So, on Thursday, I posted some background stuff, which I think I might continue to do in the future. Just in general if I want to do a real post which there’s some basic technical stuff that needs to be mentioned to do properly, I’ll post it on Thursday, but otherwise, I’m going to stick to Monday posts. So this week, I’m going to generalize the example from last week and talk about Elliptic curves in general.

To do this, we need to do a little bit of work to get the notion of genus figured out for an abstract curve (that is, a one dimensional variety with only smooth points) with no given embedding. Let $C$ be a one dimensional smooth variety, that is, a curve (we will always have curves being nonsingular and complete) and let $\Delta:C\to C\times C$ be the diagonal map. By virtue of $C$ being a curve, the diagonal map has closed image and so is a subvariety of $C\times C$. That means that there is a sheaf of ideals $\mathscr{I}$ such that for each open set $U\subset X\times X$, we have $\mathscr{O}_X(U)/\mathscr{I}(U)$ isomorphic to the structure sheaf of $\Delta(C)\cap U$. In a fashion similar to constructing tangent spaces, we look at $\mathscr{I}/\mathscr{I}^2$, and then pull it back to $C$. This gives us a sheaf called the canonical sheaf (well, in general the relative differentials, but it’s all the same for curves) and denoted by $\omega_C$. We define the genus of $C$ to be $\dim_k \omega_C(C)$, the dimension of the global sections.

Now we can give a definition of an elliptic curve. It will just be a curve of genus 1. Before we continue, we’ll need to know what a line bundle on $C$ is. For our purposes, it will be a sheaf of $\mathscr{O}_C$-modules which is locally of rank 1. It is a theorem that all of these arise in the following way: let $D$ be a divisor if it is in the free abelian group generated by the points of $C$. Then there we can construct a line bundle by taking $\mathscr{O}(D)(U)=\{f:U\to k|f$ has poles of order $\leq -n_i$ at $P_i$ if $n_iP_i$ appears in $D\}$. Of primary importance, will be the global sections of these sheaves. Now we choose $P_0\in C$ a point.

So now we want to find an equation of a plane curve isomorphic to our elliptic curve. First we show that the elliptic curve can be embedded into the plane. So look at the sheaf $\mathscr{O}(3P_0)$. That is, we want sections with poles of order at worst three at a given point. There turn out to be three linearly independent ones (we prove this with the Riemann-Roch Theorem, but we can write them down explicitly once we have coordinates) and so we get a map $\varphi:C\to\mathbb{P}^2$ by taking these sections as the coordinates. These sections will give us an embedding, as the following argument proves:

If $p,q$ are such that $\varphi(p)=\varphi(q)$, then for all $s$ global sections, we have that $s(p)=0$ if and only if $s(q)=0$. So we consider the sheaf $\mathscr{O}(3P_0-p)$, which gives a map into $\mathbb{P}^1$ and is defined everywhere. So in particular, it is defined at $q$, and so there is a section $s'$ for this sheaf with $s'(q)\neq 0$. As it happens, $\mathscr{O}(3P_0-p)\to \mathscr{O}(P_0)$ is an inclusion (any section with a pole of third order at $3P_0$ and a zero at $p$ must be in $\mathscr{O}(3P_0)$), and so $s'$ is in $\mathscr{O}(3P_0)$, and not vanishing at $q$. The contradicts our assumption, and so the map is injective (and in fact defined everywhere). To see that it is an embedding, we look at the differential of $\varphi$. The same sort of argument works here as well.

So now we have an embedding, so we can take homogeneous coordinates $(x,y,z)$ on $\mathbb{P}^2$. Let’s see if we can work out the equation of our elliptic curve, now that we know that every elliptic curve an be embedded into the plane. For simplicity, we will remove the point $P_0$ which is the point at infinity, and so only have to work with $x,y$ as coordinates in the affine plane.

Now we notice that $x$ has a pole of order 2 at infinity, and $y$ has one of order 3. Another use of Riemann-Roch (hmm, perhaps I should do a post on Riemann-Roch in the near future…) gives us that the number of linearly independent global sections of $\mathscr{O}(nP_0)=n$, at least, for positive $n$. So let’s write out what they are for small numbers. For $n=1$, we get the only section being the constant section $1$. For $n=2$, we have $\{1,x\}$, for 3, $\{1,x,y\}$, for 4 we get $\{1,x,y,x^2\}$, for 5 we get $\{1,x,y,x^2,xy\}$. So far, no nice relations. But lo and behold, for $n=6$, we get seven natural functions to look at $\{1,x,y,x^2,xy,x^3,y^2\}$, but the space has dimension six! So that means they satisfy some equation of the form $y^2+a_1 xy+a_3y=x^3+a_2x^2+a_4x+a_6$, and we can assume the coefficients of $x^3,y^2$ are 1 because we can always scale them down. Making a substitution in $y$ with the quantity $y+\frac{1}{2}(a_1 x+a_3)$ (which we can do over $\mathbb{C}$) we get $y^2=(x-a)(x-b)(x-C)$, and a linear transformation gets us the final equation, $y^2=x(x-1)(x-\lambda)$ for some $\lambda\in \mathbb{C}\setminus\{0,1\}$.

Now we are faced with another problem. Not every $\lambda$ corresponds to a different elliptic curve. There are a few $\lambda$‘s that correspond to the same one. However, from this form, we can define what is called that $j$-invariant, which does classify elliptic curves up to isomorphism. It is given by the simple formula $j=2^8\frac{(\lambda^2-\lambda+1)^2}{\lambda^2(\lambda-1)^2}$. So is that it? We’ve got a classification of elliptic curves up to isomorphism, which is really and truly spectacular. What’s left?

For one thing, there is the question of moduli spaces. So this is really a matter of classifying families of elliptic curves. There are two types of moduli spaces: coarse and fine. We have constructed a coarse moduli space for elliptic curves, because if we have a family $X\to B$ with fibers elliptic curves, we get a map $B\to \mathbb{A}^1$ taking each fiber to its $j$-invariant. However, the map $B\to \mathbb{A}^1$ doesn’t uniquely determine the family $X\to B$. This is because we have automorphisms of elliptic curves. (as they are all groups, they have infinitely many automorphisms, in fact) So there’s no hope of even finding a scheme which is a moduli space. However, this does bring us to stacks, but I’m going to back off on this point a little bit, because at some point I’m going to give stacks the time they deserve.

Instead I’m going to talk a bit about number theory. We have elliptic curves as plane curves of degree three over the complex number. And as seen last week, the rational points (in fact, the points over any subfield) form a subgroup of the elliptic curve. So one question to ask is what the structure of the group over $\mathbb{Q}$ is? The Mordell-Weil theorem in fact states that this group is finitely generated (it is abelian by geometric considerations). In fact, Siegel’s Theorem (Carl Ludwig, not me) states that there are in fact only finitely many point which have integer coordinates (for the affine curve, at least). Now, finite generation means that we can write the elliptic curve (as a group) as $\mathbb{Z}^k\oplus T$ where $T$ is a finite group. It turns out that $T$ is easy to determine, but the number $k$ is very difficult. In fact, if you can determine it for every elliptic curve, it’s worth a million dollars, because that would settle the Birch and Swinnerton-Dyer conjecture.

Additionally, elliptic curves can be used in integer factorization and in some modern cryptosystems. I think that’s all I’ve got for the moment. I might ramble more about elliptic curves at some point in the future…say, if I ever get around to reading Silverman’s book.