Timeline for If the direct image of f preserves coherent sheaves on noetherian schemes, how to show f is proper?
Current License: CC BY-SA 3.0
14 events
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| Oct 10, 2014 at 12:24 | comment | added | Filippo Alberto Edoardo | @Olivier: Ah, yes, I see - you're right. Thanks! | |
| Oct 10, 2014 at 9:42 | comment | added | Olivier Benoist | @Filippo : I think you need to extend it. The argument goes as follows. To check that $f:X\to Y$ is proper, it is sufficient to check that the $f_y:X_y\to Y_y$ are proper. For this, I want to use that $f_{y,*}$ preserves coherence. To see it, take $\mathcal{F}_y$ coherent on $X_y$, extend it to $X$, and apply that $f_*$ preserves coherence. | |
| Oct 10, 2014 at 9:35 | comment | added | Filippo Alberto Edoardo | @OlivierBenoist: Just another small question: do you really need that a coherent sheaf can be extended (which I agree is true, of course)? After all, in your situation, what you only need is that it can be restricted (preserving coherence), no? If $f: X\to Y$ is as above and you want to prove it proper (local condition), pick a point $y\in Y$, restrict $f$ to $X_0=f^{-1}(\mathrm{Spec}(\text{local containing }y))$ and get that $f\vert X_0 :X_0\to \mathrm{Spec}(\text{this local})$ is proper, hence $f$ is proper. No? | |
| Oct 10, 2014 at 8:09 | history | edited | Olivier Benoist | CC BY-SA 3.0 |
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| Oct 10, 2014 at 8:07 | comment | added | Olivier Benoist | @FilippoAlbertoEdoardo : hi ! You're of course right : I want $f$ to vanish on the point $z$ I have chosen. This ensures that $Z$ is still not empty. To prevent other degenerate situations (like $X$ being empty, or $X$ not being dense anymore in $\overline{X}$), I also require that $f$ does not vanish on various subschemes : but this I did not forget to write... | |
| Oct 10, 2014 at 7:28 | comment | added | Filippo Alberto Edoardo | @Olivier: nice proof, but I'm missing something on your reduction step towards finiteness of $Z$: do you ask something to the function $f\in\mathcal{O}_{\overline{X},z}$ (which you might call $g$, you already had an $f$ around, right? ) or simply not to vanish? What if I take $g=1$ and have $\overline{X},Z,X$ empty? Am I producing nonsense? | |
| Oct 9, 2014 at 18:05 | history | edited | Olivier Benoist | CC BY-SA 3.0 |
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| Oct 9, 2014 at 17:41 | history | edited | Olivier Benoist | CC BY-SA 3.0 |
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| Oct 8, 2014 at 22:40 | history | edited | Olivier Benoist | CC BY-SA 3.0 |
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| Oct 8, 2014 at 20:14 | history | edited | Olivier Benoist | CC BY-SA 3.0 |
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| Oct 8, 2014 at 20:11 | comment | added | Olivier Benoist | @DavidRydh Thanks ! Of course, your proof is easier in the sense that Zariski-Riemann spaces are somehow hidden in the use of Nagata compactification. (And my qs hypothesis was a consequence of my lack of knowledge of non-separated phenomena ;-) ). | |
| Oct 8, 2014 at 19:38 | comment | added | David Rydh | Nice proof. It easily extends to the case where $X$ is a separated algebraic stack by taking a proper covering by a scheme. Making the base a general stack is more difficult though (exactly as in my proof). Nitpick: why do you write "qcqs" when all schemes are noetherian? | |
| Oct 8, 2014 at 19:07 | history | edited | Olivier Benoist | CC BY-SA 3.0 |
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| Oct 8, 2014 at 15:52 | history | answered | Olivier Benoist | CC BY-SA 3.0 |