Consequences of the Birch and Swinnerton-Dyer Conjecture?

Before asking my short question I had made some research. Unfortunately I did not find a good reference with some examples. My question is the following

What are the consequences of the Birch and Swinnerton-Dyer Conjecture ?

I have read the official statement of the conjecture in Clay Institute website and followed a very nice talk by Manjul Bhargava...

Just to be clear enough, I'm not interested on the explanation of the conjecture, I would like to know what are the consequences.

How can I make this question C.W. ?

• Try to google 'congruent number bsd' to get a consequence of this conjecture. – Sylvain JULIEN Mar 30 '17 at 21:05
• This is probably not the kind of consequences you're looking for, but it is worthwhile to note that the Birch and Swinnerton-Dyer conjecture contains as a "mere" prerequisite the assertion that the Tate-Shafarevich group of elliptic curves is finite. This conjectural claim is of huge importance as it implies that global duality theory for elliptic curves (or abelian varieties in general) is well behaved. There are quite a lot of results in arithmetic geometry which are conditional on that fact. – Yonatan Harpaz Mar 30 '17 at 21:31
• Ram Murty and I recently showed that if elliptic curves over number fields have an automorphic L-function and if the rank part of twisted BSD holds, then Hilbert's tenth problem for the ring of integers of any number field is undecidable (actually we require less, but this formulation fits into a comment). Alternatively, by Mazur-Rubin one knows that finiteness of Sha would have the same consequence. – Pasten Mar 31 '17 at 1:15
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• Some computational questions about points on elliptic curves, as well as statements about elliptic curves on average, are rather easier to approach on the $L$-function side (assuming BSD is true) than on the elliptic curve side. – Greg Martin Mar 31 '17 at 19:58

There is a theorem of Michael Stoll that there are no $c\in\mathbb Q$ such that the polynomial $x^2+c$ admits a periodic 6-cycle starting at some $a\in\mathbb Q$, but the theorem is contingent on the Birch-Swinnerton-Dyer conjecture being true for the Jacobian $J$ of a certain curve $C$ appearing in Stoll's paper. The reason it's needed is because he needs to know the rank of $J(\mathbb Q)$, but the genus of $C$ is too large for him to do an explicit descent. So he's "just" using the rank part of BSwD. Here's the reference:

MR2465796 Stoll, Michael Rational 6-cycles under iteration of quadratic polynomials. LMS J. Comput. Math. 11 (2008), 367–380.

• Stoll's variety $J$ is an absolutely simple abelian 4-fold; so this is not really Birch--Swinnerton-Dyer per se, but the generalisation of BSD to abelian varieties of arbitrary dimension due to Tate. I feel rather silly pointing this out to @JoeSilverman given that one of the standard references for this generalised conjecture is his book with Hindry, but I feel it should be clarified that this is not quite the original BSD conjecture! – David Loeffler Mar 31 '17 at 8:56

Implicit in the BSD conjecture are two other basic conjectures about elliptic curves: the Parity Conjecture and the finiteness of the Tate-Shafarevich group. Most applications I know follow from the rank part, but refined BSD will allow formulas for the order of the Tate-Shafarevich group for instance.

Two specific well-known elementary applications are:

• Tunnell's solution to the Congruent Number Problem
• Rodriguez-Villegas and Zagier's solution to which primes are sums of two rational cubes

These are both contingent on BSD for some cases. Here are a couple other consequences:

• There exists an elliptic curve/modular form of analytic rank 4. (Note the existence of an elliptic curve of analytic rank at least 3 is used to solve the Gauss class number 1 problem.)
• (example) Let $E$ be an elliptic curve of conductor 17, and $-d$ a negative prime discriminant. Then the $-d$-th quadratic twist of $E$ has infinitely many rational points if and only if the Legendre symbol $(-d/17) = +1$.

It would also provide an analytic way to attack Goldfeld's conjecture on average ranks of elliptic curves.

• 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. – Noam D. Elkies Mar 31 '17 at 3:37
• @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.) – Kimball Mar 31 '17 at 4:26
• "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.) – Noam D. Elkies Mar 31 '17 at 14:55