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The Birch and Swinnerton-Dyer Conjecture is well known in the current literature

http://en.wikipedia.org/wiki/Birch_and_Swinnerton-Dyer_conjecture

My question is about the possible equivalent statements of the Birch and Swinnerton-Dyer Conjecture

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closed as unclear what you're asking by Alex B., Tilman, Stefan Kohl, Chris Godsil, Yemon Choi Jul 20 at 16:22

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

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I actually have no idea what your question is from this. Saying what your question is about is quite a bit different than asking a question directly. –  Matt Mar 20 '13 at 16:04

2 Answers 2

Weaker version of BSD (Parity Conjecture):

$$(-1)^{\mathrm{rank}(E/K)} = w(E/K),$$ where $w(E/K)$ ( +1 or -1) is the global root number of $E/K$.

Standard statement:

BSD I:

If $K$ is a number field and $E$ is an elliptic curve over $K$, then $$\mathrm{ord}_{s=1} L(E/K,s) = \mathrm{rank} (E/K), $$ where $\mathrm{rank}(E/K)$ := Analytic rank of $E$ over $K$ := Mordell-Weil rank of $E$ over $K$.

BSD II: The order of $Ш$ is finite and the leading coeffcient of $L(E/K,s)$ at $s=1$ is given by $$\lim_{s \to 1} \frac{L(E/K,s)}{ (s-1)^r} = \frac{R.|Ш|.C}{\sqrt{{\triangle}_K} {|T|}^2 },$$ where $r$ is the Mordell-Weil rank of $E/K$, $R$ is the regulator of $E/K$ (with respect to the Neron-Tate height pairing), $|Ш|$ is the order of the Tate-Shafarevich group, $|T|$ is the order of the torsion group, $\triangle_K$ is the discriminant of $K$ and $C$ = $\prod_{v} c_{v} $ is the product of the local tamagawa numbers ($v$ varies over places of $K$).

When $K$ is a function field over a finite field of +ve characteristic,

$\mathrm{ord}_{s=1} L(E/K,s) = \mathrm{rank} (E/K) \iff |Ш| < \infty \iff |Ш_l^{\infty}| < \infty$ for some $l \iff \mathrm{ord}_{s=1} L(E/K,s) \leq \mathrm{rank} (E/K)$

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