Lemma 101.18.1. Let $\mathcal{X}$ be an algebraic stack. Let $x \in |\mathcal{X}|$. The following are equivalent:
There exists a morphism $\mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ which is locally of finite type and represents $x$.
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 67.39.8 and 67.28.5). Hence we see (1) holds by Lemma 101.17.2.
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 67.25.3). By Morphisms of Spaces, Lemma 67.25.4 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. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: