scheme and let $u \in U$ be a point mapping to $x$.

Assume (1). Let $u' \leadsto u$ be a specialization in $U$. Then $f(u') = f(u) = x$. By Lemma 67.12.1 we see that $u' = u$. Hence $u$ is a generic point of an irreducible component of $U$. Thus $\dim (\mathcal{O}_{U, u}) = 0$ and we see that (4) holds.

Assume (4). The point $x$ is contained in an irreducible component $T \subset |X|$. Since $|X|$ is sober (Proposition 67.12.4) we $T$ has a generic point $x'$. Of course $x' \leadsto x$. Then we can lift this specialization to $u' \leadsto u$ in $U$ (Lemma 67.12.2). This contradicts the assumption that $\dim (\mathcal{O}_{U, u}) = 0$ unless $u' = u$, i.e., $x' = x$. $\square$

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