Geometry of a Polynomial

This week, we’re going to talk about some of the more important ideas in algebraic geometry, and in particular we will be picking a polynomial and describing its geometry in detail. This polynomial is not taken at random in any way, shape or form, but we’ll go in pretending that we have no clue how nice it is. Ready? Here’s the polynomial:

$$f(x,y)=y^2-x^3+x$$

Ok, so a lot of you might recognize it for what it is…keep quiet about it, we’re going to get to how absurdly good this polynomial is. First off, the polynomial describes a subset of the plane ${\mathbb{C}}^2$ (yes, we’re working over the complex numbers, but we’ll look at other things later if there’s time) given by $f(x,y)=0$, a set that we will denote by $E$. The first question to ask is if this can be built out of simpler subsets? To put it in the proper language this set is the variety defined by $f(x,y)=0$ and we are asking if it is irreducible. That is, can we write it as a union of proper subvarieties (what’s a subvariety? Well, we’re only concerned with the basics of things at the moment, so it’s a subset of ${\mathbb{C}}^2$ given by $g(x,y)=0$ such that it is contained in $E$ ). This corresponds precisely to $f(x,y)$ being irreducible, that is, $f(x,y)$ can’t be factored. How can we see that? Either each term will have a $y$ appearing or else one of them will have $y^2$ as a monomial. The first case is ruled out because then we would need to have a polynomial equal to the square root of $x^3-x$, which just can’t happen. The latter can’t happen as then there would be no way to cancel the cross terms.

You might be wondering why we grabbed ${\mathbb{C}}^2$, well, it’s simply because $f(x,y)$ is a polynomial in two variables. If there were more, we’d look in higher dimensional space. But there is the added complication that we could think of $f(x,y)$ as a function on ${\mathbb{C}}^3$ such that it doesn’t depend on $z$. We’ll ignore this, though, because this is just $E$ crossed with $\mathbb{C}$, and so all the geometry is already in $E$.

Moving on, when you think about a geometric object, one of the first things you should ask yourself is “what is its dimension?” So here, $E$ has dimension 1 . Why is this? Well, ${\mathbb{C}}^2$ has dimension 2 by definition, and $E$ is the solution set of a single polynomial. To be careful once again, we can associate to $E$ the ring $k[E]=k[x,y]/f(x,y)$, which is the ring of all polynomial functions on $E$. The dimension of $E$ is the Krull Dimension of this ring, which is the longest chain of prime ideals contained in it. This will turn out to be one, because ${\mathbb{C}}^2$ is associated to $k[x,y]$. This ring cannot have a chain of length more than two so we are looking for ideals containing $f(x,y)$. We can find this by $(0)\subset (f(x,y))\subset (x,y)\subset k[x,y]$ which shows that there can fit at most one ideal between our polynomial and the whole ring. So $E$ is one dimensional, so we will refer to it as a curve.

Pausing for a moment, we note that this curve has COMPLEX dimension 1. It has dimension two over the real numbers, and so is a surface. But now the question arises, is it a manifold? Because if so, it is automatically a Riemann Surface, so, by an application of the Implicit Function Theorem we get a manifold if and only if $E$ turns out to not be singular. So how can we check for singularities? Well, it just means that nowhere on the curve do the derivatives of $f$ all vanish. Rather than having to solve the possibly difficult system of equations involving $f$ (because it contains a cubic, which is not good, see this post of mine and this one of Gowers’s which show how hard a single variable cubic is), we can instead homogenize $f$ and solve the system only involving the derivatives. This also has the advantage of compactifying $E$ by sticking it into projective space. So now we denote by $F(x,y,z)=y^{2}z-x^3+x{z}^2$ the homogenization of $f$, and this is the object we care about now. So now we take the three partials and obtain the system $-3x^2+z^2=2yz=y^2+2xz=0$. Doing a bit of algebra shows that this can only happen if all three variables are zero, but that’s not allowed in the projective plane, so $E$ is nonsingular, and thus a Riemann surface. cracks knuckles Now we’re in projective space, so it’s time to really get started analyzing this thing. What points did we add? Well, anything in the standard plane can be found with $z=1$, so we shall plug in $z=0$. This gives us $-x^3=0$, which says that $x=0$, so the only point at infinity has to be $(0:1:0)$. So we only added a single point, and things are much easier! So it’s time we introduced another useful object, the Hilbert Polynomial.

