Timeline for What are non-trivial examples of non-singular blow-ups of a non-singular variety?
Current License: CC BY-SA 2.5
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| when toggle format | what | by | license | comment | |
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| Apr 22, 2010 at 21:47 | vote | accept | jlk | ||
| Apr 19, 2010 at 22:02 | comment | added | jlk | @quim, thanks for the explanation. What does it mean for J to be "complete" in this context? I think it'd be useful to people if you copied this explanation to the original question: mathoverflow.net/questions/21232/when-is-a-blow-up-non-singular | |
| Apr 19, 2010 at 8:26 | comment | added | quim | By the way, the surface singularities obtained blowing up integrally closed ideals are called "sandwiched," because they lie between two smooth surfaces, the resolution on top and the surface before blowup below. I don't know if "sandwiched" singularities of higher dimensional varieties have been studied, maybe a MathSciNet search is in order... | |
| Apr 19, 2010 at 8:23 | comment | added | quim | Let I be an integrally closed ideal, and let $J_1$, ... $J_r$ be the irreducible complete ideals associated to the base points of I (ie, blow up each point on the cosupport of I and take the strict transform; iterate until I becomes trivial). I believe that the surface obtained blowing up I an integrally closed ideal is nonsingular if and only if each of the $J_i$ appears in the Zariski factorization of I. The proof should be in some paper by Spivakovsky, maybe MR1053487. | |
| Apr 19, 2010 at 1:29 | comment | added | jlk | @Karl, if you figure out what Zariski factorization says about this question generally, then I'd be interested to hear. | |
| Apr 17, 2010 at 22:46 | comment | added | Karl Schwede | Bah, ignore the first sentence of my previous comment! The two ideals are already the SAME (no need to take the integral closure). In particular, if an ideal contains $xy$, then it also contains $x^2 y$. Sorry about that. | |
| Apr 17, 2010 at 19:51 | comment | added | Karl Schwede | If you'll notice, $(x,y)(x^2,y) = (x^3, x^2y, xy, y^2)$ which is the integral closure of the ideal you mentioned $(x^3,xy, y^2)$. Generally speaking, we should probably limit ourselves to blowing up integrally closed ideals (as ideals have the same normalized blow-up as their integral closures). Finally, blowing up products of ideals is essentially like blowing up a series of ideals in succession. In this case, blowing up something like $(x,y)(x^2,y)$ is like blowing up the origin $(x,y)$ and then a point on the exceptional $P^1$ (this is related to "Zariski factorization") | |
| Apr 17, 2010 at 17:22 | history | answered | mdeland | CC BY-SA 2.5 |