 OK. So maybe we'll start. So we were talking about products. And then, in particular, about completeness, which, as I said, was supposed to replace compactness for varieties. So the definition was that the variety x is complete if we take the second projection from x times y to y, then this is a closed map for all varieties y. So that means that the image of any closed subset in the product with any y, if I project it down to y, will be closed in y. And as we had seen, so we had shown that one thing that, if x is complete and f from x to y is a morphism, then we find that the image is always closed. So this is also property that you have for compactness. If you have a compact space which has a map to a house of space, then the image is always closed because it is compact. So now we want to give some. And we had, finally, the big theorem was that any projective variety is complete. So this was a somewhat more tricky theorem that we proved. Essentially, one had to show it only in the case of Pn. And then it was finally reduced to some kind of linear algebra statement by using the projective null-stellar sense. OK, so now we come to some corollaries. So this would be new now. So first, if let x be a projective variety, if I take the functions which are regular on the whole thing, then these are just the constants. So any regular function on the whole of x is constant. That's quite different from the affine case where we knew that the regular functions, for instance, on an are all the polynomials in k. But here it's very different. There are very few regular functions. I mean, if I assume that to be regular everywhere. The second statement is every morphism from, so say, P from x to y from a projective variety to an affine variety maps the whole of x to a point. That means there are essentially no morphisms from projective varieties to affine varieties. If I have a morphism from a projective variety to an affine variety, the only possibility is that everything is just mapped to one point. So in particular, if x is affine and projective, so affine means that it's isomorphic to a closed sub variety of affine space, and it's also projective, then it follows x is a point. So the affine and projective varieties are really something very different. So this is quite simple. So and in fact, it all follows from the first. So assume we have a regular function. So let f be regular on the whole of x, where x is a projective variety. So that means that f, a function, can be viewed as a morphism to a one. It's the same thing. So f is a morphism. So by this statement here, as x is complete, it follows that f of x is closed in a one. But so what are the closed subsets of a one? So they are finite sets of points, a one or the empty set. And f of x is irreducible because the image of something irreducible, so it follows that f of x is a point, or f of x is equal to a one. Now we have to exclude that it can be a one. But that's simple. So via the embedding of a one into p one, say a one identified with u zero or something, we get that f gives a morphism, f from x to p one, which however we know maps everything to a one, because the image is the same as before. So the image is still closed. So f of x is also closed in p one. But obviously, a one is not closed in p one. The closure of a one is p one. So it follows that f of x is a point. So this proves that f of x is a point. In other words, the function is constant. So the second statement is kind of clear. So y is isomorphic to a closed sub-variety of some an. So if I compose with this isomorphism, I can assume that y is a closed sub-variety. Well, then we know that a morphism from anything to something in a n is given by an n tuple of regular functions. So thus, p can be written as f one to f n, where the f i are functions which are regular on the whole of x. But we know then they are constant. So each of them is just constant equal to some number a i. So phi maps everything to a one to the point a one to a n. So that's also very simple. And the rest is clear. If x is a fine and projective, then the identity would be a map to which this applies. And the identity map would be supposed to map x to the point. And so x, et cetera, is a point. So as I said the other time, the fact that the image of a morphism starting from a projective variety is always closed. So a sub-variety of whatever you map to gives us a new way to construct variety. So if I have any variety and the morphism from that to some projective space, so any projective variety and morphism from that to some projective space, then the image will also be a projective variety. And so this is a way to construct many projective varieties. We want to look at one example, which is the Veronese embedding, again after some Italian mathematician Veronese. And that's rather simple. It's somewhat similar to the Sege embedding in some way. So we fixed two numbers. Why? We fixed just one number. So we fixed the number, say, d and n, so some positive integers. And we put n to be the number n plus t choose d minus 1. And so we want to construct Veronese embedding, which is new d, from pn to pn, the Veronese embedding of degree d. So how does it go? So we want that the coordinates here corresponds to all the monomials of degree d in the coordinates on the original n. And that is precisely