 So, we were talking about dimension and I had introduced finite morphisms which we wanted to use to study dimension. So if I, this was first finite morphisms were morphisms of a fine variety. So if x and y are fine varieties and f from x to y is finite is defined to be finite if and only if the, if I take the pull back. So the coordinate ring of x is a finite. So f star of a of y algebra. And I had introduced what this means. So it means that, so if f a and b k algebras then, so one is a sub-algebra of the other then we say that b is a finite as an a algebra if, well it's the analog of a finite dimensional vector space over a. So if there exist some elements finite in many b1 to be n in b such that b can be written as linear combinations of these elements with coefficients in a, we had given some characterizations of that. If b over, or some property, so if c over b is finite and b over a is finite and c over a is finite and we had some similar results and we had seen that being finite also has something to do with, so if, so one result that we had that if b over a is finite then every element of b satisfies a monic equation with coefficients, so it's a zero of a monic equation with coefficients in a. And so, yeah maybe that is, and then we finally, then we had come back to, now that we had studied this we had then come back to this definition and we had proven that finite morphisms are closed. So now we want to go forward, so maybe I, we want to, we want to study these finite morphisms to somehow study dimension, so it's somehow important that we have enough finite morphisms and so that we, for instance if we want a n to have dimension n, we will prove that in a moment and then we want to use this to compute the dimension of any variety, so if we can use finite morphisms to compare the dimensions then it would be good if we know that every, say a fine variety has a finite morphism to some a n and that's actually what is called the neutral normalization theorem, so that we can compare, so up to a finite, so there's always a finite morphism from any a fine variety to some a n, so let's, this is, and this will be a very important tool for us, so we have a theorem which is the neutral normalization theorem, so we prove a little bit more, so let z f in a n is a hypersurface, so f is a polynomial in x1 to xn, maybe irreducible, then there exists a finite subjective morphism from z of f to a n minus 1. Okay, so if later we prove that finite subjective morphisms preserve dimension and then this will hopefully show that the dimension of hypersurface in a n is the same as the dimension of a n minus 1 and that will maybe help us to show that the dimension of a n minus 1 also is n minus 1, we'll see that later today, so and more generally if x is in a fine variety, so then there exists a finite subjective morphism from say p from maybe also called p, p from x to ak for some k. Okay, so we want to prove that, it requires some work, first we will prove some very simple remark which we'll use in the proof of the first part, which is so obvious that maybe we don't even need to prove it, but anyway, so let f be a non-zero polynomial x1 to xn and so then, well then there exists a point p in a n with f of p different from 0. Now in some sense that's pretty obvious, but I still want to prove it because it actually does even take a little bit of effort. So if n is equal to 0, this is obvious because in that case we just, this is just a constant, the constant either 0 or not. Now if n is equal to 1, then we are over an algebraically closed field, so we can write f of x equal to x minus b1 to the n1, x minus say bl to the nl where the bi are some elements in k and the ni are positive and you just factors into linear polynomials because k is algebraically closed. So thus the zero set of f just consists of these points b1 to bl and so we take any point b in k without b1 to bl then f of b is different from 0. And why can we do this? S k is algebraically closed it's infinite, k is infinite because an algebraically closed field will always be infinite and so certainly if we take away these finally many points there will be still some. So now comes the actual dry one, what? Now comes the case n is greater than 1, here I assumed n is equal to 1 because the proof was in dimension 1, now we look at the general case n is bigger than 1. Now n, so it is somehow induction on n but we also we need the case n equal 1 in the induction, so that's why it's a bit strange. So we can write f to be some i f i x n to the i where f i is some polynomial k x1 x n minus y. And we assume that f is not 0 so that means that not all the f i are 0. So there is an i there's a j such that fj is not 0. I mean as a polynomial it's not 0. So by induction on n we find that there is some tuple of you know b1 to bn minus 1 such that I put it into this fj it will be known 0 such that fj of b1 to bn minus 1 is different from 0. So then we can make a new polynomial we just take g of a variable x is just defined to be f of b1 to bn minus 1 comma x. So this is not the zero polynomial because you know after all there is one coefficient which is non-zero. So this is in kx without zero it's a non-zero polynomial and so it has a zero it has a it has a there's a point