Definition 101.8.1. Let $\mathcal{X}$ be an algebraic stack. We say $\mathcal{X}$ is *Noetherian* if $\mathcal{X}$ is quasi-compact, quasi-separated and locally Noetherian.

## 101.8 Noetherian algebraic stacks

We have already defined locally Noetherian algebraic stacks in Properties of Stacks, Section 100.7.

Note that a Noetherian algebraic stack $\mathcal{X}$ is not just quasi-compact and locally Noetherian, but also quasi-separated. In the language of Section 101.6 if we denote $p : \mathcal{X} \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ the “absolute” structure morphism (i.e., the structure morphism of $\mathcal{X}$ viewed as an algebraic stack over $\mathbf{Z}$), then

This will later mean that an algebraic stack of finite type over a Noetherian algebraic stack is not automatically Noetherian.

Lemma 101.8.2. Let $j : \mathcal{X} \to \mathcal{Y}$ be an immersion of algebraic stacks.

If $\mathcal{Y}$ is locally Noetherian, then $\mathcal{X}$ is locally Noetherian and $j$ is quasi-compact.

If $\mathcal{Y}$ is Noetherian, then $\mathcal{X}$ is Noetherian.

**Proof.**
Choose a scheme $V$ and a surjective smooth morphism $V \to \mathcal{Y}$. Then $U = \mathcal{X} \times _\mathcal {Y} V$ is a scheme and $V \to U$ is an immersion, see Properties of Stacks, Definition 100.9.1. Recall that $\mathcal{Y}$ is locally Noetherian if and only if $V$ is locally Noetherian. In this case $U$ is locally Noetherian too (Morphisms, Lemmas 29.15.5 and 29.15.6) and $U \to V$ is quasi-compact (Properties, Lemma 28.5.3). This shows that $j$ is quasi-compact (Lemma 101.7.10) and that $\mathcal{X}$ is locally Noetherian. Finally, if $\mathcal{Y}$ is Noetherian, then we see from the above that $\mathcal{X}$ is quasi-compact and locally Noetherian. To finish the proof observe that $j$ is separated and hence $\mathcal{X}$ is quasi-separated because $\mathcal{Y}$ is so by Lemma 101.4.11.
$\square$

Lemma 101.8.3. Let $\mathcal{X}$ be an algebraic stack.

If $\mathcal{X}$ is locally Noetherian then $|\mathcal{X}|$ is a locally Noetherian topological space.

If $\mathcal{X}$ is quasi-compact and locally Noetherian, then $|\mathcal{X}|$ is a Noetherian topological space.

**Proof.**
Assume $\mathcal{X}$ is locally Noetherian. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. As $\mathcal{X}$ is locally Noetherian we see that $U$ is locally Noetherian. By Properties, Lemma 28.5.5 this means that $|U|$ is a locally Noetherian topological space. Since $|U| \to |\mathcal{X}|$ is open and surjective we conclude that $|\mathcal{X}|$ is locally Noetherian by Topology, Lemma 5.9.3. This proves (1). If $\mathcal{X}$ is quasi-compact and locally Noetherian, then $|\mathcal{X}|$ is quasi-compact and locally Noetherian. Hence $|\mathcal{X}|$ is Noetherian by Topology, Lemma 5.12.14.
$\square$

Lemma 101.8.4. Let $\mathcal{X}$ be a locally Noetherian algebraic stack. Then $|\mathcal{X}|$ is quasi-sober (Topology, Definition 5.8.6).

**Proof.**
We have to prove that every irreducible closed subset $T \subset |\mathcal{X}|$ has a generic point. Choose an affine scheme $U$ and a smooth morphism $f : U \to \mathcal{X}$ such that $f^{-1}(T) \subset |U|$ is nonempty. Since $U$ is Noetherian, the closed subset $f^{-1}(T)$ has finitely many irreducible components (Topology, Lemma 5.9.2). Say $f^{-1}(T) = Z_1 \cup \ldots \cup Z_ n$ is the decomposition into irreducible components. As $f$ is open, the image of $f|_{f^{-1}(T)} : f^{-1}(T) \to T$ contains a nonempty open subset of $T$. Since $T$ is irreducible, this means that $f(f^{-1}(T))$ is dense. Since $T$ is irreducible, it follows that $f(Z_ i)$ is dense for some $i$. Then if $\xi _ i \in Z_ i$ is the generic point we see that $f(\xi _ i)$ is a generic point of $T$.
$\square$

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