Lemma 105.14.1. Let $p : \mathcal{X} \to Y$ be a morphism of an algebraic stack to an algebraic space. Assume

1. $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is finite,

2. $Y$ is locally Noetherian, and

3. $p$ is locally of finite type.

Let $f : \mathcal{X} \to M$ be the moduli space constructed in Theorem 105.13.9. Then $M \to Y$ is locally of finite type.

Proof. Since $f$ is a uniform categorical moduli space we obtain the morphism $M \to Y$. It suffices to check that $M \to Y$ is locally of finite type étale locally on $M$ and $Y$. Since $f$ is a uniform categorical moduli space, we may first replace $Y$ by an affine scheme étale over $Y$. Next, we may choose $I$ and $g_ i : \mathcal{X}_ i \to \mathcal{X}$ as in Lemma 105.13.8. Then by Lemma 105.13.10 we reduce to the case $\mathcal{X} = \mathcal{X}_ i$. In other words, we may assume $\mathcal{X}$ is well-nigh affine. In this case we have $Y = \mathop{\mathrm{Spec}}(A_0)$, we have $\mathcal{X} = [U/R]$ with $U = \mathop{\mathrm{Spec}}(A)$ and $M = \mathop{\mathrm{Spec}}(C)$ where $C \subset A$ is the set of $R$-invariant functions on $U$. See Lemmas 105.13.2 and 105.13.4. Then $A_0$ is Noetherian and $A_0 \to A$ is of finite type. Moreover $A$ is integral over $C$ by Groupoids, Lemma 39.23.4, hence finite over $C$ (being of finite type over $A_0$). Thus we may finally apply Algebra, Lemma 10.51.7 to conclude. $\square$

Comment #6704 by on

Perhaps $Y = \operatorname{Spec}(A_0)$ here!

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