where it's not zero so so thus there exists b in k such that g of b is different from 0 that is f of b1 to bn minus 1 comma b is different from 0 b. Okay so it was in the end not so difficult but anyway so we have found always a point where it's non-zero. So now we come to the actual we start with the actual proof so we will just go about and start with number one. So we want to prove that this is finite subjective morphism so we have our polynomial f so we take the top degree homogeneous part of f so the it's a big f so we notice some decomposition into homogeneous components we take the one of highest degree so this will be some non-zero polynomial so this is so this is in particular f of t is in particular homogeneous so if we put the last variable xn to b1 it will not be the zero polynomial you know if you just look at it if you put take homogeneous polynomial put one variable equal to one you have precisely as many terms as before no terms cancel so if the polynomial was not zero before it's not zero afterwards you know you can get back the previous polynomial by just homogenizing so this is not zero so in particular we can use this stupid remark so thus there exists say b1 to bn minus 1 in you know a n minus 1 such that if I take fd of b1 to bn minus 1 comma 1 this is different from zero okay so this is what this stupid remark was useful for now we can change the coordinates on our an we just shift everything you know all the coordinates the first coordinate we replace x1 by x1 minus b1 x2 by x2 minus b2 and so on so by shifting the coordinates we can assume that b1 to be n minus 1 is actually zero no so by a change of coordinates assume that fd zero zero one is different from zero now when we so the f that we use to for this hyper surface is well defined only up to multiplying by non-zero constant you know the zero set does not change if we multiply by non-zero constant so this is some non-zero number we can as well assume that this number is one so by multiplying f by a constant can assume fd is equal to one but what does it mean it just means that so if you have a polynomial a homogenous polynomial in in these variables x1 for xd so these are all the monomers have degree d so there's only one monomial of degree d which will give you one if you put only the the nth variable equal to one all the other ones equal to zero all the ones that once give you zero so this statement just means equivalently that the coefficient of xn to the d in either fd or in f is one okay so this is what we and you know this should make you think of what we had about these these finite maps there was something that in algebra is finite over another one so if you if you add an element which satisfies a monic polynomial to some algebra then the extension is finite so here we it looks like we might get there let's see how it works so so let now we take this projection p from say zf so p as a projection we just take projection to the first given by the first n minus one coordinates so we just you know take the first n minus one coordinates of every point and we let say w n in the coordinate ring of zf to be the class of the last variable so so then I claim that so maybe I claim that a of zf will be the pullback of k x1 to xn minus one a joint omega n I mean the extension the k algebra extension like this because after all you know the the a of zf is just a quotient of k x1 to xn by the ideal of this thing so it's generated by the classes of x1 to xn so the the pullback just pulls back the class of x i the x i you know on the n minus one to the class of x i in here so this is okay and the the last one the xn is pulled back to this one so it's really the we find that the coordinate ring of zf is just this no because this says nothing else than it is the ring generated by the classes in eight set zf of the xi but now we see that this omega n satisfies a monic equation so so we know that the coefficient of xn to the d and f is one so f is equal to x n to the d plus sum i equals one to d minus one ai xn to the i where the ai are some polynomials in k x1 to the xn minus one no that's just what it says that they really let the coefficient is this but what does it mean if we now you know the in zf we just take maybe i want to still keep the statement so so we have that a of zf is equal to k x1 to xn divided by ideal generated by f so we have in thus in a of zf we have a relation that f is zero the class of f is zero and a of zf so that zero is equal to omega n no wn to the d plus sum i equals one to d minus one well the pullback of the ai times omega n to the i so we have a monic equation so this implies so we have precise the statement that this thing is obtained from this by adding an element which satisfies the monic equation and we had a lemma which said that under these assumptions under these if one has this then this one is finite over this that is pi from zf to a n minus one is finite okay so we have found that indeed we have a finite map like this so now we come now surprisingly the ah we are not finished because we are supposed to show that it's a finite surjective morphism okay so we have to see why this map is surjective but that's kind of easy so show that pi is surjective so we take a point we want to be n minus