 So this is the second part of the first lecture in which we will discuss some cubic curves. The first cubic curve we will discuss is the curve y squared equals x cubed plus x squared. If we draw a picture of it, it looks something like this. And what we'd like to do is to find rational or integral points on the curve. Well that's not very difficult. Suppose we've got a point x, y. What we do is we draw the line through this point and the origin. And the slope of this line will of course just be y over x, which we will call t. And obviously if x and y are rational then t is also rational. On the other hand if we choose a line through the origin with rational slope, it's going to meet this cubic in three points which will be given by the roots of some equation with rational coordinates. And two of the roots are already rational because they're just both zero. So the third intersection point must also be rational. So in other words x and y rational correspond to t being rational. So this means we almost get a one-to-one correspondence between points on this cubic and rational numbers. Well we don't quite get a one-to-one correspondence because you notice that something funny is going on at the origin. In fact t equals one, t equals minus one, both correspond to the point x, y equals zero, zero. We can of course work out x and y in terms of t. So we have y equals tx. So if we substitute that in there we find t squared x squared equals x cubed plus x squared. And if we divide through by x squared because we're not interested in the point zero, we find t squared equals one plus x. So if we find x equals t squared minus one, y equals t cubed minus t. So we see we have another birational equivalence between this curve and the line of rational points t because the map's taking x, y to t, or t to x, y are both polynomials or rational functions. You can think of this cubic curve as really just a copy of the rational line except that two of the points on the rational line are identified and mapped to the same point here. What we have is an example of a resolution of a singularity. So a singularity is roughly a bad point of the curve. So you see something is going wrong here. And a resolution means you want a nice map from some curve without singularities onto this curve here. This is the simplest example of resolving the singularity. In fact, this resolution is done by a process called blowing up. Blowing up is a very powerful method of resolving singularities. We will see a little bit later. Hiramaka showed that blowing up resolves singularities in non-zero characteristic. Sorry, in zero characteristic. The problem of resolving singularities in non-zero characteristic is a very difficult unsolved problem. Probably about the most basic unsolved problem in algebraic geometry. Finding points on curves or rational points on curves can be very difficult. For example, suppose you try to find all rational points on the curve x to the n plus y to the n equals 1. Well, if x is equal to x over z and y is equal to y over z, for some integers y is z, we just have to solve x to the n plus y to the n equals z to the n and ask whether this has any solutions other than the obvious solutions where x or y is 0. Well, this is of course Fermat's last theorem, now proved by Andrew Wiles, and it took several centuries of effort to prove it. So even a simple looking equation like this can be really hard to find the rational points on. Anyway, that's much too difficult. So let's look at a slightly simpler example. This example comes from a book called, let's see if I can show it without that light shining, The Canterbury Puzzles by this guy called Duttony who lived about a century ago and he had a newspaper column or some sort of column where he would give mathematical puzzles. And the one we want to discuss today is the one called The Problem of the Doctor of Physics, where he has a nice picture of it and the problem says there are two glass spheres, one of radius one and the other of radius two. So one has eight times the volume of the other. And the problem is to find two other glass spheres which sum up to the same volume. And you have to find spheres of rational radius. So the problem is to solve the following equation. We want to find rational numbers with x cubed plus y cubed is equal to one cubed plus two cubed, which is of course just nine. So this is a bit like Fermat's last thing for n equals three except we've got a nine here instead of a one. And there's one obvious solution which is x, y equals one, two, or for that matter two, one. And the problem is can we find some other rational points? Well, the easiest way to find rational points is to look up the answer in the back of the book. And if we do this, we find, not quite sure if it's in focus, but if you look just here, if I can do this. So just here it gives the two solutions, which are these amazingly impressively large 12 digit numbers divided by 12 digit numbers. Okay, when you look at these, you should remember that Dudney did not actually have a computer or a pocket calculator. He did this calculation by hand. So, well, it would be difficult enough to check these were a solution. So the problem is how on earth did Dudney find these numbers by hand? And the answer is he used some algebraic geometry. So what we first do is we draw a picture of this curve. Let's see if I can read this red mark. So we first draw a picture of this curve. And it looks something like this sketch it roughly sort of comes in like this. It's off like that, something like that. And here we've got these points one, two, and two, one. So we have two known points and the problem is to find others. There's a very clever idea for finding these points, which goes back to Fermat. And what you do is one way of finding points is you can take the tangent line to one of these points and see where it intersects the cubic curve. And the key point is that this intersection point is rational coordinates. And let's try and think why that is. Well, if you solve for the x coordinate of this, it's some sort of cubic equation because there's three roots because you want the intersection of a line with a cubic. And it's coefficients are rational. And two of its roots are also rational because they're both just one. So the third root must also be rational. So we get the third rational point, which is you can sit down and calculate it. Well, calculating isn't all that difficult. So the tangent curve is, if this is the point a b, the tangent to this curve is going to be y equals a squared over b squared minus a squared over b squared times x minus a plus b. And you can substitute this in and find the equation for x, which says x cubed plus minus a squared over b squared x minus a plus b all cubed is equal to nine. So as I said, this is a cubic in x and two roots are x equals a. So we can work out the third root, which must be essentially minus the constant term divided