Timeline for Consequences of the Birch and Swinnerton-Dyer Conjecture?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 31, 2017 at 14:55 | comment | added | Noam D. Elkies | "Other case" = Legendre symbol $-1$? Yes, that's what the 2-descent should give in that case. (For curves like 17B, 2-descent is basically elementary: it's the technique introduced by Fermat for $y^2 = x^3-x$. It doesn't always show the rank is zero, but in simple cases that often happens.) | |
| Mar 31, 2017 at 13:29 | history | made wiki | Post Made Community Wiki by Todd Trimble | ||
| Mar 31, 2017 at 4:26 | comment | added | Kimball | @NoamD.Elkies I haven't thought about a Heegner point construction for this example, so I'm not sure if one can avoid BSD for that or not. But I suppose one still needs to use nonvanishing of an $L$-value to get finiteness of $E(\mathbb Q)$ (i.e., a known case of BSD) for the other direction, yes? Or is there an easy way to get finiteness for such curves? (I don't work on elliptic curves.) | |
| Mar 31, 2017 at 3:37 | comment | added | Noam D. Elkies | Does this last "(example)", and others like it, actually require BSD? Presumably the primality condition forces the rank to be at most $1$ by $2$-descent (such a curve, e.g. 17B: $y^2 = x(x+1)(x-16)$, has at least one rational $2$-torsion point), and then the Legendre symbol makes the rank conjecturally odd; but at least for the "congruent number" curve $y^2 = x^3 - x$ one can prove directly that the Heegner-point construction yields a point of infinite order on the relevant quadratic twist when the discriminant is prime, and the same could be true here too. | |
| Mar 31, 2017 at 3:27 | history | answered | Kimball | CC BY-SA 3.0 |