 This algebraic geometry video will be about proper maps in algebraic geometry. So in the previous video, we talked about resultants and I want to rephrase the result about resultants in a slightly different way. So suppose F and G are homogeneous polynomials in two variables, Z0 and Z1 with coefficients in Y1 up to YK. So these are going to be polynomials in these variables here where they're going to be homogeneous in these two variables, X0 and X1. So what this means is that F and G are really hyper surfaces in the product of AK with P1 where these two variables are coordinates in P1 and these variables are just coefficients in a fine space. So I want to think of these as being polynomials in X0 and X1 with coefficients in these variables here. So suppose they define two hyper surfaces HF and HG. Then there is a projection from this product to AK and the condition for the resultant of F and G to be vanish. In other words, the condition that they have a common zero in P1 is just the condition for a point in AK to be in the image of HF into section HG. So we've got a projection from this intersection to AK and the image is equal to the zeros of the resultant. So the resultant will be a polynomial just in the variables KY1 to YK. So the key point is the image is closed. So we've got this slightly strange result that if you take two hyper surfaces, take their intersection, that's a closed set. The image in AK is also closed and it's sort of tempting to think this is a completely trivial result because it's a standard mistake to think that the image of a closed set under a projection is closed. Well, this isn't true. There are some obvious counter examples. For example, if we look at the map from A1 times A1 to A1, this is not closed. We can take the set XY equals one in A1 times A1 and if we project it to A1, this is just the set A1 minus zero, which is not closed. So this projection from A1 times A1 to A1 does not take closed sets to closed sets. By the way, there's a closely related example which is a sort of standard error that people sometimes make. The image of a polynomial from... Let's do this over the reals. Even over the reals, if you take a polynomial from Rn to R, the image need not be closed. It's rumored that proving that it is closed is sometimes set as a question for people. So from R1 to R, it's quite easy to show the image of a polynomial that's closed. If you're looking at maps from R2 to R, let's look at the set X squared plus XY minus one squared. Then the image is obviously at least equal to zero and you can easily take it to be any positive number by taking X to be the square root of that and choosing Y, so this is zero, but the image can't be zero. So the image is equal to four reals that are greater than zero, which is not closed. So even the image of the whole of R to the N projected to R under a polynomial map need not be a closed set, so maps in algebraic geometry are quite often not closed. And this is closely related to the concept of varieties being compact. So let's ask, when are varieties compact? Well, as we saw earlier, the predictive space over the reals or over the complex numbers are compact in the usual topology, usual house store topology. And we'd like to ask, what is the analog of this four varieties over fields in the Zariski topology? Well, P to the N over a field is certainly compact in the Zariski topology, but this turns out to be a really boring fact because even A to the N over a field is compact in the Zariski topology. In fact, any affine variety or any projective variety is always compact in the Zariski topology. Compactness just isn't an interesting property. It's more or less turns out to be equivalent to the fact saying that if a set of elements generate the unit ideal, then a finite number of those elements generate the ideal, which is true, but not very helpful if you want an analog of compactness. So we can ask, what is the correct analog of compactness? So we would like some property that is similar to compactness such that projective space has this property and affine space doesn't. Well, the correct analog turns out to be the concept of proper maps. So let's forget about the Zariski topology and talk about ordinary topological spaces. A map from X to Y is called proper if it's continuous and universally closed. So what does universally closed mean? Universally closed mean that X times Z to Y times Z is closed for all Z, which is a slightly unintuitive property, but it's equivalent to saying that it's continuous and closed and has compact fibers. So it's sort of saying the map has compact fibers and some sort of niceness property. It has to be closed, so it can't be too weird. For Hausdorff spaces, proper maps X to Y is equivalent to saying that it's continuous plus inverse image of a compact set is compact. Notice, by the way, it's not enough to say the inverse image of every point is compact because you would just let X be a disjoint union of compact sets mapping on to all the points of Y and this would be kind of the inverse image of the points would be compact, but the fibers wouldn't be joined together in a nice way, so it doesn't count as being proper. So in algebraic topology, we define a map of varieties X to Y is called proper if it is universally closed. This is for the Zariski topology and it's not only for the Zariski topology on X and Y, but for the Zariski topology on the product. So there's a subtlety here because the ordinary topology on the product isn't the same as the Zariski topology on the product. So calling it universally closed in this sense is not the same as saying it's universally closed as a map of topological spaces. This is a much stronger condition because the Zariski topology on the product is much finer than the product topology. So now we can discuss the analog of compactness would be to say that if a variety, if the map from a variety X to a point is closed, then that's a sort of analog of, sorry, not closed proper, then this is a sort of analog of X being compact. So saying this is proper is an analog of X compact. So what we want to do is to show that projective varieties have this property that the map from a projective variety to a point is proper. In other words, projective varieties are sort of analogs of compact spaces. So we want to show that if X is projective, then X to a point is proper. And it's easy to reduce to the case. First of all, X is P to the N for some N because if you've got a closed subset of X and the map from P to N to a point is proper, it's easy to show