 We were talking about dimension. So the last thing, so I reviewed the last few things we did. So one thing we have proven is that if x is a variety and u in x is open and stands, so that means non-empty, then the dimension of x is equal to that of u. So particular means that the dimension of variety is a birational invariance. And then if x is a variety and u is an open subset, it follows that it's irreducible. That's a relatively easy exercise, I mean, which, OK. OK. And so using this, we had shown that, say if x is a variety, x in a n is an affine variety and f is a polynomial, which does not vanish on x. Then if I take then every irreducible component of x intersected 0 set of f, its dimension precisely n minus 1, if the dimension of x is n, its dimension dnx minus 1. OK, so this was more or less the last thing we did. And now we wanted to talk about dimension of fibers, so we had this theorem. Well, I call it theorem. So let f from x to y be a morphism of varieties, morphism. Assume there's a non-empty open subset u in y, such that the fibers over all points in u have dimension n for all p in u. Then the dimension of x is equal to the dimension of y plus n. So if you have a morphism, the dimension of the source is that of the target plus the dimension of the fiber. And this also holds if the dimension of the fiber is only constant or an open subset, so a dense open subset. The rest can never be so large as to change the dimension of x. We want to, I had also stated, I think, a kind of what could be converse, I mean a stronger form of a converse. So that's the theorem on the dimension of fibers. So this says, so let f from x to y be a surjective morphism of varieties. And we take n to be the difference of dimension. We know that the dimension of x is here, so OK. So then we have first, for all points in y, the dimension of the fiber is bigger equal to the difference of dimensions. So it can never be that the dimension of the fiber is, so to speak, too small. So it can only, on an open, on some subset, it will be equal to n, but otherwise it will be bigger. But it will never be, it can never be that some fibers are smaller than most of them. And the second one is that there exists an open subset, and there exists an open, and therefore non-empty, and therefore dense subset u in y, such that the dimension is equal to n. And actually, I didn't state that before. In addition, we can assume that the fiber is irreducible. x and y are both irreducible, and the claim is that also the fiber has to be. But this, we will not prove. This is more difficult. This is something like a converse to this theorem, but a bit stronger. And it's considerably more difficult to prove than this. We will prove this, but not this, without proof. OK, maybe before we do have it. OK, so let's try to prove the first theorem. I mean, this one. So we know that the fiber will be a closed subset, so the statement makes sense. We can talk about the dimension, which will be the maximum of the dimensions of the irreducible components. So is this the wrong? OK. So we put v to be f to the minus 1 of u, which is an open subset of x. So we can now, we know that the dimension of u is equal to the dimension of y, because u is an open subset in y. At the dimension of x is equal to the dimension of v. So the statement does not change if we replace y by u and x by v. So we can assume that u is equal to y. So that means we can assume that for all the points, the dimension of the fiber's n. So replacing y by u and v x by f to the minus 1 of u, so by v, we can assume that the dimension of f to the minus 1 of p is equal to n for all points p in x. So this story that the assumption is only true for an open subset of y can be immediately removed. So we can as well assume that all fibers have this dimension. OK. So now we want to prove this somehow by induction over the dimension of y. So prove the statement. But maybe we have to first see. Prove the statement by induction over the dimension of y. So if y is a point, so if dimension is 0 is a point, there's nothing to show. Because then x is equal to f to the minus 1 of that point. It has dimension n. So that's kind of trivial. So the statement is trivial. It's actually. So we want to make induction over the dimension of y. So for this, we first make another simplification. We want to assume that both x and y are fine. Because then we can cut down by hyper planes to reduce the hyper surfaces to reduce the dimension. So replacing y by an open affine subset, which still has the same dimension as y. And x by an open affine subset of f to the minus 1 of this affine subset, we can assume that x and y are both fine. And if fine means, after all, that it's isomorphic to a close subset of some affine space. But dimension is invariant at the isomorphism. So we can, as well, assume that x and y are close subvarieties of some affine spaces. In fact, x in, I don't know how I wanted it, a l and y in a m close subvarieties. So we do this because now we can cut down with hyper surfaces and we know what happens. So given, so now f is now a morphism between subvarieties of affine space. So it's given by polynomials, as we have learned. So we can write f equal to, say, f1 to fm, as many as the dimension of the space in which y lies, where the fi are some polynomials in the variables on the space in which x lives. That was how morphisms of affine varieties were described. So now, let g be a polynomial here in kx1 to