We’ve now spoken about all sorts of invariants, the most important ones being derived from the Hilbert Polynomial, we’ve spoken about Bezout’s Theorem, and we’ve developed a nice bit of classical algebraic geometry. We’ve also mentioned algebraic groups, and it might be a good idea to refresh this stuff, because now we’ll return to it, and build up some more examples.

Our object of study for the day will be what is called an Elliptic Curve. Though technically, we should talk about Elliptic curves completely in the abstract, with no reference to an embedding in projective space, we’re going to restrict everything, including our definitions, to curves in $\mathbb{P}^2$. If you want to see a bit more of the abstract theory either wait awhile until I do it in this series, or look at my previous posts on the topic.

Anyway, we define an elliptic curve to be an irreducible smooth plane curve of degree three. That is, it is given by a homogeneous cubic polynomial that is irreducible and such that it’s three formal partial derivatives are never simultaneously zero. Let’s pause for a moment to talk about this last part. Partial derivatives are linear operators, that is, we only need to define them on monomials. So we define $\frac{\partial}{\partial x_i} x_1^{n_1}\ldots x_k^{n_k}=n_i x_1^{n_1}\ldots x_i^{n_i-1}\ldots x_k^{n_k}$. So for instance, in the projective plane, we’d have $\frac{\partial}{\partial x}x^ay^bz^c=ax^{a-1}y^bz^c$. We do allow exponents of zero, that way $\frac{\partial}{\partial x} y^bz^c=0$. We’ll prove that this condition is the same as that the tangent space has the same dimension as the variety later, and for the moment take this as the definition of smooth.

It is true that any elliptic curve can be written as $y^2z=x(x-z)(x-\lambda z)$ when working over a field of characteristic not equal to two. However, the only way I know of to do this requires the Riemann-Roch theorem, and is explained in my previous posts. For now, we’ll continue to restrict our definition and say that this is precisely what an elliptic curve is.

So now we take an arbitrary elliptic curve, call it $E$. The point $(0:1:0)$ is a triple point. By this, I mean that if we take the line given by $z=0$, it intersects the curve only in this single point. As the curve has degree three, the line should meet the curve in three points, and so we must count this single point three times. We denote this point by $P_0$.

Now we are in a position to define a group law on the cubic. It’s a bit more complicated that we would really like, but without the trick we’re going to use, it doesn’t quite work out. Late $P,Q\in E$. Then first we take the line containing both $P$ and $Q$. It intersects the elliptic curve in a third point (this point might one of $P$ or $Q$, but that’s irrelevant). If $P=Q$, then we taken the tangent line at $P$. We label the third point $R$. Now, we take the line containing $R$ and $P_0$. It intersect the elliptic curve in a third point. This point we denote by $P+Q$.

So we take this to be the group law. By being careful, you can show that this is associative, that $P_0$ is the identity element, and that there are inverses (in fact, $R=-P-Q$). And so we have a group. To check that it’s algebraic, you can write down the coordinates of $P+Q$ in terms of $P$ and $Q$, and it can be expressed as rational functions, as can the inverse. These expressions are pretty nasty though.

The point is, elliptic curves are all algebraic groups, and Bezout’s Theorem allows us to see this. There remains one big unanswered question that I’m going to answer, but leave completely unjustified. Are all elliptic curves isomorphic as algebraic groups? That is, is there an isomorphism of varieties that is also an isomorphism of groups? The answer is a very resounding no.

In fact, we can classify them completely using something called the $j$-invariant. The way to determine it is first to take an arbitrary elliptic curve. Then, place it in the form mentioned above. Then $j(E)=2^8\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2}$. It is a theorem (though not a particularly easy one) that $j$-invariant completely classifies elliptic curves. That is, $E, E'$ are isomorphic if and only if $j(E)=j(E')$.

So now what lambdas will give the same $j$? Here’s the list: $\lambda,\frac{1}{\lambda},1-\lambda,\frac{1}{1-\lambda}, \frac{\lambda}{\lambda-1},\frac{\lambda-1}{\lambda}$.

That’s it for right now, enjoy your weekends.