 This lecture is part of an online algebraic geometry course on schemes and will be about how to tell whether a morphism X to Y of schemes is separated using valuation rings. So we'll first start by looking at the case of Hausdorff spaces. So a space is Hausdorff means informally that sequences have at most one limit point. That's not precisely what it means, but in practice it's kind of similar. So you can think of it as following suppose we've got a space X. And suppose we've got a map from the open unit interval 01 to X. We can ask how can this be extended to a map from the closed unit interval to X. And if X is Hausdorff, there is at most one way of doing this. So there's at most one extension if X is Hausdorff. On the other hand, if X isn't Hausdorff, there may be more than one extension. For example, we may take X to be a line with two origins. And then if we've got an open interval 0 to 1, we might map it to here in the obvious way. And if we extend it to a closed interval by adding a point here, then this point can go to either there or there. So there are two possible extensions. Well, it's also useful to have a sort of relative case of this. So suppose we've got a map of spaces from X to Y. Then we can ask, suppose we've got an open interval mapping to X. And we want to extend it to a map from the closed interval to X. Well, we suppose we're given a mapping from this to Y. We've got an obvious inclusion homomorphism there. Then we can ask, is there more than one way to extend this? So we can ask how many extensions of this map are there? So if there's only one extension, we might think of X as being kind of relatively Hausdorff over Y. For example, we might take X to be a plane with two Y axes. So X is going to be a plane, and you kind of duplicate the Y axis by gluing two copies of the plane together. And we might map this to a line with two origins. And now, if you've got an open interval mapping to X, say it maps it to this, and we're given an extension of this to the half open interval to Y, well, the green point can go to either here or here. So there are two ways to extend this to Y. Y isn't separable. However, if we choose one of these extensions, there's only one way to lift it to a map here. So there's only one way to lift this. So there's only one lift. Well, I guess in general, there's at most one lift. So let's say at most one lift. So although X and Y are both non-Hausdorff, in some sense this map from X to Y is relatively Hausdorff. All the non-Hausdorffness of X comes from the fact that Y is non-Hausdorff in some sense. And what we want to have is an analog of this in algebraic geometry. So this property here will turn out to correspond an algebraic geometry to the property that morphism X to Y is separated. And as we see, X and Y can be non-separated, but this morphism might be separated. So we need the analog of the open interval contained in the closed interval. So this is our analysis. This will be the open interval 0 to 1. And this will be the point 1. And the analog of this is the spectrum of R, where R is a discrete valuation ring. So you know the spectrum of R has two points. It has a generic point and it has a closed point. So this is the generic point given by the ideal 0. And this is the closed point given by the ideal P, where P has valuation 1. If you would like an explicit example, we could take R to be, say, the ring of formal power series over the reals. The quotient field K is then the ring of Laurent series and the spectrum of R. Well, the elements of this ring are a sort of functions on the real axis given by power series, except they're not quite because the power series is only formal and doesn't converge everywhere. But you can think of it as being converging in an infinitesimally small part of the real line. So it converges at 0. It doesn't really converge anywhere else at 0. But we sort of pretend that if you're infinitely close to the point 0, then this power series converges. So its spectrum can be thought of informally as looking something like this. And if you look at these two pictures and sort of squint at them and don't look too closely, they look very similar. Of course, they're not similar at all because this is two points and is none howst off and this space is howst off and has an uncountable number of points. But if you don't worry about that too much, these two things are very, very similar. So if we go back to this diagram, here we had a diagram we had earlier where we were trying to have a map from X to Y. We had maps here and we were trying to ask if you can extend this in at most one way. Now let's translate this into the language of schemes. What we do is we take schemes X and Y. And well, what we have down here is the spectrum of a discrete valuation ring R. And here we have the spectrum of K where K is the field of quotients of R. And we ask if we're given these four maps, we can ask how many lifts are there. And we might expect that most one lift should be somehow related to the property that X to Y is separable. So we can ask, is this related to X to Y being separable? And the answer is sometimes. So there's a theorem by growth and it says that if X is a finite type over Y, that means the morphism from X to Y is a finite type. And Y is notarian. Then the two properties here that there's always at most one lift and X, Y are separable. So these