 So we were still talking about dimension. So we had finished by theorem about the dimension of an. So theorem, so we have the dimension of an is equal to n. And if f in an be a polynomial of positive degree, then the dimension here is maybe say reducible. Then the dimension of the zero set of f is equal to n minus 1. So the dimension of hypersurface is 1 less. And that of the ambient space. And the third one was that if I have a sub-variety of an, which has dimension n minus 1, it is a hypersurface. So if x in an is a closed sub-variety of dimension n minus 1, then it follows that x is equal to set f for f reducible. Okay, so we had proven the first two. And the last one is very simple. So we can maybe do it. So we just prove three. So we have x somehow is a closed sub-variety of an. And it's not equal to n because after all it has dimension n minus 1. So there exists some polynomial f which lies in the ideal of x. There will be some such polynomial. And again, we can choose f irreducible. Again, so actually f is also not constant because x is also not the empty set. So again, if we have an f here in the ideal of x, x is a prime ideal. So one irreducible component is so whenever we find there is an f here. And so therefore, there is also an irreducible f in the ideal of x. And so thus x is contained in the zero set of f. Both are irreducible and they are of the same dimension. So we know that if we have one variety contained within another, which are both irreducible of the same dimension, then they must be equal. Because the statement was that if a variety is strictly contained in another, then the dimension is smaller. So then it does it follows that x is equal to zero of f. So I should maybe notice that so in two and three, I can drop the assumption that it's irreducible. So that either f for two or x is irreducible. Because by definition, if you have a dimension of an algebraic set, so is equal to the maximum of the dimensions of the irreducible components. So if f is a polynomial which is not irreducible, then its zero set is a union of the zero sets of irreducible components. And each of them has dimension n minus one. So the thing also has dimension n minus one. So we don't have to worry about that. And even here, if x is a close subset of an, which has, well, no, that's not true. So if every irreducible component has dimension n minus one, then it is a hypersurface. But that's not the same as that the dimension is n minus one. Okay, so I'll leave it maybe like that. Okay, so another corollary is, what, nothing in the mark? So in part two, we can drop the assumption that f is irreducible. And the statement is still true. And we also find that every fine variety is finite dimension. Because we have seen that an has dimension n. If you have a defined variety, it's isomorphic to a closed sub-variety of some en. So its dimension is smaller equal to n, I mean for that n. So we just write it so if x subset n is a closed sub-variety, then it follows the dimension of x smaller equal to n. So that, okay, you don't have to. So now we want to in some sense have some kind of generalization of this part two. Namely, instead of saying we take the dimension of a zero set in an of such a polynomial, we take n a fine variety and we take its intersection with the zero set of polynomial. And we want to see that the dimension goes down precisely by one, okay? So let x in say a large n in a fine variety of dimension small n. So let f be a polynomial in this fine space. Which does not lie in the ideal of x, it doesn't vanish identically on x. Then if I have that the zero set of f intersected with x is not equal to the empty set, it could happen. But if it doesn't, then it follows that the dimension of zf intersected x is equal to n minus one. So the dimension, if you intersect with a hyper surface dimension goes down precisely by one, okay? And this is, you know, could view this part two here as a special case. So I should remark, I'm not saying here, so this intersection might very well be reducible. It's not said that it's irreducible, okay? So by definition, the dimension of something reducible is the maximum of the dimensions of the irreducible components. So thus, it means, the statement means that the dimension, so for all irreducible components of zf x, so I may be called this component y of zf, so yi of zf intersected x. We have that the dimension of yi intersected with, so yi is smaller or equal to n minus one, and there exists an i, there exists a component yi with dimension of yi is equal to n minus one. That's what it means that the dimension of this is n minus one because it says the dimension of something reducible is the maximum of the dimension of the components. In fact, we will later prove something stronger is true. So later we'll show that all irreducible components have dimension n minus one. Okay, but for the moment we prove the weaker statement. We are not yet quite ready to prove what is really, I mean what is really the truth here. So now again, this will be, we take some