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I decided to start this blog by writing about my favourite proof that can be fit in a blog post and which put me on the track to learn algebraic geometry. I was originally fascinated by the connections that algebraic geometry makes between branches of mathematics and uses these to solve actual problems, which is what I saw nicely demonstrated in the geometric derivation of the formula for the Pythagorean triples. Not only does this geometric method give you a formula for the Pythagorean triples, but the proof also teaches you a novel, general technique for solving Diophantine equations. The proof is familiar to most people who have studied algebraic geometry, as it is presented in many books such as Undergraduate Algebraic Geometry, Miles Reid and Foundations of Algebraic Geometry, Ravi Vakil, and in YouTube videos such as the Abel lecture by Ravi Vakil and this video by Richard Borcherds. This post is aimed at students who have yet to learn algebraic geometry, but in the next post I dive deeper into the modern theory related to this proof.

The idea of the proof is to translate the problem of finding integer triples $(X,Y,Z)\in\mathbb{Z}^3$ satisfying $$X^2+Y^2=Z^2$$ into a geometric problem in such a way that the translation can be reversed once we have solved the geometric problem. This is rather simple: First note that there are no non-trivial solutions with $Z=0$, so we can divide by $Z^2$ to get the relation $$\left(\frac{X}Z\right)^2+\left(\frac{Y}Z\right)^2=1.$$ You should now be able to convince yourself that the problem of finding (non-zero) Pythagorean triples $(X,Y,Z)\in\mathbb{Z}^3$ is equivalent to the problem of finding rational pairs $(x,y)\in\mathbb{Q}^2$ satisfying $$x^2+y^2=1.$$ This is our geometric translation! Why? Because the above equation is the equation of the unit circle and we are asking for the rational points on the circle. But how do we list these rational points?

Note that it is easy to find a solution, for example $(-1,0)$. We now use the following trick to find the rest of the rational points. If $(x,y)\in\mathbb{Q}^2$ is some other rational point satisfying $x^2+y^2=1$, draw a line passing through $(-1,0)$ and $(x,y)$. Note that the line does not intersect any other points of the unit circle. Now, a line through a given point is completely determined by its slope, and one can see that in this case the slope is: $$\frac{y}{x+1}.$$ As we chose $(x,y)$ to be different from $(-1,0)$, we have $x\neq -1$, so the quotient is defined. Moreover, since $x$ and $y$ are rational, so is the slope. Thus, this construction defines a function $$\Phi\colon\left\{\,\text{Rational points }(x,y)\text{ on the unit circle}\,\right\}\setminus\{(-1,0)\}\to\mathbb{Q}.$$ taking a point on the unit circle to the slope of the line through it and $(-1,0)$. I claim $\Phi$ is bijective, which means we can parametrise the rational points on the unit circle by the rational points on the number line. Thus, let us try constructing an inverse $\Psi$.

For each rational number $\lambda\in\mathbb{Q}$, there is a line through $(-1,0)$ with slope $\lambda$. We know that the line should intersect the unit circle at some point $(x,y)$, and we should define $\Psi(\lambda)$ as the intersection point, but what is this point? We already noted that $$\lambda=\frac{y}{x+1}$$ and since $(x,y)$ lies on the unit circle, it also satisfies the identity $x^2+y^2=1$. Thus, we can try solving this system of equations. From the first equation, we derive $$\begin{gather*}\lambda=\frac{y}{x+1}\implies x+1=\frac{y}\lambda\\\implies x=y/\lambda-1.\end{gather*}$$ Plugging this into the quadratic equation gives $$\begin{gather*}1=\left(y/\lambda-1\right)^2+y^2=y^2/\lambda^2-2y/\lambda+1+y^2\\\implies 0=y\left(\left(1+1/\lambda^2\right)y-(2/\lambda)\right).\end{gather*}$$ From this factorisation we get two solutions for $y$, where $y=0$ corresponds to the point $(-1,0)$. The $y$-coordinate of the intersection point is then $$y=\frac{2/\lambda}{1+1/\lambda^2}=\frac{2\lambda}{1+\lambda^2}.$$ Finally, we can also solve for the $x$-coordinate: $$\begin{align*}x&=y/\lambda-1=\frac2{1+\lambda^2}-1\\&=\frac{2-(1+\lambda^2)}{1+\lambda^2}=\frac{1-\lambda^2}{1+\lambda^2}.\end{align*}$$ Therefore, we define $$\Psi(\lambda)=\left(\,\frac{1-\lambda^2}{1+\lambda^2},\frac{2\lambda}{1+\lambda^2}\,\right)$$ and it is easy to check that $\Phi\circ\Psi=\mathrm{id}$ and $\Psi\circ\Phi=\mathrm{id}$. I leave it as an exercise for the reader to figure out how this can be used to generate all Pythagorean triples.

