 OK, so maybe we'll just start. So we were talking about rational maps. So I will brief recall, and then we finish there. So rational maps were morphisms defined on an open subset. So if phi from x to y, a rational map between varieties will be an equivalence class of pairs phi. So where u is open in x, and phi is a morphism from u to y. And we had then seen that we also have the largest open set, the largest such u, u will be called the domain. And then we had to find what it means for a rational map to be dominant. So phi is dominant if its image is dense in y. I know, and as I mentioned, this is necessary because we can only compose dominant rational maps, because otherwise it might happen that the image of the first map lands into the part where the second map is not defined. And so if the first map is dominant, so see, and the second we have another rational map, then we can make the composition. Just find that if you take the image of the intersection of the domain of this with the image, then this will be an open subset in x, and we can define the map there. And in particular, we can define, so if f is an element in kx, this is the same as saying this is equivalent to f being a rational map to a1. So we can pull back rational functions. Namely, we can say that phi star of f, so maybe now for this, it would have to be in y, is just the composition. So this is if phi is for phi is 1, so this is a rational function. So we can pull back rational functions. And this is more or less, so we also had defined what it means to be a birational morphism, a birational map. So this is a rational map which has a rational map as inverse. But now we want to talk about the relation of rational maps and maps on the function fields. This is the following theorem. Now we start with the new thing. So let x and y be varieties. So we have a bijection between the dominant rational maps from x to y to the k-algebra morphisms from ky to kx. Well, and the map is given by the pullback. So phi is sent to phi star. So we have in particular that two rational maps are equal if they give the same pullback on rational functions. So in some sense, and you know, so in particular, I will see this later also. We have that x and y are birational, even though near kx is isomorphic to ky. But we will see this later. So if one talks about varieties up to birational isomorphism, it's the same as just talking about the function fields. OK, let's try to prove this. I don't give a completely complete proof. I will give the inverse map to this map. So we have a map which associates to a dominant rational map, the pullback. And now we want, if we are given a k-algebra homomorphism, we are supposed to find a dominant rational map for which it is the pullback. So we construct the inverse to this map phi, maps to phi star. And then one has to check in detail that it is indeed the inverse. So we start with the k-algebra homomorphism. So let theta from ky to kx be a k-algebra homomorphism. And we have to find a rational map. So we want to construct a map phi from x to y, which corresponds to it, so of which it is the pullback. Now, if we replace y by an open affine subset, so we take an open subset here, which is smaller than y, and we can even find the phi instead of from x to y, instead to this open subset. We certainly have found it with the inclusion into y. We also have found a rational map to y. So we can, as well, assume that y is affine. And once it sets, it's isomorphic to some closed subset of some affine space. We can, as well, assume that it is actually a closed subset of affine space, because we just compose with an isomorphism. So we assume that y is closed. OK? So then, obviously, the advantage is that we know how we can characterize morphisms to y. But anyway, so let's see. So anyway, we are in an, so we have the coordinates on an, which I call y1 to yn. We take them as functions. I mean, the restrictions to y are the coordinate functions. So let y1 to yn be the, in Ay, be the coordinate function. Now we basically do the same thing we did before, when we, you know, this is similar to previous statement about morphisms between affine varieties. We do the same as there, that we use the, you know, the thing, when we apply it to the coordinates, we will get the coordinates of the map. So at least that's the hope. So let, so therefore, does theta y1 until theta of yn. So if these elements in Ay, they are certainly regular functions on y. And so the pullback, so regular functions are certainly rational functions. So the pullback will be a rational function. So these will be some elements in kx. But if they are rational functions, then for each of them, there is an open subset on which they are regular. And if I take the intersection of all these open subsets, we find an open subset where all of them are regular. So there exists an open, non-empty in x, such that theta of yi is regular for all i from 1 to n. So then I claim that this proves that theta y1 to yn from u. Do you want to do