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Mar 24, 2017 at 12:15 history edited Tomasz Lenarcik CC BY-SA 3.0
Fix grammar a little bit
Nov 1, 2013 at 9:27 vote accept Tomasz Lenarcik
Nov 1, 2013 at 9:26 history edited Tomasz Lenarcik CC BY-SA 3.0
added summary
Nov 1, 2013 at 6:28 answer added Pete L. Clark timeline score: 3
Nov 1, 2013 at 3:05 comment added Tomasz Lenarcik Actually, I only need this lemma in case when $L$ is a pure transcendental extension of $k$.
Nov 1, 2013 at 3:03 comment added Tomasz Lenarcik That's what I was afraid of :) Hopefully, in my particular use-case I can prove that $C$ has enough $k$-rational points.
Nov 1, 2013 at 1:59 comment added Qiaochu Yuan The argument fails at the last step: you need a $k$-rational point on $\overline{C}$ to carry this step out (if I understand you correctly) and such a point may not exist.
Nov 1, 2013 at 1:51 answer added Abhinav Kumar timeline score: 18
Nov 1, 2013 at 1:43 history asked Tomasz Lenarcik CC BY-SA 3.0