Lemma 100.18.1. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. The following are equivalent:

1. There exists a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ which is locally of finite type and represents $x$.

2. There exists a scheme $U$, a closed point $u \in U$, and a smooth morphism $\varphi : U \to \mathcal{X}$ such that $\varphi (u) = x$.

Proof. Let $u \in U$ and $U \to \mathcal{X}$ be as in (2). Then $\mathop{\mathrm{Spec}}(\kappa (u)) \to U$ is of finite type, and $U \to \mathcal{X}$ is representable and locally of finite type (by Morphisms of Spaces, Lemmas 66.39.8 and 66.28.5). Hence we see (1) holds by Lemma 100.17.2.

Conversely, assume $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ is locally of finite type and represents $x$. Let $U \to \mathcal{X}$ be a surjective smooth morphism where $U$ is a scheme. By assumption $U \times _\mathcal {X} \mathop{\mathrm{Spec}}(k) \to U$ is a morphism of algebraic spaces which is locally of finite type. Pick a finite type point $v$ of $U \times _\mathcal {X} \mathop{\mathrm{Spec}}(k)$ (there exists at least one, see Morphisms of Spaces, Lemma 66.25.3). By Morphisms of Spaces, Lemma 66.25.4 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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