one in n minus one we have to show that the fiber of it is non-empty so you know to see it's not empty so we put it's kind of similar to what we had before we put say g to be the polynomial where we so we take f and we replace the first coordinates by these so g of x so some polynomial and we know so we know that the coefficient of x to the xn to the d of f is is one so that implies that this polynomial is not zero because it's not affected by okay so the coefficient of x to the x to the d of this polynomial is one so and is well is not zero and actually it's non-constant because the coefficient as a set of x to the d is non-zero so therefore g has a zero say b in k well and then obviously if this has a zero then this is a zero of this thing so we have a thus we have that the inverse image of this point this was b so we call it a so the inverse image of no we have a point we have to show you know it's surjective so we have to show for every point the inverse image is non-zero but it was certainly not written very prettily so so this inverse image prior to the minus one of b is just a set of all b1 to the n minus 1 comma c such that f of b1 to the n minus 1 c is equal to zero which is nothing else then you know this b times the zero set of g which is non-empty okay so the map is indeed surjective so a finite surjective map here and now the surprising thing is that we have done most of the work now we kind of do some basically reduce the the main statement to this is it clear you well I said you know actually it has a zero it has it might have many zeros no no so I know it has a zero so z of g is non-empty and then this will be non-empty but in particular it contains if you want if it contains the point b times a okay now we come to the second part so so if say x is just equal to an then you know it has a the identity is finite then it's clear so assume it's not equal to an so assume that so assume x is in an is a closed up variety and okay so and we want to prove that we want to prove it by induction on a x should be non-empty obviously because otherwise the statement would also be for be false but okay so let's see so we want to somehow use use this case so you know x is not equal to an so it will be there will be some f in the ideal of x so so let f be an element in the ideal of x without zero so it may be equal to assume irreducible so if all right you know as f is as x is not equal to an it follows that the ideal of x does not consist just of zero so there is an f there and we can then also assume it's irreducible because ix is a prime ideal so whenever we write f as a product of two factors one of the two factors must be in the ideal of x okay so we can apply the first part so there exists a finite subjective morphism say now here it's also called pi whatever maybe what i call pi from the zero set of f to an minus one and we know that x is contained in the zero set of f is closed we know that the embedding of a closed subset into something into of a closed sub variety into a variety is a finite map so the map the the embedding of x into the f is finite so thus if i take say pi tilde which is equal to pi composed with i from x to n minus one this is a finite morphism so we have a finite morphism from x to n minus one but obviously there's no reason to assume it's subjective i mean the point is we can we want to be able to find once possibly smaller k such that the map becomes subjective because you know by definition of finite so to be finite we need that what is it if we look at the map x to y this will be finite if a x is a finite f upper star of a y module or something is finite over this okay and now if the map is injective then it turns out that the pullback is surjective that's actually a homework exercise and so therefore you know if they are equal certainly one is finite over the other and then the composition of finite morphisms is finite so this is finite so where am i so this is not necessarily finite but we can look at the image where we now we can use the induction hypothesis y lies in a n minus one by induction on n we know that there's a finite morphism from y to some ak okay how would i would call it finite morphism finite subjective morphism so phi from y to ak for some k now this thing is a finite morphism as a map from x to a n minus one but actually maps to y so then we know that if we viewed as a map to its image it's also finite morphism so that was one of the things we proved so we have that pi tilde from x to y is finite and so if i put now maybe this should have if i not now put the thing that we are talking about to be um phi composed with pi tilde this is a map from x to ak this is certainly finite as a composition of finite maps and it's also subjective as a composition of subjective maps because this map you know this map was just uh you know if we have restricted here on this side to the image so by definition it's finite and the other is subjective and we also have um we have a finite subjective morphism here so n subjective and so that proves it okay so this is this Neuter normalization theorem so we want to prove so we will see uh in a moment how one can apply this for instance to prove that the dimension of a n is equal to n if we find the time before that i