 So this talk will be about the question, what is an etal morphism from X to Y? So in previous lectures, I was talking about the etal topology and giving some idea of how to calculate, but I never actually defined what an etal morphism actually was. So I want to try and repair this. The first question is, what does etal mean? Well, it's a French word, which means slack. And this is not really very helpful. You sometimes get the impression that when people like Seren Grothendick were inventing all this terminology, they opened a dictionary completely at random and chose a random adjective. So I've no idea why these are called slack morphisms. Anyway, what I'm going to do first is define etal morphisms for the usual topology of complex manifolds. And then we will use that to motivate the definition for etal morphisms of schemes. So an etal morphism, F from X to Y is one such that, one that is a local homeomorphism. What this means is that given the point X in X, it has a neighborhood U, so that F is a homeomorphism from U to the image. So a picture of this is as follows, typical etal morphism, X might be some sort of double cover of a circle, so Y might be a circle. And what it's saying is that if you choose any point in X, like this one, there's some little neighborhood such that the map to its image in Y is a homeomorphism. And so here are some typical examples. So first of all, if we have an open set U contained in X, then the inclusion is obviously etal. Secondly, if we take say the non-zero complex numbers and map it to the non-zero complex numbers by mapping Z, goes to Z to the N, this is etal. But you can see if we allow the zero complex number, that's not an etal map because it's not a local. I mean, you can't find a neighborhood on which it's an isomorphism. The third example is if we take the line with two points and map it to the line, then that's also etal. And in fact, you can see if you take this point here, it's got a perfectly good neighborhood that maps homeomorphically to a neighborhood of there. So etal maps are allowed to be non-house dwarf. In fact, they quite often are. Etal maps from X to Y are more or less the same as sheaves over Y. In fact, in the early days of sheaf theory, sheaves were often defined as etal maps rather than maps from open sets to something or other. And the way you get from one to the other is, if you've got an etal map, the corresponding sheaf is just given by sections of FX to Y. In other words, for each open set in Y, you just take the maps from U to X that are sections and that gives you the value of the sheaf on Y. On the other hand, if you've got a sheaf, then we can define the etal space X over Y to be the union of the fibres of the sheaf over points of Y. And you can put a topology on these and check it's an etal map. So the next question is, what are sheaves for the etal topology? So we remember we said there was something called an etal topology in algebraic geometry and you could define some rather funny looking sheaves for it. So let's think about sheaves for the usual topology. Here we're talking about a complex manifold Y with the usual complex topology. So sheaf for the usual topology means for each open set in Y, we're given a set and this has to be funcorial and satisfy some sort of covering condition. So a sheaf for the etal topology means that for each etal map from X to Y, we're given a set and this is funcorial and satisfies covering conditions and whatever. And the point is that sheaves for the usual topology are more or less the same as sheaves for the etal topology. And that's because if you've got an etal map from X to Y, say draw a typical etal map looks something like this. The point is it's covered by sets that are isomorphic to open sets of Y. That's more or less the definition of an etal covering. So the point is X is covered by open sets of Y. Well, when I say it's covered by open sets of Y, I mean it's covered by sets that are isomorphic to open sets of Y. So this means so any sheaf on X, so any etal sheaf on X is determined by its values on open sets U in Y because there's a sort of covering condition for sheaves that says that if an open set is covered by other open sets and the sheaf on this is determined. So what this means is that sheaves for the usual topology are more or less the same as sheaves for the etal topology. So in this particular case, the etal topology isn't any different from the usual one for at least for most practical purposes. So that covers etal maps for complex manifolds with the usual topology. Now we want to look at etal maps in algebraic geometry where we're going to use the Zariski topology and let's see what happens. So let's try attempt number one, which will fail. Let's consider maps F, X to Y that are local isomorphisms. What this means is that any X in X has an open neighborhood U such that F on U to the image is an isomorphism, say an isomorphism of schemes. So this is the obvious analog of the definition of etal map we gave. And well, it's a perfectly good definition of a local isomorphism, but it fails as an analog of etal maps. And it fails in two ways. First of all, if we take the map from the non-zero elements of a field to itself taking X to X squared, say, this is not a local isomorphism in the Zariski topology. So you see if we take this map here and we map it to there, if we take a point here and we take an open neighborhood in the usual complex topology, that's quite easy. It's a small set there, but an open neighborhood in the Zariski topology is huge. It's sort of everything except a finite number of points and it certainly doesn't map isomorphically to anything in there. So the map taking X to X squared is not a local isomorphism, but we want it to be etal. The second reason it fails is that sheaves are the same as Zariski sheaves. So you remember we were introducing etal sheaves and etal cohomology in order to get different cohomology groups because the cohomology groups constructed using the Zariski topology were much too small in general. So using local isomorphisms is just not going to give us any better cohomology groups. We're gonna get the same sheaves and the same cohomology groups as for the Zariski topology. So