Skip to main content
5 events
when toggle format what by license comment
May 14, 2018 at 3:36 comment added Ashvin Swaminathan Qing Liu's proof is great, but I think you can prove this without any algebraic geometry. If you add a "useless" generator $f_{n+1}$ to the given list, then the corresponding row of the Jacobian is a linear combination of the other rows, so adding such "useless" generators doesn't change the rank. Thus, if $f_{r+1}, \dots, f_n$ and $g_{r+1}, \dots, g_m$ are two different lists of generators, the associated Jacobians have the same rank as the Jacobian associated to the combined list $f_{r+1}, \dots, f_n, g_{r+1}, \dots, g_m$.
Nov 3, 2011 at 23:27 vote accept Nicolás
Nov 3, 2011 at 20:56 answer added Qing Liu timeline score: 13
Nov 3, 2011 at 18:19 answer added Greg Muller timeline score: 3
Nov 3, 2011 at 18:09 history asked Nicolás CC BY-SA 3.0