the correct number. And so let mi from, say, of x0 to xn for 0 to i to n be all the monomials of degree d, while in x0 to xn. And then we define w vd, this mu d, I suppose mu d d4, from n to n as given by all these monomials, m0, mn. So just the ith coordinate of an image point is just the, they are somehow numbered in some way. It's the ith monomial of degree d in these coordinates, evaluated on that point. So I claim this is a morphism. So to define a morphism, this should be all homogeneous polynomials of the same degree, that they obviously are, they are polynomials of degree d. And they should have no common zeros on pn. So that's only other condition. So and that's easy to see, because we can just choose a few one of those, which already have no common zeros. Because already x0 to the d, until x1 to the d, xn to the d, which certainly are some of these, belong to these, have no common zeros. Because x0 to the d is 0, x0 is 0 and so on. So these have no common zeros. So this defines us a well-defined morphism. And so we get nu d of pn is a closed sub-variety. So in this case, actually, one can show that this map is an isomorphism onto the image. Maybe I will not claim nu d as a morphism from pn, so its image is an isomorphism. And this is basically because one can define the inverse morphism. I just do it locally. So if we look at, so the open subset ui, so this is the points of all a0 to an. In pn, such that ai is different from 0, they form an open cover of pn. So maybe I will not, the other end. So maybe I give a name to the coordinates on p large n. So say, for any monomial of dvd in p0 of pn, we may be right, so this monomial is called m. Maybe I write zm for the corresponding coordinates on p large n. Now you see that according to this, the coordinates on the large pn correspond and 1, 1 correspondents with the monomials of degree d in x0 to xn. They're just, as a notation, I index them by the monomials. It's just the way, the only thing that I know is that they are large n such coordinates. I can call them whatever I want. I call them zm, where m is the monomial. So then I can form, say, ui tilde, which is the set of all points, which is pn minus the 0 set of z xi to the d. What is it? Yeah, that's a bit so there are two different z's. So there's a large z and a small z. So this is the set where the coordinate corresponding to xi to the d is non-zero. So this is obviously an open subset, not pn, nu d, open subset of nu d of pn. And this ui tilde form a cover of nu d of pn. Because we have kind of used this already here, we know that on pn, at least one of the coordinates is non-zero. And if we end the image, then at least one of the xi to the d is non-zero. Because what we started with, one of the coordinates was not zero, the corresponding coordinate xi to the d on the p large n will be non-zero. So this form an open cover. And we see that nu d, if I restrict it to ui, is a map from ui to u tilde i. If the i-th coordinate here was non-zero, then the coordinate with xi to the d will be non-zero there. And so we just want to give always the inverse of this map. So and this nu d, so give inverse of nu d restricted to ui. And this is very simple. This is just, so it's given by, well, I say, you just take, say, zx0 to the d zx0 to the d minus 1x1. So I assume here, for simplicity, assume that i is equal to 0. You can look for any other i, obviously. And then this is given by zx0 to the d minus 1xn. So why is this the inverse? So if we start from ui, we map x0 to xn to x0 to the d, I mean to all these things. And then if we go back, we just have to, so this inverse map. So if I look at this, this would just map, so if we would start with a0 to en, then it maps to all these two quotes. And then, so the cm, and then we project this just to a0 to the d, a0 to the d minus 1 a1, and so on, a0 to the d minus 1 an. But this is the same as a0. So to an, because we just have to divide by a0 to the d minus 1. And you can also check that it works the other round. If I start, that's a bit more what? Yeah, yeah, that's just a misprint. Thank you. OK, so you have to check that the other composition gives you that. OK, so we find this. So in particular, let's just give some. So one thing which occurs often is the case that n is equal to 1. So the simplest example is the map nu d from p1 to pd. If you work it out with the numbers for n equal 1, you get just pd. And the map now is just, if you look at it, to take all monomials of degree d in the variable. So we have, so a point a0 a1 is mapped to a0 to the d, a0 to d minus 1 a1, and so on, until we finally get to a1 to the d. And so the image curve is usually called nu of d of p1 is called a rational normal curve. So this is just a curve which is isomorphic to p1, but somehow it lies in pd. And this is considered because it often gives a counter example to some things for months, might think to be true. And another case is if you take nu2 from p2 to p5, then the image is called the veronizosurface. Again, it is considered because it often gives an example of something one would not expect. OK, so this is this example. There's one kind of application. So we have seen here that in some sense, the coordinates on this large pn correspond to monomials of degree d in the coordinates on the original pn. And this will lead to