we need to do something else to find a good analog of etal maps. So let's think about why does it not work? So why does it fail? And the answer is Zariski open sets are too big. We saw this, the Zariski open set was everything but a finite number of points on the circle and so it couldn't be a local isomorphism. So we want to find a way to kind of say something as a local isomorphism in a more local way than for Zariski sets. Well, let's take a look. First of all, if we've got a local isomorphism and we've got a point here and its image might be y which is f of x, then we get an isomorphism from the local ring of x. Sorry, it goes from the local ring of y to the local ring of x. So this would be an isomorphism. If this map is a local isomorphism it means all the maps of local rings are isomorphisms. And local rings see too much of x in algebraic geometry. I mean, it sees all of x but except for co-dimension one subsets in some sense. And there's a variation of the local ring that sees far less that we can take the completion of our x hat and our y hat. And these are also going to be isomorphism. There's a local isomorphism but the completion sees much less. For example, if we take a curve that looks like that and a curve that's just the intersection of two lines the local rings at this point is going to be quite different from the local ring at that point but the completion of the local ring at this point is going to be the same as the completion of the local ring at this point. So the completion of a local ring is seeing far less of the variety. So we can just say this is an isomorphism for all points x where y is the image of x. And this gives a reasonably good definition of a tile maps. This works for varieties over an algebraically closed fields. So if all you want to do is mess around with varieties over algebraically closed fields you can use this as the definition of an etal map. It just has to be an isomorphism on local rings. Well, there are however some problems with this definition. First of all, it's a little bit clumsy to use because completions of local rings are not exactly an elementary concept. Secondly, we want to work with varieties that are not over algebraically closed fields and this definition just goes wrong. It also goes wrong even if we're just looking at the spectrum of field, in other words, single points. I mean, we saw last lecture that etal cohomology is more or less equivalent to Galois cohomology in the case of fields and it just isn't if you use this definition. So we need to modify this a bit and let's look at it a bit more closely. So what do we need to get an isomorphism of local rings, Ry to Rx? Well, let's think about what the completion of Ry is. Well, it's an inverse limit where we take Ry over the maximum ideal or Ry over the maximum ideal squared or Ry over the maximum ideal cubed and so on. So what we want to do is to, in order to define a map from the local ring to the local ring of X, we need to define compatible maps for all of these. And the way we can do this is we start with a map from this ring and then we lift it to a map from this ring and then we lift it to a map from this ring and so on. In other words, we have to solve the following problem. We want the spectrum of Ry over M to the N plus one mapping to X, mapping to the spectrum of Ry over M to the N, mapping to Y, and we want to find a lifting of this map. So this says that, sorry, that should be an N and that should be an N plus one. In other words, if we've managed to lift this ring to a map of X, we want to lift it to a map of this ring to a map of X. And if we do this for all N and stick them together, we've got a map from the local ring of Y to X. Well, this suggests the following problem. Suppose you've got X and Y and we've got the spectrum of some ring, mapping to Y and we've got the spectrum of some ring modulo, an ideal, well, we could take N squared equals north, or we could, if we want, take N to be null potent. And we want to know, is there a, given this configuration where C is some ring and N is a null potent ideal, we can ask, does there exist a lifting here? And we say, this condition is almost the definition of et al. We say X to Y is formally et al. If this map G, if there is a unique map G, making this diagram commute but if any choice of C and N. By the way, there are two other very similar conditions. We say it's formally smooth if there's at least one G and we say it's formally unrammified if there's at most one G. So something is formally et al if and only if it's both formally smooth and formally unrammified. Formally smooth maps turn out to be very close in relation to the notion of non-singular varieties. So if Y is the spectrum of an algebraic closed field, then a map from a variety to there is formally smooth if and only if this is a regular variety in the sense of algebraic geometry. Anyway, we're mainly interested in the formally et al ones. Well, that almost gives us the definition for et al. Formally et al ones, turns out there's some sort of pathological cases and Grozendick found the right condition for et al map. So a map X to Y is et al if and only if it's formally et al plus locally or finite presentation. So in practice, this condition being locally finite presentation is almost always satisfied, trivially satisfied in any map you're considering. So formally et al is in practice more or less the same as et al. Incidentally smooth and unrammified are defined in the same way. They're smooth if they're formally smooth and locally finite presentation and the same for unrammified. So we should give some examples. First of all, the trouble with this definition is a bit hard to get to a head round it. So let's first of all look at a very simple example. So first we take X to Y and suppose this is et al and suppose we take a point and a point with a little tangent vector like this. So this would be the spectrum of some field K that you're working over and this might be the spectrum of K of X over X squared which is really a point with a little bit sticking out in some sense. And this map being et al means we can always find a unique lifting. So let's draw a picture of this. What it means is if we choose