the map from that closed subset to a point is proper. Secondly, we've got to show that X times Y to Y is closed for all Y. And it's enough to show that X times A to the N goes to A to the N is closed. Essentially because Y is covered by affine subsets, so we can reduce the case, sorry, affine open subsets so we can reduce the case of Y's affine open subset. And for an affine open, for an affine variety, if this map here is closed, then it's also closed if we replace A, N by some closed subset. So what we want to do is to show that P to the N times A to the M goes to A to the M is closed because that will reduce to this case here. Well, we're first going to show that P1 times A, M goes to A, M is closed. And we will use this using resultants. So now let's show that P1 times A, M to A, M is closed. So suppose S is closed in P1 times A, M. So S is the zeros of F1, F2 and so on, some collection of polynomials where these are polynomials in, well, they've got to be polynomials in X and Y and then let's take Z1 up to ZM. So these are going to be coordinates of A, M and X and Y will be the coordinates of P1. So we have X, Y, it'll be a point here. And now what we do is we look at S1, F1 plus S2, F2 and so on, and T1, F1 plus T2, F2 and so on where the SI and the TI are just new variables. So these are going to be homogenous polynomials in X and Y with coordinates. They're going to be polynomials in all the S's, the T's and Z's. Then the projection of S to A, M is given by the points where these two polynomials have a common zero. So this will just be given by the vanishing of all the coefficients of all the polynomials in S and T of the resultant. So what we do is we take the resultant of these two polynomials, it'll be some polynomial in ST and Z and then we take all the coordinates of all the polynomials in ST and that will give us a very large number of polynomials in all the ZIs and the zeros of all those polynomials, you can check is the projection of S to A to the M. So the projection that P1 to a point is proper. Well, what about mapping Pn to a point? So we can ask, is Pn to a point? Well, this would be very easy if Pn equals P1 times P1 times P1 and so on. Because we could just do it by induction on N, we would know that Pn minus 1 times P1 to a point is proper by induction because Pn minus 1 times P1 times X to X is closed because the map from Pn minus 1 times X to X is closed, so we can just multiply by P1. But there's a bit of a problem. Pn is not equal to Pn minus 1 times P1, so we can't apply induction going down like this. However, it nearly is and it turns out to be close enough to this product that we can reduce to the proving properness of P to the N by induction. So the correct version says that blow-up of P to the N at a point is equal to a non-trivial P1 bundle over P to the N minus 1. So a P1 bundle over P to the N minus 1 is something that sort of looks locally like a product of P1 times something, but might not be globally a product because there might be some sort of twisting here. So this shows that P to the N is pretty similar to P1 times P to the N minus 1. The difference is we have to blow up a point which is a fairly harmless operation and the other difference is that the bundle might be a bit twisted and not a product and as we will see, this doesn't really matter either. Well, first of all, we have to show why this is true. Well, we've got a map from P to the N minus 1 to P, sorry, Pn to P to the N minus 1, which isn't quite defined everywhere. So, however, it's a sort of correspondence because some points in Pn will have image that's more than one point in P to the N minus 1. So what we do is we look at the graph C contained in P to the N times P to the N minus 1 and Z isn't a sort of attempt to form the graph of this function and it won't quite be a graph of this function because this isn't a function. So Z is going to be the set of pairs X0, X1 up to Xn. Those are coordinates for P to the N times Y1 up to Yn. So these are going to be coordinates of P to the N minus 1 such that XIYJ equals XJYI for all IJ for which this makes sense. So the map Z goes to P to the N is an isomorphism except the fact that it maps a whole copy of P to the N minus 1 to the point 1, 0, 0 and so on. So in other words, Z is just a blow up of P to the N at this point. So it's pretty close to being PN. On the other hand, Z maps to P to the N minus 1 is locally a product of P1 times some affine set to the same affine set. And to see this, you just take this affine set to be say that the points YI equals 1 and then we see that over this hyperplane we're just mapping X0 up to Xn to X0, X1 times X1 up to XI minus 1, XI plus 1 up to Xn. So locally this map from Z to P to the N minus 1 looks like just a product of P1 times something. So Z maps to P to the N minus 1 is proper. That's because it's proper over an open cover of P to the N minus 1 and you can easily check that being proper is a sort of local property. So if this map is proper for an open cover of P to the N minus 1, then it's proper. And now we can show that the map from P to the N minus a point is proper by induction because we just look at this map Z goes to P to the N minus 1. Z also maps to P to the N. If we map that to a point, we map that to a point and this map is proper and this map is proper. So this map here is proper by induction and this map is proper because it's locally a map from P1 times something to something and this map here is on to and very nearly an isomorphism. So the composition of two maps is proper. So this map is proper and now it follows easily that map from P to the N to a point is also proper, which is what we were trying to show. So this illustrates the technique of using blow-up to turn a rational map into a regular map. So we start off with a rational map from P to the N minus 1 times P1 goes to P to the N, which wasn't defined everywhere and we just blow this up to a map Z and the map Z now has a regular map. Sorry, this error is in the wrong direction. We now have a regular map from Z to P to the N and to this product here. By the way, for N equals 2, the surface Z we had above is something called a ruled surface over P1. What this means is it's a surface Z mapping to P1 whose fibres are all copies of P1. Another obvious ruled surface is P1 times P1 to P1, but you can also have twisted versions of this and Z is the simplest example of a non-trivial ruled surface over P1. Ruled surface is over P1 or sometimes called here it's a brook surfaces. Here it's a brook. Study them a bit.