this? I make it here, so different thing. So this thing is a y is a morphism. But I can also say it differently. So I could also say that if I want to refer to a previous theorem, we have that theta defines an injective, but maybe I leave it like this. So this is a morphism. Well, maybe I don't. So this is true, but I can say it in different ways. So theta defines an injective homomorphism of k-arge bars from a of y to ox of u. So if I have any polynomial in the yi, then theta of it is the corresponding polynomial in the theta of yi, which lies in ox of u. And then we had this theorem that if we have a homomorphism of k-arge bars, this defines us a morphism like here. But we also have, so which gives us this morphism by the theorem gives the morphism u to y. And now it is an exercise, which however this time I did not give you, but it's an easy exercise which you can check, that the fact that this morphism is injective implies, and in fact is equivalent to the statement that the image of, so that the fact that this homomorphism is injective is equivalent to the fact that this morphism here has dense image. So to show that theta is injective implies that the image of this map from u to y is dense. So first, why is this map this theta injective? Well, we have a homomorphism of fields from ky to kx. And it's not the zero homomorphism, so therefore it is injective. Because every homomorphism of fields is either the zero homomorphism or it is injective. Because the kernel is an ideal. And there are no ideals. And so we take the restrict. So this ay is just a subring of ky. So certainly if it was injective before, if I restrict it, it's still injective. And this fact is quite simple. You can just see that somehow the kernel, I mean it's quite, it's really an easy exercise to check that if this homomorphism is injective, then the image will be dense. Because basically, how shall I put it? So you find somehow that the closure of the image is precisely the locals where the kernel here vanishes, something like that. So anyway, it's very easy to check. And you can also easily try to do it yourself. So now, what this means, therefore, thus we find we have a morphism phi from u to y, which is dominant. Image is dense. And so it's a rational map phi from x to y. And the statement was also before here when you say that theta defines an injective homomorphism of k algebras in this way. So if I call that maybe phi, which by the theorem gives a morphism phi from u to y. So how does it give the morphism? This is the statement was precisely it gives a morphism with the property that the pullback of it is the theta. So when we had this theorem about morphisms of a fine varieties, how it is related to homomorphisms of the k algebras, the statement was that there were bijection between them precisely in such a way that the pullback is the corresponding homomorphism of k algebras. And so we find that at this level, the pullback is this. And so it's easy to reduce from that that also here, the phi star is equal to theta. Anyway, so the details you can check yourself. And then one has to, I mean, it's an exercise to check also the converse that if you compose them the other way, you get also the identity. OK, so this is this statement. So basically, we have kind of reduced it to the previous result for a fine varieties. Now I want to see what this has to do, how this works for birationalities. So I had already given the statement that it follows from this that x and y are birational, if and only if kx is isomorphic to ky as k algebras. And now we want to do slightly more, slightly more. So let x and y are, again, varieties. Then the following are equivalent. First, x and y are birational. Second one is that x and y contain open subsets, which are isomorphic to each other. So there are non-empty open subsets. u and x, v and y such that use isomorphic to v. And the last one is that, indeed, kx is isomorphic to ky as k algebras. So in some sense, it seems kind of obvious that these three should be equivalent in more or less what we just proved will imply that 1 and 3 are equivalent, more or less. And this statement is kind of also clear, because a birational map is a morphism on an open subset. And it's birational if it has an inverse which is a rational map. So you can imagine that it will be two open subsets which are isomorphic. So let's carry this out. So we assume that x and y are birational. So now we want to show that there are such two non-empty open subsets. So let p from x to y, a birational map with inverse, I write it c, but it's just p to the minus 1, where it is defined from y to x. And so some we take, for instance, the domain of phi. So we know that if I take the composition psi composed with phi, this will be, where will this at least be defined? I mean, what is an open set where it is defined? For one thing, we must have that phi is defined. So it's defined on the domain of phi, but domain of phi. So we need that. And we also need that then once we map from there, we