the fact that if you look at the intersection of p small n with a hyper surface of degree d in the image under the veronizosurface, this will correspond to the intersection with a hyper plane. And that somehow simplifies things sometimes. So let me write this down. So let f equal to sum ai mi. So mii again our monomials, i equals 0 to n, the homogeneous polynomial of degree d in x0 to xn. So now let x subset pn be a closed sub-variety. Then I claim, so if I look at nu d to the image under the veronizomap of x intersected the 0 set of f, then I claim this is equal to the image of x under the veronizomap intersected 0 set of the hyper plane sum i equals 0 to n ai z. If you think of it, this is completely obvious because the image of, so the map is just given by sending by all the monomials of degree d. And so you find precisely this. If I take the intersection of x with a hyper surface of degree d, it's mapped to the intersection of the image of x with a hyper plane. These are just the coordinates, it's a linear combination of the coordinates. So this is essentially directly from the definition, even though it maybe looks strange. So we know that nu d is an isomorphism onto its image. So this allows us sometimes to restrict to reduce questions about intersections with hyper surfaces to intersections with hyper planes. You can use this, reduce questions of about intersections with hyper surfaces to intersections with hyper planes. Because after taking the veroniza embedding, intersection with a hyper surface becomes intersection with a hyper plane. So we give an application, so that's not just theoretical nonsense. So let x in pn be a projective variety. So let f be a non-constant homogenous polynomial of degree, so homogenous polynomial of degree d, degree d bigger than 0. So then first, so f is obviously in x0 kx0 to xn. Then the first statement is that if I take x minus a 0 set of f, this is n a fine variety. And the second statement which actually follows, if x is not a point, then the intersection of x with a 0 set of f is non-empty. So that means if you have any sub-variety and you intersect it, and it doesn't happen to be just a point. Well, then the intersection with any hyper surface will be non-zero. So it's kind of a very strong strengthening of the fact that in projective space, any two lines meet. So we have something much more. Anything which is not a point with a hyper surface will always intersect. And we will later make use of it when we talk about dimension. So let's prove this. This is a relatively straightforward application of the above. So first, so if f is a hyper plane, so degree is 1, we can apply a projective transformation such that the image of f under the projective transformation is just the hyper plane at infinity. I could say maybe whatever, it doesn't matter. So you could write a to the set of f. This is equal to the 0 set of x 0. And then, obviously, we know that x minus 0 set of x 0 is just the open subset u 0, which is isomorphic to an. Or we can view it as an. So pn without and x intersected x without the 0 set of x 0 is equal to x intersected an, which is certainly a fine. And this projective transformation is an isomorphism of pn to itself. So it follows that it's also true before that. So by the projective transformation reduced to the case that the hyper plane was just the hyper plane at infinity and then it's open. So this is the. And now, if it's not a hyper plane, we apply the Veronese map. If f has degree v, so we see that x without the 0 set of, so via the Veronese map, we have that x without the 0 set of f is isomorphic to x minus a hyper plane to mu d of x minus a hyper plane, which is just the hyper plane which was obtained in this way here. And therefore, mu d of x minus a hyper plane is a fine by what we already saw. So this is a fine. So for the second statement, it's just an application of the first. So assume it's not the case. So if x intersected the 0 set of f is empty, we have just seen. So what do we have then? If we take x minus the 0 set of x, this is, after all, just like equal to x, then this is a fine. And it is projective because x is projective. Thus, it is a point. And so this is precisely the statement here by a contraposition. If we say that if the conclusion is false, then the assumption is false. It's the same as saying that the statement is correct. OK, so this proves this. OK, so this is everything I wanted to say for the moment about completeness. So what we, for the moment, say about the product. And now we come to one more topic about, in some sense, morphisms, which is rational maps. So we want to do this relatively fast. Maybe a bit more than one lecture of the lectures over, so we're not expecting to finish today. So rational maps. So I mean, we have these morphisms. We have introduced these morphisms because we want to use them to study varieties. So they are somehow the tool that we use to study them. Under we apply the morphism, we see what happens, and we find out something about the varieties. So the problem that we, in some sense, have with this is that there are not very many morphisms. So for instance, I mean, for instance, we have this. We have just proven that from a projective to a defined variety, there's basically no morphism at all. It maps everything to a point. But in