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Apr 6, 2020 at 8:51 vote accept Joey van Langen
Mar 31, 2020 at 10:22 comment added Chris Wuthrich You are absolutely right. This is up to automorphisms that fix the kernel. Luckily our curves only have $\pm 1$ as automorphisms.
Mar 31, 2020 at 8:43 comment added Joey van Langen You seem to use that every isogeny is uniquely determined by its kernel, but I can multiply any isogeny with a degree 1 endomorphism to get an isogeny with the same kernel, but maybe different eigenvalues. For example replace $ \phi $ by $ - \phi $.
Mar 31, 2020 at 7:13 vote accept Joey van Langen
Mar 31, 2020 at 8:43
Mar 30, 2020 at 15:34 history edited Chris Wuthrich CC BY-SA 4.0
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Mar 30, 2020 at 13:36 history answered Chris Wuthrich CC BY-SA 4.0