Timeline for Noetherian property in the small fppf site of an algebraic stack
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 13, 2017 at 19:35 | comment | added | Sam | @MarcHoyois: Ok...then it's fine. | |
| May 13, 2017 at 12:49 | comment | added | Marc Hoyois | @Sam $X\to Y$ is still finitely presented. | |
| May 13, 2017 at 11:48 | comment | added | Sam | @Jason Starr: Thanks! Just to clarify: Since $X \times_{Y}U$ is an algebraic space (by the definition of an algebraic stack), we can consider the affine scheme from where there exists an etale surjection on the algebraic space and then show that this affine scheme is Noetherian. Now using your first comment the argument can be completed. I think this is what you meant. I wrote this just to be sure, since I am a beginner in algebraic stacks. | |
| May 13, 2017 at 11:29 | comment | added | Sam | I guess I must mention this simple example was pointed out to me in stackexchange. | |
| May 13, 2017 at 11:27 | comment | added | Sam | @Marc Hoyois: I don't see how is that true without any condition! For any non Noetherian ring $A$, consider the morphism $Spec(A) \longrightarrow Spec(k)$ where $k$ is a field. $Spec(k)$ is certainly Noetherian but $Spec(A)$ is not. | |
| May 12, 2017 at 13:34 | comment | added | Jason Starr | For an algebraic stack, there is a smooth, faithfully flat morphism $U\to Y$ from an affine scheme $U$ to the stack $Y$. Thus, also $X\times_Y U \to X$ is smooth and faithfully flat. By my previous comment, it suffices to check that $X\times_Y U$ is Noetherian. | |
| May 12, 2017 at 13:34 | comment | added | Marc Hoyois | The answer is yes even if $X\to Y$ is not flat and $X$ is not a scheme. This follows from stacks.math.columbia.edu/tag/04YF (1) $\Leftrightarrow$ (2): X is locally noetherian iff there exists a smooth surjective map $U\to X$ where U is a locally noetherian scheme. | |
| May 12, 2017 at 13:27 | comment | added | Jason Starr | For a ring $R$, being Noetherian can be checked after faithful base change, $R\to S$, in the following sense. For a set $\{J_\alpha\}_{\alpha\in A}$ of ideals of $R$, if the associated set of ideals$\{J_\alpha S\}_{\alpha \in A}$ has a maximal element $J_\beta S$, then also $J_\beta$ is a maximal element of the original set. Indeed, suppose that $J_\beta \subset J_\alpha$ is a proper inclusion, i.e., $J_\alpha/J_\beta$ is nonzero. Since $S$ is faithful, also $J_\alpha S/J_\beta S$ is nonzero. | |
| May 12, 2017 at 11:30 | history | asked | Sam | CC BY-SA 3.0 |