 Okay, this is the third part of lecture one on algebraic geometry where we will cover the theorems of Beizu, Papus and Pascal. So we start with the theorem of Beizu. In fact, this theorem was originally stated by Newton in his Principia. It's book one, section six, Lemma 28, if you want to see it. Newton didn't really prove it, or at least his proof was kind of a bit informal and difficult to follow. So what does Beizu's theorem says? Well, it says, roughly speaking, says two curves of degrees M and N in the plane have at most M, N points of intersection. Well, that's not quite true because the two curves of degrees M and N might have an entire component in common. For instance, they might even be the same. So it should say if there are no components in common. Well, there's a slightly stronger version of Beizu's theorem that says the number of intersection points should be exactly M, N. Except to get the number exactly equal to M, N, we need to work over an algebraically closed field such as the complex numbers. Secondly, we must somehow count points at infinity because we might have say two parallel lines of degree one that don't meet anywhere on the plane, but they sort of meet at infinity in some sense. And third, we have to worry about intersection multiplicities. So if we've got a curve, a straight line, meeting a parabola, then you need to count that as meeting at two points rather than one point to get it right. And it's actually quite tricky to get even a correct statement of Beizu's theorem. And the problem is defining intersection multiplicities isn't all that easy. So people used to define intersection multiplicities by saying you perturb both curves slightly and count the intersection multiplicities. And the problem is it's actually quite difficult to make sense of this. So you can see Beizu's theorem is plausible. For example, if we take two ellipses, like this, it's kind of plausible. They have four intersection points and so on. We can give a very informal proof of Beizu's theorem, as I haven't even stated Beizu's theorem correctly. I certainly can't prove it correctly, but here is an informal proof. Suppose the curves are f, x, y equals zero and g, x, y equals zero, where we assume both of these curves have degree m and degree n, we assume there. Then what we can do is we assume that the number of intersection points doesn't vary if we perturb f and g. So we can just vary their coefficients. And what we can do is f is a product of linear factors and g is also a product of linear factors. Well, then Beizu's theorem is now obvious because f is just a union of m lines. For instance, it might be a union of lines like this. And g is also a union of n lines, so it might be a union of lines like that. So if f and g both are degree three, then f might look like the green thing and g like the purple thing. And it's completely obvious that two lines have a single point of intersection, so the total number of intersection points will be mn. Well, there are several things rather doubtful about this proof. The main one being how do we know the number of intersection points doesn't change if we perturb f and g. And we run into all sorts of problems because as we perturb f and g, we might get multiple points and so on. So this is really unclear. Well, this sort of style of proof was very common in algebraic geometry in the early 20th century. So there was a big school of algebraic geometry called the Italian school of algebraic geometry, which had some spectacular results. But it just sort of peated out in the middle of the 20th century. The problem was they used a lot of this sort of informal reasoning. And the number of results they produced that were incorrect just kept going up and up. And by about 1950 or 1940 or 1950, it had just become impossible to tell whether the results they proved were correct or not. And this was eventually fixed by people like Zariski and Andre Ve, who finally put all of algebraic geometry on a much firmer fitting, much firmer foundations. The problem was the proof became a lot more complicated. I mean, this proof has the disadvantage it's wrong and meaningless. It doesn't have the advantage that it's very simple and easy to understand. So these sort of informal proofs do have a use there. They're good for guessing what the right answer is. For example, if you understand the reasoning of this proof, you can easily figure out what happens to Bezou's Theorem in higher dimensions. For instance, if we worked in n dimensions and had n hypersurfaces, sorry, n, then you could perturb each hypersurface. It was a union of hyperplanes and you get the analog of Bezou's Theorem in n dimensions, which says that if you take an intersection of hypersurfaces of the right number, the total number of intersection points is the product of their degrees. Anyway, that's Bezou's Theorem. Next we go on to Pappus's Theorem. So Pappus was one of the last of the great classical Greek mathematicians. He lived around 300 to 350. And at the time he lived, classical Greek mathematics had almost come to an end. So there were a few people like Pappus and Diophantus who did spectacular results and the trouble was, as it was coming to an end, very few people knew what he was doing. So he wasn't all that well known and very little is known about his life. Anyway, one of his theorems is a really spectacular theorem because, and what is so spectacular about it is that it's just a theorem about straight lines in the plane. And it's very difficult to think of anything interesting to say about lines in the plane because they seem so simple and basic. Anyway, his theorem goes as follows. Suppose you take two lines in the plane. So here I'm going to take two straight lines and on each line I'm going to choose any three points. So here I'm going to choose three green points. And here I'm going to choose any three red points. I actually want to number these as one, two, three. And now I join up every point to the points with a different number on the other lines. So here I'm going