two properties are equivalent. Okay, this is slightly different from the version in Hart-Shorn's textbook. Hart-Shorn has slightly less restrictive conditions on X and Y. And it's also a bit different because Hart-Shorn doesn't just require R to be a valuation discrete valuation ring, but he uses the case when R is allowed to be any valuation ring. So the solution here comes from, if you want to look at top, it comes from growth index elements of algebraic geometry, chapter two, part 7.2 or whatever. And which version you use is really a kind of matter of preference. If you don't like general valuation rings as growth index didn't, then you put on slightly stronger conditions on X and Y so you can get away with just using discrete valuation rings. And one of the other problems in scheme theory, every theorem not only has a large number of small conditions, you need to make it work, but there are always half a dozen versions of every theorem with slightly different variations in which conditions you have and slight variations in how strong the theorem is. So it's a real headache trying to remember all these. So I'm not going to prove either version of this theorem. There's a perfect proof written out in Hart-Shorn's book. Instead I'm going to give some examples of how to use it. So the first problem is, before using this, is what is a morphism from spectrum of K or the spectrum of R to a scheme X. So, well, first of all, morphism from the spectrum of a field K to X, easy to describe. So here K is a field. Well, first of all, the spectrum of K has just one point. So it's image in the spectrum of X is going to be a point. So we need to choose a point in X. Let's call this point P. And this point will have a local ring. Let's call it XP. So XP equals local ring of X at P. Informally it's even think of it as being something like functions on X that are defined near P. And in order for this to be a homomorphism of schemes, we need to give a homomorphism of rings from X to the local ring of spec K at this, which is just K. And this must be a local homomorphism. And you remember the local homomorphism means the inverse image of the maximal ideal of K must be the maximal ideal of this ring here. So if we let K of P be the X of P modulates maximal ideal, you can think of K of P as being the sort of field associated to the point P. Then we have a homomorphism from K of P to K. So a homomorphism of schemes from spectrum of K to X is equivalent to choosing a point of X plus a homomorphism of fields from the local ring, not the field at P to the field K. So morphisms from spec K to schemes are easy to control. Now we need to know what is a homomorphism from the spectrum of R to X where R is a discrete valuation ring. Well, the spectrum of R has two points and we can call these points X0 and let's call the generic point X1. And we've got a homomorphism from the spectrum of K to this ring. It's always difficult to remember which were around these wretched arrows go. So we need to choose two points X0 and X1 and X, which are the images of these. And in order for this to be a continuous map, we must do X0 is containing the closure of X1. And we notice that if X0 is in some open affine, then X1 is in the open affine set. So this makes it particularly easy to work at homomorphism from the spectrum of R to X because if X is covered by open affines, then this is just given by homomorphism to one of these open affines. We don't have this problem that the image of the spectrum might be in several different open affines, which makes it rather tricky to work at homomorphisms. And we get a homomorphism of the spectrum of fields as before. So we get a map from K of X1 to K where K of X1 is the field of this point here. And we also need a map from the local ring of X at X0 to R. And this must be a homomorphism of local rings. Sorry, it must be a local homomorphism of local rings. So the inverse image of the maximal ideal of here must be the maximal ideal of K here. Well, since R is a subring of K, we may as well quotient out X, the local ring here by the inverse image of zero in K. So we can take the image of this local ring in KX1. And what we get is a homomorphism of the image of X, X0 in KX1 to R. So this allows you to describe homomorphism from the spectrum of a discrete valuation ring to any scheme. So let's work out two examples. So the first example, let's check the line with two origins is not separated using this method using valuation rings. So the picture we get here is let's take the discrete valuation ring to be just the localization of polynomials at the point zero. So this is all quotients of polynomials such that the denominator is none zero at the origin. And our diagram that we've got to investigate looks like this. So here we've got the bug-eyed line or line with two origins and we're given maps like this and we want to investigate how many ways that are of lifting this. So let's translate this into homomorphisms of rings. So here we've got KX and here we've got the localization as a subring of it. And here we've just got the morphism from K to there and here we've got two copies of KX kind of glued together. So we've got two different maps like that. And we're going to take the homomorphism here that's just the obvious inclusion of K of X, the ring of polynomials to the field of rational functions. So we just take these to