Timeline for How do you compute the primes of bad reduction?
Current License: CC BY-SA 3.0
8 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 13, 2012 at 17:57 | answer | added | François Brunault | timeline score: 5 | |
| Dec 13, 2012 at 17:31 | history | edited | R.P. | CC BY-SA 3.0 |
added 84 characters in body
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| Dec 13, 2012 at 16:49 | comment | added | R.P. | That makes a lot of sense, thanks! Your second remark agrees with what I thought myself, I just thought it made the notation simpler if I restricted to $k=1$. | |
| Dec 13, 2012 at 16:18 | comment | added | François Brunault | In the case of a subscheme Y of arbitrary pure dimension d, then I think you should consider the ideal generated by the F_i's together with all (n-d)-minors of the jacobian matrix. | |
| Dec 13, 2012 at 15:55 | comment | added | François Brunault | Geometrically, taking the resultant of two polynomials corresponds to compute the projection of an intersection. This approach only gives an upper bound for S, because two projections might intersect while the original objects don't (but I don't have a counter-example where all orderings are taking into account). Anyway, a possibly better method would be to use Gröbner bases over the integers, see magma.maths.usyd.edu.au/magma/handbook/text/1112#12186 At least for hypersurfaces you will see the integer N you're interested in as the first member of your Gröbner basis. | |
| Dec 13, 2012 at 13:08 | history | asked | R.P. | CC BY-SA 3.0 |