effort, so let's see. This I called proposition. So, okay, so now let's start with the proof. So somehow we want in some way or other to reduce this to the previous theorem, I mean to the part two. So to the case of hypersurfaces of A n, we know it in that special case. You want to somehow relate it to that. But it's not completely trivial to see how this should work. But we have one tool to relate the variety to A n. It is Neutral Normalization Theorem. We always know that if you have a variety of a finite subjective morphism to some A n. And in fact, if x has dimension n, it has a finite subjective morphism to A n, to this, no? So we, I write just so that I have it. y equal to x intersect with that f. And then as z, f, as f does not lie in the ideal of x. It's not equal to x, so it's strictly contained. So by what we know is that we know therefore the dimension of y is smaller than that of x. So it follows that the dimension of y is smaller equal to n minus 1. So one kind of direction of this equality we know. And now to prove the other inequality, we will use the Neutral Normalization Theorem. So by the Neutral Normalization Theorem, there exists we have a finite subjective morphism, say pi from x to A n, no? Because there is a finite subjective morphism to some a fine space and preserves the dimension. So it must be this one. So we want to make use of the fact that this is finite, again using this thing with the monic polynomials. So we identify kx1 to xn. So the coordinate ring here with its pullback, which is a suffering of Ax. So I will somehow view now the xi as elements in Ax, just identifying them, just dropping the p up a star in the notation. So by the finiteness, pi is finite, we know that in particular, remember that f. So let f be the class of our polynomial f in Ax. So in particular, so this is an element in Ax. So as Ax is finite over kx1 to xn, it follows that f satisfies a monic equation. So then we can write there exists a monic polynomial, say xn plus 1 to the d plus sum i equals 0 to d minus 1, ai xn to the i, where the ai are elements in kx1 to xn, such that this polynomial is satisfied by f. So if I take h of x1 to xn, f, this is equal to 0. So this is just the fact that the f satisfies a monic equation with coefficients in kx1 to xn. So we know that. So replacing h by an irreducible factor, if necessary, we can assume h is irreducible. If we have a monic polynomial, we write it as a product of two other polynomials, then up to maybe multiplying one by a constant and the other one by the inverse of the constant, we find that these two polynomials, again, have to be monic so that the product can be monic. Just this term x to the, it can only come from one factor from the one side and one factor from the other side. So therefore, if h is not irreducible, I can just write it as a product of factors. And if this polynomial is 0, if I put f into it, then one of the irreducible factors must have the property. And so therefore, I can just replace h by that. I can always assume that h is irreducible. So now I use this. So I define a morphism. So let phi pi comma f from x to an plus 1. So just on the first, I just take the map which maps here into an. It's the map pi. And then the last component of the map is f. So we know that pi is obviously just equal to this map phi composed with the projection to the first few factors, so x1 to xn. So if I take the map from an plus 1 projecting to the first n coordinate, then I get just pi. If I start, the point is this map is just given by adding an x coordinate where it's f. And so I get rid of that. So that's clear. We know that this is finite. So therefore, we know that if a composition of morphisms is finite, then the first one is finite. So phi is finite. So it's always the same jigs. And by definition, we have that the image of x is contained in the 0 set of this polynomial h. Because what does it say? So if I replace to, this is just what is written here. If I have the, so this is just, so if I restrict this f to x, it becomes just the function f. And here it precisely says that this expression, if I apply h to this, I get 0. So that means precisely that the image of x lies in the 0 set of h. So as this is a finite morphism, this is a close subset. So thus, phi of x is a close subset of dimension n in, so close sub-variety of dimension n in z of h. Because it's a finite morphism, so it preserves the dimension. So it's also still n. And now, again, z of h is a 0 set of a polynomial, of an irreducible polynomial in a n plus 1. It has dimension n. It contains as a closed sub-variety another variety of dimension n, so they are equal. So thus, phi of x is equal to z of h. So in other words, phi from x to z of h is a finite subjective morphism. And now, although it doesn't seem so evident, we are almost done. So in some sense, we see now that we have somehow, at least we have replaced x by a hypersurface in the n plus 1. We will now see, but it's actually better. So if we now, we have to keep this, where's the formula here. So by definition, if