Timeline for Deformation of Line bundles over dual numbers
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2011 at 9:11 | comment | added | Veen | You helped me a lot, Mattia. Thanks. | |
| Nov 10, 2011 at 9:11 | vote | accept | Veen | ||
| Nov 9, 2011 at 13:00 | comment | added | Mattia Talpo | Once you know that it is locally free, it must be of rank one, since its pullback to $X$ is.. | |
| Nov 9, 2011 at 12:35 | comment | added | Veen | Well, but one needs locally free of rank one as one wants it to be a line bundle on $X'$. But I think it really is, due to local splittings of the sequence $0\rightarrow L \rightarrow M \rightarrow L \rightarrow 0$ ($L$ is locally just the structure sheaf and so you get a retraction). Does this sound reasonable? | |
| Nov 8, 2011 at 20:27 | comment | added | Mattia Talpo | I think so, if say $X$ is loc noetherian. Take an open affine $U=Spec A \subseteq X$ (so the corresponding open of $X'$ is the spectrum of $A'=A[\varepsilon]$) and apply the local criterion of flatness for the nilpotent ideal $(\varepsilon)\subseteq A[\varepsilon]$ to the restriction of $M$ to $U$, call $N$ the corresponding $A'$-module. Namely, since $N/(\varepsilon N)$ is flat over $A$ (loc free even) and $Tor_1^{A'}(N,A)=0$ (check that $\cdot \varepsilon: A\to A'$ stays injective after tensoring with $N$), then $N$ is flat over $A'$, so it is locally free. (maybe there's an easier way..) | |
| Nov 8, 2011 at 18:09 | comment | added | Veen | @Mattia: is this $M$ with the $\mathcal O_{X'}$-structure you defined a line bundle on $X'$? It should be given the above definition of deformation. | |
| Nov 8, 2011 at 17:41 | history | answered | Mattia Talpo | CC BY-SA 3.0 |