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Mar 9, 2010 at 5:26 vote accept Anonymous
Mar 8, 2010 at 23:02 comment added Dror Speiser That won't be enough, specifically in Kevin's example. Taking all the 2-torsion points amounts to taking the splitting field of f(x), in which the maximal order can still not be generated by the coordinates of the points. Finding the maximal order of a cubic field (and its normal closure) is at least as hard as finding the squarefree factorization of the discriminant, and that is considered a hard problem. Hence, I believe, that at least for the 2-torsion case, you cannot find a nice description for the ring of integers.
Mar 8, 2010 at 21:46 comment added Anonymous @Kevin: thanks for your answer. Obviously I don't know anything, but maybe it will be better if I adjoin all the n-torsion points to Q and find the ring of integers in that field instead?
Mar 8, 2010 at 21:34 comment added Pete L. Clark @Kevin: I think that comment may be worthy of an answer in and of itself, possibly augmented by some remarks about rings of integers in cubic fields (they can already be complicated, right?).
Mar 8, 2010 at 21:26 comment added Kevin Buzzard To expand a little on Pete's answer: imagine a 2-torsion point in the curve y^2=f(x). It's of the form (a,0) with a a root of f(x). Now f(x) can be any cubic with distinct roots, so Q(a) can be (for example) any degree 3 field, and in this case a can be any element of it that isn't in Q.
Mar 8, 2010 at 21:25 comment added Pete L. Clark I think people will probably see this answer and respond to it accordingly. Let's wait a few hours and see if that's actually the case.
Mar 8, 2010 at 21:21 comment added Anonymous Thanks for your enlightening input. If you don't mind, could you paste your response to my question? or make it an entirely new question? so other people can respond to "the correct question"?
Mar 8, 2010 at 21:01 history edited Pete L. Clark CC BY-SA 2.5
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Mar 8, 2010 at 20:56 history answered Pete L. Clark CC BY-SA 2.5