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Jan 21, 2010 at 22:10 vote accept natura
Jan 21, 2010 at 22:10 vote accept natura
Jan 21, 2010 at 22:10
Jan 21, 2010 at 12:37 comment added Kevin Buzzard @basic: if your $y^2=x^3+tx$ were a trivial family, then it would be equal to the family y^2=x^3-x. The latter curve has all its 2-torsion, so if your family were trivial then the 2-torsion in your curve over S:=Spec(k[t][t^-1]) (considered as a global object) would be four copies of S. But the two-torsion in the t-curve isn't four copies of S, as you can easily check (you are taking a square root of -t).
Jan 21, 2010 at 8:44 comment added Pete L. Clark @basic: Your example uses "quartic twists" rather than quadratic twists (for that, see Chapter 10 of Silverman's Arithmetic of Elliptic Curves) but it is otherwise the same (and there are still infinitely many isomorphism classes of fibers over k, if k is any number field). So yes, that works too.
Jan 21, 2010 at 7:49 comment added Kevin Buzzard There are infinitely many kinds of fibres! Q^x/(Q^x)^2 is an infinite group: it's a vector space over Z/2Z with basis -1 and the prime numbers.
Jan 21, 2010 at 7:18 comment added natura I think I get what you mean. Those fibers are isomorphic over C but not over Q, right? My example seems to also work out, except that the fibers seem to be more than 2 kinds (over Q)?? Thank you!
Jan 21, 2010 at 7:03 comment added natura Is this right? Consider the family of elliptic curve $y^2=x^3+tx$ defined over Spec(k[t]), every fiber (except for t=0) is an elliptic curve with the same j-invariant 1728. but then how to demonstrate that this family is not a trivial family (i.e. tensor product of Spec(k[t]) and a single elliptic curve)?
Jan 21, 2010 at 6:38 history edited Pete L. Clark CC BY-SA 2.5
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Jan 21, 2010 at 5:44 history answered Pete L. Clark CC BY-SA 2.5