want to show one more kind of topological property of finite morphisms that is you have the statement that if you have a finite morphism and you have a sub to have two sub varieties two irreducible closed subsets in the source which are one strictly you know one contained in the other but strictly contained they are not equal then the same holds in the image it's not it cannot happen that if you have two sub varieties one contained in the other which are not equal that they become equal by mapping with the finite map and this will in particular imply that the fibers of a finite morphism are finite which is you know a good reason to call it finite okay so let's see that so lemma let say phi from x to y be a finite subjective morphism so let so z strictly contained in w we close sub varieties of x then I have that the image of z is strictly contained in the image of w so you don't have the two things become equal if they were not equal to before so somehow things cannot be contracted so first we can so we can restrict the map phi to w no so therefore we can assume that x is equal to w and we can in the target we can restrict the map to the image of w it still is a finite map if it was one before so we can assume that y is equal to f of w okay so this simply this simplifies a little bit so in other words we the statement would be that so in other words so if so if z is strictly contained in x then f of z is strictly contained in y so where z in x is a close sub variety so I always use variety so this only makes sense if things are irreducible okay so we have this tiny simplification so now we want we have to check this so we have that z is strictly contained in x so there will be so this was actually I changed it from phi to f but now I may be so we take there's no f you know here it was called phi and here it can move fast itself as an f but it's still a phi so let g be an element in the x which is not zero such that g restricted to z is zero that means that g is in the ideal of z in in ax and you know we know that if z is not equal to ax it's a close sub variety there will be some function which vanishes on it no that's one of the things we always know so we take such a thing we really want to finally use the finiteness whatever to show that then there will be something similar in the image so now we again we have to use the finiteness and we again use this by having this thing with a monic polynomial so this g will satisfy monic polynomial over ay so g satisfies a monic equation say g to the n plus sum i equals zero to n minus one say phi star of ai times g to the i is equal to zero no where the ai are elements in a of y no that's what the finiteness tells us so what do we so now if we have such a monic equation we can assume that the degree is as small as possible so there might be more than one monic equation which is satisfied we take the one with the smallest possible degree so if this is the case in particular it cannot be true that the whole thing is divisible by g so the coefficient of g to the zero must be non-zero because otherwise we could just divide the thing by g so we have phi star of a zero is non-zero so it follows that phi star of a zero it's different from zero no otherwise divide by g and we get a monic equation of smaller degree because g is after all non-zero we can certainly divide by it okay so but now we can bring this thing to the other side so it means that phi star of a zero if we bring it now to the other side so this is non-zero this we just if we take it out we can divide by g and so if i'm not mistaken this is just equal to minus g times g to the n minus one plus sum i equals one to n minus one phi star of a i g to the i minus one so we have just you know brought you know this phi star of c zero to the other side and then we obviously can divide by g and so we get this but you know we see that this thing is divisible by g so if this thing is divisible by g it means that it vanishes on that because g vanishes on that but if you know phi star is just obtained by taking a zero composed with phi if a zero composed with phi vanishes on that it means that a zero vanishes on phi of that but so what would have we found we have found a non-zero element in so this is a non-zero element a zero is a non-zero element in ay which vanishes on phi of of z so it means that phi of z is not equal to y and that proves it so we somehow so we find just that the constant coefficient of this thing must vanish on the image so the thing is not equal to okay and now as a corollary we get the thing that I mentioned so if say again phi from x to y is a finite subjective morphism then all the fibers of phi are finite are finite okay this is quite simple it follows basically directly from the statement here so you know that the fibers are finite is equivalent to the statement that every irreducible component of a fiber is a point you know because you know you know in algebraic set has finitely many irreducible components and if each of them is a point then it's finitely many points okay so it's enough to show so every irreducible component of say phi to the minus one of y where y is a point in y is a point well that's very similar it's very simple so I call this component maybe large z so we take a