by the leading term divided by the other two roots. And you find the third root is given by x equals a cubed plus b cubed or cubed minus nine b to the six or divided by b to the six minus a to the six times a squared. And see these coefficients can rapidly get rather big because you're taking sixth powers of something. So, and this point here turns out to be minus 17 over seven 20 over seven. So that is one solution minus 17 cube plus 20 cubed is nine times seven cubed. Well, that needs original problem. Asked for diameters of gloss spheres. So obviously these numbers have to be positive. Well, here we found negative number. How do we find a positive number? Well, we can use another idea or actually really the same idea. What we can do is we would take this point and join it to this point and look at the intersection point of this line with a cubic curve. Now, just as before this line will be given. You can write down the equation of this line and its intersection points with a cubic will be roots of a cubic equation of rational coefficients. Two of the coefficients of rational the third must be so we can work at this point here. And turns out to be I can find it something like minus 271 over 438 919 over 438. So already it's getting a bit tedious to work out. And again, one of the coefficients is still negative. Well, we can keep on repeating this process. We can take another two points and take the intersection with the cubic curve and keep going until sooner or later. If we're lucky with one whose two coordinates are both positive. And this is roughly how Dudley came up with this spectacularly large example. The first two examples we had a circle or a cubic with a node were both birational to the line which makes them particularly simple to deal with because finding points on the first two curves is more or less the same as finding points on a line. This curve is not birational to a line. It's easiest to see if you look at it in over complex numbers. So if a curve is birational to a line, it must be essentially just a complex fear. If you ignore a few points, so curves that are birational to a line, meaning they're rational curves, sort of look like a Riemann sphere, except you might have to remove a few points or add a few points. If you draw a picture of x cube plus y cube equals nine of the complex numbers, it doesn't look like that at all. It looks like a torus. Or more precisely a torus with one point removed because we haven't yet done projective geometry, so we've actually missed out one point at infinity. And taking birational transformations of curves only modifies a finite number of points, and there's no way you can modify this by a finite number of points, removing them or sticking them together or whatever to make it look like a sphere. So this example is fundamentally harder than the first two examples. In fact, we've got a sort of algebraic operation. So on our curve, I draw our curve again, we've got an algebraic operation where if you've given any two points at that point, you can form a third point by drawing the line through them. So if this is A and B, we can form a third point. Let's call it A quiddle B. So we've got a binary operation on points. There are two minor problems. For instance, what happens if A is equal to B? Well, then you have to take the tangent line. The other problem is what happens is if you choose two points, say there and there, and you might find that line through them doesn't actually intersect this qubit curve in any other points. These two only intersect in two points. Well, you can see it does sort of intersect the qubit curve at infinity. So in order for this operation to be well defined, we have to add a sort of pointed infinity, which is what you get if you go off in this direction and infinitely long distance. To make sense of that, we need to use projective coordinates, which we will discuss later. Anyway, we've got a sort of operation and is it a group operation? Well, first of all, it's not a group operation because we haven't chosen an identity yet. So you might choose an identity and the simplest identity is the point at infinity. It's the most canonical point. And it's not actually a group operation because however, what happens is it turns out we do get a group operation if we say that A, B and C lie on a line is equivalent to being A plus B plus C equals naught in the group. So we're going to define a group law by saying that if three points on a line, then their sum is zero. So we see this operation here is not actually a group operation. It's really minus a group operation. So it counts as minus A plus B. And this operation is obviously commutative and it obviously has inverses and the problem is, is it associative? So we can ask, is A plus B plus C equal to A plus B plus C? And this is really rather a pain to check if you do it in coordinates. So if you actually write out the formula for A star B, it's a bit of a mess. And when you start having two group operations, it's really rather painful to check that law is associative. Fortunately, it turns out there's a much easier way to check it's associative. The point is that A1 plus B1 plus A2 plus A3 and so on turns out to be equal to B1 plus B2 plus B3 and so on. It's equivalent to saying there is a function, a rational function with poles at all the Ai and zeros at numbers Bi. And using this fact, it's quite easy to check that the group operation is associative. So we get a group operation on this elliptic and this cubic curve. This turns out to be something you can do with any reasonable cubic curve. You can usually make it into a group. There are a few minor exceptions. If the cubic curve has singularities, as in the previous example, then you sort of, if it's only got one singularity, you can sort of sometimes get a group law if you throw out the singular point. But if it's got too many singularities, you don't get a group law on it at all. So these groups are for rather strange historical reason called elliptic curves, which is a bit unfortunate because these cubic curves have almost nothing whatsoever to do with ellipses, but the name has stuck and there's nothing we can do about it. So elliptic curves are the one-dimensional case of things called abelian varieties. So abelian varieties are algebraic groups that are projective, which means roughly they don't have any missing points. There's one thing I should give a slight warning about. Abelian linear groups are not abelian. The reason for this is that abelian linear group is an old name for a symplectic group. And it was rather embarrassing that abelian linear groups are non-abelians, so Hermann Weier changed the name of abelian linear group to the symplectic group. So if you ever come across the name abelian linear group in an old book, be rather careful, they're not abelian. In the next part of the lecture, we will discuss Bezut's theorem, Pappas's theorem and Pascal's theorem.