end in the domain of psi. So if we apply, and in the same way, to take phi composed with psi, this is defined on the domain of psi intersected with psi to the minus 1 times the domain of phi. And we can see if we apply now, if we take phi and apply it from the domain of phi, intersected phi to the minus 1 domain of psi, this will map this to this one. Because if I apply phi to it, then this gets mapped into the domain of phi. And so I claim these maps to the domain of psi, intersected psi to the minus 1 domain of phi, which is the same as psi to the minus 1. OK, let's see. No, no, now we have to see why it's true. And if not, then it will be, we shouldn't. OK, so if you apply phi to this, we certainly are mapped to this one. If you apply phi, so the idea is that this should be the, so now I'm not quite sure. So I want to say that it maps to this one. Because that is, well, that certainly, but that is not what I want. That certainly is true. But you know, so here's psi, so what do I have here? Let me just see. So I apply phi to this. If I apply to this, I get the domain of psi. And then psi to the, and now we apply, so here we would have just intersected phi of the domain of phi. But OK, and I want to say that phi is just, if we are, you know, phi is just psi to the minus 1. Where, you know, so the map, so where the inverse, so where phi is defined, it's the inverse of psi. So I think it is, you know, this is the same. Anyway, so if we are given this, so then it means we have these two sets. We have the map phi between them. And we have, obviously, psi goes the other way around to the other one. So then what is written here says precisely that phi and psi are inverse to each other as mapped from here to here. What? Yeah, yeah, yeah, so that's a misprint. No, no, that's actually false. Yeah, yeah, OK. What was the misprint, but not the same. OK, so this statement, just this fact, just says, so this is defined here, and this is defined here, and they go like this. And, you know, where both, where it is defined, this composition is the identity, you know? And they are the identity here. Because, after all, this was supposed to be the inverse. Wherever the composition is defined, it is the identity. So we find that phi and psi go from one set to the other. And the compositions are the identity. So these two sets are isomorphic. That's, well, this is the isomorphic. So we have that, this intersection, the domain of psi intersected to psi to the minus 1 domain of phi is isomorphic to the corresponding thing with phi and psi exchanged. So this was the only slightly non-trivial part. So we have found this open subset. And now, so this proves 2, so 2 to 3 is obvious. Because, by definition, if I take an open subset of a variety, it has the same function field. In this case, so kx is isomorphic to ku. And with our definition, they are even equal. And ky is isomorphic to kv. And obviously, if u and v isomorphic, then ku is isomorphic to kv. And the last one, 3 goes to 1, is basically what we proved in the previous theorem. So if kx and ky are isomorphic as k algebras, then we know that x and y are birational. Because if we are given an isomorphism, we have for that a birational map from x to y. So OK. So we had last time this example of the Cuspital Cubic, which you maybe recall that we take a curve C equal to the 0 set of y squared minus x to the 3. That this is not isomorphic. C is not isomorphic to a1, but they are birational. And in particular, you find that the function field of this is just the kx, the field of rational functions in one variable. OK. But I mean this example we briefly did last time. So now I want to come to one specific example of birational morphisms, actually. So the birational morphism is a morphism, which is also birational, but it doesn't necessarily have an actual inverse, which is defined everywhere. And the most standard, most typical example of that is a so-called blow-up. So I will briefly introduce the blow-up. And then I will, so maybe just in a small special case, it's a very general thing, the blow-up. And we will look what it does to basically just do it for blowing up a point in A2. And then we'll briefly see what it does to curves in the point, which pass to that point. So this example, this is an example. So it's one of the main example of birational morphism. So this is a, can maybe say a few words about that. So blow-ups are quite general construction. So you have, for instance, the notion of if x, so these are just things I say. So I will not make it precise. So if x is in a fine or protective variety, one can blow up and say I is an ideal either in the coordinate ring of x or in the homogeneous coordinate ring of x or in the final protective case. You can say you can blow up x along i. But I will not maybe define this. But somehow you can, if you want to, so blow up, you can either, so if you want to have a picture in your mind, you can either think that you have really some kind of pump and you have a point and you