we take this z of f intersected with x, if we look at this map, I claim that this is equal to phi to the minus 1 of the 0 set of h and xn plus 1. So we are here in an. We know that phi of x is equal to c of h. So if I take its inverse image of z of h, this would be x. And this f, the last component was just f. So the inverse image of xn plus 1 would just be f. And so this z of f with this is just the common 0 set of h, the inverse image of this. Because the inverse image of the 0 set of xn plus 1 is the 0 set of f, the inverse image of h. The 0 set of h is x. So this is really OK. But if you look at this, we have here our wonderful h. I hope you, I mean, I don't know why. So I mean, I was all the time talking about h. I forgot to write that it was called h. Now, either you knew it anyway, I mean, either you understood it, which I hope. Or otherwise it is very strange that you wouldn't say anything. Because I mean, obviously it's a mistake of me to not write. But on the other hand, here it says this and that h of this is equal to 0. So I mean, it wouldn't have made very much sense at this point. But still, anyway, this is h. And it was all over the place. So I can only hope that you were aware of that. So anyway, what? No, I'm sorry, too. But in order not to feel too guilty, I have to give the fault to you. OK, so here we are. So this is fine. But now we look at this h. So we have the 0 set of h of xn plus 1. So then we can also, for looking at this common 0 set, we can put xn plus 1 equal to 0 in this thing. It's the same as a 0 set of xn plus 1. And whatever remains of h when we put xn plus 1 equal to 0. And what remains, the constant term. So this is the same as p to the minus 1 of a 0 and xn plus 1. So in other words, this is the same as the inverse image of the 0 set of a 0. A 0 depends on the first few coordinates, only on the coordinates 1 to n times 0. The last coordinate is 0. So now this means this now, so if I take something times 0, it's obviously isomorphic to this without taking it times 0. So this thing has the same dimension as a 0 set of a 0. And a 0 is a hyper surface. And a z of a, I mean, so thus the dimension of z of f intersected x. Yes? Yeah, of a 0 and so these are two different times. A 0 is the constant term of this. But that means it's a polynomial in x1 to xn. And if I put xn plus 1 equal to 0, then the only thing that remains of h is a 0. And therefore, if I take the common 0 of this polynomial and x0, it's the same as the common 0 of, so if I take the common 0s of h and x0, that's the same as the common 0s of x0 and whatever I get from h if I put x0 equal to 0. That is the common 0 of x0 and a 0. So that's, and that gives you this. And now, but this just mean it's the, this is z of a 0 times 0. z of a 0 is a closed subset in an. So the dimension of this will therefore be equal to the dimension of the 0 set of a 0. Because phi is a finite morph, isn't it? Preserves the dimension. And now, this is a hyper surface in an. So we know for a hyper surface in an that, so either a 0 is constant, then it follows that zf intersected with x is equal to the empty set. Or otherwise, the dimension of 0 of a 0 is equal to n minus 1. Because that was the theorem that we had, or the part of the theorem that we proved for a hyper surface if the intersection is not empty. So if you have a non-zero, if you have a hyper surface, so a hyper surface is always a 0 set of a non-constant polynomial where that depends on, I kind of have changed what I, how I called it a few times. But so an irreducible hyper surface is 0 set of an irreducible polynomial. Now, if you have a polynomial which is not irreducible, then it is a product of irreducible components. And the 0 set is the union of the 0 sets of those. And so the dimension of the 0 set is still equal to n minus 1, because each of them has dimension n minus 1. But if the polynomial is constant, then the 0 set is either everything if it's 0, or it is nothing if it's a non-zero constant. So therefore, the 0 set of this will either be empty, or it has dimension n minus 1. And so this, well, and that proves it. No, because we wanted to prove that dimension, that the statement was precisely that the intersect, that the 0 set of, that the intersection of a hyper surface with a fine variety was dimension 1 less, it was empty. So now, where am I actually? No, in this case, it is injective, because we have a fine-net surjective morphism. And so it is, I mean, it was only for notational reasons that I made the identification. But it is OK, because it is injective, because we have a surjective morphism. And then, so therefore, the pullback is injective. So there's no problem with that. Yeah, otherwise it would be most likely would allow me to prove things which are not true if I would make identifications via maps which are not injective. OK, so now we come to something, to another result with this approach to dimension is kind of one of the more difficult theorems. If there's other ways how one can define dimension, then this would be very