xm. Such that if I take the 0 set of this polynomial intersected with y, we want that this is neither empty nor the whole of y. So z of g does not lie in the ideal of y. We can certainly do that. And this shot, and we can also assume that this will not vanish on this. We can always find the polynomial which doesn't vanish at any given point, and so we find that. So now we know that this intersection has dimension 1 less than y, so we want to use that to prove our result. So let y prime be, say, equal of the 0 set of g intersected x, and we put x prime equal to the f to the minus 1 of y prime. Then we know, how do we get this? So this is just f to the minus 1 of the 0 set of g intersected with y. So that means we obtain it by putting f, it is just the 0 set of g composed with f, which means x prime is equal to, say, x intersected the 0 set of g of f1 to fm, the coordinates replaced by these polynomials. But this itself is, again, a polynomial in kx1. So this thing, so g of f1 to fm is a polynomial in kx1 to xl. So we take the intersection of x with the 0 set of such a polynomial. And this polynomial does not lie in the ideal of x, because otherwise, already here, we would have that this is equal to the whole of y. Because I mean, the 0 set is not equal to the whole of x, because that would mean that here the 0 set is equal to the whole of y. And so we know that by our previous theorem that the dimension of this thing is the dimension of y minus 1. The dimension of this thing is the dimension of x minus 1. So we find that the dimension, which we, how would we want to write it here? So the dimension of x is equal to the dimension of x prime plus 1 by this previous theorem, which is equal by induction. Now we can apply the induction hypothesis to the map. So we apply induction hypothesis. So for any irreducible component of x prime, so x tilde of x prime mapping to corresponding irreducible component y prime. So every irreducible component of x, so y tilde of y prime, every irreducible component of x, we map to some irreducible component of y, because they are irreducible. So it cannot map to more than one. So for any such, we know that the dimension of, so that by induction, the dimension of x tilde is equal to the dimension of y tilde plus n, because now the dimension of this thing is smaller, it's n minus 1. And so we have this. And so thus we can do this. This dimension of x prime plus 1, which is the dimension of y prime plus n plus 1. The dimension of y prime plus 1 is the dimension of y. And so this proves it. So this is relatively simple. We just, I mean, we make this number of reduction steps, but then afterwards we just have to cut down by a hypersurface, and then the thing is, and then we know it for the hypersurface, and by induction we get it. OK, so we give a number of simple applications, which it's nice to know that this is true. For instance, the dimension of the product is the sum of the dimensions of the factors, which by definition does not follow. If you look at these chains of things, there's no reason why. So the example dimension of x times y for varieties x and y is equal to dimension of x plus dimension of y. And for this, we just apply this result to the projection to one of the factors. So we take p from x times y to y, say, then the dimension of the fiber is the dimension of x, and the dimension of the target is the dimension of y, and then the theorem applies. Similarly, we can look at the affine cone of a projective variety. So let x in the n be a projective variety. So we can look at the affine cone, so cx. So you maybe remember this is the set of all points in the n plus 1 whose class in pn lies in x and plus the 0.0. And so then the claim is dimension of cx is equal dimension of x plus 1. Well, if I take cx without 0, this is certainly an open subset, an open-dense subset in x. At least I've assumed that x is not empty. It's an open-dense subset of cx. And we have a morphism from cx without 0 to x, namely just we take a point x0 to xn in a n plus 1, and we map it to its class x0 to xn in pn. And you can show, so it doesn't follow directly from the criteria for what we have, for what the morphism is. But you can also just prove it directly from the definition that this is straightforward to see that this is a continuous map, inverse image of something closed is closed. And you can also easily see if you pull back a regular function, then it becomes a regular function here. I mean, in particular, if instead of x you take here, say, pn and here an minus the origin, and then the general thing will be the restriction of that. What? Is this a homework? Oh, you got this as a homework. OK, I forgot that. But anyway, so then you certainly know how to do it. OK. And so you have a, so there's no worry about this. I also didn't attempt to prove it. OK. And so, and then obviously the fibers, all fibers. So what are the fibers? These are all the n tuples, say, a0 to an, which have the same class as 0 to an. But the equivalence class is by multiplying by non-zero constant. So the fibers are c with, you know, k without 0. The fibers are isomorphic to a1 without 0. No, I mean, you can check this. No, the fibers, if you just look at what the fibers are, you see an isomorphism with a1 without 0. And so, again, the theorem applies and tells you that this thing has dimension of x plus 1. And this is an open subset. So the dimension of cx is dimension of x plus 1. So now we want to get some