Timeline for Hartshorne's associated scheme for a variety
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 3, 2011 at 16:48 | vote | accept | abourdon | ||
| Jul 2, 2011 at 16:04 | comment | added | Ramsey | rather, not contained in, $U^c$, then $X\cap U$ is nonempty irreducible closed in $U$. Now show that this $X$ is the closure of $X\cap U$ and you're done! | |
| Jul 2, 2011 at 16:02 | comment | added | Ramsey | The image is as your describe. To see this, I suggest that you first try to prove that the inclusion of $t(U_i)$ into $t(V)$ I described (because I was working in the context of a more general argument to show that $t(V)$ is a scheme) is more simply described as "taking the closure". From this description it's not difficult to show that the image of $t(U)$ is $t(U^c)^c$. Basically, if $X$ is a nonempty irreducible closed subset of $U$, its closure cannot be contained in $U^c$, so it is an element of $t(U^c)^c$ by definition. Conversely, if $X$ is closed irreducible in $V$ not containing ... | |
| Jul 2, 2011 at 12:47 | comment | added | abourdon | Thanks so much for your answer! I think I now understand your map, but I'm still trying to show the image of $t(U_i)$ is open. By saying this map addresses the issue I was having, do you mean that the image will equal $[t(U_{i}^c)]^c$? Or do I need a different approach? | |
| Jul 1, 2011 at 18:22 | history | answered | Ramsey | CC BY-SA 3.0 |