 Okay, so we continue with our discussion about general varieties namely projective spaces and projective varieties. So if you recall from the last lecture what I had told you was that if you take projective space which is you take affine space of dimension n plus 1 over an algebraically closed field k. So this is as I said just Cartesian product of k n plus 1 times but given the Zariski topology. And then you take away the origin and look at the space of lines then you get the projective space over k and there is this map which is the natural projection map you can think of it as a quotient map and you can also think of it as associating every point in projective space to the line through the origin that it corresponds to okay. And what we saw last time is that this projective space is it can be given the Zariski topology in 3 ways one is to take the quotient topology given by this map pi which makes use of the topology above the Zariski topology above. The second way of giving the Zariski topology and projective space is to declare closed sets to be of the form given by 0 common 0 loci of a bunch of homogenous polynomials in n plus 1 variables okay. And the third way is to consider an open cover of projective space by open sets which are homomorphic to affine space and then or which are which are bijective to affine space and then transport the variety structure. So it is a third aspect namely the affine cover of projective space by n plus 1 copies of affine space which is what I explained last time. So you know you define the subset ui which is the locus where Xi does not vanish so basically you have so the coordinates here are x0 through xn that is how a point here looks like and well it goes down to a point here with which will now have homogenous coordinates x0 semicolon etc xn and I am using the same Xi to denote both the coordinate functions as well as the coordinates general varying coordinates. And ui corresponds to Xi not equal to 0 this is an open subset because it is complement is the 0 set of Xi which is the 0 set of homogenous polynomial Xi is the homogenous polynomial degree 1 so it is a complement of a closed set so it is an open set but the point is that you have this map phi i from ui to an n dimensional affine space which is gotten by sending a point with coordinates x0 etc xn to what you do is you divide every coordinate by Xi and then at the entry Xi you will get Xi by Xi which is 1 and you forget it so you just send it to the point x0 by Xi, xn by Xi and here you omit Xi by Xi. So what we saw last time was that this phi i is actually an isomorphism of varieties okay so phi i is an isomorphism of varieties and what it will induce is pull back by phi i induces a k-algebra isomorphism so the k-algebra isomorphism will be say a pulling back regular functions so given a regular function here you pull it back to a regular function here pulling back means you given a given a function here you compose it with this to get a function here. So you get a k-algebra isomorphism from the regular functions on an to the regular functions on ui and the fact is that this is the same as the coordinate affine coordinate ring of affine space and this right here is can be identified with the homogenous localization of the polynomial ring corresponding to this affine space at Xi which means that you take you call you call s as the affine coordinate ring of affine n plus 1 space and is also called as the homogenous coordinate ring of the n dimensional projective space and what you do is you take you take that that is just the polynomial ring in the variables x0 to xn you localize it with respect to Xi ok and s has a grading there is a natural grading on x on s which is given by degrees of polynomials homogenous polynomials every polynomial in s breaks up into uniquely into its homogenous components and each homogenous component has a fixed degree that is the grading on s and this that grading will induce a grading on this localization by simply by very simple and obvious manner namely an element here is going to be just of the form some element of s divided by a power of Xi ok and then what you do is you define the degree of a homogenous a homogenous element here is of the form homogenous element of s divided by some power of Xi and its degree will be the degree of the numerator minus the degree of the denominator ok that makes this integrated ring and then you take the degree 0 part ok. So, this is the this is the ring of this is the ring of regular functions on this open subset of projective space mind you on projective space a regular function is defined as something that is locally a quotient of homogenous polynomials of the same degree ok whereas on affine in affine space it is simply a regular function simply a quotient of two polynomials of course it is if supposed to be defined where the denominator polynomial does not vanish right. So, so this is this is isomorphism and in fact so you know if I call the if I call the coordinates on this a n as t1 t2 etc up to tn then this isomorphism which is given by pull back of maps. So the map is very simple you give me a regular function here its pull back is just composition with phi i. So, it is just f going to first apply phi i then apply f. So, this is the so this is the this is the pull back map and this is an isomorphism ok. So, moral of the story is that this each of these UIs there are n plus 1 of them ok each of these UIs is just an affine space affine n dimensional space and these cover these n plus 1 affine spaces they cover projective space and the fact that this is an isomorphism actually tells you that the the projective variety structure on projective space is given by gluing the affine variety structures on each of these affine spaces why are these why are these isomorphisms of varieties ok. So, yeah so there are I had asked you to check that I had asked you to check that this equality holds probably it is not so difficult to check that. So, so let me check let me check this and let me also tell you that in terms of rings this isomorphism is given in a very nice way this way it is given by a homogenization map and this way it is given by a de homogenization map ok. So, let me just explain that. So, so first thing that I want to tell you is that as to why the regular functions on Ui are given by this by this ring ok. So, you know so the situation is like this I have so let me draw a diagram. So, I have I have affine space so, here is a n plus 1 minus the origin this is the punctured n plus 1 dimensional affine space and this is the projection on to the projective space and here I have Ui ok. And well the if you take if you take pi inverse of Ui if you take its inverse image under this map under this under this map then you get pi inverse Ui which is actually which actually is if you think about it for a