kind of, maybe that's the last result we want to prove in this form about dimension. We will still go on about dimension but about the kind of what the dimension of things are. Namely, a formula, a statement about the dimension of closed sub varieties in an, of the dimension of intersections of sub varieties in an and pn. So how the dimension behaves when you intersect. And the statement is that the dimension will go down, well, I write it down, then I can say. So theorem, so what happened? So let x, so there's two parts, first the fine case. x and y in an closed sub varieties. Then every irreducible component of the intersection has dimension at most, at least, say, bigger equal to the dimension of x plus the dimension of y minus n. We'll kind of say something about that in a moment. And the second statement is the one for projective space. So let x and y in pn be closed sub varieties. So the first statement is the same as before. So every irreducible component, x intersected y has dimension bigger equal to the dimension of x plus the dimension of y minus n. But there's more, namely, if this number here is positive, is non-negative. So if dimension of x plus dimension of y is bigger equal to n, then the intersection is non-empty. Here, there's no prediction about the intersection being empty or not, because I just say something about every irreducible component. If the intersection is empty, there is no irreducible component. Therefore, the statement is completely void. And in fact, you can certainly have two hyperplanes in an, which are parallel, and the intersection is empty. So this can certainly happen regardless of how big the dimension of n is. But here, you have the same, first you have the same set about dimensions, but if there's not too much room for them, if the dimensions add up to at least the dimension of the ambient projective space, you get it's non-empty. OK, so first, this statement, I mean, it's not so complicated, but it's somehow slightly easier to remember if instead of talking about dimension, you talk about co-dimension. So if x in n, so this is a remark, has a dimension, say, n minus k, we also can say the dimension, the co-dimension of x in pn, but I just said co-dimension of x, is equal to k. Now, the definition of co-dimension is just the dimension of the ambient space minus the dimension of the thing. So then in this form, the statement in this form, the statement is the co-dimension of x intersected y is smaller equal to the co-dimension of x plus the co-dimension of y. So when you intersect things, the co-dimension is at most the sum. We had seen the special case for hyperplanes. If you intersect the hyper-surface, if you just intersect the sub-variety of the hyper-surface, the dimension drops at most by 1. So and this statement here, that the intersection is non-empty, is something which is very special for projective space. Where is it? Did I actually write this? So the fact that x intersected y is not empty if the co-dimension of x plus the co-dimension of y is small or equal to n. So it's the same as that dimension. So it's not so exciting anyway. So this means this is special for projective space. So we know it's false for a fine space because we can have parallel lines in A2 and we can have parallel hypersurfaces in NEAN. But even we, for instance, have as an example, we find that, for instance, we can use this to prove that p1 times p1 is not isomorphic to p2. So it's not. And this is just very simple. So if we take, so if they were, so I maybe write numbers up, if p1 times p1 is isomorphic to p2, then we would have that this statement holds. So then for any one-dimensional sub-variety, any sub-varieties say x and y in p1 times p1, we have that the intersection is non-empty. We get x intersected y is different from the empty set because this is true in p2 and this statement about intersection in isomorphism will preserve that. But this is obviously not true because if p and q are two different points in p1, then we can look at p times p1 and certainly and q times p1. And this is just the product. So p is different from q. So p times p1 is disjoint from q times p1. You have just two different fibers. This is empty. Obviously, these are isomorphic to p1, each of them. So have dimension 1 and this has dimension 2. So this is the case where this does not hold. So now let us try to prove the theorem. Well, it's also you can also do so if you take a fine space, you can take a point. You can do it like that. It's more than many more examples in the fine space where things don't intersect. But you can also make this. I mean, a fine space is an open subset of that and you can restrict to that. That is also OK. So now we come to the proof of the theorem. So it's not so clear how to do it. But we will use a trick. We know how to intersect with hypersurfaces. So we want to reduce to the case of intersection with hypersurfaces. That's not quite clear. It's not true that if x has co-dimension n, it can be taken as the intersection of n hypersurfaces. This is not true. But we can still reduce it to intersection of hypersurfaces by bringing in the diagonal. So let's see how that works. So use the diagonal. So the trick, so we do one trick, use intersection, use the diagonal. So reduce to intersection, actually with hypersurfaces, but even more especially with hyperplanes. So with linear hypersurfaces. And so how we use the diagonal, so we are in the first case. So let delta in an