obvious, and other things would be much more difficult. But anyway, the theorem is that the dimension of a variety and a non-empty open subset are equal. So that somehow the dimension is something which only depends on how the variety looks like almost everywhere. And from the definition, it almost looks obvious, but then if one looks a bit more carefully, one actually has to work. So let x be a variety, and u in x, a non-empty open subset, then the dimension of x is equal to the dimension of u. So anyway, this is quite a fundamental property. I mean, if you have many folds, you have an open subset of a manifold, and it's also manifold of the same dimension. So if you really want this to be true. So one direction is kind of easy, but if I want to write it the way I wrote it, I have to do like this. So I want to show that as u is contained in x, one would expect the dimension of u is smaller or equal to that of x, and that is actually the direction. So let u1, u0 contained in u1 and so on until un equal to u be a chain in u. Then we want to make a chain in x of the same length, and we do this by taking the closures. So let xi equal to ui, so the closure of ui in x. Note that by definition, we have that ui is equal to u intersected with xi. You know that this u carries the induced topology as usual, and then it is an intersection of u with a closed subset. In particular, ui is the intersection of closed subset of x, and certainly it is the intersection. Then it must be the intersection of the smallest closed subset of x, which contains it. So that's clear. So therefore, thus, the xi are not equal, because otherwise, and so we get a chain. Is a chain in x. Also, if it's a chain, the ui is supposed to be reducible, then we know that the closure is also irreducible. So it's fine. So we get a chain in x, and thus the dimension of x is bigger equal to the dimension of u. So that is the easy part. Now, if you want to do the other direction, it looks in some sense even simpler. So you could just say you take a chain in x, and then you make a chain in u by taking all the intersections called ui, xi intersected with u. And this way, you get a chain in u, which has the same length. But there is an error in this argument, because there's no reason to believe that, for instance, x0 has to lie in u. So we might not be able to get a chain like this by intersecting just with u. So we have to somehow make sure that we can do this. And so we have to take a bit more effort. But still, let's start, let me see. So the first statement that we first want to reduce to the case that x is a fine. So how do we do this? So we, again, start with a chain in x. And now we take any open affine subset of x. Let w in x be an open affine subset with x0 in w. If you have such a chain, it starts with a point. And well, we know that any variety has an open cover by affine subsets. So certainly, we can find an open affine subset which contains the point x0. So we should maybe, I have to be slightly more careful. So I want a chain which starts with a point. In principle, we can always, so here I was not assuming it's a maximal chain. But I can always, if the lowest one is not a point, I can also, if this has, in some sense, a positive dimension, I can make it longer by putting a point here. So we can take such an open affine subset. And then the argument that I just sketched in words will work. I can just take the intersection with the xi with w. And this gives me a chain in w of the same length. So let w i equal to xi intersect w for i. So then we know that w i plus 1 is dense in xi plus 1. So and xi is not. So it follows that w i plus 1 strictly contains w i for i. So thus, w 0, which is of course equal to x0, contained in w1, wn, is a chain in w. Actually, anyway, it doesn't matter. I don't want to confuse the issue more. OK, so we get in this way a chain in w. So for any chain in x, what? Yeah, yeah. OK, thanks. So for any chain in x, I find a chain in w of the same length. So the dimension of w is bigger equal to that of x. We know already the other direction. So it follows that the dimension of x is equal to the dimension of w. So if I have for every variety x, there is an open defined subset which has the same dimension. So therefore, thus, you can replace, say, x by w and u by w intersected u. And so if we prove that the dimension of w is equal to the dimension of w intersected u, then we have proven the statement that dimension of x is equal to dimension of u. So to reduce to the case, x is a fine. So now we assume that x is a fine. So first, so x is a fine. Well, this is a moment still not easy enough for us. So we first do it in the case that x is a fine space. So now we deal with the case that x is an. So if x is equal to an, so we can take let x0 be a point in you and I put, say, xi, whatever, xi, so large xi, the space, I put this to be, say, in a fine linear subspace containing xi minus 1 for all i. So for instance, if the point x0, for instance, if the point 0 lies in u, if the point 0, 0, 0 lies in u, we can