a point of X and we choose a point of Y and a little tangent vector at this point in some sense then we can lift this tangent vector to a tangent vector in X uniquely. So in particular et al means you can uniquely lift tangent vectors and that's obviously closely related to the idea of this being in some sense a local isomorphism. So let's give some examples of maps that are or aren't et al. So let's look at the spectrum of K of X over X squared mapping to spectrum of K. So this looks like a point and a point with a little bit sticking out from mapping to it. And we can ask is this et al? Well, for this we need to check the following conditions. So we've got this map spectrum of C over N mapping to the spectrum of C mapping to the spectrum of K mapping to spectrum of KX over X squared. And we want to know if there's a unique map like this and let's turn this into a question about rings by reversing all the arrows. So we've got a map from K to K of X over X squared and this maps to some algebra C and this maps to some algebra C over N and we want to know if we can find a lifting like that. And the answer is you can't always. For instance, we can take this element C to be K of X over X cubed and you can take N to be the ideal generated by X squared. So this is K of X over X squared and you can take this to be the identity map and you can see there's no way of lifting this map here because this must map it to X plus something and X plus something squared is never going to be zero in this ring, although it can be zero in that ring. So this map is not et al and that's very plausible geometrically because this doesn't sort of look infinitesimally like this. So a way to think of et al maps is as follows. Suppose you've got a space X and a space Y. What et al means is that if you've got some something in Y so this might be the spectrum of C and suppose we've got spectrum of C here. Sorry, spectrum of C over N. Then suppose you extend this infinitesimally by making it so going from this scheme to this scheme is kind of sort of making it infinitesimally bigger. So it says that if you extend this infinitesimally then we can lift this to the scheme X. So we can sort of lift infinitesimal thickenings of a sub-scheme and you can see you can't do that here if you sort of, no, never mind. So for another example, I want to look at some cases over a field that isn't algebraically closed. So the next example is going to be, let's take a field which is not algebraically closed. I'm going to take L to be a finite, separable extension. So you remember this means L is of the form K of X, modulo F of X where F is irreducible and has no multiple factors over the algebraic closure of K. This is equivalent to saying that F and its derivative are co-prime. In particular, this means AF prime plus BF equals one for some A and B. So AF prime equals one in L. So the derivative of F at X is, so the derivative of F is now invertible. Now we've got to ask the question, is spectrum of L to spectrum of K et al? So we need to look at the following diagram. We have C over N and spectrum of C goes to here. It's very difficult remembering which around all these arrows going. We want to know if we can lift it like that. And as usually, it's easier to write this in terms of rings. So let's take this and we want to know, is there a lifting of L to C? Well, we have to map X to something in C and X maps to something in C over N and we can lift that to an element C of C. But if we map X to that, that's not necessarily a homomorphism of rings. So what we want to do is we want to find N in this ideal N, so that F of C plus N equals zero because then we can map X to this element C plus N. And we're allowed to add N to C because that will still have the same image in here. So we know F of C is in the ideal N. So we want to solve F of C plus N which is F of C plus F prime of C times N because N is null potent. And we want this to be equal to zero and we can solve this equation because F prime of C is invertible. So we can choose N so that F of C plus N equals zero and that gives us our mapping from L to C. So finite separable extensions are etal. And this is really useful because knowing finite separable extensions is more or less the same as knowing the absolute Galois group of a field K. So what this turns out to mean is that etal cohomology of the spectrum of a field is more or less the same as the Galois cohomology of that field. And if you remember we used that several times in working out the cohomology of a curve. Incidentally, notice that we were working out the cohomology of a curve over an algebraically closed field but in doing that, we needed to work out the Galois cohomology of a non-algebraically closed field, the field of rational functions on the curve. So this is why you need to know etal maps for non-algebraically closed fields even if you're working with varieties over algebraically closed fields. We should finish off just by showing that inseparable extensions are not etal. So here's an example of an inseparable extension. So let's take a field K of characteristic P and let's map it to some inseparable extension K of X, modulo X to the P minus A where A is some element in K that's not a piece power. And now we have the following lifting problem. We're going to take this map to be the identity map. So we just map K of X over X to the P minus A. And now we want to show that we can't solve this lifting problem for something here. So we've got to find a suitable something and we can just take K of X epsilon and mod out by their ideal generation by epsilon squared and X to the P minus A minus epsilon. And now what we need to do to lift this map is to find a Pth root of X to the A minus epsilon. So A plus epsilon has no Pth root. If we try to find a Pth root of the form X plus something times epsilon to the P, which is the obvious thing to do. This is just X to the P plus P times something. And this is just zero because we're in characteristic P and this is just X and this is just equal to A. So this is not equal to A plus epsilon. So we can't lift, so we can't find a lifting of this map here and inseparable extensions are not etal. Okay, that's all about etal maps at the moment.