Skip to main content
7 events
when toggle format what by license comment
Sep 29, 2014 at 8:25 vote accept jsm
Sep 21, 2014 at 18:38 answer added Matthias Wendt timeline score: 2
Sep 21, 2014 at 17:52 history edited Matthias Wendt CC BY-SA 3.0
fixed LaTeX
Sep 19, 2012 at 14:08 comment added jsm You're right, but if $C$ has good reduction, then $K_1(\mathcal{C}_p)$ is torsion. I'm not claiming that $t_p=0$ implies $K^'_1(\mathcal{C}_p)=0$.
Sep 17, 2012 at 11:16 comment added Damian Rössler (sorry: "$C$ has good reduction", not "$C^p$ has good reduction")
Sep 17, 2012 at 11:15 comment added Damian Rössler I don't understand (2). If $C^p$ has good reduction, then this would imply that $K_1(C_p)=0$. But $K_1(C_p)$ does not vanish because it contains at least the non-zero elements of ${\bf F}_p$.
Sep 17, 2012 at 10:59 history asked jsm CC BY-SA 3.0