take as xi just the 0 set of the coordinates from xi plus 1 until xn, so that we have like we did before. So these are just all linear subspaces. And then we put ui equal to xi intersected u. And we get again a wonderful chain. Un is a chain in u. And so we get the dimension of an open subset in en is n. So to finally deal with the case that we have that x is a fine, we need to use again the neutral normalization theorem to reduce to the case of a n. So now it's all a bit, and this is again a little bit of this trick. So if x is a fine, so then we have there exists a finite subjective morphism, say phi from x to en, where I assume that n is the dimension of x. So now we take the complement of x. So we have x minus u is a closed subset of x. And we can take phi of x and say call this z. This is a closed subset. This is lies, obviously, in en. It's closed because we have a finite morphism. And as this is a closed subset of x, so every irreducible component of it has dimensions smaller than that of x. And this finite morphism preserves the dimension. So therefore, z has dimensions smaller than n. And in particular, they are certainly not equal. Or I could also say this is the case, because as a finite morphism, it preserves strict inclusions of closed, of irreducible closed subsets. Anyway, it will be called phi is finite. So let f be the polynomial which vanishes on z. And we put v to be the complement of this 0 set. Then this v is open, dense, so open and non-empty in en. So we have the dimension of v is equal to n. On the other hand, we can put w to be the inverse image of v. This lies of v. This lies in x. And in fact, if you look at it, it lies actually in x minus z. But maybe we don't need it. So for the moment, it certainly is an open subset in x. So phi is a surjective closed morphism, because it's finite. So if I restrict it to the pre-image of this thing, it's still surjective. And it's also closed, because we have induced apology on both sides. So we have that phi restricted to w from w to v is surjective and closed. So therefore, we know if you have a surjective closed morphism in the dimension of the source, it's bigger equal to that of the target. So we have the dimension of w is bigger equal to the dimension of v, which we knew to be equal to n. But now, by definition, we have that u contains w. Because we have taken v by taking the complement of z of f. And f is something which lies in the ideal of z. So the inverse image of this thing will certainly contain the whole of u. So u contains w. So therefore, the dimension of u is bigger equal to the dimension of w, which we have just seen, is bigger equal to n. And now this proves it, because the other direction we had. But you see it's a bit. So the basic idea is somehow quite simple that in some sense what you want is the argument that I gave in the beginning, that if you have a chain in x in u, you get a chain in x by taking the closures and to get a chain in u by taking the sections with u. But that doesn't quite work. This only goes in the works under some, because you would need that x0 lies in u. But you can still use the argument to prove that you can assume it's a fine. And then you can assume that x is a fine. And if x is itself an, it's easy to see. It's kind of straightforward. You just intersect with linear spaces. And otherwise, you somehow use the neutral normalization theorem to reduce to the case of an. But then there's a little bit of kind of slightly complicated argument in the end to see how you reduce to the case of an, which is somehow by taking the image, the pre-image, and so on. But anyway, but the basic thing is that, you have to make, it's a little bit, it's not just you say, I want to reduce. And then you reduce. But then you have to see it's, how do you go back and forth? And so one way to do this like this. What if you use close? If you use close. Well, if you use close, we know that the dimension is smaller. So you know that open subsets are dense, so the close subset is always much smaller. And we know that we had this theorem that if you have a variety, then every close sub-variety has a smaller dimension. And so if it's not a sub-variety, but just a close subset, it's a union of sub-varieties. And each of them has smaller dimensions, so the dimension is smaller. So the relation there is quite, so as I get some, so some corollaries, so first corollary. So all quasi-projective varieties are finite dimensional. So all what we had called varieties are finite dimensional, because we had seen that every variety has an open cover by the fine varieties, and the fine varieties are finite dimensional. Oh, all. Yeah, I don't know. I started saying something, and then, so all varieties are finite dimension. So as I said, every variety has an open subset, which is in a fine variety. And you know that that has a certain dimension, a certain finite dimension, and the variety has the same dimension. Another corollary is that if we have varieties, if x and y are two varieties which are birational, then they have