Timeline for Can I conclude that a morphism of vector bundles is zero if it is so fiberwise?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 10, 2013 at 18:25 | vote | accept | Falter | ||
| Aug 10, 2013 at 18:26 | |||||
| Aug 10, 2013 at 9:11 | comment | added | rita | Assume $X=Y$. Then locally the map of vector bundles is given by a matrix with entries regular functions and the question becomes: can a nonzero regular function be zero at every point? if this happens, then $X$ has an irreducible component all of whose points are non-reduced. | |
| Aug 10, 2013 at 8:13 | answer | added | Sasha | timeline score: 2 | |
| Aug 10, 2013 at 6:41 | comment | added | Falter | Yes, but this is indeed the question. So Angelo's example shows that the claim is false for schemes $X$ which are not reduced. The question remains if it holds for reduced $X$. | |
| Aug 10, 2013 at 0:13 | comment | added | Piotr Achinger | I think Angelo's comment dealt with the slightly different situation when you assume that the map is zero on the fibers of the two vector bundles. Then indeed $X=Y=Spec\, k[x]/(x^2)$, $\phi=x: O_X\to O_X$ is zero on the fiber i.e. after diving by $x$. | |
| Aug 9, 2013 at 19:03 | history | edited | Falter | CC BY-SA 3.0 |
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| Aug 9, 2013 at 19:01 | comment | added | Falter | Dear Angelo, thank you for the comment. It would be nice if you could slightly elaborate it. I also think that the assumption of $X$ reduced is not bad for me, so the question remains if the claim holds for $X$ reduced (I edited the question at this point). | |
| Aug 9, 2013 at 18:50 | comment | added | Angelo | Consider the case that $X = Y$, $f = \mathrm{id}_X$, $\cal U = \cal V = \cal O$, and $X$ is not reduced. | |
| Aug 9, 2013 at 18:46 | history | edited | Falter | CC BY-SA 3.0 |
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| Aug 9, 2013 at 18:18 | review | First posts | |||
| Aug 9, 2013 at 18:39 | |||||
| Aug 9, 2013 at 18:02 | history | asked | Falter | CC BY-SA 3.0 |