Timeline for Control on the locus of bad reduction for divisors
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 25, 2022 at 22:35 | comment | added | Jason Starr | One example of this is plane conics in $\mathbb{P}^2_K$. | |
| Jul 25, 2022 at 22:32 | comment | added | Dori Bejleri | @JasonStarr Yes that makes sense thanks! | |
| Jul 25, 2022 at 22:21 | comment | added | Jason Starr | @DoriBejleri To get such an upper bound, you also need to bound the height of the "moduli point" $[D]$ inside its complete linear system (such a height is sometimes called a "moduli height"). | |
| Jul 25, 2022 at 21:05 | comment | added | Dori Bejleri | I would guess that the best you can hope for is an upper bound which depends on the degree of $D$ measured with respect to a fixed ample line bundle on $\mathcal{X}$. Of course fixing the degree gives an upper bound on $f_\mathfrak{p}(D)$ but the trickier part is to use the fact that $D$ varies within a finite type Hilbert scheme to show there exists an upper bound on the number of primes of bad reduction. Also making this effective sounds very hard. | |
| Jul 25, 2022 at 15:41 | comment | added | Jason Starr | There is no upper bound depending only on $\mathcal{X}$ but not on $D$. So long as the dimension of $X$ is at least $2$, a union $\mathcal{D}_p$ of $s$ (pairwise distinct) general very ample divisors in a single fiber $\mathcal{X}_p$ will admit a "lift" $D$ that is geometrically irreducible. | |
| Jul 25, 2022 at 15:31 | comment | added | manifold | yes, $D$ has to be geometrically irreducible. | |
| Jul 25, 2022 at 14:43 | comment | added | Daniel Loughran | Do you want $D$ to be geometrically irreducible, rather than just irreducible? Otherwise $S_D$ can be infinite. | |
| Jul 25, 2022 at 12:58 | history | edited | manifold | CC BY-SA 4.0 |
added 65 characters in body
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| Jul 25, 2022 at 12:53 | history | asked | manifold | CC BY-SA 4.0 |