general, it's difficult to construct morphisms. You have this condition. If you have a projective variety, you want to map to another projective variety, you have to find so-and-so many polynomials, homogenous of the same degree with no common zero, and whatever. And that's just not easy to find. And so we want to try to make do with something which is not as good as a morphism, but it's easier to construct. And this is a rational map. And that's something quite simple. It's kind of almost a morphism. Namely, it is a morphism which is not defined everywhere. So there's just an open subset of the source where it is defined, a non-empty open subset. And then the morphism we had before is just a special case where this open subset happens to be the whole of the variety. And it actually, we will not really come to that. But if one is doing more advanced later, the way how one actually usually construct morphisms is to construct them as rational maps and then prove that they are defined everywhere. So it's somehow, it is actually the way how one does it. Well, so a function for us is always a map or a morphism to K to the ground field. And we will prove that a rational map to the ground field is the same as a rational function. But I always call, it's maybe my personal viewpoint, but for me and also in the usual language, now we're talking about the function is always thought as a map to the ground field. And a map is always to another variety or a morphism. But so it means that rational functions are a special case of rational maps. But that we'll have to see. And they're actually quite, we will find out that they're quite closely connected to rational maps and rational functions are closely connected. So lemma. So first before starting, we need, I want to, so that everything is reasonably well defined, we prove a small lemma, which in principle we have already done, but I want to repeat it. So if phi and c are two morphisms from x to y of varieties, and we assume if there's a non-empty open subset in x such that phi restricted to u is equal to psi restricted to u, then it follows that phi is equal to psi. And if you remember, we have in some sense already proven this, we have, namely we have proven that if we have two morphisms from x to y, then the set of points in x where phi of p is equal to psi of p is a closed subset of x. And now we are looking here at this closed subset in terms of the condition that it contains a non-empty open subset. Now we know that any non-empty open subset in a variety is dense. So it follows that the closure of it is the whole of x and that proves this, okay? So this we already have proven it, but it's anyway. So now we want to define a rational map and it's slightly more complicated than what I made you believe. So a rational map. So maybe I can basically say what the idea is. So I said a rational map is a morphism which is defined on an open subset of the target. But we don't want to care about which precisely what open subset that is. So therefore a rational map will be an equivalence class. So where we call two such things equivalent, so where a representative will be an open subset plus a morphism from that open subset and we call it two of them equivalent if on the intersection of the open subsets they are the same map, okay? But this is just, it basically just says it's a morphism on an open subset and we don't care about which open subset. So a rational map f from x and it's usually done with a broken arrow is an equivalence class. So I write u comma phi of pairs. So u and phi where u and x is non-empty and open there's an open subset in x and phi from u to y is a morphism. And as I said, the equivalence relation is that two of them are equivalent if only the section, they are the same map. So here say u comma phi is equivalent to v comma psi if and only if this is restricted to u intersected v is equal to psi restricted to u intersected v, okay? So obviously when I say I'm talking about equivalence class one has to see that this is an equivalent relation to exercise this is an equivalent relation. If you, you know, you have to see symmetry which is trivial, reflexivity which is trivial and transitivity which is not trivial and to prove that the transitivity you use this lemma and then it's actually quite simple then. So I mean I just, so we say that say this map the insertion of that psi is defined by say v psi if this is representative. So if it's one of the things equivalent to it. So now that I have made this complicated thing with the equivalence classes I want to tell you that I could have also done it in a different way which is in some sense easier. Namely instead of saying I say it's a morphism on an open subset and I don't want to say which one I can instead say it's a morphism on an open subset which I make as large as possible on a maximal open subset. And so that cannot be further extended. So there's kind of one best representative always. And so this is as follows. So let v from x to y the original map say defined by you know phi on an open subset then the claim is phi defines a morphism from an open subset. So phi I could still call it phi from the what I call the domain of phi to y. So with where the domain of phi will be the largest open subset on which I can define it. And it's just what? So the domain of phi. So