times an be the diagonal. We have seen that delta is isomorphic to an. We have delta, small delta, from an times an, the diagonal morphism, which maps an to delta. So it maps the point p to the point pp. It maps and it is an isomorphism onto its image. OK. And now we see, well, I expect the statement you can remember. It's not so complicated. So the remarkable thing that you want to use is if you take the inverse image under delta of, say, x times y, or if you want of, which is the same as the inverse image under delta of x times y intersected delta, so now with this delta, which is an isomorphism, this thing, I claim, is just x intersected y. So if you just think of it, this is the set of all points p in an such that it gets mapped to x times y under the diagonal map. So such that p comma p is an element in x times y. But that means that p is an x and p is also in y. OK, so this is just obviously the case. And so we see, so thus, intersected y is isomorphic to x times y intersected delta under the diagonal morphism. And this intersection takes place in A to N. So well, now I claim that delta can be written as an intersection of hyperplanes. So let's see how. So let's see. So let x1 to xn be the coordinates on the first factor, first an, and y1 to yn the coordinates on the second factor an. So how do I describe the diagonal? Well, I claim that then delta is a 0 set of x1 minus y1, x2 minus y2, and so on, until xn minus yn. Because if I have a point p equal to a1 to an, then the ai are just the value of this. And the statement that you have a point pp means precisely that the ith coordinate in the first thing is equal to the ith in the second. So that's just obviously the case. So what you see, this is the intersection of N hyperplanes, so N polynomials of degree 1. Now we had this theorem. So by a previous theorem, we have that every irreducible component of, say, a variety, whatever, c in an with a hypersurface, say, z of f in an has dimension in big n, at most one less than the dimension of c, bigger equal to the dimension of c minus 1. Namely, there are two possibilities. Either z is contained in the 0 set of f, in which case the dimension is equal to the dimension of z, or it's not contained in 0 set of f, then we know the dimension is one less. So if z, you have dimension of c is equal to the dimension of z set at cf. If z is not contained, big dimension of c, I don't know why I do it right in this way around, but anyway it's OK. So we can do induction. So if we now would intersect our z with n different hypersurfaces, the dimension would drop at most by n, because at each step it drops at most by 1. So if you apply this to our situation, so by induction, we have that dimension of x intersected y, which is equal to a dimension of x times y intersected delta, is bigger equal to the dimension of x times y, which is the dimension of x plus the dimension of y, minus the number of equations, which are n. And this is actually true not only for this, but for every irreducible component, because the statement says every irreducible component has that dimension. So you see that by induction, every irreducible component, say z of x intersected y, has dimension of z is equal to the dimension of the corresponding irreducible component here, has at least, OK, the notation is not so great. But anyway, so which is this, OK? Well, it doesn't always have to be that on the, I think it's obvious, even though it's true. OK. So now we come to part two. So we want to, now this argument to the diagonal doesn't work for projective space. You cannot say that it's like that. But instead you can reduce the state. So we want to prove this by reducing to the case of a fine space, which is a bit funny because the statement that you want to prove in the case of projective space is actually stronger. It's kind of, it's a bit amazing that you can prove something stronger to reduce to a case where that stronger statement is false. But we will see it can be done all the same. So reduce to 1 by using a fine cone. So we have if x and y in pn are closed sub varieties. And by definition, if I take this section of Cx with Cy, this is the same as the cone. Just look at the definition. This is obvious. So everything is compatible. We know that the dimension of Cx is equal to the dimension of x plus 1. Same for x, same for y, and x intersected y. So we have this very simple relation. And so therefore we can just say, so let Z be an irreducible component of x intersected y. And then C of x of Z is an irreducible component of C of x intersected y. And we know that this thing has the correct kind of co-dimension. We know, and the dimension of Z is by what we had seen as an example, equal to dimension of the cone over Z minus 1. And we had, on the other hand, seen that this is bigger equal to the dimension of the ambient of the dimension of Cx plus dimension of Cy minus the dimension of the ambient defined space, which is now an n plus 1. And this is the dimension of x plus 1, this dimension of y plus 1. And here however I forgot to subtract 1. The dimension of Z is equal to the dimension of Cx plus the dimension of Cy minus the dimension of the ambient space. And I have to subtract the 1. Otherwise it doesn't work out. So minus n, so minus, well I can say minus n plus 2. And then you see this is just equal to the dimension of x plus dimension of y minus n. So besides the state S k. So this is kind of trivial. Now for the other statement, we