make something much bigger there. Or you can even view it much more drastically that there's some place in x where you put a bump and you make it explode and you look where the pieces fly. And so somehow it's a way to look at some small part of the variety like under a very big microscope. So what happens very near to the point you can see very large, or it becomes very large. But we will only consider the blow up of A2 in the origin. That's very simple. So how does that go? Where do we have it actually? So this I can just write down, so definition. So on A2 take with the coordinates x and y. And on A2, P2, we say we take the homogeneous coordinates P1. Say I call them t and u. And then the blow up is something which lives in the product. So the blow up A2 hat of A2 at 0 is the following. So this is 0 set of xu minus yt in A2 times P1. So what? A2 hat, yes. So this is defined to be A2 hat. And this is A2 hat, yes. So I should maybe say we didn't precisely have that this is a closed subset of A2 times P1. This doesn't directly follow from what we said, but it's an easy exercise to prove. We proved something very similar for closed subsets of Pn times Pm. So the closed subsets of An times Pm are the 0 sets. So z of f1 to say fr by the fi. So if the coordinates on An, I call x1 to xn on Pm y1, y0 to ym, then these are polynomials in x1 to xn, y0 to ym, which are homogeneous in the variables which live on some projective space and no condition on these variables, so homogeneous in the yi. And the proof is basically the same as what we did when we proved that what the closed subsets of A Pn times Pm is. And so in particular, we find that this thing, as it's a 0 set of polynomial with this property, is closed in A2 times P1. So we take pi, which is just the first projection from A2 hat to A2. It's the blow-up map, and we call the inverse image of the origin, so I just write 0 for the 0.00. And it's called the exceptional divisor. That's how it's called. I mean, in the moment that doesn't make much sense, but that's how it is called. So we want to see that pi is a birational morphism. If I take pi, restrict it to A2 hat minus E is an isomorphism. And well, it's obvious from the definition that E is just equal. If you look at it, it's a 0 set of x times u, yt, where now x and y are both supposed to be 0, because we take, we are over the origin. But then it's a 0 set of nothing, so it is just the whole of P1. So E is just equal to 0 times P1. So in particular, here the map is not an isomorphism on E. So everything kind of stays unchanged outside 0, but over the point 0 we have a whole line P1. And one should think of this P1 as all the directions that you can have in the origin. As I said, if you want to have the slightly violent viewpoint on the blow-up, it's really, if you put a bomb in the origin, these are all the directions into which the pieces can fly. So this is therefore a birational morphism, but not an isomorphism. So let's see in more detail that happened. So now we want to first see the state. So it's clear that E is equal to 0 times P1, but you might want to see now that this map is indeed an isomorphism. So first, so we take a point. So I just write the coordinates in x and y for the coordinates of the points, even though I could call them AB and so on. So we take a point in A2 without the exceptional divisor. So that means that this point is not the point, this x and y are not both 0. So somehow by symmetry, it somehow will look the same whether it is x or y. So not both of x and y are 0, so we can assume that the x in which we want, we can assume that x is non-zero. We can assume. So if we are on this A2 hat, it means that this equation is fulfilled. So we have x u is equal to y t. And if x is not 0, we can also say that u is equal to y divided by x times t. Or equivalently, if I take the point in projective space, which is given by the pair x y, this is the same as this point in projective space. So x u, this holds anyway independent of whether it is x which is non-zero. So we see from this, we have indeed a morphism. So A2 hat is the graph of the morphism, say from A2. So I don't know how it's from A2 without 0 to p1 point x y. So it's just given by x, so we have that x y is mapped to x y. Now, because we see precisely that the points are precisely the pairs of x y and t u, such that if I take the image point, x y is equal to the second coordinate. So it's precisely the graph of this morphism. So in fact, we would see that A2 hat is irreducible, thus A2 hat is the closure of the graph. In fact, you can always view the blow up of anything like that, that you blow up an ideal or whatever. You can always define a rational map. So if you haven't defined to you what general the blow up is, but you can always define it in such a way that you have some rational map, which is defined somewhere and you take the closure of its graph. That is the blow up. So now we want to, so I'm just describing this. It's not a big, so now I want to see that this