be the obvious inclusions. Now, we want to work out, sorry, I've got this arrow the wrong way round again. Probably whenever you switch from schemes to rings, you always have to change the direction of the arrows and it's incredibly easy to get them the wrong way round. So if we fix one of these rings K of X, there's obviously exactly one way of extending this map because the only way is obviously the natural embedding of this ring into here. However, there are two possibilities because we can map the closed point here to either one of these two points here. So this corresponds to either mapping this ring to here or this ring to here. And these give us two different homomorphisms from this scheme to this scheme here. Informally, if you think of a spectrum of this as being an open set followed by point, then the two maps, the open point always goes to the generic point here. But there are two possibilities for where the green point it can go to, it can go to either of these. But once you've chosen the image of this green point, then there's that then the rest of the morphism of the schemes is uniquely determined. So there are exactly two morphisms here. Anyway, there's more than one morphism. So this map here is not separated. So the second example, let's show that the project of line mapping to let's work over the integers to the spec project of line over the integers. We want to show that this is separated. So the project of line and the line with two origins. So P1 and the line with two origins are both obtained by gluing two copies of A1 along A1 minus the origin in some sense. So these two ways differ in that one of them is not separated and the other one is separated. So being separated is a fairly subtle property of exactly how you glue things together. So first of all, we need to recall what are the points of P1 of Z with values in a local ring. So we're only going to use a discrete valuation ring, but we may as well do it for all local rings. We need a map from the spectrum of R to P1 over Z. Now P1 over Z consists of two copies of the affine line over Z, which have been glued together. And spectrum of Z has a spectrum of R has a closed point and then it's got various other points that we don't know very much about. The point is that if the image of the closed point is in some open affine, then the image of the spectrum of R is in the same open affine. So this makes it particularly easy to find maps from the spectrum of R to P1 of Z because the image is going to line one of these open sets. We don't have to worry about it maybe being partly outside the first one and partly outside the second. So morphisms from the spectrum of any ring to the affine line over Z are very easy. The affine line over Z is just spectrum of Z of X and morphisms from Z of X to a ring just correspond to elements of the ring. So morphisms from spectrum of R to the affine line just correspond to elements of R. So morphisms to the first copy of the affine line correspond to elements say R0 of R. And morphisms to the second copy of the affine line correspond to some element R1 in R. And we have to glue these together that some morphisms to the first A1 will correspond to certain morphisms of the second. And the way they glued together is that R0 is the same, so R0 is the same as R1 if R0 times R1 is equal to 1 because we're gluing these two copies of the affine line by sort of changing something to its inverse. So the points can be identified as pairs X0, X1 such that X0 or X1 is a unit and X0, X1 is the same as lambda X0, lambda X1 if lambda is a unit. And here the point R0 that we've got there might just correspond to the point R0 colon 1 and R1 might just correspond to 1 colon R1. And you can see that this description here is the same as this description here. So points of the projective line just are roughly what you would expect. So now let's check the lifting property. So you remember we've got this map spectrum of K goes to P1 over Z goes to spectrum of R goes to spectrum of Z. And we're checking maps that go in, I think it's this direction. And now let's check this in terms of rings, sorry no we don't need to do it in terms of rings. So maps here, so these sorts of maps just consist of pairs K0, K1 with not both 0. And of course K0, K1 is equivalent to lambda K0, lambda K1. And points here correspond to pairs R0 colon R1 with R0 or R1 a unit. And again this is equivalent to lambda R0, lambda R1. So the question is, if we're given a point K0, K1 like this. Is there at most one point R0, R1 like this and given K0, K1, there is exactly one R0, R1 up to equivalence. That corresponding to it, which obviously is K0, R1 equals K1, R0. And this is obvious if K0 is not 0, then we can just take the point 1, K1 over K0. And if K0 is 0, we just take the point 0, K1, 1. And you can check that if R0 and R1, if you take R0 and R1, one of them is a unit. Say R0 is a unit, you may as well divide by R0. So you can assume R0 is 1 and then there's only one point correspond, then that point is unique given the ratio between R1 and R0. So the projective line over the integers satisfies the condition for being separable given by valuation rings. In fact, you can easily extend this to n-dimensional projective space over the integers or that any field. Okay, next lecture we're going to talk about proper maps in general topology and algebraic topology.