Timeline for Cohomology and proper base change
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 4, 2014 at 18:27 | comment | added | Piotr Achinger | That's right, $\pi$ is of course not projective, but we can consider the blow-up of $B\times \mathbb{P}^1$ instead. | |
| May 4, 2014 at 13:46 | comment | added | user45397 | @Achinger: Thanks for the answer. Could you also tell me why $\pi$ is projective? I understand the morphism from $\mathcal{X}$ to $\mathbb{A}^2$ is projective. | |
| May 3, 2014 at 20:41 | comment | added | Piotr Achinger | Good question. That's because $\mathcal{X}$ is integral and the base $B$ is a smooth curve (so "flat" means "torsion free"). See e.g. III 9.7 in Hartshorne's "Algebraic Geometry". | |
| May 3, 2014 at 20:09 | comment | added | user45397 | Why is $\pi$ flat? | |
| May 3, 2014 at 19:29 | history | answered | Piotr Achinger | CC BY-SA 3.0 |