Skip to main content
8 events
when toggle format what by license comment
Oct 5, 2011 at 17:44 comment added David Zureick-Brown @Felipe Right. I was taking K = Q.
Oct 5, 2011 at 17:36 comment added Felipe Voloch @David I don't think there is a Frey-Mazur conjecture over an arbitrary field.
Oct 5, 2011 at 15:42 comment added David Zureick-Brown Phrased this way, I think this is equivalent to the Frey-Mazur conjecture, which is open.
Oct 5, 2011 at 14:34 comment added Adam Harris Yes - sorry I meant non-isogenous over $\bar{K}$.
Oct 5, 2011 at 14:24 comment added Alex B. Adam, quadratic twists are usually not isogenous. Certainly, for any $E$, there exist infinitely many quadratic twists of $E$ that are not isogenous to $E$.
Oct 5, 2011 at 14:22 comment added David Zureick-Brown To nit-pick -- the answer is still no, for the same reason. Quadratic twists are isomorphic over an extension, but aren't necessarily isogenous over their field of definition.
Oct 5, 2011 at 14:02 comment added Adam Harris Thanks David. I have edited the question so that they are non-isogenous now
Oct 5, 2011 at 13:49 history answered David Zureick-Brown CC BY-SA 3.0