Timeline for "un-nil-ifying" ideals via deformation
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| Dec 5, 2010 at 11:13 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
edited tags; added 30 characters in body
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| Dec 5, 2010 at 4:58 | comment | added | Sándor Kovács | Francesco, I agree. | |
| Dec 5, 2010 at 2:50 | comment | added | Francesco Polizzi | Sandor, I think one of the problems is, as you said, the scheme structure on "$2L$", and this obviously depends on the ambient space where you see $L$. Results on rigidity on Hartshorne's book apply only to $L$ a line on a smooth cubic surface (the arithmetic genus is $-2$). In this case, it is sure that no possible smoothing exist, neither over the dual numbers nor over a curve. Probably your construction provides another nilpotent structure, which is smoothable. The arithmatic genus must be $-1$, so one possibility is the scheme "$2L$" where $L$ is a fibre of a ruled surface... | |
| Dec 5, 2010 at 2:00 | comment | added | Sándor Kovács |
Also, how about taking the example above and replacing $X$ with $L\times {\rm Spec} \, k[\varepsilon]/\varepsilon^2$?
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| Dec 5, 2010 at 1:58 | comment | added | Sándor Kovács | You are right, I just erased the part about the irreducibility as you posted your comment. But, then, what's wrong with my argument (see below). | |
| Dec 5, 2010 at 1:57 | comment | added | Sándor Kovács | Francesco, if the base is a smooth curve, then flatness only requires that all associated points (in this case irreducible components) dominate the base. I think this is OK. | |
| Dec 5, 2010 at 1:56 | comment | added | Francesco Polizzi | I'm a bit confused. The double projective line cannot be the central element of a flat family whose general element is the union of two disjoint lines (as your example seems to suggest), because the arithmetic genus is different (0 instead of -1). It seems to me that this argument does not involve the irreducibility of the total space. Or am I missing something? | |
| Dec 5, 2010 at 1:51 | answer | added | Sándor Kovács | timeline score: 2 | |
| Dec 5, 2010 at 0:49 | comment | added | Francesco Polizzi | Sandor, are you sure that you obtain a flat family in this way? If so, where is the mistake in my example of double projective line? | |
| Dec 5, 2010 at 0:35 | comment | added | Sándor Kovács | So, you don't really have a scheme to start with, but a reduced scheme and you want a family for which that reduced scheme is the support of the special fiber with whatever scheme structure, right? Otherwise, for instance if your $X$ is reduced, then this is not going to happen. In the case when you don't care about the scheme structure, take two trivial families of your reduced scheme over the same base and glue them at $t_0$ like in your example. | |
| Dec 4, 2010 at 23:16 | answer | added | Francesco Polizzi | timeline score: 3 | |
| Dec 4, 2010 at 22:44 | history | asked | anon | CC BY-SA 2.5 |