thing, A2 hat, has a cover by two open subsets, each of which is isomorphic to A2. So you have kind of two charts. So we want to see A2 hat has a cover by two open subsets, isomorphic to A2. And I can just, so I still recall that A2 hat that we want to write, so we have it. So we keep in mind. So we can take, as a one open subset, we take VT to be equal to A2 hat minus the zero set of T. So T is one of the coordinates of P1. And I claim this is isomorphic to A2 in the most obvious way. So how can I write this? Well, if the T coordinate is non-zero, I can put it equal to 1. So this can be written as a set of all x, y, 1 comma u in A2 times P1, such that this relation, x, u is equal to y, T is 1, so that y is equal to x, u. So I can write it like that. But that means, what does? If I take the map from VT to A2, which sends such a thing to just x comma u, then it's given a nice way by coordinate. It is actually a morphism. And it is an isomorphism. Why is that? Because we can write down the inverse map. So if we know x and u, we know what y is. And they are given just by, again, by polynomials with inverse. So x, u is mapped to what is it, x comma x, u, 1 comma u. And in the same way, obviously, when u is non-zero, in the same way, the u is isomorphic to A2 via the corresponding map. So to take x, y, so point here, we look as x, y, and T comma 1. Because the u is 1, would be mapped to x comma T. And then you can reconstruct in the same way as before. So it's also an isomorphism via this map. So we have a cover by, so A2 hat is equal to V, u, VT, u, and each of them is isomorphic to A2. And it's an exercise. So if x has an open cover, x equal to u1, union u2, where the ui are irreducible, and u1 intersected u2 is non-empty, then it follows that x is irreducible. So this is very easy to prove. You just use the definitions and you find that. So this applies here. So we find that A2 hat is irreducible. Anyway, so we have these nice charts. So this is how one will want to use the blow-up to somehow change the original charts x and y on A2 by the charts x and u, which somehow is finer. And we need to just be able to describe what the exception divisor is in these charts. So if we look at this, so in these coordinates, so in. So I want to say it again. So A2 is the union of two open subsets isomorphic to A2. And in fact, one is A2, which is so open subsets Vt and Vu, which are isomorphic to A2. Namely, the first one, Vt, is isomorphic to A2, where I have the coordinates with coordinates. What are the coordinates here? x and u. And Vu is isomorphic to A2, where the coordinates correspond to the coordinates x and t, y and t. I didn't do it here, but the coordinates come out y and t. And so we can work with these coordinates. So we find, for instance, in Vt, the exceptional divisor, E. So the exceptional divisor is all the points where both x and y are 0. It's just the 0 set. And in Vu, it will be the 0 set of y. So I mean, that somehow, if one knows a little bit more, this is the reason why this thing is called divisor. Because later, a divisor will be essentially something like the 0 set of one equation, one polynomial, in some other variety. And we see that out of a point, which is not the 0 set of one thing, we get the exceptional divisor is the 0 set of one coordinate. The exceptional divisor is this. And so now we want to, so we have somehow the description. And so finally, if one looks at, I just now look in the code in Vt, the projection from A2 hat, so rather from Vt, to A2 is given. It's just that all the stuff is mapped to this tuple. So it means this is given by x u. Now the coordinates on this A2 is mapped to x comma x u. So one, we find an open subset which isomorphic to A2. And its projection to the actual A2 is given by this map. So we locally describe this thing also as an A2. But the coordinates are different. And the map is not the identity. And so we can use this to, now we want to see, so that one can have some big idea why this could be useful, what the blow up does to curves in the plane. So what does, so example, what does the blow up do to curve C in A2 through the origin? So it contains this point. So how does it change? So what do I mean by what does it do to it? So we want to compare the curve C to the closure of, we take the inverse image here, but we take only the closure of the points which lie outside the exceptional divisor. So we call, so let C upset A2, the curve. So the strict transform of C is C hat, which actually can also be used with some blow up, which is the closure of P to the minus 1 of C without 0 in A2 hat. So I mean, so we want to see what happens for some curves. We just look at two examples. So the first one we take C, so let C be the 0 set of F, where F is a nodal cubic. So F is equal to, say, y squared minus x squared times x plus 1. So you can see we have a double 0 here at 0 and another 0 at 1. This will lead to the following picture. If I just look at the real points, then, so here, if x is equal to minus 1, so