Timeline for Does every ample divisor "span" a hyperplane?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 7, 2013 at 8:48 | vote | accept | Tomasz Lenarcik | ||
| Oct 7, 2013 at 8:46 | comment | added | Tomasz Lenarcik | Ok, now I see that this example is in fact pretty general and can be modified so that the intersection set is almost arbitrary. If $\gamma(x)=0$, $\deg\gamma=d$ is any curve, then consider Veronese embedding of degree $2d$. The hypersurfaces $\gamma(x)^2$ and $\gamma(x)\delta(x)$ (sopposing that $\deg\delta=d$) both contain the curve $\gamma(x)=0$. | |
| Oct 6, 2013 at 23:41 | comment | added | Lev Borisov | If you consider any $X$ and any ample divisor $D$ on it, then you can take an embedding by $kD$. One of the hyperplanes will be just $kD$, and so on... | |
| Oct 6, 2013 at 22:48 | comment | added | Tomasz Lenarcik | You're right! It wasn't that hard after all :) It also made me realize that I forgot about some assumptions about the geometry of $X\cap H$. Please check out the updated question. | |
| Oct 6, 2013 at 22:39 | history | answered | Lev Borisov | CC BY-SA 3.0 |