moment it is it is D of D of Xi minus the origin ok. So, Ui in the projective space corresponds to the ith homogenous coordinate Xi not vanishing and its inverse image above will correspond to ith coordinate not vanishing that means the equation. So, you are looking at Xi not equal to 0 and Xi not equal to 0 is a basic open set D of Xi you know D of f always denotes the locus where f does not vanish it is a complement of Z of f which is the locus where f vanishes. And we have already seen that D of f is itself an affine variety isomorphic to an affine variety because of the Rabinovish trick ok it can be D so this D of Xi can be embedded in an affine space of one dimension more and we can embedded in A n plus 2 as a close sub variety ok. So, this D of Xi in the whole affine space is certainly it is an affine variety and it is and since I am looking I am since I have since I am looking at its intersection with this I have to throw away the origin ok and that is the inverse image of this UI ok. And the point is that this map Pi so first observation is that this map Pi itself is a morphism of varieties ok see this map this natural projection from the punctured affine n plus 1 dimensional space to the projective space this itself is a morphism of varieties and the reason is I mean the reason is obvious because it is continuous ok and a regular function here if you give me a by definition a regular function here is a quotient of two homogenous polynomials. So, if you compose it with Pi I will simply get a regular function here so not every regular function here on an open set goes down to a regular function below ok but regular function this map Pi of course pulls back regular functions and open sets to regular functions and open sets. So, it is obvious that Pi is a morphism ok so continuity is obvious and in one of the definitions you give this the quotient topology from that ok. So this is certainly continuous map and if there is an open set here what is a regular function an open set here locally it is of the form f by g where f and g are homogenous polynomials the same degree but if I take that function and compose it with Pi you know if you if you pull it back it will correspond to it will correspond to a function there ok and that function is of course it is going to be a function that is constant on lines it is going to be a function that is constant on lines passing through the origin because that is what that is what will happen any function any function on this set when you compose it with the projection map will give you a function above which is constant on lines through the origin and conversely any function above which is constant on lines through the origin will go down to define a function below ok. So, if you take a function of the form f by g f and g are homogenous polynomials of course define an open set contained in the locus where g does not vanish then it is inverse image will be simply the same function f by g which is which will which is again a quotient of polynomials and that is certainly quotient of polynomials a quotient of homogenous polynomials is also quotient of polynomials and any quotient of polynomials defines a regular function locally on a fine space. So, it is very clear that pi pulls back regular functions to regular functions. So, pi is a morphism ok. So, pi is a morphism of varieties and you know therefore pull back of pi pull back via pi will give me a map from you know regular functions here to regular functions there. So, I am looking at pi restricted to this. So, I am going to get O ui O of ui to O of pi inverse ui this is what I am going to get and so namely you give me a regular function on ui you compose it with pi I will get a regular function on pi inverse ui this is the pull back map alright. So, this is and of course this is a this is a k algebra homomorphism any morphism of varieties will induce a k algebra homomorphism which corresponds to the pull back of regular functions ok and of course mind you ui is a variety now as far as our definition is concerned because our definition allows varieties to be either affine or quasi affine or projective or quasi projective ui is a quasi projective variety it is an open subset of a projective variety in this case projective space ok. And the point is that pi inverse ui is also a variety pi inverse ui pi inverse ui is actually you know it is the it is just the this variety dxi it is basic open set first thing is certainly if you take an element here an element here is of the form f it is of the form f by xi to the power of degree f where f is homogenous of certain degree right. So, an element an element of a sxi 0 is of the form f by xi to the degree f ok. So, sxi a s localized at xi is just inverting xi. So, elements there will be of the form simply f by some power of xi but then if I want degree 0 part then I want the degree of this element to be 0 and the degree of this degree of a quotient is defined as degree of the numerator minus degree of the denominator. So, if you want that to be 0 then the power of the xi in the denominator should be equal to the degree of the numerator polynomial which is supposed to be homogenous ok. So, this is how an element looks like and this element is if you think of this element this element is of course a function which is constant on lines passing through the origin because it is both the numerator and denominator of homogenous of the same degree. In other words it goes down to define a function on the projective space and where it will define a function it will define precisely a function on ui because that is where xi is not 0. So, it is very clear that clearly it is the pull back via pi of the regular function f by xi to the degree f which is in o of ui ok. This is clearly a regular function on ui because a regular function on projective space is supposed to be on an open set is supposed to be a quotient of polynomials homogenous polynomials of the same degree. So, in this case it is certainly a quotient of homogenous polynomials of the same degree that is because you have taken it in the degree 0 part the graded localization and therefore it is a regular function ok. So, what I am trying to say is that you know for the moment so you know I have so if I want to draw a more particular diagram I have o so I have o of T xi in the affine space above this is an affine variety and you know that this can be identified with s sub xi ok. If you take T xi in the affine space that is the locus where xi does not vanish and it is an affine variety it is affine co-ordinate ring is s xi. In fact this is the same as af can write af dxi. The reason is because d of xi basic open set defined by xi is of the form d of f and every element of the form d of f is an affine variety right and its co-ordinate ring is just given by localization at f. So, in this