Ciro Santilli @cirosantilli 34

The BSD conjecture states that if your name is long enough, it will always count as two letters on a famous conjecture.

Maybe also insert a joke about BSD Operating Systems if you're into that kind of stuff.

The conjecture states that Equation 1. "BSD conjecture" holds for every elliptic curve over the rational numbers (which is defined by its constants $a$ and $b$)

The conjecture, if true, provides a (possibly inefficient) way to calculate the rank of an elliptic curve over the rational numbers, since we can calculate the number of elements of an elliptic curve over a finite field by Schoof's algorithm in polynomial time. So it is just a matter of calculating $N_p$ like that up to some point at which we are quite certain about $r$.

The Wikipedia page of the this conecture is the perfect example of why it is not possible to teach natural sciences on Wikipedia. A million dollar problem, and the page is thoroughly incomprehensible unless you already know everything!

The elliptic curve group of all elliptic curve over the rational numbers is always a finitely generated group.

The number of points may be either finite or infinite. But when infinite, it is still a finitely generated group.

For this reason, the rank of an elliptic curve over the rational numbers is always defined.

TODO example.

One major difference between the elliptic curve over a finite field or the elliptic curve over the rational numbers the elliptic curve over the real numbers is that not every possible $x$ generates a member of the curve.

This is because on the Equation "Definition of the elliptic curves" we see that given an $x$, we calculate $x^3+ax+b$, which always produces an element $y^2$.

But then we are not necessarily able to find an $y$ for the $y^2$, because not all fields are not quadratically closed fields.

For example: with $a=1$ and $b=1$, taking $x=1$ gives:and therefore there is no $y\in \mathbb{Q}$ that satisfies the equation. So $x=1$ is not on the curve if we consider this elliptic curve over the rational numbers.

That $x$ would also not belong to Elliptic curve over the finite field ${\mathbb{F}}_4$, because doing everything $\mathrm{m}\mathrm{o}\mathrm{d}4$ we have:Therefore, there is no element $y\in {\mathbb{F}}_4$ such that $y\times y=2$ or $y\times y=3$, i.e. $2$ and $3$ don't have a multiplicative inverse.

For the real numbers, it would work however, because the real numbers are a quadratically closed field, and $\sqrt{3}\in \mathbb{R}$.

For this reason, it is not necessarily trivial to determine the number of elements of an elliptic curve.

Mordell's theorem guarantees that the rank (number of elements in the generating set of the group) is always well defined for an elliptic curve over the rational numbers. But as of 2023 there is no known algorithm which calculates the rank of any curve!

It is not even known if there are elliptic curves of every rank or not: Largest known ranks of an elliptic curve over the rational numbers, and it has proven extremely hard to find new ones over time.

TODO list of known values and algorithms? The Birch and Swinnerton-Dyer conjecture would immediately provide a stupid algorithm for it.

This construction takes as input:and it produces an elliptic curve over a finite field of order $p$ as output.

The constructions is used in the Birch and Swinnerton-Dyer conjecture.

To do it, we just convert the coefficients $a$ and $b$ from the Equation "Definition of the elliptic curves" from rational numbers to elements of the finite field.

For example, suppose we have $a=3/4$ and we are using $p=11$.

For the denominator $4$, we just use the multiplicative inverse, e.g. supposing we havewhere $4^{-1}=3\mathrm{m}\mathrm{o}\mathrm{d}11$ because $4\times 3=1\mathrm{m}\mathrm{o}\mathrm{d}11$, related: math.stackexchange.com/questions/1204034/elliptic-curve-reduction-modulo-p