I take the union over all. So I take another one which is equivalent which is in the equivalence class. So we take all v psi which are equivalent to u phi. So that means that psi restricted to intersection is equal to that of v. So it's just, so I take any function which on the intersection of u intersects so which is defined on some open subset such that on the intersection with u it is equal to phi and I just take the union of all these open subsets. This is an open subset. And how is this morphism defined? Well just whenever I am in a v I take the corresponding map. So by, so we define phi of a point p to be equal to be say psi of p if we have that v psi is equivalent to u phi and p is in v. And obviously you have to see to show that this is well defined and the morphism. Well defined morphism. And it's kind of clear because to be a morphism means there is a neighborhood of the you know that for every point where it's defined it must be morphism in the neighborhood but it's given by morphisms. And so we have to see it's well defined but so if we have a point p which lies in two different such open subsets then it means it lies in the intersection of the, so on it lies maybe in v one and v two and we have c one and c two. So then it's a point in this intersection of v one and v two and we know that c one and c two will be equal to each other on the intersection. Now that is will again follow from the transitivity of the equivalence relation. So it is well defined morphism. So we can instead of saying it's an equivalence class we can also always extend the rational map to the largest possible open subset. And if this largest possible open subset happens to be the whole of x then psi is actually morphism. Well yeah so it's somehow glued. Yeah I mean you can say yeah but yeah you could call it we kind of glue it yes but you know in some sense it's just that the equivalence relation in such a way that you just but yeah you sometimes call this gluing in more complicated situations. Okay so let me see. So first then so I have some remarks. So the first one is what you asked in the beginning. So the rational maps from x so f from x for a one are the same as the are the rational functions. K x namely so in O x U for some open subset. So if f is an element in K x then it follows that f is regular on some open subset. So namely you can always write it as a quotient of two elements say in the coordinate ring and then the if where the denominator is non zero it is a non zero. And so as f is a map function or a map and so it's a regular it is a rational map to a one and you also find that the equivalence relation you have on both sides are the same. You know we had seen that if two rational functions coincide on some open subset then you have a non zero function two rational functions coincide on some open subset then they are equal. So it's somehow the same story. Conversely if f from if say U f from x to a one is a rational function or rational map well then why we see you can just reverse steps. So we have f from U to a one is a morphism thus f is an element in O x of U and so f is also an element in K x. As I said the equivalence relations are the same. So you know the two things are the same as elements in K x if and only if they are the same as rational maps. So it's really that the rational maps are the same as rational functions. So you can use this to look at some other examples. So if x is a variety and you take any elements f one to f n in K x then we get a morphism then say f one to f n from x to p n so maybe a n is a morphism is a rational map and so these f i will be regular on some open subsets which I maybe call the domain of f i so this is the set of all points where you can write f i as a quotient of two regular functions so that the nominate is non-zero and so it is a morphism so from the intersection i equals one to n the domain of the f i which is some open subset of x to a n and we can also do the same in the objective case so let x in n be the quasi-projective variety and we take some say n f zero to f m some rational functions then say i write f zero f m from p m is a rational map which is defined on some open subset namely the open subset will be the points where all of these functions are regular so each of them is regular on some open subset so there will be the intersection of them is still a non-empty open subset and they have no common zeros so therefore I must have that not all as zero but just very different before it was very difficult to for instance to have morphisms but now you can just take any rational functions, anything this will always give you a rational map which is defined somewhere and here it's in some sense even more extreme this could be a projective variety so you know there's no morphism but you have an enormous supply of rational maps there are always lots and lots of rational functions the rational functions are the same as that on any open subset and for any tuple of them you get a morphism for a n whereas you know that there's basically no morphism at all and we can do this also in a different way with four polynomials so we have the same so you have the same assumption that x in p and part of the variety and say f0 to fm the polynomials homogene is of the same degree so and they are not all in the ideal of x so not all