to join up points one and two like that. And I'm going to look at their intersection point. And then I'm going to do the same with points two and three. Join them up and mark the intersection point. And then I'm going to join up points one and three. And then I'm going to mark the intersection point. And now what you do is you look at these three intersection points. And Pappas's rather astonishing theorem says that if I've drawn this accurately enough is all night on a straight line. And if you try and prove it using methods of Euclidean geometry, you'll find it's not actually all that easy. It's not too difficult to prove using coordinate geometry. Coordinate geometry wasn't invented until about a thousand years after Pappas lived though. So I guess he had some other way of proving it. Pappas's theorem incidentally turns out to be essentially equivalent to commutativity of multiplication. What this means is you can define a plane over any division ring. So we can say Pappas's theorem is equivalent to multiplication commutative. Because if you look at the analogous results in a plane over a division ring, you find that Pappas's theorem is true if and only if the division ring is commutative, in other words, a field. Well, we're going to deduce Pappas's theorem as a special case of Pascal's theorem. So Pascal was a French mathematician. He's famous for having Pascal's triangle named after him although he was nowhere near the first person to think of it. Anyway, Pascal's theorem was discovered when he was a teenager and it goes like this. Suppose you take an ellipse and I'm going to choose six points on the ellipse and I'm going to number them as one, two, three and I'll number them as one, two, three down here. So you would choose any six points. And now I'm going to join up these points in pairs like this. So I join up one and two and I join up points two and three. I join up the points three and one like so. And then Pascal's theorem says that if you look at these three intersection points, they all lie on a straight line and they probably won't because I'm not terribly good at drawing accurate ellipses. Well, that seems to work out quite well. So you get a line like this, which is called Pascal's line. And it should be pretty obvious from the way I've drawn them that Pascal's theorem is very similar to Papas's theorem. In fact, you can see that Papas's theorem is really the special case of Pascal's theorem. So Pascal's theorem holds where this is any degree two curve and one rather special case of a degree two curve is just two straight lines. It's a sort of degenerate case. So here we have a degree two curve consisting of two straight lines. And you can see that Papas's line is just the same as Pascal's line in this special case. So how do we prove Pascal's theorem? Well, we don't actually know Pascal's original proof because it's kind of been lost. There are some plausible reconstructions of it. So you can prove Pascal's theorem by first of all proving it for a circle using Euclidean geometry, although it's not all that easy to think of a proof. And then you notice that theorem is invariant under various projective transformations and you can use a projective transformation to turn any circle into a conic. So Pascal's theorem holds for conics. Well, we're going to use a different proof of Pascal's theorem using algebraic geometry that uses Bezut's theorem. And what I'm going to do is I'm going to let's number these lines in order as line number one, two, three, four, five, and six. So six is that one, two is that one, and so on. Incidentally, you notice these six lines really form a hexagon, although it's a kind of funny hexagon that intersects itself. So the hexagon goes one, two, three, four, five, six. I've drawn a funny hexagon that intersects itself rather than an honest one that just rather than an honest convex one, partly to show the connection with Papas's theorem and partly because if I drew a proper hexagon, these three intersection points would be a long way away and would be even less likely to line up in a line. Anyway, so we've got a hexagon and I'm now going to choose six linear polynomials, p1, p2, p3, p4, p5, p6, where pi equals naught on line i. And now what I'm going to do is I'm going to look at the polynomial p1, p3, p5, and the polynomial p2, p4, p6. And now these polynomials obviously vanish on all six points on the hexagon. I think I've gone off the bottom of the piece of paper, so let me try again. So these vanish on all six points. So what we can do is we can consider the first polynomial minus lambda times the second polynomial, and this will again vanish on all six points. And now we can adjust lambda. So p1, p3, p5 minus lambda, p2, p4, p6 vanishes on the seventh point of the conic. And we can do that because we can just choose any point on the conic. These will have some values at the point and we can just choose lambda so that this thing vanishes. Now we notice this is degree 3 and the conic has degree 2. So by Bezut's theorem, there are at most 2 times 3 equals 6 intersection points. Okay, well we seem to have a contradiction because we've got 7 intersection points and Bezut's theorem says there are at most 6 intersection points. Well that's not a contradiction because if you remember Bezut's theorem says that there are at most 6 intersection points unless they have a common component. Well the conic doesn't split as a product of smaller degree polynomials, so the only possible way they can have a common component is if the conic is actually contained in the degree 3 curve like this. So this degree 3 curve is equal to the union of a conic and a line. Well this line will obviously be Pascal's line because, for example, if we go back to Pascal's theorem, we notice that p1, p3, p5 and p2, p4, p6 both vanish on all these 3 points. So these 3 points must be on the degree 3 curve and since they're not usually on the conic, they must be on the line. So this gives a proof of Pascal's theorem using Bezut's theorem from algebraic geometry.