want to show that if this number is non-negative, then the intersection is non-empty, which is something which is not true in the affine case. So assume the dimension of x plus dimension of y is bigger equal n. We want to show that the intersection is non-empty. Well, it's not true such a thing if you take affine varieties, but it's always if you take any such cones, the intersection will always be non-empty because every cone contains a 0.0. So we know that Cx intersected Cy is not empty because the origin lies in Cx intersected Cy. So every cone, by definition, contains the origin. So we know that every irreducible component of Cx intersected Cy has dimension. What is it? Bigger equal to 1. No, the dimension of Cx plus dimension of Cy minus n plus 1, which is one more, the dimension of x plus dimension of y minus n. And so this is at least bigger equal to 1. So it follows that Cx. So we know from this story that the intersection is non-empty. So there is an irreducible component. So the statement that every irreducible component has dimension bigger equal to 1 is a non-empty statement. We know it has this dimension. So it's somehow. But the dimension is not just bigger equal to 0, which would mean that it contains 0.0, but it actually is bigger equal to 1. So it does not only consist of the 0.0. So thus Cx intersected Cy is different from the 0.0. And we know that, which, after all, is just C of x intersected y. So if the cone of something is not just the 0.0, then the something is not empty. So thus, x intersected y is different from the empty. So it's this trick with a cone that they always have to intersect. So the thing that in general was kind of not such a big deal that every irreducible component has to have a certain dimension does not necessarily mean that any such irreducible component exists. But if you are talking about cones, then it will have to exist. And so it always is a non-empty statement. OK, so this was as much as I wanted to say about this way, about dimension. But there is another way of looking at dimension, which is also about as commonly studied, used as the one that I used. I mean, a different definition of dimension, which is equivalent to the one we use. And I want to introduce this dimension and briefly explain this new other definition of dimension and briefly explain why it's equivalent to the one that we have used. But this will be very sketchy. In particular, we introduce some concepts using some theorems, which I do not prove. But anyway, it's maybe also a nice, brief repetition of some of the field theory that you had or something. But anyway, it's dimension and transcendence degree. So I mean, just as a motivation, so we know that the dimension of x is equal to the dimension of y if x and y are birational. We know that because we have seen that the dimension of a variety is equal to the dimension of any open subset. And from this it follows. So on the other hand, we know that x is birational to y if and only if the function field, the field of rational functions on x is isomorphic as a k-algebra to the function field of y. So therefore, it must be possible to express the dimension of x in terms of the function field. So thus, time x must be computable or determined by kx. And in fact, the answer is that it is equal to the transcendence degree of kx over k. We will see that the dimension of x is the transcendence degree of kx over k. Now, I don't expect that you know what that is, so I have to introduce it first. So maybe you know what it is, but I have to tell you what the transcendence degree is and then to show you that this will be the case. So I want to introduce this transcendence degree. So this is, I have to say a bit about field extensions. So if say k, let's let k over k be a field extension. So that just means that large k and small k are both fields and small k is a subring of the big k. And so if say a1 to an are some elements of k, we can look at the field, the extension of k generated by these elements. So denote k a1 to an with round brackets, the smallest subfield of k of the large k, which contains the small k and these elements. It's just what if you want its intersection over all these, but it's just can be obtained by taking, OK, we can look at this. So this is the field extension of k generated by these elements. And so if there are, finally, many such elements such that k is equal to this thing, we say that k over k is a finitely generated field extension. So if a1 to an k such that k is just equal to this, then we say the field extension is finitely generated. For instance, complex numbers over q is not finitely generated field extension. OK, so let's see. And now we want to come to the transcendence degree. We only do this for finitely generated field extension. So let's assume we have such a thing. So let k over k be a finitely generated field extension. So it means it's obtained from the smaller field by adding finitely many elements. And now we take some elements. So elements b1, b1 to bn, and k are called algebraically independent. Well, if you cannot get them, b1 to bn cannot be a 0 of a polynomial with coefficients, so over k, of course, more k, of a polynomial with coefficients in small k. So if there is no polynomial, say f in n variables, except for the 0 polynomial, because that always would work, such that if I take f and I replace the variables by these elements, you get 0. So there's no polynomial in several variables that