of them vanish identically on x then f0 to fm from x to pn m is a rational map so we don't have to worry whether these things have common zeroes as long as they are not all identically zero on x we always get a rational map okay so for instance in just a stupid example so say we take projection from a point so we have p equal to the point zero one in pn then we have the projection from we can look at we had talked about projection from a point if we have a sub variety which does not contain the point but we can instead also define the projection from a point as a map from pn to pn minus one is a rational map so pp so projection from p so this is p equal to x0 to xn minus one this is a map from pn to pn minus one and in this case we had seen this was just the map was just given by taking any given point in pn and taking the line from that which connects that point with this point and intersecting the line with the hyperplane so the map is actually defined everywhere on pn except the point p so the domain of pp okay so now we have these rational maps but usually if you have maps you want to compose them and now the problem with these rational maps is that they cannot always be composed and that's kind of obvious because a rational map is not defined everywhere so if you have two maps and the first map maps everything to the locals where the second map is not defined then you don't have a composition so you want to avoid this and you avoid this by only considering maps where the image is dense in the target so then this problem cannot happen okay so definition a rational map so phi from x to y is dominant well if in a suitable sense its image is dense in y so you can say this if I could call it phi of the domain of phi is dense in y and you can check as an exercise or it's also in the notes equivalent is so if say u phi is representative so u is any open subset on which phi is defined only non-empty open subset then this is equivalent to the fact that the image of the domain is dense is equivalent to the statement that the image of u is this is equivalent to so the statement that the image is dense if it holds for one open subset holds for all open subsets so now I want to define a composition so let phi from x to y be a dominant rational map so assume that it is defined on u an open subset u and psi a rational map from y to z rational map then we can define the composition so say defined by defined on some open subset v of y then it follows that if I take the inverse image of v under this map if I take the map from u to x to y then this is a non-empty open subset of u then u intersected phi to the minus one of phi is a non-empty open subset of u and therefore of x now obviously q to minus one of v if it's the only question is whether it's non-empty it's certainly open because the inverse image of an open set by a morphism I mean the morphism on u so this is a non-empty open subset and we can define our rational map by just this as a representative so if I take the pair u intersected phi to the minus one of v and on this I define psi composed phi this is from x, z is the composition so I get the rational map which I have now defined to be I have given it some open subset on which it is defined we don't know precisely how big the maximum open subset might be might be much larger but there is an open subset on which it is defined and that's what we call the composition and then the domain of this would be the maximum open subset on which I can extend it in many cases this might be the whole of x but even if it's not true for the individual pieces so in particular we can do this if the second one was a rational function so we can use rational maps to pull back rational functions so this allows to define the pull back of rational functions so let f into y be a rational function again and let phi from x to y be a rational map so then we have again the pull back so we have phi star so this here I would call psi composed of phi star of f is defined to be phi f composed with phi which is a rational function so a map from x to a1 so one would do the same thing one finds some open subset which phi is defined one takes the inverse image of that open subset intersects it with the set on which f is defined and then from this to that gives a rational function so we have this pull back and again we can easily see that this behaves nicely so if we or something so easy to see if we take phi star from ky to kx so I should say that it's not a rational map it's a dominant rational map so for dominant rational maps we can define the pull back so phi star is a homomorphism so if we take a constant function it pulls back so actually homomorphism of k algebras if you pull back the product it pulls back the product of the pull backs the sum as the sum okay so I'm a bit faster than I thought so now you know we want to so we had before defined what an isomorphism between varieties is this is a morphism which is an inverse and if two varieties isomorphic then in many ways we kind of view them as more or less the same thing you know I mean they have all the same properties now we have this rational map which is a bit less than a morphism and we have so therefore we can also have something which is a little bit less than an isomorphism which is a bi-rational map so that's a rational map which