Lemma 67.25.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. The following are equivalent:
There exists a morphism $\mathop{\mathrm{Spec}}(k) \to X$ which is locally of finite type and represents $x$.
There exists a scheme $U$, a closed point $u \in U$, and an étale morphism $\varphi : U \to X$ such that $\varphi (u) = x$.
Proof. Let $u \in U$ and $U \to X$ be as in (2). Then $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$ is of finite type, and $U \to X$ is representable and locally of finite type (by the general principle Spaces, Lemma 65.5.8 and Morphisms, Lemmas 29.36.11 and 29.21.8). Hence we see (1) holds by Lemma 67.23.2.
Conversely, assume $\mathop{\mathrm{Spec}}(k) \to X$ is locally of finite type and represents $x$. Let $U \to X$ be a surjective étale morphism where $U$ is a scheme. By assumption $U \times _ X \mathop{\mathrm{Spec}}(k) \to U$ is locally of finite type. Pick a finite type point $v$ of $U \times _ X \mathop{\mathrm{Spec}}(k)$ (there exists at least one, see Morphisms, Lemma 29.16.4). By Morphisms, Lemma 29.16.5 the image $u \in U$ of $v$ is a finite type point of $U$. Hence by Morphisms, Lemma 29.16.4 after shrinking $U$ we may assume that $u$ is a closed point of $U$, i.e., (2) holds. $\square$
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