Let $X = \text{Spec}A$ be an affine scheme. It is well known that if $U$ is an open subset which contains $\text{SpecMax}A$, then $U\supseteq X$. The previous statement generalizes to arbitrary schemes, due to the nature of the Zariski topology of a scheme, being locally affine.

When talking about stacks, one typically replaces the notion of closed points with points of finite type. Again, say $\mathcal{X}$ is a stack, and assume $U\subseteq \mathcal{X}$ is an open substack which contains all points of finite type in $\mathcal{X}$. Is it true that $U\supseteq \mathcal{X}$?

I am not really sure how to approach such a question, as my intuition from the Zariski topology, having a basis of basic open subsets of the topology doesn't generalize at all.

Thanks in advance!

  • 2
    $\begingroup$ There are schemes without closed points, so I'm not sure what you mean by 'the previous statement generalizes to arbitrary schemes'. $\endgroup$ Commented Jun 14, 2023 at 11:50
  • $\begingroup$ Every scheme is locally affine, and every ring has a maximal ideal. $\endgroup$
    – kindasorta
    Commented Jun 14, 2023 at 11:53
  • $\begingroup$ So do you mean then that there is a cover by affine pieces such that intersection of your open with those affine pieces contains all closed points of these affine pieces? Or that this holds for any cover by affine pieces? $\endgroup$ Commented Jun 14, 2023 at 13:01
  • $\begingroup$ I think that this is one of those cases where if something holds for one cover then it holds for any cover, doesn't it? $\endgroup$
    – kindasorta
    Commented Jun 14, 2023 at 13:02
  • 3
    $\begingroup$ The type of example to think about is when $R$ is a DVR with closed point $s$ and generic point $\eta$. Then the affine open $U = \{\eta\}$ contains $\eta$ as a closed point, but $\eta$ is not closed in all of $\operatorname{Spec} R$. In fact, one way to see that a ring is Jacobson is exactly that its locally closed points are closed [Tag 01TB]. $\endgroup$ Commented Jun 14, 2023 at 13:07

1 Answer 1


Assuming you mean that $\mathscr X$ is an algebraic stack (e.g. in the sense of [Tag 026N]), then this is true, and follows relatively straightforwardly from the case of schemes.

Although you did not make a precise statement in the scheme case, it is indeed true that if $U \subseteq X$ is an open subscheme containing all locally closed points, then $U = X$. Moreover, locally closed points are the same thing as points that are (locally) of finite type [Tag 01TA].

Now let $\mathscr U \subseteq \mathscr X$ be an open substack, i.e. $j\colon\mathscr U \to \mathscr X$ is representable by open immersions. In particular, if $x \colon \operatorname{Spec} k \to \mathscr X$ is a point of finite type, the pullback along $\mathscr U \to \mathscr X$ gives an open immersion $U \hookrightarrow \operatorname{Spec} k$, and we say $\mathscr U$ contains $x$ if $U \neq \varnothing$. If this holds for all finite type points, the question is whether $\mathscr U \to \mathscr X$ is an isomorphism.

By assumption, there exists a smooth surjection $f \colon X \to \mathscr X$ where $X$ is a scheme. Checking that a morphism is an isomorphism is smooth-local, so it suffices to check that the base change $U \to X$ of $j$ along $f$ is an isomorphism. Note that $U \to X$ is again an open immersion, so by the above it suffices to check that $U$ contains all points of finite type of $X$. If $x \colon \operatorname{Spec} k \to X$ is a point that is (locally) of finite type, then so is the composition $\operatorname{Spec} k \to X \to \mathscr X$, since $f$ is locally of finite type [Tags 06MH and 03YJ]. By assumption, the pullback of $U \to X$ along $x$ is an isomorphism, i.e. $x \in U$. $\square$

  • $\begingroup$ Many thanks, I will try to understand it first and accept it after I understood it. $\endgroup$
    – kindasorta
    Commented Jun 14, 2023 at 13:03

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