vanishes on them. So this is the generalization of the concept of transcendent, because you say an element b in k would be called algebraic if there is a non-zero polynomial, such that f of b is equal to 0. And you call it transcendence if it's not algebraic. So you call an element b in k transcendent if there's no polynomial f in kx, such that f of b is equal to 0. And now this is the generalization for more elements, called them algebraically independent. And for the intuition, it's quite good to think of algebraically independent behaving very similar to linearly independent and vector spaces. I mean, so the formal properties are similar. Obviously, it's a different concept. But there are some, in particular, so let me, we will be able to talk about something which you call a transcendence basis. So just to say it again, if b in k is algebraically independent over k, if it's just one element, if and only if b is transcendent. So there's no polynomial which vanishes on it. So if b1 to bn are algebraically independent, transcendent. You know the word transcendent? Oh. Well, OK, so I'm not so sure about the spelling. No, I don't think so. I would write it like that. But it might, you know, you have to, you can check it. I would write a z. But it might be that I'm confused by the fact that I'm German. I cannot guarantee. Anyway, it's a, you know, there's no, at least there are no two different concepts where once it's with z and once with z. OK, so I will, OK, I'm willing to believe you. OK, OK. So if these elements b1 to bn, these are algebraically independent, then it's easy to see that if I take this field k, so algebraically independent over small k, if I take the field generated by them, that this is isomorphic to the field of rational functions in n variables. So whose elements would just be quotients of polynomials in the xi. I mean, basically, for some of the definition, it would be isomorphic to this. So, and then finally, a maximal set of algebraically independent elements over small k would be called the transcendent spaces. OK, so this is this definition. And now I get to do this theorem, which is basically a statement that the transcendent spaces will always exist, which is analogous to the statement that every vector space is a basis. And also that if you have a set of generators for the field extension, you can choose the transcendent spaces from the set of generators, which is also a fact that you have for field extensions, for vector spaces. If you have a set of elements which generate the vector space, then you can take a subset of it, which will be a basis. So as you see, now I have made some compromise. So this power, I will do this without proof. Although it's not particularly difficult, but it takes some effort. It's not what we are talking about. So let k over k be a finitely generated field extension. So then first, there is a transcendent spaces. So there exists a transcendent spaces of k over k. And I can choose it as a subset of the set of generating elements. It's supposed to be set to the end. OK, that's now the main thing. Then the second one is that every transcendent spaces has the same number of elements. So that every base, analogous to the state and world vector space, that every basis has the same number of vector space, which is called dimension of the vector space. So every transcendent spaces of k has the same number of elements. And this number is called the transcendence degree. As I said, the last one is that if you have a transcendence basis for field extension, the rest is a finite algebraic extension. So let b1 to br be a transcendence basis of k over k. Remember that k over k is a finite, finite degenerated field extension. Then if I take k over this thing, this is a finite algebraic extension. OK, so every other element will be algebraic over this. And so I mean, it's just that you have a finite degenerated algebraic extension, so it's a finite algebraic extension. OK, so this is this thing. I mean, I said I will not prove it, so I don't have to worry about the proof. But it's not very difficult. But as I said, it's not what we are concerned with. So now we want to use this. So we want to now start relating this to our thing of dimension. So we want to say that, as I said, the dimension of x is equal to the transcendence degree of the function field of x over k. So let's first see. So let now x be an algebraic variety over k. So then x contains an open affine subset, which I call v. And as the function field for x and open subset is the same, we have that kx is equal to kv. So we can assume that v is a closed sub-variety of some n by applying its isomorphic to closed sub-variety, so we can find isomorphism. So we want to describe kx. So kx is equal to kv, which is this. So let x1 to xn be the coordinates on a n, and say y1 to yn be the coordinate functions. So just the restrictions of the coordinates to v. Well, then we know that kx is equal to kv is equal to the quotient field of av. And so that's the, and av will be the extension of the ring k by the elements y1 to, so by this I mean that I'm not saying it's a polynomial ring generated by these, you just have k and you add to it all polynomials in y1 to yn. They don't have to be all different from each other, like we had also before. And so this will then be k y1 to yn. So kx, so we see that this is indeed a finitely generated field extension of k. So now let say the transcendence degree of kx over k