we will have somehow it looks like this, where this is the origin, this is the point minus 1, 0, and then it goes like this. So it will somehow look like this. This is the point 0, 0. So the real points look like this, and you can somehow see that something kind of not so nice happens at this point. And so we can try to, we want to compute what is the strict transform. So we look just, look at this draw in the chart, T different from 0 that is in VT. We see that the preimage, so pi to the minus 1 of C, is just the 0 set of this F composed with T. And so what does it give me? I just have to, the map was given by this. So it just means I put y equal to x u and see what I get. So this is just equal to the 0 set of x squared times u squared minus x plus 1, if I'm not mistaken. So just, the point is just to put y equal to x u, and then hopefully it comes out correctly, no? It comes out correct. So it looks like this, and you know that the 0 set, so this is equal to the 0 set of x squared union, the 0 set of u squared minus x plus 1. And this thing, as we know, is the exceptional divisor. And so we see, and this thing is obviously irreducible, if you look at it, it's just a quadratic polynomial like that. So this is the strict transform of C. And you can see that this thing is actually isomorphic to A1. Namely, where am I? You just, I didn't write it. Oh yes. So I claim this is isomorphic to A1. So if I have the map from C hat to A1, which sends a point with coordinate x u to u, then I claim this is an isomorphism with inverse. So we just take the projection to this u factor with inverse. Well, we have to reconstruct it. So t maps to t squared minus 1 comma t. If u is equal to, or if you want u, is mapped to u squared minus 1 comma u, because the x will be equal to u squared minus 1, obviously. And so therefore, this thing is isomorphic to A1. So we somehow see that C has become simpler. This thing is certainly not isomorphic to A1. It has this point, which is special. In fact, one would call this a singular point when one knows what that is. And here, it just becomes A1 after blowing up. So we have used this blowing up to simplify the curve to make it nicer. And one can even kind of make a picture if we have them. So this would be the curve. And in the blow up, we get one point for every tension direction at this point. So we find that there's a whole line, which is the exception divisor. And this intersection in the curve somehow will look like this, lying over it, intersecting in two points. And so this is just an A1. This thing is the exception divisor, which we throw away. So out of this thing, we have kind of pulled this apart to make it smooth. And we can have another small example, which is the hospital cubic, which is very similar. So I should maybe mention, it's easy to check. So now everything done, everything in this chart, t is different from 0. In principle, we have to also look at the other chart. But it's easy to check that the strict transform is contained in vt. So it will not happen that the u equal to 0 is not allowed. Because you can also see, I mean, no. So t equal to 0 will not be, there will be no points in the, except those on the exception divisor, which lie in u equal to 0. So therefore, it's enough to just look at this chart. But in general, it could be different for other curves. And so if you have the hospital cubic, which I've seen many times. So c, we call it again c, is the 0 set of y squared minus x to the 3. Then we can do the same thing again. We again look in the chart, t different from 0. So on vt, we have now the coordinates t and x again and u. And we find that, so the inverse image will be 0 set of x squared times, say, x, well, times u squared minus x. No? We just, yeah, precisely. And that's what it says. And so, yeah, yeah, no, certainly. That should be obvious. It's not supposed to be a miracle. And so we again have that the strict transform. So you can see it's just a parabola. This is just, is obviously, again, isomorphic to a1. In this case, as you know, if you take the real points, picture downstairs looks like this. And here we have the exceptional divisor is like this. And the curve will be a parabola, which is tangent to it. But again, it's a smooth curve, and we can see that. So I mean, it goes to map pi. OK, so in general, so we have seen in these two cases that the curve has somehow become simpler by blowing up. In fact, what you, I mean, if one knows what it means in general to blow up in any kind of variety also after having blown up, one can prove, for instance, that one can make any curve into a smooth curve, whatever that means, but that really locally, that is a manifold in the complex topology by repeatedly blowing up. And more generally, that is a very big theorem. I haven't told you what it means. But I told you one can blow up ideals in a variety. And there's a general theorem that if you have any variety. Now I've not told you what it means for a variety to be non-singular, but over the complex