Lemma 100.23.4. Let $\mathcal{X} \to \mathop{\mathrm{Spec}}(k)$ be a locally quasi-finite morphism where $\mathcal{X}$ is an algebraic stack and $k$ is a field. Let $f : V \to \mathcal{X}$ be a locally quasi-finite morphism where $V$ is a scheme. Then $V \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite.

**Proof.**
By Lemma 100.17.2 we see that $V \to \mathop{\mathrm{Spec}}(k)$ is locally of finite type. Assume, to get a contradiction, that $V \to \mathop{\mathrm{Spec}}(k)$ is not locally quasi-finite. Then there exists a nontrivial specialization $v \leadsto v'$ of points of $V$, see Morphisms, Lemma 29.20.6. In particular $\text{trdeg}_ k(\kappa (v)) > \text{trdeg}_ k(\kappa (v'))$, see Morphisms, Lemma 29.28.7. Because $|\mathcal{X}|$ is discrete we see that $|f|(v) = |f|(v')$. Consider $R = V \times _\mathcal {X} V$. Then $R$ is an algebraic space and the projections $s, t : R \to V$ are locally quasi-finite as base changes of $V \to \mathcal{X}$ (which is representable by algebraic spaces so this follows from the discussion in Properties of Stacks, Section 99.3). By Properties of Stacks, Lemma 99.4.3 we see that there exists an $r \in |R|$ such that $s(r) = v$ and $t(r) = v'$. By Morphisms of Spaces, Lemma 66.33.3 we see that the transcendence degree of $v/k$ is equal to the transcendence degree of $r/k$ is equal to the transcendence degree of $v'/k$. This contradiction proves the lemma.
$\square$

## Post a comment

Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: