 So this algebraic geometry video will be about Bezou's theorem. So Bezou's theorem in its simplest form says that if we have two curves of degree M and N, they intersect in M-N points. Well, the trouble is this version of the theorem is actually incorrect and there are quite a lot of problems with it. So let's just run through the problems. First of all, we may have two parallel lines. So these both have degree 1 and they intersect in 0 points and the way around that is we need to include points at infinity. So we need to work in projective space, or the projective plane rather than affine plane. Secondly, if we take two curves like y equals 0, y equals x squared plus 1, these intersect in no real points, although they do intersect in two complex points. So we need to use the complex numbers, not the real numbers, or more generally an algebraically closed field. A third problem is the two curves may have a component in common. In fact, they may even be the same. For instance, we could have a curve here and if we take its intersection with this curve, which is exactly the same curve, then they have an infinite number of points in common. So we should say that they need to have no components in common. A fourth problem is that points may have multiplicities and there are several ways this can happen. For example, if we take a line and try and intersect it with this parabola, then it only has one intersection point even though the parabolas degree 2 and the line has degree 1. But you have to think of this as being an intersection point of multiplicity 2, whatever that means. You see, if you deform this parabola very slightly, it meets the green line at two points. So you should think of this as being two points that just happen to be in the same place. And things get even more complicated if the curves themselves have multiple points. For instance, if you look at y cubed minus x to the 4 equals 0, it's got some sort of multiple point at the origin and suppose you then take the curve y5 minus x to the 6 equals 0, this is going to be some other curve which also has a multiple point at the origin. And you know, what is the intersection of these two curves going to be? And we have to count some sort of multiplicity at this point. It's not all clear. What the multiplicity of that point should be. So there are quite a lot of problems in even stating Bezou's theorem correctly. So one version of it would be that two distinct irreducible curves in projective space over the complex numbers of degrees mn intersect in mn points countered with multiplicities. And that's not really a complete statement of the theorem because you notice we haven't actually said what we mean by the multiplicity. And one of the problems is to try and figure out what the multiplicity of an intersection actually is. So we were saying they distinct, this eliminates the problem that they're the same curve. We're saying they're irreducible, which eliminates the problem. They might have a component in common, and we're working over projective plane over complex numbers to eliminate the problem of points not really existing and so on. There are various other different versions of this. We can ask for an n-dimensional version. So we might want to take n distinct hypersurfaces in p to the n over the complex numbers of degrees d1 up to dn should usually intersect in d1, d2 up to dn points. And usually means, roughly speaking, it means that the intersection shouldn't have the wrong dimension. So if you intersect n hypersurfaces in dn, you usually expect the intersection should just consist of a finite number of points. Sometimes it might consist of an infinite number of points. For example, you could take three hyperplanes. You could take a hyperplane there and another hyperplane here. So they might intersect along a line like this. And there might be a third hyperplane containing the same line. And then you see any two of these hyperplanes intersect very nicely. Their intersection is just a line. But when you intersect three of them, then instead of getting a finite number of points, you get an entire line. So you also want to exclude cases like that. And as you can imagine, it's it's not at all easy to correct, to find a correct statement of this theorem. Another variation of the theorem you might try and say is that if you take two algebraic sets of degrees m and n, well, what's the degree of an algebraic set? Well, we we don't know yet. And then is the intersection of degree mn? And sometimes it will be and sometimes it won't be. It quite often will be. You see, this actually generalizes Bezos's theorem about intersecting two curves because there we're intersecting two curves of degree m and n and saying they should have mn intersection points. And if you think of a point as having degree one, then this just says that the intersection has degree mn. And again, we've got the usual problems that the intersection might have the wrong degree. And then who knows what happens and so on. So we expect the theorem like this is quite often true, but might sometimes not be true. There's an informal proof of Bezos's theorem. Well, an informal proof isn't actually a proof at all, but this is so this is a sort of justification for it. What we might do is if we've got two curves given by fxy equals nought and gxy equals nought, we can deform fg. So they're both a union of a finite number of lines. In other words, we just vary the coefficients gradually until we've deformed them so that they're just the product of linear functions of x and y. And if we do that, then we see that they're obviously going to be mn intersection points because f will be the union of m lines and g will be the union of n lines where m and n are their degrees. And if we take mn lines, they will usually intersect in, sorry, if we take m lines intersect with n lines, then there will generally be mn points of intersection. And that's obviously not a proof because we're sort of assuming that a number of intersection points doesn't change as we deform f and g. And this is by no means clear. I mean, we still have this usual problem of what do we mean by multiplicity of an intersection and so on. But this sort of informal proof at least makes you see that the theorem is probably generally correct that the right number is the degree of f times the degree of g. And the same informal argument works when we take n hyper planes in sorry, n hyper surfaces in n-dimensional projective space. Again, we can deform them to be a union of hyper planes and then just check that easy case. So how do we actually state and prove Bezu's theorem correctly? Well, as I said, one of the biggest problems is