the same dimension. Because we have seen that so this, if x is birational to y, then we know there are open subsets of x and of y which are isomorphic. Non-empty open subsets, say u and x and v and y, such that u is isomorphic to v. So then, by what we have just seen, as u is an open subset in x, a non-empty open subsets in x, we have the dimension of x is equal to dimension of u. As u is isomorphic to v, the dimension is equal to that of u. And then v is open in y, so they have the same dimension. So this is quite simple. And then, finally, we have the analog of what we had in the fine case for projective varieties. So this is corollary. So the dimension of projective n space is equal to n. This is because an isomorphic to an open subset in pn. So they have the same dimension. Then, if f in k x0 to xn is a homogeneous polynomial of positive degree, then we have the dimension of the 0 set of f in pn is equal to n minus 1. So here may be, anyway, the last one is that if x in pn is a close sub-variety of dimension n minus 1, then x is a hypersurface, homogeneous and irreducible. OK, so I will not say so much to that. The last one, three, is precisely, you just copy the same proof as in the fine case. And two is also more or less the same. So by a projective transformation, so we just take a linear change of coordinates, transformation, can assume that the 0 set of f is not contained in the hyperplane at infinity. This anyway would be, if we have some polynomial, we can change the coordinates so that it's not the hyperplane at infinity. Well, and then if I take the 0 set of f intersected with, in some sense with u0, but I could say, I just write with an, this is equal to the 0 set of the polynomial f, which we have de-homogenized. So it's just a hypersurface in a n. So this has dimension n, n minus 1. And it stands, it's an open subset of the 0 set of f. So also the 0 set of f has dimension n minus 1. And I just say, for the last one, I just say you now do the same proof as in the fine case. We just say that x, you take an element in the ideal, some f in the ideal of x, some irreducible element in the ideal of x. Then x is contained in the 0 set of this polynomial. And as both are irreducible of the same dimension, they are equal. And that's it. We did this just at the beginning of this lecture. Now we can prove the statement that I had said before when I proved this fact about the intersection of a variety with a hypersurface. Namely, we can prove that now if we take the intersection of a fine variety with a hypersurface, that every irreducible component of the intersection has dimension 1 less. And this is actually very easy. Namely, what we do is just we consider the open subset where we throw away all the irreducible components of the intersection except for 1. And then this thing has the same dimension. And then if you argue for this, it means the intersection has just one irreducible component. And the dimension of the intersection is n minus 1. And so this irreducible component has dimension n minus 1. So somehow we use this previous theorem to turn this into some kind of triviality, whereas before it seemed impossible. So let's see. So theorem, I still call it theorem because it's important. So let x in a n be in a fine variety. And let f be a polynomial, which does not vanish on x. Then every irreducible component of the zero set of f intersects x has dimension n minus 1. Yeah, maybe I write still of dimension n. And then this must be like this. Do I want it like this or no? For some reason, maybe it will stick to my dimension. So this irreducible component of this thing has the dimension of x minus 1. So it's actually funny because in the previous case, we said that the weaker statement was if the intersection is not empty, then it has dimension n minus 1. So this somehow looks like this statement is stronger also in another way that says something about non-emptiness. But that's not the case. If the intersection is empty, this means there's no irreducible component. And so this theorem says nothing. So it just says that every irreducible component, if there's any, has dimension n minus 1, dimension of x minus 1. Now let's come to the proof. So let we take any irreducible component of this thing. Let z be an irreducible component of the zero set of f intersected with x. And we take w to be the union of all the other irreducible components. So we take a polynomial g, which lies in the ideal of w, but not in the ideal of x. So this g vanishes on all these irreducible components of zf intersected x, except for the z. So this is not I wanted this. We have chosen one irreducible component. We want to show that this irreducible component has dimension of x minus 1. And we do it by throwing away all the other irreducible components by restricting to an open subset. And so we have here the union of the other irreducible components. We take a polynomial, which vanishes on all other irreducible components, but not on the one we are interested in. And we let u be equal to x without the zero set