Timeline for descent of nef divisors
Current License: CC BY-SA 4.0
14 events
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| Jul 4, 2021 at 16:31 | history | edited | anonymous | CC BY-SA 4.0 |
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| Jun 8, 2021 at 17:13 | comment | added | Jason Starr | You do not need to use alterations or resolution of singularities. Because of $S2$ extension / Hartog's phenomenon, you only need to prove triviality of an invertible sheaf away from codimension $2$. The normalization of a scheme is regular away from codimension $2$. That is why the normalization of a finite base change suffices (combined with the simple analysis of ramification in characteristic $0$ at codimension $1$ points). | |
| Jun 7, 2021 at 21:02 | comment | added | anonymous | Thanks for your answer. I'm not exactly sure what you mean by normalized base change. I was thinking of making an alteration of $Y$ to make the morphism semistable (though it doesn't seem like nefness is needed here) | |
| Jun 7, 2021 at 18:34 | comment | added | Jason Starr | You can use normalized base change and norms to prove that your invertible sheaf is "relatively $n$-torsion" for some positive integer $n$, i.e., the $n$-fold self tensor product is a pullback from the base. | |
| Jun 5, 2021 at 12:51 | comment | added | Damian Rössler | Note that the problem you are facing is the possible non-separatedness of the Picard scheme. If $f$ is cohomologically flat in degree $0$, the answer to your question is positive if the relative Picard scheme is separated (nothing new here but it might help research the problem). | |
| Jun 4, 2021 at 21:43 | history | edited | anonymous | CC BY-SA 4.0 |
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| Jun 4, 2021 at 17:50 | comment | added | Damian Rössler | Why does Grauert's theorem imply that $f_*(L)$ is a line bundle? To apply Grauert's theorem, you need to know that $H^0(X_y,{\cal O}_{X_y})\simeq\kappa(y)$ for all $y\in Y$ (in your situation). Why is this so? | |
| Jun 3, 2021 at 21:34 | history | edited | anonymous | CC BY-SA 4.0 |
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| Jun 3, 2021 at 21:33 | comment | added | anonymous | Thanks for your comment. I'm interested in the the char 0 case here (So I've edited the question). You're right, even if all the fibers are connected and reduced, then also the proof works. But I don't know what to do in case non-reduced fibers show up (for example, pencil of smooth conics degenerating to a double line). | |
| Jun 3, 2021 at 19:49 | comment | added | R. van Dobben de Bruyn | Are you assuming the fibres are geometrically reduced? In positive characteristic, it is not necessarily true that $f_*\mathscr L$ is invertible (think about the relative Frobenius morphism and $\mathscr L = \mathcal O_X)$. But if the geometric fibres are reduced and connected, then $f_*\mathcal O_X = \mathcal O_Y$ holds universally, and this still gives $\mathscr L \cong f^*f_* \mathscr L$. | |
| Jun 3, 2021 at 19:26 | history | edited | anonymous | CC BY-SA 4.0 |
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| Jun 3, 2021 at 17:06 | history | edited | anonymous | CC BY-SA 4.0 |
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| Jun 3, 2021 at 15:46 | history | edited | anonymous | CC BY-SA 4.0 |
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| Jun 3, 2021 at 15:39 | history | asked | anonymous | CC BY-SA 4.0 |