be the transcendence degree of this thing. And now the claim is that, so what one can easily see, so if x is equal to an, or x is equal to pn, then we know that kx is equal to kx1 to xn. So the field of rational functions in these, the field of rational functions in these variables. And so the transcendence degree is precisely n. So which happens to be the same as the dimension of an or pn. And the claim is that this always holds. So we want to claim the following theorem. Let x be a variety, then the dimension of x is equal to the transcendence degree of kx over k. And in many books you find that also as a dimension, as a definition of the dimension. For some things, this definition of the dimension is easier for other things, our definition is easier. So anyway, we have chosen. Now we want to see, at least to sketch a proof, why this is true, why these two definitions of dimension are equal. So we will prove the next time. So we'll show another theorem. Every variety x is birational to a hypersurface in a fine space. So in, say, a to the dimension of x plus 1, OK? So it's actually somewhat surprising. So it shows that this equivalence relation by rationality actually identifies many things because obviously there's a big difference between hypersurfaces and varieties of higher core dimension. But every variety has an open dense subset, which is isomorphic to an open dense subset in a hypersurface in a n plus 1. So that's a surprising fact, but that's how it is. And now we will prove that the next time, or if you're very fast even today, but I don't think so. And now we use this to prove this theorem. So this is next time. So now comes the proof of theorem 1. So if this is the case, we can assume that our variety is a hypersurface. So by this previous theorem, we can assume x in a n, x is 0 for f in a n, is a hypersurface. So where f in k x 1 to x n is an irreducible polynomial, OK? So we know, therefore, that the dimension of x is equal to n minus 1. Because we know the dimension of hypersurface is 1 less than the dimension of the ambient space. So now we want to show that also the transcendence degree of the function field of x is n minus 1. So to show k x over k is equal to n minus 1. OK, so we take, just to check, let x 1, x n be the coordinates on a n and say y 1 to y n in a x, the coordinate functions. We have just seen that in that case, we can say that k x is the field generated by the field extension of small k generated by these coordinate functions. And now, first we want to see that these n elements cannot be algebraically independent, which then implies that this thing is smaller equal to n minus 1. And that's very simple. If we take f of y 1 to y n, this is just the class of f in the ideal of x in a x, which is, after all, k x 1, x n divided by f. So in other words, this is equal to 0. Because obviously, the class of f is 0. And so it does y 1 to y n are algebraically dependent. And so we know that the transcendence basis can be taken as a subset of these elements. And we cannot take all of them. So we can take at most n minus 1. So it follows that the transcendence degree of k x over k is smaller equal to n minus 1. But we want to prove it's equal to n minus 1, so we are not done. So we have to show the opposite inequality to show equality. We can assume that the last variable x n occurs in f. I mean, we have a polynomial in the variable x 1 to x n. It will depend on some. It might not necessarily have to depend on all the variables, but it will depend on some. It's not a constant polynomial. That then you wouldn't have a hyper surface. So we could, in the worst case, remember the coordinates in such a way that the last one actually occurs in the polynomial. So that's kind of trivial. And then we want to show that the previous ones, y 1 to y n minus 1 are algebraically independent. Want to show y 1 to y n minus 1 are algebraically independent. So then that means that the transcendence degree is at least n minus 1, and so they're equal. And this is quite simple. So assume they are algebraically independent. So otherwise, we would have a polynomial in these variables which vanishes on them. So there exists an element g in k x 1 to x n minus 1 with g of y 1, so a non-zero element, with g applied to these elements is equal to 0. But these are just the classes of the xi in ax. So this means, in other words, this just means that if you take the polynomial and divide by f, you get 0. I mean, you take the polynomial and take its class in the polynomials divided by f. In the quotient ring, you get 0. So in other words, we have that g is in the ideal generated by f. But this is not possible because if you take the ideal generated by f, you take f and multiply it by any polynomial. You know that f itself depends on x n. So every polynomial, if you multiply f by any non-zero polynomial, you get a polynomial which still depends on x n. So the only way how a polynomial which depends on less variables would lie in the ideal is if it's 0. So thus, we have a contradiction to the choice. And so we find that, therefore, in fact, these elements are algebraic, independent, n to the 10 sense degree is equal to n minus 1. So thus, OK, so maybe I will stop here. So next time, we will have to prove this other result, which actually is not so difficult, but looks in the moment rather surprising that every variety is vibrational to a hyper surface. OK, so that's it for today. We meet again on Wednesday, I think.