The Birch and Swinnerton-Dyer Conjecture, Another $1,000,000 Problem

31.01.2018 |

Episode #7 of the course Great math problems for the 21st-century mind by John Robin


Like Fermat’s Last Theorem, the Riemann Hypothesis looms on the horizon as the behemoth that interests almost every mathematician. Even non-mathematicians who enjoy learning about and appreciating the beauty of math can now explore it deeper, thanks to a great deal of public information that researchers have shared to turn the hunt for a proof into a more open-source endeavor.

Now, the Riemann Hypothesis might be our current day’s new Fermat’s Last Theorem, but there are five other millennium problems still, also worth $1,000,000. I will spend the next few days telling you about my other favorite one.

This is the Birch and Swinnerton-Dyer Conjecture, and we’ve already touched briefly on its domain. Do you remember Lesson 3 when I explored how Andrew Wiles proved Fermat’s Last Theorem? The beautiful curves he worked hard to understand so he could show that the special one of the form y2 = x(x − an)(x + bn) could not have a modular form?

I introduced you to elliptic curves for a reason. As a reminder, they have the form:

y2 = x3 + ax + b

Now, I also mentioned that every proper elliptic curve has a modular form. I said this is found by counting the number of solutions in number fields of limited size, growing progressively and counting again, and correlating the number of solutions, in a field of a given size, to a polynomial whose coefficients match the size of the field.

In the next two days, we will turn to looking more precisely at what is happening here.

Let’s take our friend from Lesson 3, y2 = x3 − x.

Start with the number field of size 3. So, this means we only have the numbers 0, 1, and 2. That gives us only the (x,y) pairs (0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2).

If we try every one of these out (plugging them in for (x,y) in the equation y2 = x3 − x), we will find that only three of them work:

(0,0), (1,0), and (2,0)

I should note that when I say our number field is limited to size 3, that means when you hit 3, you go back to zero. So for example, 4 = 1, 5 = 2, 6 = 0, and so on. Think of a clock, where when you hit 12, that’s 0 again. Here, the “12” is the number “3.”

Now, we count three solutions. We call the number of solutions Np, where p is a prime number, and we say here for field of size p = 3 that N3 = 3.

We can do this for field sizes of all prime numbers, 3, 5, 7, 11, 13, etc. Each time, we count the number of solutions, Np, and keep track of them.

This is hard to do by hand. For our friend y2 = x3 − x, for example, you might be okay with checking all possible pairs (0,0), (0,1), … (4,4) for p = 5, but for p = 7, 11, and on, you’ll need a computer to calculate all these. (When I was a research student, I had to write a program to calculate all these numbers and look for a pattern.)

The point is, this search for various proper elliptic curves unearth a pattern, and this is what the Birch and Swinnerton-Dyer Conjecture is all about. Tomorrow, we’ll get some more context so in Lesson 9, we can dig in with more detail.


Recommended book

A Friendly Introduction to Number Theory by Joseph Silverman


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