It turns out that the Hilbert polynomial can be used to define dimension, so we’d better hope that we extract the number one out of it when we calculate it. I’ve taken the liberty of finding it when no one was looking, and the Hilbert polynomial of $E$ is $p(t)=3t$, and the degree of the Hilbert polynomial gives the dimension, so our two definitions agree! Wonderful! The Hilbert polynomial also defines two other invariants of the curve $E$, the degree and the arithmetic genus. The degree will just be the lead coefficient multiplied by $(\dim E)!$, which in our case is 3 , and the arithmetic genus is given by $(-1{)}^{\dim E}(p(0)-1)$, which gives us 1 . This notion of genus agrees with the topological one, and so this tells us that the Riemann surface we get is a torus. In particular, this shows that $E$ isn’t the same thing as a sphere, which is the (unique) nonsingular genus o curve.

We mentioned that the degree is three, which is the same as the degree of the polynomial (which makes it a good word for it) but what does this really mean, geometrically speaking? In my previous post about the history of the subject, I mentioned Bezout’s Theorem which tells us that in the projective plane, two curves intersect in a number of points equal to the product of their degrees. In particular, this means that a line intersects $E$ at three points.

What can we do with this information? Well, we can make $E$ into a group! (isn’t this exciting?) Adding on this structure, which I’ll describe in a moment, makes $E$ into what is called an Elliptic Curve, which has deep connections to number theory. On to the group structure. Let $P,Q\in E$ be distinct points. Then we can take a line that passes through $P$ and $Q$, and it must hit a third point on $E$ (which may be $P$ or $Q$ again) and we can call that point $R$. (If $P=Q$, then take the tangent line to $P$.) Now, this seems like a great way to define a group operation, but it has a few problems. For one, there isn’t an identity, that is, there’s no point on $E$ such that the line between it and any other point of $P$ is tangent to $P$ . We can correct this though, by not taking $R$ to be the sum of $P$ and $Q$, but rather take the unique third point on the line connecting $R$ and $(0:1:0)=e$. It is not too hard to check that this actually gives us a group structure on $E$ and that $e$ is the identity element. In fact, if you wrote out a formula for $P+Q$ in terms of the coordinates of $P$ and $Q$, they turn out to be rational functions, and in fact to be defined everywhere on the curve, so this group structure is algebraic, which makes it particularly nice.

Now, we’ve actually discussed quite a lot of the geometry of $E$, in particular with respect to its embedding into ${\mathbb{P}}^2$, but there’s also a lot of number theory in this curve (and others like it). I’ll just post some links to additional material talking about these relationships, and either come back to it some future week, or else move on and look at the geometry of another polynomial (perhaps a surface? Does anyone have a preference?) next time. So anyway, on to some number theory.

As it happens, a theorem of Mordell and Weil says that an elliptic curve is a finitely generated group if you only look at the points with rational coordinates (why is this subset a subgroup? The formulas for sums are rational functions of rational numbers, and so give you new rational numbers), and then one of the big $1,000,000 questions is the Birch and Swinnerton-Dyer Conjecture about the rank of this group. For more on that, see the Clay Math website.

Additionally, a theorem of Ken Ribet’s is that the Modularity Theorem (then the Taniyama-Shimura Conjecture) implied Fermat’s Last Theorem, one of the big long standing open problems in mathematics. This theorem tied modular forms (a number theory gadget that I know almost nothing about) to elliptic curves, and Andrew Wiles started to prove the theorem and got the biggest and most important case, and so he gets the credit for solving Fermat.