point small z in large z then obviously the image of the small z is y which is the same as the image of the large z if you want you can also put brackets around to make it into a set so by the statement if they're equal if the images are equal the things you started with are equal that's all okay so this was the whole story okay so that was as much as I wanted to do for now about finite morphisms and now we want to apply it to study dimension so I maybe should recall so I should recall the definition of dimension so so if x is a variety then so if I have a chain that empty set different from x0 contained in x1 and so on until xn the chain of irreducible closed subsets in x so we just call this a chain in x and n we would call the length of the chain in safety speaking they're n plus one elements but we call the length n okay so then so actually I call this not call I will write just x0 x1 and so on until xn the chain in x and n will be its length and then the dimension of x is it's not precisely correct in the notes so it's called the dimension of x so this will be either the maximum the maximal n that's such a such that we have a chain in x if such a maximum exists and infinity otherwise so if one you know in in analysis want to say it's a supremum of the length of the chain now it's a bit strange and that's why I also wrote it slightly incorrectly in the notes that we know that we have it if we have a descending chain of closed subsets it will be finite so in that sense one would think there would be always such a maximum but in theory in Iparora this is not obvious because it could be that every chain is finite but there might be chains of bigger and bigger length for one given set we will see that this is actually not true so we will we will prove that all quasi-projective varieties have a finite dimension but until we prove that we cannot use it in our arguments so we for a while have to pretend we don't know that because we after all don't know it yet yes what in no no no infinite I don't know why I I don't know why I wrote empty set yeah no certainly not zero I mean another question is what kind of dimension you want to this I didn't dwell on what kind of dimension one wants to give the empty set yeah I think the most reasonable thing to say that for the empty set it should be minus one okay that that fits best with with the formulas so and I mean to deal with this infinite for a while we say that obviously infinity is smaller equal to infinity and if n is in z then n is smaller and in particular smaller equal to infinity okay no if we want to write down and so note that so remark so if x well no I mean we don't write I will use it so now let's look at the following lemma so we want to somehow show some basic things about dimension the first one is we would like to so to show that if we have a close sub variety in a variety then its dimension is smaller okay this is however will a priority not be entirely true because we might have some trouble with infinity but apart from that it is so let y so first let y in x be a close sub variety then the dimension of y is smaller equal to that of x and if the inclusion is strict and the dimension of y is not infinite then the dimension of y is strictly smaller than that of x okay later we will prove that all varieties have finer dimensions so that so if we have a strict include this just says that if you have a strict inclusion then dimension of the smaller one is smaller no but for the moment we are not yet there the second statement is let f from x to y be a surjective closed morphism of varieties then the dimension of x is bigger equal to the dimension of y okay so so let if y0 so to say y k is a chain in y then we can then it's also a chain in x and thus the dimension of x is bigger equal to that of y because the fact to prove that the dimension of x is bigger equal to that of y we need that for every chain in y we find one in x which is at least as long and we can just take the same no because it says and this this holds independent of whether some of them is infinite or not so on the other hand so if now but if now y is strictly contained in x we can even get another chain then y0 y1 and so on yk strictly contained in x is a chain in x so that means for every chain in y we can find a chain in x which is longer so if the dimension of y would be infinite this wouldn't really prove anything but if the dimension is finite then you know we take the chain which is you know such that the k is the dimension and we find the dimension of x is larger this is one yeah and second is a bit more it takes slightly more effort but it's not big deal so again so we want to show that the dimension of y if x is bigger equal to that of y so that is we want to find for every chain in y a chain in x which is at least as long that's enough so let say yn be a chain in y so what we want to do is show that there exists a chain in x so x0 strictly contained in x1 and so on in x such that well just if I take f of xn is equal to yn so I mean the fact that f of xn is equal to yn will imply that this is always strict you know because they are different and then we have found a chain of at least the same length as the corresponding chain in y so the dimension of x is bigger equal