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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jan 15, 2017 at 13:53 comment added user19475 I think Mordell is a consequence of Shafarevich, not vice versa.
Mar 2, 2014 at 13:35 history edited Colin McLarty CC BY-SA 3.0
Clarified an issue.
Mar 2, 2014 at 2:49 vote accept Colin McLarty
Mar 2, 2014 at 2:43 comment added Qiaochu Yuan Yes, Sage can do it. First, it can find a basis of the appropriate space of forms (sagemath.org/doc/reference/modfrm/sage/modular/modform/…). Second, it can compute the action of the Hecke operators on this basis (sagemath.org/doc/reference/modfrm/sage/modular/modform/…), so you can diagonalize this action.
Mar 2, 2014 at 2:40 comment added Colin McLarty Yes of course searching a finite list is effective. Do we have an effective way to create the finite list?
Mar 2, 2014 at 2:39 comment added Qiaochu Yuan In this case there's no obstacle to searching the space, since the list of eigenforms is finite and we know how to find it. (Right?)
Mar 2, 2014 at 2:38 answer added Noam D. Elkies timeline score: 31
Mar 2, 2014 at 2:37 comment added Colin McLarty Non-effectiveness is no obstacle to validity of the proof. But I wonder if current proofs are effective.
Mar 2, 2014 at 2:33 comment added Qiaochu Yuan The non-effectiveness of the proof is not necessarily an obstacle. The proof guarantees that a certain search (namely, if I'm not horribly mistaken, the search for a cuspidal eigenform with the right Fourier coefficients) will return a result, but one can perform this search knowing that it's going to return a result without worrying about why one knows it's going to return a result (because the appropriate space of forms, for fixed $N$, is finite-dimensional). Right?
Mar 2, 2014 at 2:18 history asked Colin McLarty CC BY-SA 3.0