 This lecture is part of an online algebraic geometry course on schemes, and we will be about separable schemes and morphisms. So, let's start with some background. In analysis, we have two really basic topological properties, that have been house dwarf and that have been compact. The trouble is in algebraic geometry, these two properties are not really terribly useful, because if we look at say varieties in the Zariski topology, then they're almost never house dwarf. I mean points are house dwarf, but almost anything other than that is not going to be house dwarf. In fact, in a variety in the Zariski topology, any two non-empty open sets intersect, so it's about as far as possible from being house dwarf. Similarly, with compactness, varieties are almost always compact unless you sort of allow funny sorts of abstract varieties with infinite numbers of open sets or whatever. So, compactness is not terribly useful either, and we want to know what is the correct analogue of these for schemes. For example, n-dimensional projective space over C is compact in the usual complex topology, and A to the Nc is not compact again in the usual complex topology. So we would like some property of the abstract varieties, which holds for projective varieties but not for affine varieties, and similarly we'd like to find some analogue of house dwarf for them. So the analogue for schemes or varieties, house dwarf is replaced by the property of being separable. In fact, the word separable in general topology sometimes means house dwarf, especially in French-speaking countries. And the analogue of being compact, well it has two analogues, this corresponds to the notion of completeness for varieties, so we discussed the fact that projective varieties were completed in an earlier course. However, we also like relative versions of these, so if we've got a map from a scheme S to a base scheme S, then we can also talk about separable morphisms. And instead of talking about complete morphisms, you usually talk about proper morphisms. So the correct analogue of compactness turns out to be related to the notion of a proper morphism, and the correct analogue of being house dwarf turns out to be related to the notion of a separable morphism. Very informally, a morphism is separable if and only if it's all its fibres at house dwarf, and a morphism is proper if and only if its fibres are all satisfied, whatever the correct analogue of compactness is. So we have to find a good way to define the analogue of house dwarf, so in topology, so in general topology, a space X is house dwarf if and only if the diagonal of X times X is closed. And this is kind of almost obvious if you draw a picture, so here's a copy of X and a copy of X, and here's the diagonal, and if we've got a point X, Y with X not equal Y, what we can do is we can choose a little neighbourhood of X and a little neighbourhood of Y, and we end up with a little neighbourhood of X times Y, so this neighbourhood is U and this is V, then U times V is the neighbourhood of X times Y, and if the diagonal is closed, it means we can find a neighbourhood U times V of X, Y disjoint from the diagonal, so U and V are disjoint, and conversely if you can find disjoint neighbourhoods of U and V, then the diagonal is closed. So we can use this idea to define separability or scheme, so we say a scheme X is separated if the map from X to X times X is closed immersion. In fact, it's not difficult to prove this is equivalent to the image being closed. It's obvious that closed immersion implies the image is closed, the other implication requires a little bit more thought but isn't very difficult. Well, more generally, we want to define the concept of a separated morphism, so X to S is a separated morphism if the diagonal map from X to X times over S X, here we've got these two maps here, if this thing here is a closed immersion or has closed image, as before these two conditions turn out to be equivalent. So informally, a morphism is, if a morphism is separated, you expect all its fibers to be separated, it's not quite equivalent but it's very close. So X is separated as a scheme is equivalent to saying the morphism from X to the spectrum of Z is a separated morphism, because this is the terminal object in the family of schemes. So that's some examples. The first example is the traditional example of a non-separated or non-house store space, you just take the line with two origins. So in general topology, you can take two copies of the real line and stick them together by identifying everything except the origin, so you get something that looks like this. There are two copies of the lines and there are two origins and you identify all the other points. And of course we can do the same thing in algebraic geometry except we take the affine line which is say spectrum of K of X and this is spectrum K of X and they're glued over the affine line minus the origin which is the spectrum of K of X X minus one. Now of course when you take the product, you don't give this the product topology that you use in general topology, if you did that all you would do is get exactly the definition of house store which doesn't work. So this thing here should of course have the Zariski topology which is finer than the product topology. So if you draw a picture of this in the Zariski topology, the X times, well I guess we're working over the point spectrum of K sort of looks like this. What it does is it has a sort of double X axis and a double Y axis. Sorry, got the X and Y the wrong way around so this is a sort of double X axis and if you look carefully at it, you'll see that there are four points. I mean it's got a sort of quadruple origin and if you look at the diagonal, you see it contains two of these four points and these two red points are in the closure of the diagonal but not in the diagonal. So the diagonal is not closed and this is an example of a non house dwarf scheme or a non house dwarf morphism from extra spectrum of K. Another example. Suppose we've got any morphism from the spectrum of a to the spectrum of B, where A and B are of rings. And this is always separated and what we have to do is if we call this X and we call this Y, then we're looking at X to X times over Y X. So we're looking at this map of schemes here and we have to unwind it and figure out what this is in terms of algebras. Well, that's just a tensor over B a mapping to a sort of map goes the other way, very confusing