Timeline for Blow up along codimension one closed subscheme
Current License: CC BY-SA 2.5
16 events
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| Jan 7, 2010 at 21:57 | history | edited | Olivier Benoist | CC BY-SA 2.5 |
added 54 characters in body
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| Jan 7, 2010 at 21:54 | comment | added | Olivier Benoist | @Matt Yes, you're so right ! It does'nt change the fact that the fiber over the origin is a projective line. However, and you were right about that too, the total space is smooth, and should coincide with the blow-up of the origin. Sorry about that. I will edit my post. | |
| Jan 7, 2010 at 21:45 | comment | added | mdeland | I think the correct blow-up algebra should be $A[t,u]/(zt - (x+y)u, zu - (x - y)t)$ | |
| Jan 7, 2010 at 20:12 | comment | added | mdeland | I'm confused - is there something wrong with your presentation of the blowup algebra? When $u \neq 0$, I calculate the open affine of your projective variety to be $k[x,y,z,t] / (x^2 - y^2 - z^2, zt - (x + y))$ (here I abuse notation and continue to use $t$ as the affine coordinate). This is singular away from the line $Y$ I think. Also I don't see the relevant calculation exactly by Taisong Jing. | |
| Jan 7, 2010 at 16:09 | comment | added | Olivier Benoist | I don't think so ! In fact, the total space of the blow-up considered here is singular along the exceptional line, as you can see in the chart calculated by Taisong Jing, whereas the blow-up of the cone at the origin is smooth (it is the traditionnal resolution of singularities of the cone) ! Hence, these two blow-ups do not coincide. | |
| Jan 7, 2010 at 15:38 | comment | added | mdeland | Another way to see this example: since $Y$ in $X$ is Cartier away from the singular point, the blow up of $X$ along $Y$ is the same as the blow up of $X$ along the intersection of the line with the singular locus - namely, the vertex of the cone. Then you again see a projective line sitting over the vertex. | |
| Jan 6, 2010 at 22:50 | vote | accept | TJCM | ||
| Jan 6, 2010 at 22:29 | comment | added | TJCM | I see, you are right. I should not have localized on D(z). It is a subalgebra of A_z, but not need to localize. Thus this statement is indeed not true. Thank you! | |
| Jan 6, 2010 at 22:23 | comment | added | Olivier Benoist | If you want to take the fiber over the origin of $k[x,y,z,t]/(z^2-x^2-y^2, zt-(x-y)))$, you just let $x=y=z=0$ and get $k[t]$, which is the line I'm talking about. | |
| Jan 6, 2010 at 22:17 | comment | added | Olivier Benoist | No, it's not. It is the union of two lines, the strict transform of $Y$, and another contracted to the origin by the blow-up as you can see from : $$Proj(\sum_{n\geq 0}I^n)=Proj (A/I)[t,u]/((x+y)u).$$ | |
| Jan 6, 2010 at 22:07 | comment | added | TJCM | No, unfortunately. After localizing on the open set D(u), one would get (k[x,y,z,t]/(z^2-x^2-y^2, zt-(x-y)))_z = (k[y,z,t]/(z^2-y^2-(zt+y)^2))_z = k[z,z^{-1},t], and the fiber of the origin point is indeed finite. As you have seen, the point is that you need to take localization at D(z) if you take the definition as $Proj (\sum_{n>=0}I^n)$, but you don't localize if you take definition as $Proj_A A[t,u](zt−(x+y)u)$. | |
| Jan 6, 2010 at 22:00 | comment | added | TJCM | And after calculation, if you take $Proj (\sum_{n>=0}I^n)$ as the definition of blowing up, then you will see it's indeed quasi-finite in this example. | |
| Jan 6, 2010 at 21:59 | comment | added | Olivier Benoist | We do have the same definition ! And $Proj_A A[t,u](zt−(x+y)u)$ is just the concrete expression of $Proj(\sum_{n\geq 0} I^n)$ in my example ! | |
| Jan 6, 2010 at 21:58 | history | edited | Olivier Benoist | CC BY-SA 2.5 |
edited body; edited body
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| Jan 6, 2010 at 21:53 | comment | added | TJCM | Here I take $Proj (\sum_{n>=0}I^n)$ as the definition of blowing up along V(I) on Spec A. So it is not always equal to $Proj_{A}A[t,u]/(zt-(x+y)u)$. More precisely, it is equal to $Proj_{A}A[t,u]/(zt-(x+y)u)$ when $Proj_{A}A[t,u]/(zt-(x+y)u)$ is integral. But here this expression is not an integral scheme. So this counterexample does not work under our definition of blowing up! | |
| Jan 6, 2010 at 21:46 | history | answered | Olivier Benoist | CC BY-SA 2.5 |