of g. Now we want to, u has now the wonderful property that u is in a fine variety. We have this theorem that if we take the complement of a polynomial in the fine variety, it's again in the fine variety. So we can use the results we proved for fine varieties for this thing, although it is an open subset of x. So then u is in a fine variety. And if I take u intersected the zero set of f, well, we have just done it in such a way that the only part of the zero set of f that intersects u is z, because we have thrown away all the other ones. So this is just u intersected with z. And now this, if you, so this is, however, still a hyper surface in this refined variety u. If you also remember how it was shown that this is a fine, it's actually a closed sub-variety of a n plus 1. And the first n coordinates are the same. So if f is a polynomial in kx1 to xn, it's also a polynomial wherever u lives. So therefore, I mean, I hope you remember that or whatever. We'll look it up. So therefore, f is a polynomial function on u. So we find that u, the dimension of z intersected u, is equal to the dimension of x, of dimension of u minus 1, because it's a hyper surface in this thing. And u is an open subset in x. So this is equal to the dimension of x minus 1. And this is an open subset in z. And so I find that my given irreducible component of this intersection, any given irreducible component of the intersection has dimension, dimension of x minus 1. So it's somehow, it looks a bit like cheating, but because you just prove it by throwing away the others and still having the same theorem, but that's the way it works. So now we come to somewhat an application, which tells us in a kind of more precise way how the dimension changes under morphisms. So we had, for instance, I had claimed at some point. So we had seen that if you have a finite subjective morphism, then it preserves the dimension. And I had said in the beginning, this comes somehow is supposed to be related to the fact that if you have a morphism with finite fibers, then it preserves the dimension. But we haven't proven that. Now we'll prove something which actually implies that. Namely, we prove that if the fibers all have a given dimension, then the dimension of the source is by that number bigger than that of the target. And actually, it's enough that this just holds over an open subset. So let me state it. So here, I don't want to call it corollary proposition. Let f from x to y, the morphism of varieties, assume there exists a non-empty open subset, u in y, such that all the fibers over points in u have the same dimension. For all p in u, the dimension of f to the minus 1 of p is equal to some number n. So there's an open subset where the fibers all have dimension n. We assume that, then it follows that the dimension of x is equal to the dimension of y plus n. So this is, and in particular, for instance, if all the fibers are finite, then it means the dimension of the fiber is 0, then the x and y have the same dimension. But if you're not quite sure, and I don't quite know how to manage with time because it's somewhat tricky, I wanted to also state, without proof, kind of converse to this, which is kind of the more powerful theorem. This is called a theorem without proof. So I will not prove it. So this is often called theorem on dimension of fibers. The statement is basically that this is if and only if, and actually slightly more than that it's if and only if, namely, let f from x to y be a morphism. And assume that the dimension of x is equal to the dimension of y plus n. So maybe I take a subjective morphism. I mean, otherwise, I will restrict to the image. Then there are two statements. So first, there's the converse of this. I know, first I have something else. So for all points p in x in y, we have that the dimension of the fiber is at least equal to the difference of the dimensions. So the dimension of the fibers cannot be too small. It can only be too big. And secondly, there's an open dense subset. There is a non-empty open subset u in y, such that for all p in y, the dimension of f to the minus 1 of p is equal to n. Yeah, indeed. OK, so this first thing is actually not so difficult. It follows by some modification of the proof I would be giving for the proposition. It would be essentially, I could call this an exercise. So if you understand the proof well enough, you should be able to figure it out, although it is a somewhat hard exercise. The second one is actually much more tricky. I think in order to prove this, one needs to prove a better version of the neutral normalization theorem. Anyway, so now, however, I have the problem that I cannot seriously try to prove. So I will not prove this theorem, because it's a bit too complicated. As I said, the first one I could prove, or if I'm in a particularly nasty mood, I might ask you to prove it. But I would prove next time this. But I don't think I want to start now, because I don't think I can get, I mean, I can rush the proof. Then just next time, I will take five minutes more.