defining multiplicities of intersections correctly. And in order to do this, we need to use some theory about modules. These are going to be finitely generated modules over notarian rings. And what I'm going to do is I'm just going to very quickly recall the theory of these modules since this course sort of assumes that commutative algebra is already known. So suppose you've got a finitely generated module m over some notarian ring r. Let's call this ring r. Then what we can do is we can filter m what contains so m nought is nought and this is contained in m1, which is contained in m2, which is contained in m n for some n, so that each mi over mi minus 1 is isomorphic to r over p for some prime ideal p. And this is quite easy to prove. So the proof, we choose a maximal, so we choose i maximal among the ideals so that r over i is isomorphic to a sub module of m. And then it's easier to check that i must be prime. So we we put m1 equals isomorphic to the sub module r over i and then apply induction on m over m1. So it's not very easy, so it's not difficult to show that finitely generated modules over notarian rings have this property. And the problem is we want to define the multiplicity of r over i in m to be the number of times r over i so this should be a prime ideal p. r over p is isomorphic to mi over mi plus 1. The problem is that this definition doesn't actually work in general. Let's see why it doesn't work. Well, there are some cases when it works just far. So the first example is if m is let's take r to be the integers just for simplicity and let's take m to be a finite group. Then r over p is going to be z over 2z, z over 3z, and so on. And the multiplicity of z over pz in m is obviously well defined. So the number of times z over pz is in m is equal to well, if you take the order of m to be p1 to the n1, p2 to the n2, and so on, then the order of number of times z over p1z occurs in m is obviously just n1. You can read it off from the order of m. So for finite modules of the finite number of points over the integers, the multiplicity is well defined. However, if m is finitely generated, it might not be. For instance, we could just take m to be the module z and then we've got this inclusion nought contained in z. We could take nought to be m0 and this to be m1 and then we see z occurs once and z over 2z occurs nought times. Well, what's the problem? Well, we can take another filtration of m. We can take nought contained 2z contained in z and then this quotient here is isomorphic to z over 0 and this quotient here is isomorphic to z over 2z. So z over 2z now occurs once. So we've got this problem. We've got two different filtrations of z and the number of times z over 2z occurs is different. So we cannot define the multiplicity of z over 2z in z. It varies depending which filtration we choose and there's there's just no really good way to define it. For example, 3 let's take m to be the finitely generated module over z. The multiplicity of z over 0 in m is defined. So this is now just the rank of m as a module over z and you can work out the rank by say tension with the rational numbers and just taking the dimension of the vector space. So we see sometimes the multiplicity of a module in m is not defined in this case here and sometimes it is defined and what's the difference? Well, the answer is we get a well-defined multiplicity if we're looking at minimal prime. So we look at the primes. So look at the prime ideals p1, p2 and so on with r over pi in a filtration of n. If p1, pi is minimal in this set, we get a well-behaved multiplicity. And if you look at the previous examples, you can see that this is happening. So if m is finite, then the only prime ideals that occur are 2, 3 and so on and 0 doesn't occur and 2 and 3 are all minimal so their multiplicity is defined. In this case, if m is finitely generated, we see 0 is minimal among the primes such that z over the prime occurs in m, but 2 is not minimal. So that's the difference. We get a well-defined multiplicity for some of the primes more specifically for the primes that are minimal. It's not too difficult to see that the multiplicity is well-defined for minimal prime. So if pi minimal, the multiplicity is equal to the length of mp over rp, where this is just the localization of r at the prime p and this is just the localization of m. And roughly speaking, localization is a way of turning an ideal into a maximal ideal because you just kill off all the ideals bigger than it. So if pi is minimal in this set and we localize, we're sort of turning p into a maximal ideal and it's also a minimal ideal so it's really the only ideal that now occurs in here and now has a well defined multiplicity just by taking its length as a module over this. So we're not going to prove that. We're just going to assume it from some commutative algebra course. There's a geometric way of looking at this which is kind of useful. Here I'm now going to take r to be say ring of polynomials and two variables and primes and prime ideals of this correspond to points and curves and the whole plane. So if we've got the plane here, then we can imagine prime ideals since there might be some prime ideal here and another prime ideal here and the third prime ideal here and the minimal prime ideals. So suppose we draw the prime ideals that occur in some filtration of a module m and suppose we get these three prime ideals, then we end up with two minimal prime ideals and this prime ideal is non-minimal. It sort of looks at this point as smaller than this line so why are we calling it non-minimal? Well you have to remember that when you change prime ideals to the corresponding varieties order gets reversed so minimal prime ideals correspond to maximal varieties. So we can sort of picture the prime ideals occurring in module m like this and what we're interested in are the maximal components so that would be this red one and this green one. So here the multiplicity of the red prime and the green prime would be well defined and the multiplicity of the blue prime would not be well defined. So you can think of this set here as being something like the annihilator of m if you take the annihilator of m that's things that kill everything and m that would be an ideal and it will correspond to some algebraic set and it will split up into irreducible subsets which the primes we're interested in, the minimal primes we're interested in. So to summarize if we've got a module over ring r we can define the multiplicity for some primes of r they're the minimal ones that sort of appear in the decomposition of m. So that's for notarian rings there's a similar theorem for graded modules over