Timeline for How do you compute the primes of bad reduction?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2012 at 19:56 | vote | accept | R.P. | ||
| Dec 16, 2012 at 19:55 | vote | accept | R.P. | ||
| Dec 16, 2012 at 19:56 | |||||
| Dec 14, 2012 at 8:48 | comment | added | François Brunault | (...) which vanish on the potential singular points. Then you have a finite number of matrices over the algebraic closure of $\mathbf{F}_p$ whose rank you need to check, which is probably much easier from a computational point of view. | |
| Dec 14, 2012 at 8:46 | comment | added | François Brunault | Dear René: With the Gröbner basis computation, there are a lot of minors to compute, which is indeed an annoying thing. But I think we can combine both approaches: put some minors in $I$ until $I \cap \mathbf{Z} \neq 0$ (probably a generic choice of $d+1$ such minors will do). Once you get an integer $N$ which gives a reasonable upper bound for $S$ (and assuming you can factor $N$!), you can check whether $X_{\mathbf{F}_p}$ is smooth as you suggest. But for this the Gröbner basis computation is useful, because it will give you, I think, univariate polynomials in each variable (...) | |
| Dec 14, 2012 at 3:11 | comment | added | R.P. | Dear François: Thank you very much. I must say though I wonder: in terms of complexity, one might be worse off with the Gröbner basis calculation than if you would upper-bound $S$ using my method and then for each $p \in S$ would check if $X_{\mathbf{F}_p}$ is smooth. I can't prove this, of course, after all you do have to factor an integer $N$ over whose size you don't really seem to have much control | |
| Dec 13, 2012 at 17:57 | history | answered | François Brunault | CC BY-SA 3.0 |