# What is the Birch and Swinnerton-Dyer Conjecture – Part 1

Yesterday, I watched a talk by Manjul Bhargava called What is the Birch and Swinnerton-Dyer Conjecture?. I think he did a really good job in explaining the conjecture on a level that can be understood by a non-expert yet mathematically inclined person.

In the caption, there is a link to these lecture notes (not written by Bhargava himself). I thought it might be interesting to go through these lecture notes and explore the territory of topics mentioned in the talk (especially in the introductory part of the talk). Therefore, I will walk through the lecture notes in this blogpost (and one or two following posts) and try to look into the mentioned theorems in a bit more detail.

The first two paragraphs are:

Manjul Bhargava spoke about elliptic curves and the Birch and Swinnerton-Dyer
Conjecture (BSD), one of the most fundamental probelms in number theory. He
explained its origin and its statement in elementary terms, as well as summarising
progress towards a proof, including results that were emerging in the workshop taking place during week of the Research Conference.
A central theme of number theory is the search for rational solutions of polynomial
equations. For a polynomial in one variable with integer coefficients, the problem is
easily solved by using the Rational Root Theorem.

That sound interesting. Let’s see what the Rational Root Theorem tells us. I’m taking the following material from Jared Weinstein’s Number Theory course.

Rational Root Theorem: Suppose the polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0$ has coefficients $a_i \in \mathbb{Z}$. If $p/q$ is a fraction in lowest terms which is a root of $f(x)$, then $q|a_n$ and $p|a_0$.

Proof: The fact that $p/q$ is a root of $f(x)$ means that $f(p/q) = 0$. After
clearing away denominators, this becomes $a_n p^n + a_{n-1} p^{n-1} q + \ldots + a_1 p q^{n-1} + a_0 q^n = 0$. Now, we move the last term to the right and factor out $p$: $p(a_n p^{n-1} + a_{n-1} p^{n-2} q + \ldots + a_1 q^{n-1}) = -a_0 q^n$.
Since the left hand side is obviously divisible by $p$ the right hand side must be as well, so $p|a_0 q^n$. But since $p$ is coprime with $q$ but divides a product containing $q$, it must divide $a_0$ (by the generalized form of Euclid’s Lemma).
The proof that $q|a_n$ is similar. $\Box$

The main topic of Bhargava’s talk is the question whether certain polynomial equations have rational roots. Using the Rational Root Theorem, we can now answer this question for, e.g., the polynomial $x^2 - 2 = 0$. We know, by taking the square root of this equation that the solution over the real numbers is $\pm \sqrt{2}$. But with the RRT we can now show independently that there cannot be a rational solution. Observe that a rational solution $p/q$ would have to fullfill $p|-2$ and $q|1$. Therefore, the only possible solutions would be $\pm 1, \pm 2$. However, by plugging them into the equation, we see immediately that they are no solutions.

In the next post we will look at the next paragraph of the lecture notes that discuss polynomials of degree 2 in 2 variables.