graded rings. The only difference is that if you've got a graded prime you may have to shift its degrees a bit as well because if you take the ideal r modular or graded prime you might have to shift the degree of everything by some integer l which is sometimes denoted like this. So for every graded module you get a collection of primes and some of them have been sort of had their degrees shifted a bit. So now we get back to Bezou's theorem so the version of it we're going to prove following Hawthorne we take y to be a variety in pn and h is a hyper surface given by the equation f of x0 up to xn equals zero and we're going to assume that y is not contained in h so this is going to exclude all those funny things where y has a component h as a component or something like that and what we're going to prove is that sum over j of iyhzj times the degree of zj is equal to the degree of y times the degree of h. You remember we defined degrees of algebraic subsets earlier. Here y intersection h is a union of irreducible components zj and this thing here is going to be the multiplicity of the intersection of y and h at zj. For example if we take y to be this and we take h to be this then we might have z1 might be the intersection here and the multiplicity in this particular case i and yhz1 well in this case be two because there are sort of two this is a sort of double point I guess and what we do is we look at the following exact sequence we take nought goes to kx nought up to xn modulo the ideal of y and then we want to multiply by f and we take kx nought up to xn and we quote out by the ideal of y again and this maps to kx nought up to xn now we're going to quote out by the ideal of y and f. So we get an exact sequence of of modules where we don't quite because when we multiply by f we have to shift the degree so here we have to shift the degree it's going to be shifted by minus the degree of f on this bit here and now we're going to look at the Hilbert polynomials of these three and Hilbert polynomials of exact sequences behave very nicely so the Hilbert polynomial of this minus the Hilbert polynomial of this plus the Hilbert polynomial of this is going to be zero so what's the Hilbert polynomial of this well it will be the Hilbert polynomial of this piece here minus the Hilbert polynomial of this piece here so it's equal to p y of z minus p y of z minus d so we have to subtract d here because we've shifted the degrees by minus the degree of f sorry I'm saying the degree of f equal to d so this bit here comes because we shifted the degree and now we can work out the leading term of this as follows um so the Hilbert polynomial of y is degree of y times z to the dimension of y divided by the dimension of y factorial plus lower terms and the Hilbert polynomial of this is again degree of y times z minus d to the dimension of y divided by the dimension of y factorial plus lower terms and if we work out what this is it's um um actually I don't want to write out the whole thing so let me just take this bit here and figure out what it is well we've got z to the dimension of y minus the dimension of y times um d times z to the dimension of y minus one so if we subtract these the leading term is going to be d which is the degree of f times the degree of y um times z to the dimension of y minus one divided by the dimension of y minus one or factorial um so um point is this is the leading term of this is now the degree of our hypersurface times the degree of y plus the standard fudge factor where z to the something divided by something factorial so the conclusion is um the Hilbert polynomial of k x naught up to xn divided by i y f um has leading term um degree of h times the degree of y times the dimension of y intersection h sorry z to the power of that over dimension of y intersection h factorial so that should be minus one so um this bit isn't very interesting it just says the dimension is equal to the dimension of y minus one sorry that shouldn't have been a minus one there what it's saying is the dimension of y intersection h is the dimension of y minus one I was just getting them a bit muddled up um so um we've got this term here which is what we want and as our answer the trouble is we now have to interpret it so we want to show that this um is this equal to the sum over z i of the intersection multiplicity of y and h at zj and to do this we need to look at a filtration of um the module so um so the module k x naught up to xn modulo i uh f i y f will have a filtration um um naught equals m naught contained in m one contained in m two contained in m k equals m and all these quotients um are of the form r over q i for some prime i prime ideal um some greater prime ideal of this and they should of course be twisted by some number l i and um now um shifting the degree by l i doesn't actually change the leading coefficient because the late leading coefficient is the dominant term in the asymptotic behavior of the graded pieces and just shifting by a constant isn't going to change that so so the the Hilbert polynomial of each of these has leading term degree of zj over the dimension of zj factorial times z to the dimension of zj and the dimension of zj will be the dimension of y minus one by quoting some dimension theory um so um so we see that the degree of h times the degree of y is equal to the degree of sum over j of the degree of zj times the multiplicity of um um r over q i um twisted by various things in this graded module um k x north up to x n modulo i f so all we have to do is to show this term here is the multiplicity of the intersection so how do we prove that that the multiplicity of these um prime ideals possibly twisted by various l's in this module here is the multiplicity of the intersection well there's an old joke about a target shooter who got a hundred percent in his target shooting and he did this very easily what he did was he would first fire at the target and then whenever one of his bullets hit he would draw a circle around it to indicate that that was the target he had been firing at and we're going to use the same technique in order to show that this is the multiplicity of the intersection it's very easy we just define the multiplicity of the intersection to be this multiplicity here in fact if you are paying careful attention you will notice that i never actually defined the multiplicity of an intersection of y and h along zj so now we're going to turn around and actually define it to be given by the number of times um um these graded primes possibly shifted occur in this graded module here so if that more or less gives the proof of bezzou's theorem and finishes the first chapter of hot