to that of y okay and so we do it so we use induction on n so anyway if n is equal to to 0 well we can just take inverse image of x0 or something then it's anyway it's easy um anyway we'll see that but now let's we assume otherwise um what yeah yeah so n equals 0 is clear so otherwise so we do induction on n so let z1 so whatever cr be the irreducible components of the inverse image of yn minus 1 so the point is that I mean in theory we could just think we take as the xi the inverse image of the yi but that wouldn't be allowed because we need these things are always irreducible so we have to take irreducible components but you know you know that the union of the xi of the f of xi is equal to yn minus 1 and they are closed and yn minus 1 is irreducible so it follows that one of them must already be equal to to this so thus one of the f of xi is equal so we can just take this thing so then so maybe I so for for some i it's equal to that then for this i we have a chain so so this so that means f of xi so if I take the map now if I take f from xi to uh yn minus 1 this is now a subjective closed morphism and so we do indeed um anyway now we can apply the induction hypothesis and see that there is you know we we are in the situation we just start now with yn minus 1 make induction over n now we have a length of chains so thus by induction there are so we get a chain x0 it's contained in x1 and so on and as xn minus 1 we take this zi is a chain in x with f of xi is equal to yi well okay and then so here I write f xn but I what I meant obviously yeah so I'm sorry but this is not written correctly no I want that f of xi is equal to yi for all i because then it means we really have a chain here okay so now we have this under n minus 1 and then for the last one we just have then you know we have that if we take x0 the last one xn is always x so the last one is is a chain with so f of xi is equal to yi for all i so I mean here we just have taken the previous chain we have left out the last one and now we put the last one back and over definition we have that the map from x to y is a finite subjective morphism okay and now this will allow us to prove not a finite subjective a subjective closed morphism now finally we can use this to prove that finite subjective morphisms preserve the dimension so we see that if we have a closed subjective morphism the dimension of the source is bigger equal to that of the target and finite morphisms are closed and finite subjective morphisms are closed and subjective and so we have to show that also the other inclusion holds so let so this is I call theorem that f from x to y be a finite subjective morphism morphism of varieties then the dimension of x is equal to the dimension of y so we know already that the dimension of x is bigger equal to the dimension of y because f is subjective and closed so we have to show the other inequality the dimension of y is bigger equal to that of x so that is for every chain in x we must find the chain in y which has the same length so let x0 xn be a chain in x so so let for all i let yi to be just the image of xi and the claim is y0 contained y1 to yn is a chain in y which then proves the dimension of y is bigger equal to that of x so the main point is that you have to show that these inequalities are strict no that none of them become equal but this is true that yi is always strictly contained in yi plus 1 because because this is true for the xi and we had proven that finite morphisms preserve strict inclusion of closed sub varieties so indeed the inclusions here are all strict and the yi are irreducible as images of something irreducible so this is indeed a chain in y of length n and so this um proves this yeah now my task which i don't have very much time the next task is indeed to prove that the dimension of an is equal to n so let's see how i can go about this so this is now you want to apply what we know so far to to prove that which then will also prove that at least all the fine varieties have finite dimension so theorem the first statement is the dimension of a n is equal to n somehow obviously that's a very basic thing if this were not true then we would have chosen the wrong definition because obviously it makes no sense to have a notion of dimension which doesn't give the obvious dimension to the obvious thing okay so we will again also compare with hypersurfaces so let f in k x1 to xn be an irreducible polynomial then the dimension of the zero set of f well we want it to be 1 less is n minus 1 so zero f okay so if we take the zero set of one equation it says dimension one less and the third one which is rather simple conversely any sub variety x in a n of dimension n minus one is a hypersurface so if i have a sub variety of the n which has dimension n minus one it's a zero of one polynomial well because okay you see well i will just start so first i do something which i call implication of one to two but we will actually also need it in the proof of part one which is that we which is which will be the statement that we want to prove that the dimension of a hypersurface is the same as the dimension of a n minus one okay which one are equivalent no these are not these are things that we prove three statements there's nothing to do