has a rational map which is inverse to it and in that case we call the two varieties bi-rational and that means equivalent to the statement that they contain open subsets which are isomorphic so we know that in a variety an open subset is dense so if two varieties are bi-rational to each other then they contain open subsets non-empty open subsets which are isomorphic now if you know the point is that in Zariski topology open subsets are very large so it is a non-trivial statement that they contain open subsets which are isomorphic if you are in differential geometry if two manifolds have open subsets which are isomorphic this says that they have the same dimension so that's not very exciting but you know here it is a much more interesting notion I mean it means really that they are almost everywhere isomorphic and so we'll introduce this so definitions so a dominant rational map say phi from x to y is called a bi-rational map so it's again if it has an inverse so if there exists another dominant rational map say if and only if there exists another dominant rational map which I just call phi to the minus one phi to the minus one from y to x such that the composition of them is in one direction identity of x and the other one the identity of y so such that phi to the minus one composed of phi is equal to identity of x and phi composed of phi to the minus one is equal to the identity of y and here so I should say this is they are equal as rational maps so what this means is that wherever this is defined you know this is so when phi is defined and maps into where phi to the minus one is defined the composition will be the identity but that means that if I extend this composition to its maximal domain it will be the identity on x defined everywhere and the same here so it is not necessary that phi is defined everywhere or that phi to the minus one is defined everywhere but the composition is automatically defined everywhere if it's the identity so x and y are called by rational or by rational equivalent if there exists a by rational map phi from x to y okay so let me see what I can do now just 5 minutes the next thing would be some theorem which is a bit tricky so one special case which so in particular a variety will be called rational if it's by rational to an for some n or the same to pn the variety is called rational if it is by rational an for some n and so you can see that example we have that well obviously an is rational and pn is rational no because we have just the embedding of an to pn is a by rational map if I take say the caspital cubic which is the 0 set of y squared minus x to 3 subset a2 this is rational because we had seen that there's an open subset of it so we have an isomorphism from an open subset to another open subset if you remember a1 so because if I take the map from a1 to c which sends so it's what I could say it's given by t squared t to 3 so this is a morphism and the a bijective morphism actually and the inverse is a morphism on c without the point 0 0 so we see that we have an inverse rational map no that we had seen before and finally if you have pn times pm this is also rational because we have an open subset we have seen that it contains open subsets which are isomorphic to a n times a m so at this point you might wonder whether maybe all varieties are rational which would be a little bit disappointing no I mean we cannot be sure but in fact in a very strong sense most of our varieties are not rational it's a kind of a special thing to be rational so I think for instance I cannot prove it because that's actually would actually be difficult but I can just say if I have I cannot now write down the equation but if you have a a general so so if f c in k x0 x1 x2 is general I don't say what general is but somehow if you write down one at random polynomial of degree homogeneous polynomial of degree d bigger equal to 3 then the 0 set of f is not rational I haven't told you what general means it means for instance non-singular we have to see so if we take the so we will not explain this now but anyway so we have one can there is a certain but anyway maybe it's enough to say for any d bigger equal to 3 there is a polynomial in fact many of degree d such that the 0 set of f is not rational and actually much more is true I mean there are very many different kinds how things cannot be rational I mean usually by rational to each other okay so anyway maybe that's enough now and then next time we will say something more what being by rationally equivalent has to do with the function field and then we will talk about particular briefly talk about particular rational morphisms by rational morphisms so these are morphisms such that the inverse is that they have a morphism as a rational map which are is the blow up which is a very important construction I mean in general we are not going to use it very much but I just want to introduce it briefly and then this will finish the story about the rational maps and we will start talking start a completely new topic which is dimension so you know how to define dimension how to compute dimension and then we will find that now we actually are able to say something about it and make some theory because we can use these what we have learned about morphisms to study study the dimension okay so see you on Wednesday