Timeline for When is an algebraic variety $\mathbb{Q}$-factorial?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 29, 2011 at 5:48 | comment | added | naf | Any curve $C$ is quasi-projective. So assuming $C \subset \mathbb{P}^n$ you can find a hyperplane $D$ which intersects $C$ at a given singular point $x$ and no other singular point. Since any smooth point is a Cartier divisor and $D$ is also Cartier, it follows that some multiple of $x$ is the Weil divisor associated to a Cartier divisor. | |
| Aug 28, 2011 at 19:57 | answer | added | Remke Kloosterman | timeline score: 9 | |
| Aug 28, 2011 at 15:55 | vote | accept | Fei YE | ||
| Aug 28, 2011 at 15:41 | comment | added | Fei YE | Thanks. I was thinking of lifting a non-zero multiple of a Weil divisor to a Cartier divisor. Can you explain how to show that a singular curve is $\mathbb{Q}$-Cartier? I was thinking to lift a Weil divisor to the normalization and then maybe push forward to get something useful. But I don't have a clear picture in my mind. | |
| Aug 28, 2011 at 14:46 | comment | added | naf | It depends on how you define $\mathbb{Q}$−factorial in the non−normal case. If you just say that a non−zero multiple of every Weil divisor lifts to a Cartier divisor then singular curves will be $\mathbb{Q}$-factorial. | |
| Aug 28, 2011 at 13:20 | answer | added | Donu Arapura | timeline score: 10 | |
| Aug 24, 2011 at 5:38 | comment | added | Fei YE | Thanks. But is it possible that a nodal-cuspidal plane curve is $\mathbb{Q}$-factorial? | |
| Aug 23, 2011 at 15:32 | comment | added | naf | $\mathbb{Q}$-factoriality is usually defined only for normal varieties since one wants the group of Cartier divisors to be a subgroup of the group of Weil divisors; a singular curve is never normal. | |
| Aug 23, 2011 at 9:19 | history | asked | Fei YE | CC BY-SA 3.0 |