Timeline for Image of a projective variety is closed
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2021 at 21:48 | vote | accept | LAGC | ||
| Jun 16, 2021 at 4:34 | history | edited | LAGC | CC BY-SA 4.0 |
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| Jun 16, 2021 at 4:33 | comment | added | LAGC | @DoriBejleri If we assume that $Y$ is a scheme, then is the answer to my question affirmative? | |
| Jun 16, 2021 at 4:02 | answer | added | David Benjamin Lim | timeline score: 2 | |
| Jun 16, 2021 at 4:01 | comment | added | Dori Bejleri | Here I was implicitly assuming that the action of $\mathbb{G}_m$ is the usual one by scaling with all weights equal to $1$. | |
| Jun 16, 2021 at 3:59 | comment | added | Dori Bejleri | No, $Y$ is the Artin stack $[\mathbb{A}^n/\mathbb{G}_m]$. If you unravel the definition of the quotient stack, a map $T \to Y$ is the same as a a tuple $(L, s_1, ...., s_n)$ where $L$ is a line bundle on $T$ and $s_i$ are sections. On the other hand, a map $T \to X$ is the same data but with the condition that the $s_i$ don't simultaneously vanish. Dropping this condition induces the embedding $f : X \to Y$ which you can equivalently think of as identifying $X$ with the quotient $[(\mathbb{A}^n\setminus 0)/\mathbb{G}_m]$ which is an Artin stack which happens to be isomorphic to a scheme. | |
| Jun 16, 2021 at 3:36 | comment | added | Dori Bejleri | This is false. Take for example $Y = [\mathbb{A}^n/\mathbb{G}_m]$ and $X = \mathbb{P}^{n-1}$. Then $X$ is projective and the natural inclusion $f : X \to Y$ is an open embedding with dense image but $f(X)$ is not closed. The issue is that $Y$ is not separated so $f : X \to Y$ is not proper. | |
| Jun 16, 2021 at 3:14 | history | edited | LAGC | CC BY-SA 4.0 |
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| Jun 16, 2021 at 2:49 | history | edited | LAGC | CC BY-SA 4.0 |
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| Jun 16, 2021 at 2:19 | history | asked | LAGC | CC BY-SA 4.0 |