Let us zoom out a little and try to see how this proof fits into the greater context of algebraic geometry. In the proof, we construct the function $\Phi$ taking in rational points on the unit circle and give "points on the rational number line". We call the unit circle and the number line affine varieties in algebraic geometry, because they are subsets of $\mathbb{Q}$-vector spaces defined by polynomial equations. Note that the function $\Phi$ can be expressed as a rational function in the coordinates of the point in the sense that given a point $(x,y)$ on the unit circle, the number $\Phi(x,y)$ is the result of inputting $x$ and $y$ into a quotient of two fixed polynomials. In the same way, the inverse $\Psi$ can be written as a rational function (or a pair of rational function, each specifying one coordinate of the resulting point). When there are rational functions between two varieties that are inverses to each other, we say the varieties are birationally equivalent. So, if we start with a Diophantine equation and can translate the problem into a geometric one, then one way of solving the geometric problem is to find a new variety that is simpler than the variety we want to understand but which is equivalent to it. Using this equivalence, we may be able to solve the geometric problem, and hence, the original Diophantine equation. Recall that the birational equivalence $\Phi$ is not defined at $(-1,0)$, because the line intersecting the unit circle only at $(-1,0)$ has "infinite slope". A birational equivalence which is defined at all points is called an isomorphism.

Now, we can broaden our picture further by expanding the domain. So far we have concentrated on the rational solutions to polynomials, but we could also consider all complex solutions. At first look this seems like a crazy idea which unnecessarily complicates everything, but it will all make sense soon. So, let us consider all complex points in $\mathbb{C}^2$ on the "unit circle" defined by $z^2+w^2=1$. The space $\mathbb{C}^2$ is somewhat tricky due to it having four real dimensions. But note that the solution set of $z^2+w^2=1$ in $\mathbb{C}^2$ has one complex dimension, which is equivalent to two real dimensions. Thus, it is just a surface! But even though the unit circle looks like a surface from the point of view of the real numbers, we call it an algebraic curve. For the sake of consistency, algebraic geometers call $\mathbb{C}$ the complex line. Since the unit circle is embedded in a four-dimensional space, we cannot visualise its geometry exactly, but we can visualise its topology, because we could in principle move the surface continuously from four-dimensions down to three, making it possible to visualise.

The first thing we want to do with the circle is to make it compact. To understand the point of compactification, consider first the complex number line $\mathbb{C}$. The complex line is not compact, because one can move infinitely far in one direction without approaching any one point. To solve this, we add a "point at infinity" (Concretely, this is done by stereographic projection). This "compactification" of the complex line is called the projective line in algebraic geometry or the Riemann sphere in complex geometry. The circle can also be compactified by adding a point at infinity, resulting in the projective circle. Now, recall that we defined the birational equivalence $\Phi$ between the rational circle and the "rational number line". It turns out that this equivalence extends to an isomorphism between the projective circle and the projective line, which is defined also on the point $(-1,0)$ of the circle: it is mapped to the point at infinity on the projective line.

Since we are now looking at compact surfaces, we can study their topology more closely. Compact (orientable) surfaces have the nice property that they have a well-defined notion of genus, which counts the "number of holes" on the surface. It is then possible to check that the genera of the projective circle and the projective line are both 0. This is obvious in the case of the projective line, since we know it is topologically a sphere, and requires a bit more thought in the case of the circle. With this topological information at hand, one can actually prove more!

Any compact (and smooth) algebraic curve (over $\mathbb{C}$) of genus 0 is isomorphic to the projective line.

Furthermore, the isomorphism can be constructed by choosing a point on the curve and drawing lines through it exactly the same way we did with the circle. To connect this back to the original problem of solving Diophantine equations, this tells us that if the solutions to a Diophantine equation are in correspondence with the rational points of a complex surface of genus 0 with a rational point, then we can always parametrise the solutions to the Diophantine equation by rational numbers. With this geometric language, we can go on to study curves with higher genera, which will give us information about more Diophantine equations. The (smooth) genus 1 algebraic curves are the infamous elliptic curves. The points on an elliptic curve cannot be parametrised easily as in the case of a genus 0 curve, but the points form an Abelian group and one can construct more points on the curve by "adding" points together. Elliptic curves are extremely important in number theory and related areas. For example, they lie at the heart of the proof of Fermat's Last Theorem. For algebraic curves with higher genus, there is no simple algorithm for producing points on them, but Falting's theorem states that every algebraic curve with genus $\geq 2$ has only finitely many rational points.

If this got you interested in algebraic geometry, here are some resources to look at. Algebraic geometry can be approached—as one might expect—from two different angles, from geometry or algebra. As a general piece of advice, it helps a lot to know some differential geometry and algebraic topology to understand the geometric aspects of algebraic geometry, and knowing commutative algebra is necessary for understanding the algebraic aspects. For the geometrically inclined students, the first five chapters of Complex Algebraic Curves by F. Kirwan provide a nice, gentle introduction to the geometric side of algebraic geometry. I also recommend looking at the lecture notes by Andreas Gathmann. Students interested mainly in algebra might enjoy starting directly with commutative algebra by reading chapter 1 of Commutative Algebra by Atiyah and MacDonald. After reading the chapter, one should do all the exercises in the chapter, because they lay the groundwork for building the modern dictionary between algebra and geometry, the theory of schemes. For a deeper dive into commutative algebra, I recommend Commutative Algebra by Eisenbud. Regardless of whether one is interested more in geometry or in algebra, one should ultimately learn scheme theory. For me, the best reference for scheme theory is Vakil's Foundations of Algebraic Geometry. Getting a good intuition for schemes requires some effort, and to help with this, I recommend looking at Geometry of Schemes by Eisenbud and Harris and the other set of lecture notes by Gathmann. I wish you a good luck with your journey to learning this exciting field of mathematics and hope you get to appreciate all the bridges it builds between so many other fields.