Timeline for on the ``generic" modules of finite length (skyscrapers)
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 30, 2016 at 20:10 | comment | added | Dmitry Kerner | @Mohan: see the update. Say $R=k[[x_1,..,x_p]]$. If $m=1$, i.e the case of ideals, then $dim(R)\le n$. For $dim(R)=n$ then genericity means: $\phi$ is a tuple of mutually generic linear forms, then $coker(\phi)=k$, hence the ideal defines the reduced point. | |
| Jan 30, 2016 at 19:53 | history | edited | Dmitry Kerner | CC BY-SA 3.0 |
added 517 characters in body; edited tags
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| Jan 28, 2016 at 16:35 | comment | added | Mohan | The only natural choice I can think of is the closure of the ones of the form $k[[t]]/t^n$, but it does not have the largest dimension. In fact, if I remember correctly, the way one proves lack of irreducibilty is by showing a component of dimension larger than the natural one. | |
| Jan 28, 2016 at 16:13 | comment | added | Dmitry Kerner | @Mohan: right, but I guess among the components of the Hilbert scheme there is one of the biggest dimension? (or some other "natural choice" of the component) | |
| Jan 28, 2016 at 14:29 | comment | added | Mohan | If $p\geq 3$, in general the Hilbert scheme is not irreducible and so can have many generic modules and I do not think much is known. If $p=1$, little to be said and if $p=2$, the Hilbert scheme is irreducible and the generic ones, say of length $n$ are isomorphic to $k[[t]]/t^n$. | |
| Jan 28, 2016 at 10:44 | history | asked | Dmitry Kerner | CC BY-SA 3.0 |