getting these all the way around. And we have a map going this way to a and the point is the map from a tensor over B a to a is surjective. In other words, a is just a tensor over B a modulo sum ideal. So this is this is just more or less by definition a closed immersion. So all morphisms between affine varieties are automatically separated from this. It's fairly easy to deduce the fact I mentioned earlier that if the image of the diagonal map is closed, then the diagonal map is in fact a closed immersion. So so to check something is separated you just need to check the image of the diagonal map is is closed and there's a risky apology. So let's have a third example. Let's take X to be equal to Y to be the line with two origins. So here we've got X looking like this. And we're just mapping it to Y. It looks exactly the same. And you notice that X and Y are both not separated and Y is not separated. But the morphism X to Y is separated. In fact, since X equals Y the diagonal map from X to X times over Y X is just the identity because X times over Y X is just equal to X. And any identity map X to X is a closed immersion. So the fact that a morphism is from X to Y is separable doesn't imply anything about X or Y individually being separable. Roughly speaking, what it means is that if you've got an open affine subset of Y, then the inverse image of that is a separable scheme. So there is actually that has actually going to change in terminology in the old terminology. Schemes are called pre schemes and separable schemes were just called schemes. So, so these things here are separable and these things here are not necessarily separable. So things like if you read old copies of growth index elements of algebraic geometry initially started off by using the terms pre scheme and scheme instead of scheme and separable scheme. So that's one minor thing to watch out for. There's a variation of being separated called quasi separated. You sometimes come across a morphism X to S is quasi separated means the diagonal map from X to X times over S X is quasi compact. Any closed immersion is quasi compact so separated implies quasi separated. And actually the notion of being quasi separated is not used all that much in scheme theory. It's much more important than the theory of algebraic stacks and algebraic spaces. So it's actually quite difficult to find examples of morphisms that are not quasi separated. For instance, if you take the line with two origins X and you map it to spectrum of K, then X times over spectrum of K X map from X to this is quasi compact. In fact X is notarian and if you've got any notarian scheme then any map from it to another space is automatically quasi compact because every open set is quasi compact. So any map from X to anything else is is automatically quasi separated. So if you just want to stick to notarian schemes then every map whatsoever is automatically quasi separated. As I said, you really have to think a bit to find an example of a morphism that is not quasi separated. Well, here's an example. Let's take X1 equals X2 to be copies of spectrum of K X1 X2 X3. So this is infinite dimensional affine space. And what we glue, what we do is we glue X1 and X2 along this infinite dimensional space minus the origin. So the origin is either the point north north north and coordinates where it's the ideal X1 X2 and so on. And if you look at this, we then get a map from if we call X the result of gluing these, then if you look at X times over S X to X X to S where where S is just the spectrum of K. So we're working here for point map from X to this. What we do is you can take the image of X1 times over S X to which is containing X times over S X and this is an affine scheme. So it's quasi compact. And the inverse image of this in X is is is just is just affine space, infinite dimensional affine space minus the origin. So we saw earlier that infinite dimensional affine space minus the origin was a was an example of a scheme that wasn't quasi compact. So here we have an example where the diagonal map from X to X times X is not quasi compact. There's an open affine quasi compact set there whose inverse image is not quasi compact. As you see this example is a bit convoluted and not the sort of thing you're likely to encounter in practice. So we'll finish with an application of separability. We have the following problem is the intersection of two open affine sets and open affine set. So the open affine sets form a base of the topology of any scheme and it's really convenient if if this base is closed under taking into sections it avoids a certain amount of technical mess at times. And the answer is no in general. So here's an example of an intersection of two open affine sets that's not open affine. Let's take X to be the plane with two origins. So X is equal to X1 union X2 where X1 is isomorphic to X2 is isomorphic to the affine plane and X1 intersection X2 is equal to the plane minus the origin. And we saw earlier that this is not affine. So you definitely can get cases when the intersection of two open affine sets isn't affine. However, you notice this space here is very definitely a non separable scheme. And in fact, if you if you don't allow this sort of non separable behavior, then the intersection of two open affine sets is indeed affine. So suppose we have a morphism F from X to S with F separated and S affine. Then the intersection is open affine for UV open affine subsets of X. You can check that if you drop the condition of S is affine or F is separated, then the intersection of two open affines need not be open affine in general. And to do this, it's fairly easy. We just notice that U times over SV is open affine because UV and S are all affine. And if you if UV and S are all affine, then then the product is just affine because it's the tensor product of the corresponding algebras. So this is open affine in X times over SX. And we know that U intersection V is the inverse image of U times over SV under the diagonal map from X to X times over SX. And now we notice that for a closed immersion, the inverse image of an open affine is open affine. So if X to S is separated, then this diagonal map is a closed immersion. So the inverse image of U times V is open affine. So we find U intersection V is an affine open set. Okay. So next lecture, what we're going to do is review valuation rings in order to characterize separated and proper morphisms using valuation rings.