numbers, it means it's a manifold, a complex manifold. And there's a general theorem, which is extremely difficult to prove, that if you have any variety over the complex numbers, you can blow it up repeatedly in some ideals and it becomes a manifold. And this is a very useful theorem. Somebody by Yonaka won the Fields Medal for it. So it's really difficult. But anyway, so you see that it is a big thing. And yeah, again, it's a subset of the T. I mean, I was implicitly using it. So again, well, then you have to look at both charts. There would be, so if C hat does not lie in VT, it means that you have to describe C hat by both charts. So you find a part which is over U, which is in VU and one in VT, and both of which describes an open subset of the curve. No? But so I mean, I don't know whether I can get one. So you have to see where it comes from. So this would correspond to kind of remember everything. So I don't know whether. Yeah, maybe I will do it next time. Because I don't like to kind of think when there are witnesses who see me trying to think. So C hat was supposed to be the closure of the inverse image minus the exceptional divide. Now, here we have something irreducible. All the points, except for actually these finitely many, in this case precisely 2, lie out. You have this C hat is a close subset of A2. And it contains, so all the points, except for 2, are actually outside the exceptional divisor. So therefore, it is the closure of that set. I mean, this open subset is dense. So this is an open subset of. So if I take C of U squared minus x plus 1, and I assume that with and intersected, so without the 0 set of x, this is equal to the inverse image of C without, so intersected with, so of C without 0. And this is an open subset of the 0 set. And so if this thing is irreducible, then the closure of this is obviously this. So this is the closure in VT. And the claim is that if I take the closure in, yes. But the point is that, let me see. OK, so this is by itself the closure in VT, that is true. But now if I take the 0 set of this, this will be entirely contained also in VT and close there. And it will, even if I take the closure of this set in the other set, it will. So if I take the closure of this thing in, so the claim is also that if I take the closure of this, so I mean the inverse image of that, so the thing in, so if I take the closure of that in A2 hat. So I claim that this thing is also closed in A2 hat. And now you can just see, if you look in the other chart, what it means, you'll find that there's no, it does not contain, if you take the closure, if you look at the 0 set of that in the other chart, you find that it does not contain any point which does not lie in VT. But there's many small things to check, so I have only done part of it. So therefore, where was I? OK, so it doesn't, now you can also ask more questions, because now I can definitely not start anything else. OK, so this, otherwise I just say that, so this finishes this chapter, so with this I have talked enough about morphisms and rational maps and so on. And now we will, in the next chapter, talk about dimension. So we had already defined dimension, which was somehow, if you remember, so very briefly we defined it, if you have a chain of, say irreducible, if you, say, have a variety, then the dimension of the variety will be the biggest number, so that there's a chain of irreducible closed subsets of that length. And now we want to somehow prove something about dimension. For instance, we want to prove that dimension of n is n, or many other such things. And so we couldn't, to do this, we want to use morphisms. So we want to somehow, basically the point, what you can imagine is, if you have a morphism, then you should expect that the, if you have a surjective morphism from a to b, from x to y, you would expect that the dimension of x is bigger equal to that of y, that you would think. Not even, or if the morphism has finite fibers and is surjective, you would expect that they have the same dimension. I mean, that's what you would think, what dimension is. And so we will actually prove something slightly weaker. So if you have something which is called a finite morphism, which is, turns out to be a surjective morphism with finite fibers, but there might be more surjective morphisms with finite fibers, then in fact, the dimension is preserved. And therefore, these finite morphisms will be our main tool to study dimension. And so I will first have to tell you what the finite morphism is. It turns out that something about commutative algebra somehow, it's defined in terms of some algebra. And so I have to introduce that. And then we will prove something about finite morphisms, also that there are enough finite morphisms to serve our purposes. So there's a Neutral Normalization theorem, which tells us that. And then we will work with it. But so next time, we will introduce these finite morphisms by doing the algebra. And then we will really start with the dimension.