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
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 16, 2014 at 8:49 | vote | accept | user41650 | ||
| Oct 9, 2014 at 10:39 | comment | added | David Rydh | @DavidSpeyer I phrased it a bit sloppy. It should be something like "the pairwise intersections of the sections are all equal to $U$". This would be an approximation from the case you mention. The valuative criterion only gives a curve with an "n-fold" point so one has to study this case. I'll try to incorporate this in my answer at some point. | |
| Oct 8, 2014 at 23:16 | comment | added | David E Speyer | @DavidRydh That specific version doesn't work: Take $Y = \mathbb{A}^2$ and $X$ to be $\mathbb{A}^2$ with a doubled origin. The two obvious sections agree on the affine open $U=\{ x \neq 0 \}$ (and on the larger non-affine $\{ (x,y) \neq (0,0) \}$), but $R^1 f_{\ast} \mathcal{O}_X = \mathcal{O}_Y$. In this case, $R^2 f_{\ast} \mathcal{O}_X$ is non-coherent. But I think an easier approach would probably be to use the valuative criterion to find a curve with doubled point in $X$ and use the structure sheaf of that curve. | |
| Oct 8, 2014 at 20:17 | comment | added | David Rydh | @Olivier If you proceed as in my proof, I think one easily reduces the question to: if $Y$ is affine, $f\colon X\to Y$ is universally closed and there are n sections (given by open immersions) covering $Y$, all equal on a dense open affine $U$ of $Y$, then $R^if_*\mathcal{O}_X$ is not coherent for some $i$. If there are $2$ sections, then $R^1f_*\mathcal{O}_X$ is not coherent (a Cech calculation). I haven't bothered to calculate the general case. | |
| Oct 8, 2014 at 19:54 | comment | added | Olivier Benoist | @TomGraber I am very curious about the statement you hint at. Do you have an idea about how to prove it ? | |
| Oct 8, 2014 at 17:43 | comment | added | Karl Schwede | Allen, if you take a line through that missing point then the pushforward of the structure sheaf of that line won't be coherent. Right? | |
| S Oct 8, 2014 at 16:20 | history | suggested | David Steinberg | CC BY-SA 3.0 |
fixed typo, formatting
|
| Oct 8, 2014 at 16:12 | review | Suggested edits | |||
| S Oct 8, 2014 at 16:20 | |||||
| Oct 8, 2014 at 15:52 | answer | added | Olivier Benoist | timeline score: 10 | |
| Oct 8, 2014 at 11:09 | answer | added | David Rydh | timeline score: 23 | |
| Sep 16, 2014 at 19:38 | comment | added | Tom Graber | Vivek is right, although I think if you also demand that the higher direct images are coherent you can deduce separatedness of f. | |
| Sep 15, 2014 at 13:59 | comment | added | Allen Knutson | What about the map from $\mathbb P^2 \setminus pt$ to a point? That's not proper, but takes $\mathcal O_X \mapsto \mathcal O_{pt}$, I think; maybe there's some other coherent sheaf that doesn't stay coherent. | |
| Sep 15, 2014 at 7:16 | comment | added | Vivek Shende | why should it even be separated? the map from the affine line with the doubled origin to the affine line seems like it ought to preserve coherent sheaves | |
| Sep 14, 2014 at 22:23 | comment | added | Karl Schwede | Or maybe even just do this. Take a compactification $X'$ of $X$, choose a curve going through a point of $X' \setminus X$. That should do the job... | |
| Sep 14, 2014 at 20:58 | comment | added | Piotr Achinger | A good example of such would be the structure sheaf of an incomplete (so affine) closed curve on this non-proper variety. You get such curves thanks to the valuative criterion of properness. | |
| Sep 14, 2014 at 20:41 | comment | added | Daniel Barter | i guess that the first thing to do is prove that if you have a non proper variety over a field k, then it has a coherent sheaf whose global sections are infinite dimensional | |
| Sep 14, 2014 at 20:26 | history | asked | user41650 | CC BY-SA 3.0 |