 This lecture is part of an online algebraic geometry course about schemes, and we will be discussing proper morphisms. So for background, the problem is to find a good analog of compact spaces in topology. So the problem is that pretty much every scheme you encounter is compact in the usual topology, and so both the affine line and the projective line are compact, but in the classical complex topology, the affine line definitely isn't compact. So we want to find what is the real analog of compactness. So we'll start by looking at general topology. So in general topology, you can see that x is compact if and only if the map from x times z to z is closed for all z, means all topological spaces z. So saying a map is closed just means it takes closed sets to closed sets. Now already, this gives you a possible way to define compactness in algebraic geometry. You could just say a scheme is compact if the corresponding map here is closed, where here you would take the topology of the scheme x times y, not the product topology on the spaces x and y. Well, we could do this, but in practice it's much more useful to have a relative version of this. Well, there is a relative version in topology. We say that continuous map from x to y is called proper if the map from x times z to y times z is closed for all z. So we're working with topological spaces and tacitly assuming that all maps are continuous. For example, x is compact if and only if the map from x to a point is proper. So this sort of property is sometimes called being universally closed. So we're not only insisting that the map from x to y should be closed, but all these product maps should be closed as well. And the difference is illustrated by the following basic example. So the map from r1 to a point is obviously closed. However, if we take r1 times r1 and map this to a point times r1, this is just the projection from the plane to the x-axis. And it obviously isn't closed because we can take a closed set, say, x, y equals 1. This is closed in r1 times r1. And its projection on the x-axis is just the x-axis minus the origin. So it's obviously not closed. So just because a map is closed, it may not remain closed if you sort of multiply the top and the bottom by something. And you can see the problem here is that somehow this curve has zoomed off to a point at infinity. And this is possible because the y-axis isn't compact. And if the y-axis were compact, you wouldn't have this problem. In fact, in general topology book, you can see that x to y is proper. It's equivalent to saying that the fibers are closed, so the fibers are compact, and x to y is closed. If you don't have the condition that x to y is closed, the fibers can be compact without the map being proper. For example, x could be r1 minus the origin, and y could be r, and the map could be the obvious inclusion. And the fibers of this are all compact. They're either empty or a point, but this map obviously isn't closed because it takes a closed set to a non-closed set. So the condition that the map be closed is sort of saying that points don't suddenly disappear as you move around in the base somehow. So I just mentioned there's another criterion for properness. If y is locally compact, and let's add house store because I can't remember if this is necessary or not, then x to y is proper. It's equivalent to saying that the inverse image of any compact set is compact. So the condition of the fibers be compact just says the inverse image of a point is compact. And more generally, this condition that x to y is closed sort of says that not only is the inverse image of a point compact, but the inverse image of any compact set is compact. So that's what happens in general topology. Now we want to kind of find the analog of this in algebraic geometry. So now we're going to forget about topological spaces and work with schemes. So we say x to y is universally closed if the map from x times over z times over y z to y. Here we've got x to y. So that's z down there. So if this map here is closed for any morphism from z to y. Now this is actually a site extension of the condition that when you multiply x and y by z it gets closed because if you look at x goes to y. And you just look at y times z to y, then the pullback of this is just x times z. So in particular, if z is any topological space, the map from x times z to y times z is closed. So this is a sort of generalization of the previous condition. And growth index defined a map to be proper if it is separated of finite type and universally closed. So that's a long list of extra conditions. And of these conditions, the conditions about being separated and the finite type are usually obviously satisfied. And the important condition is this one here. So proper basically just means universally closed plus a much rather minor conditions that are in practice usually satisfied just to make sure that things behave smooth nicely. So there are two basic examples of proper maps, proper morphisms. So proper morphisms come from finite morphisms and projective morphisms. So we'll first discuss finite morphisms. So any finite morphism is proper. And what I mean to do is just sort of sketch roughly why this is true. Notice by the way, quasi-finite does not imply proper. So having finite fibers isn't good enough. And this is a counter example. This is very similar to the example that having compact fibers doesn't imply properness for topological spaces. You could just take the affine line minus the origin and map it to the affine line. And this map is certainly not proper. It's not even closed, but it's quasi-finite. Anyway, so finite morphisms are more or less obviously separated as they sort of look like affine morphisms. And they are obviously a finite type. So we just have to prove, show that they're universally closed. So step one. And step two, we notice that if X to Y is finite, then any pullback X times over YZ is also finite. So if this is finite, this is also finite. So it's enough to show that finite morphisms are closed. So we don't need to worry about being universally closed because finite morphisms are closed under pullback. Thirdly, we can reduce the affine case by covering Y with open affine subsets. And this means we want to show that if A goes to B is a homomorphism of rings and B is a finite A module, then the map from the spectrum of B to the spectrum of A is closed. So we've reduced our problem to a problem about commutative algebra. And now you can factor this into A goes to A over I, which is contained in B, not quite contained in B, but isomorphic to something contained in B. And this corresponds to a closed immersion of schemes, which is closed. Here I is of course the kernel of the map from A to B. So we can assume that A is contained in B because the composition of two closed morphisms is closed. Next, a closed set of B is given by an ideal of B. It's called this ideal J of B, and it would be something like the spectrum of B over J, which is contained in the spectrum of B. And now we get a map A modulo A intersection F minus 1 J is contained in B over J, where F is the map from A to B. So the spectrum of A over A in F minus 1 J is closed in the spectrum of A. So it's enough to show that if A is contained in B and B is a finite A module, then the map from the spectrum of B to the spectrum of A is onto. So if we've got a closed set in B, we can just apply this result to the map from the spectrum of B over J to this space here. So we've reduced our theorem to this question of commutative algebra. This is still quite a tricky theorem to prove, but we're not going to prove it, we're just going to quote it. This is a theorem by Cohen and Seidenberg, which just says that exactly this, that if A is a subring of B and B is a finite A module, then this is onto. In fact, they proved a little bit more than that. They proved something called the going up theorem, which says that not only is this map onto, but you can add some extra conditions about the primes of B and A, which we don't need for this result. So that's final maps. The other main example of projective maps, projective morphisms, we've had a special case of this before. We saw that if X is a projective variety over field K and Y is any variety, then X times Y goes to Y is closed. So this is a result of elimination theory. So this is, it's well on the way to proving that the map from extra point is universally closed. We've done it for varieties and you need to do it for arbitrary schemes. So we'll explain roughly how to do this next lecture when we discuss the relation between valuation rings and proper morphisms.