December 24, 2023: Introduction to arithmetic geometry: plane quadrics (for a general audience)

This is my first blog entry for the mathematically interested public.

Basic questions in arithmetic geometry. My research interests lie in the field of number theory, especially arithmetic geometry. One studies properties of the natural numbers 𝐍={0,1,2,}\mathbf{N}= \{0,1,2,\ldots\}. One basic question is to ask for integral or rational solutions to systems of equations fi(X1,,Xn)=0f_i(X_1, \ldots, X_n) = 0 with fi𝐙[X1,,Xn]f_i \in \mathbf{Z}[X_1,\ldots,X_n] polynomials in several variables. These are called Diophantine equations after Diophantus of Alexandria (maybe 250 AD). They define geometric objects, called varieties or schemes. These algebraic objects are investigated using both algebraic and geometric methods.

Why is it interesting? One of the reasons why we are interested in them is that finding their solutions is an easy-to-state problem, but solving them can be extremely hard and lead to significant and deep theoretical advances in number theory and related fields. An example of this is the solution of the 17th century Fermat equation an+bn=cna^n + b^n = c^n, which required late 20th century mathematics. Number theory uses almost every other area of pure mathematics. Many mathematicians developed deep theories to tackle diophantine questions, and often one cannot easily see anymore what their connection to the original questions is. The abstract objects may even become more interesting than the originating equations.

Undecidable in general. However, the question whether a diophantine equation has an integral solution is undecidable (Hilbert’s 10th problem, theorem of Matiyasevich–Robinson–Davis–Putnam 1970). (There is no algorithm that enumerates the set of parameters of a parameterized diophantine equation for which there exists a solution.) Restricting the class of diophantine equations considered, the problem can or might be solvable, though. For example, it is solvable for (at most) quadratic equations, which follows from the Hasse–Minkowski theorem, a local-global principle. The class of curves, i.e., equations in only two variables is still open.

Curves. Aside from 00-dimensional objects, which are algebraically also very interesting, the easiest objects are curves, i.e., 11-dimensional objects. There is a discrete invariant of curves, called the genus. There are infinitely many “different” (i.e., non-isomorphic) curves of any genus g0g \geq 0. In this blog entry, I will talk about the simplest ones, those of genus 00.

Pythagorean triples. A classical example is the (projective) equation X2+Y2=Z2.% \label{eq:PythagoreanTriples} X^2 + Y^2 = Z^2. By the Theorem of Pythagoras, its positive integer solutions correspond to right triangles with integral side lengths.

Excluding the trivial case Z=0Z = 0 (because it implies X=Y=0X = Y = 0 in integers) and dividing by ZZ, we can equivalently determine the rational solutions to x2+y2=1x^2 + y^2 = 1 (with x=X/Z,y=Y/Zx = X/Z, y = Y/Z). Geometrically, the solutions to this equation form a circle:

The circle x2+y2=1x^2 + y^2 = 1 of rational points of the quadric X2+Y2=Z2X^2 + Y^2 = Z^2.

The integral solutions to the projective equation correspond to rational solutions to the affine equation (with the exception of the points at infinity). The algebraic question of finding solutions to the defining equation translates into the geometric question of finding points on the curve.

Describing all solutions. Now we do some geometry: There is one rational solution (1,0)(1,0), and if one draws a line through (1,0)(1,0) with slope t𝐐t \in \mathbf{Q}, this line intersects the circle in another point with rational coordinates (Bézout’s theorem—here we use that the equation has degree 22), and all rational points can be obtained in this way. By calculating the coordinates of the intersection point, one gets an algebraic parameterization of the solutions: (x,y)=(1t21+t2,2t1+t2),t𝐐\begin{aligned} %\label{eq:Pythagorean parameterization} (x,y) = \Big(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\Big), \quad t \in \mathbf{Q} \end{aligned} Reverting the process of dividing by ZZ again yields a parameterization of the original equation (X,Y,Z)=k(m2n2,2mn,m2+n2)(X,Y,Z) = k(m^2 - n^2, 2mn, m^2 + n^2) with m,n,k𝐙m,n,k \in \mathbf{Z}. For example, letting m=2,n=1,k=1m = 2, n = 1, k = 1 yields the smallest and most well-known Pythagorean triple 32+42=523^2 + 4^2 = 5^2.

Plane quadrics. More generally, plane quadrics (i.e., curves described by quadratic polynomials in two variables) ax2+by2=cax^2 + by^2 = c belong to the simplest group of curves, those of genus 00. They have either no or infinitely many solutions, and there is a practical algorithm to decide what is the case. If there are infinitely many of them, one can parameterize them with one parameter t𝐐t \in \mathbf{Q}, as above.

An elliptic curve. A more complicated example of a curve is the (affine) elliptic curve y2=x3xy^2 = x^3 - x with real picture:

The real points of the elliptic curve y2=x3xy^2 = x^3 - x.

It is a curve of genus 11. Its properties are better described by the complex points, which form a complex torus.

One can show that the only rational solutions are (0,0),(±1,0)(0,0),(\pm1,0). There are also elliptic curves with infinitely many points. Elliptic curves are much more complicated than plane quadrics, and several extremely deep conjectures, most notably the Birch–Swinnerton-Dyer (BSD) conjecture, have been formulated about them.

There is also a method, motivated by geometry, to generate new solutions out of old solutions. However, there is no simple geometric parameterization of all solutions anymore.

Next time. I will write more about elliptic curves in my next blog entry for the public.