Timeline for Morphism of schemes with non-sober image
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2019 at 15:11 | vote | accept | CommunityBot | ||
| Apr 22, 2019 at 6:40 | comment | added | Laurent Moret-Bailly | @schematic_boi Your question would be more appropriate for math.stackexchange so this is a welcoming present: you easily reduce to observing that if a space is noetherian, irreducible, and a finite union of finitely many irreducible locally closed subspaces, then one of them must be dense, hence open. | |
| Apr 21, 2019 at 20:26 | comment | added | user138661 | @LaurentMoret-Bailly but is a non-locally closed constructible subset of a Noetherian sober space necessarily sober? Do you have some reference for that? | |
| Apr 21, 2019 at 20:02 | comment | added | Laurent Moret-Bailly | @schematic_boi In the case of $\mathbb{C}$-schemes of finite type, assuming that $f$ is a $\mathbb{C}$-morphism, I suggest you use Chevalley's theorem that the image of $f$ is constructible. | |
| Apr 21, 2019 at 19:51 | comment | added | Laurent Moret-Bailly | @Wojowu You can describe it as the open subscheme of $\mathrm{Spec}(\prod_p\mathbb{F}_p)$ whose complement is defined by the ideal $\bigoplus_p\mathbb{F}_p$. I do think your definition is better! | |
| Apr 21, 2019 at 19:25 | comment | added | Wojowu | This scheme is not a spectrum of any ring (since it's not quasicompact), if that's what you are talking about. I honestly don't know how to describe this scheme any better than just as a disjoint union of ringed topological spaces. | |
| Apr 21, 2019 at 19:19 | comment | added | მამუკა ჯიბლაძე | Which scheme do you mean? As a space, the disjoint union of the $\operatorname{Spec}{\mathbb F}_p$ is discrete. It is the underlying space of which scheme? But it is not e. g. $\operatorname{Spec}$ of the product of all ${\mathbb F}_p$, is it? Because not seeing what is $X$ I do not see why $(0)$ is not in the image. | |
| Apr 21, 2019 at 18:02 | comment | added | user138661 | can this happen if both source and target are integral schemes of finite type over $\mathbb{C}$? | |
| Apr 21, 2019 at 18:00 | history | answered | Wojowu | CC BY-SA 4.0 |