Lemma 101.23.4. 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 101.17.2 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 100.3). By Properties of Stacks, Lemma 100.4.3 we see that there exists an $r \in |R|$ such that $s(r) = v$ and $t(r) = v'$. By Morphisms of Spaces, Lemma 67.33.3 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.
$\square$

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