Timeline for Two different resolutions of a three fold
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 19 at 17:00 | comment | added | Sasha | It is easy to see that the Picard group of $Y_i$ has rank 1; therefore the morphism $Y_i \to X$ has no factorizations into a composition of two nontrivial projective morphisms. | |
| Apr 18 at 15:35 | comment | added | George | Is it obvious that $Y_i\rightarrow Z$ must be isomorphism? Could you tell me why? | |
| Apr 18 at 15:13 | comment | added | Lazzaro Campeotti | @George: for $i=0,1$, if $Y_i \rightarrow Z \rightarrow X$ is a factorisation with $Z$ smooth then $Y_i \rightarrow Z$ must be an isomorphism over $X$. Since $Y_0$ and $Y_1$ are not isomorphic over $X$, this means that the two morphisms $Y_i \rightarrow X$ cannot both factor through a morphism $Z \rightarrow X$ for some smooth $Z$. | |
| Apr 17 at 11:24 | comment | added | George | Thanks. I understand why these blowup are resolutions. How to show this three fold has no minimal resolution by using these fact. | |
| Apr 17 at 4:43 | comment | added | Sasha | This is a computation. Over the complement of the singularity, the blowup center is a Cartier divisor, so the blowup is the identity. | |
| Apr 17 at 0:26 | comment | added | George | Why are these blow up resolutions? Even if we restrict these maps to the complement of the fiber of singular locus , we cannot get bijection on to it's image | |
| Apr 17 at 0:20 | comment | added | George | Thanks. I would like to ask you a one more question. | |
| Apr 16 at 9:01 | history | answered | Sasha | CC BY-SA 4.0 |