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Mar 17, 2015 at 4:00 vote accept raynor14
Mar 11, 2015 at 10:18 comment added Chris Wuthrich Lie$(A^{\vee})$ is defined to be the $O_K$-dual of the differentials on the Néron model. So if that is not free, neither will Lie$(A^{\vee})$. But a basis of Lie${}_K(A^{\vee})$ is all that you need.
Mar 11, 2015 at 3:18 comment added raynor14 Thank you very much for your answer. Does that mean, there is no integral basis for $Lie(A)$ in the general case then ?
Mar 10, 2015 at 16:43 history answered Chris Wuthrich CC BY-SA 3.0