with the equivalent no we prove that dimension of a n is n we prove that yeah yeah but at the point is just i shouldn't maybe call it like that you know we want to prove all these statements and you know we we will prove something that once we have proven one we'll prove two okay but it's a bit more tricky because in the proof of one we will use this statement okay so what i really mean is just so we first so so first proof that the dimension of a zero set of f of such a polynomial is equal to the dimension of a n minus one okay so once we have proven one this will prove two but you know first we have to prove one so why is that so this is quite simple so let f be a non-zero polynomial then there exists a finite subjective morphism from the zero set of f to a n minus one this was the Newton normalization theorem so thus it follows that the dimension of that f is equal to the dimension of a n minus one maybe i don't call it one implies two this is what we use but it means that now we don't have to prove two anymore once you once we prove one two is automatic no non-constant otherwise not you know if it's constant it will be empty and then you know hypersurface you always need that that it's not constant so so we can now we want to prove one which is the main step and we cannot maybe quite do it now but i can prove at least the trivial direction so let say i take z i to be the zero set of say you know f x i have the variables are called x i so x i plus one until x n in a n so for instance x z zero z zero will be just the zero set of all the variables so it will be the point zero and z n will be this there will be nothing here this will be a n so so then z i is isomorphic to a to the i and thus so in particular it's irreducible and thus we have that z zero strictly contained in z one strictly contained in z two and so on is a chain in a n of length n so it follows that the dimension of a n is bigger equal to n well maybe now we'll try even the converse direction is not very long it's only slightly surprising so we want to prove the opposite inequality no so i mean if i put here from x n so it's maybe you don't like the notation but it's you know it's like you would do if you write a sum in in a program or something you know it's going up from this to this so if this one is bigger than this there's nothing so so x x n minus one would be z of x n and z and so i mean anyway z n is supposed to be just zero set of nothing yeah and then z n minus one is z of x n z n minus two z of x n minus one x n and so on i mean this is if ever you have written a a program in which there are sums you will notice that you are allowed to do it that way if the lower index is bigger than the upper index then the sum is empty and the same here okay so we prove the opposite inequality so by induction on n it's always induction so for n equal to zero it is trivial because a zero is a point and it certainly has dimension zero we can only find the choice a chain of length zero so now let x zero strictly contained in x one and so on say x k minus one which we call x strictly contained in a n be a chain in n power we don't know what this number k is that we want to find out you want to find out that this number would have to be smaller equal to n minus one anyway now so we know that then obviously x in strictly contained in n minus one is a closed subset closed sub variety so the ideal of x will be non-empty so i can find the polynomial f which is non-zero which lies in the ideal of x so let f in the ideal of x be irreducible so first i can find a non-zero element and then i can maybe write it if it's not irreducible right and it's written as a product of two factors then one of the two factors must lie in this prime ideal so i can find an irreducible element as i used it before so then x is contained in the zero set of f now we somehow have to use the part i mean the part two know if thus you know the dimension of x is smaller equal than that of the zero set of f so um it follows that k minus one is smaller equal to the dimension of the zero set of f which is equal to the dimension of a n minus one you know but we um which is a small let me try again um you know we had found that the dimension of the zero set of f is equal to the dimension of a n minus one no that was what we had said that is what's written here okay so i'm just using this and this is equal to n minus one by induction okay so this makes induction quite powerful because we may we assume that the dimension of a n minus one is n minus one by induction okay so now um so this is f such so thus so we have that so k plus one k minus one is equal to n minus one thus the dimension so for any chain in x you know the length of the chain the length of the chain is actually equal to k the length of the chain is at most k so the dimension of the n is smaller equal to k no because we have written any an arbitrary chain and its length is k no k minus one plus one which is equal which is smaller equal to to n and this is what we wanted to prove it's not a contradiction okay okay so um so i will the third part i will prove next time this also will give me occasion to review the statement and then it will be tomorrow yeah so you can you don't have to look forward it for too much time