Timeline for Locally free sheaves trivial on fibers of a flat projective morphism
Current License: CC BY-SA 3.0
13 events
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| Nov 13, 2012 at 9:05 | history | edited | Qing Liu | CC BY-SA 3.0 |
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| Nov 13, 2012 at 9:01 | comment | added | Qing Liu | @Sándor: Thanks. Hope the edited answer is more clear on the isomorphism $O_Y\simeq \pi_\star O_X$ (under rather weak hypothesis) and the equality $h^0=1$ which holds under OP's hypothesis. Of course $h^1(O_F)=0$ is somehow a strong condition. On the other hands, I belieave that if $X,Y$ are both smooth, then $h^0(O_F)=1$ holds (over $\mathbb C$). Note that by Bertini, one can reduce to the case $\dim Y=1$. Then this is true in relative dimension 1 case. | |
| Nov 13, 2012 at 8:44 | history | edited | Qing Liu | CC BY-SA 3.0 |
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| Nov 13, 2012 at 7:13 | comment | added | Sándor Kovács | @Qing: The issue is not whether $X$ is normal. Mohan's point is that for your argument you need that $X_y$ is reduced. This is a property of $\pi$ not of $X$. Otherwise you do not get $h^0=1$ on the fibers. Also, what does it mean that "$\pi_*\mathscr O_X$ injects into its stalk at the generic point" ? Another issue is that your statement does not answer when the required isomorphism happens, just reformulates it. Finally, just to be nitpicking, it's called the "structure" sheaf. | |
| Nov 13, 2012 at 6:24 | history | edited | Qing Liu | CC BY-SA 3.0 |
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| Nov 13, 2012 at 6:22 | comment | added | Qing Liu | @Mohan, I don't know why $X$ was normal in my mind. Thanks ! | |
| Nov 13, 2012 at 6:17 | history | edited | Qing Liu | CC BY-SA 3.0 |
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| Nov 13, 2012 at 6:07 | comment | added | Qing Liu | @Mohan, I will add explanations. @aglearner: I read trivial as isomorphic to the structural sheaf, sorry. | |
| Nov 13, 2012 at 0:08 | comment | added | aglearner | (I am sorry if I misunderstood your answer) | |
| Nov 13, 2012 at 0:04 | comment | added | aglearner | Dear Liu thank you for the answer! You write "Therefore if $\pi^* \pi_* E\cong E$, then $E$ it the pull-back by $\pi$ of some invertible sheaf". It seems to me that this is the logic opposite to the one that I need? I need to prove that $\pi^* \pi_* E\cong E$. Also, in my case $E$ is not necessarily invertible, it can have rank $>1$ (i.e. it is a vector bundle). | |
| Nov 12, 2012 at 23:24 | comment | added | Mohan | Connectedness is not enough for $\pi__O_X=O_y$. For example, take $X=Y\times \text{Spec} k[\epsilon]$ with $\epsilon^2=0$. | |
| Nov 12, 2012 at 22:17 | history | edited | Qing Liu | CC BY-SA 3.0 |
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| Nov 12, 2012 at 22:02 | history | answered | Qing Liu | CC BY-SA 3.0 |