case instead of f I have got xi ok do not confuse it with the f here ok. So and you know you can see that certainly this is a sub of this ok and what is happening is it is very clear that if I take an element of this form it is certainly going to be a regular function on pine on pine was ui ok is going in fact be constant on lines. So, it is very clear that this is going to sit inside this alright. I want to say that you know if you go there is a map like this ok you start with you start with an element here and element here is certainly a regular function here ok and it comes from a regular function here ok. So, what I want to say is that you know so there is a map like this ok. So, let me explain. So, you know you start with an element here this element is of this form now this element certainly defines it defines a regular function here ok which goes down to a regular function there and the pull back of that regular function is this regular function ok. So, you have a map like this mind you this is also an element of pi inverse ui this is also an element it is also a regular function there and the same thing goes down to a regular function below and if you pull back this regular function you will get this regular function above. So, when I write like this here I am thinking of it as a regular function above on pi inverse ui. In fact this makes sense as a regular function on dx i this certainly makes sense as a regular function on dx i ok. So, when I write like this this can be thought of as a regular function above ok and it is since it is constant on lines passing through the origin it also goes down to define a function and that regular function is also given by the same expression it is of the form one homogeneous polynomial by other of the same degree ok. So, it is the same regular function that it defines below also ok and the and if you pull back this function below via pi you will get this regular function above alright. Here dx i dx i-0 is the same as dx i because I do not have to I do not have to remove already origin is not there certainly and the other thing is of course so, pi inverse ui is just dx i and what I will have here is and this dx i will of course have affine co-ordinator ring sxi of which if you consider it as a graded ring the degree 0 part is sxi localize at 0 and the claim is that this map from so, this pull back map which goes from you for regular functions on ui to regular functions on dx i this map the claim is that it factors through this ok and it factors through this and it is an isomorphism that is the claim alright. So, this is the claim this is what one has to prove so, the argument is as follows so, any regular function on ui is locally of the form g by h where g and h are homogeneous of same degree and with and defined and defined in defined where h and defined where h is not 0 ok inside ui this is how see any regular function on ui looks like this locally it is of the form coefficient of two homogeneous polynomials of the same degree and it is defined where h is not 0 and it is inside ui ok see it is pulled back via pi gives a regular function on ui on dxi and hence it has to be of the form g by xi to the power of degree g ok because regular functions the regular functions on dxi they have to be of the form g by xi some power of xi and if they go down the this power if they go down to a function on projective space this the power of xi below should be equal to the homogeneous degree of the polynomial above it has to be home the first of all any function any regular function on dxi will look like some f by some power of xi ok but if it has to go down to a function on projective space ok then the numerator polynomial f has to be homogeneous and the denominator polynomial will be a power of xi the power being equal to the degree of the numerator polynomial. Therefore if since I am pulling back a regular function below I will get a regular function above the fact that it has come from below will tell you that when I pull it back it has to look like this ok and therefore this will also be the expression for the regular function below ok because this below g by h is on some open subset ok and this coincides with this on some open subset alright and that will tell you that it has to coincide on the largest possible open set ok. So it has to be of this form so maybe it may happen that there are some common factors but so if you want I can put g1 here I can put g1 by degree g1 ok and this means that g1 by xi to the power of degree g1 is the same as g by h on all of you ok. So I am using the fact that you know on a variety if two regular functions defined on a variety you know if they coincide on an open non-empty open subset then they coincide everywhere ok therefore I start with the regular function here I start with an element here it is locally of the form g by h ok it may have locally many representations but I am just taking one such representation g by h ok where g and h have to be homogeneous of the same degree. Now if I pull this function there locally to the affine space above puncture affine space above of course I will land in dxi ok and locally on dxi I will get the regular function that I get by pulling back this regular function on ui will be a regular function on dxi which has to be this ok. Therefore the function I pulled from below has to be also of the same form it is given globally in this form ok and so what I am just I mean this is just the statement that you know this map this mapping is surjected ok. This shows this shows that oh ui to odxi factors through which factors through sxi 0 is both surjective and surjective I mean it is surjective first of all the map from here to here actually lands in here ok. The second thing is that it is surjective because if you take anything here it certainly defines regular function below that is the first line that I wrote ok and it is injective for obvious reasons because if a function if a function above goes down to function below and if the function is below is 0 then the function above you started with must be 0 and conversely if the function below is 0 if the function above is 0 of course the function it goes down to is 0 ok. So it is very clear that this is both that this map is injective and that this map is surjective is very very clear ok. So here what I am doing is I am using the fact that this is a morphism and this is and I am pulling back the functions ok but you can also think of it in another way by taking any element here which is an element of this form and that defines a regular function below. So you have also a map like that ok and then you can check that if you do it like that then you are just defining the inverse map ok. You start with an element here it defines a regular function below then you will get a map like this ok you can check that that map is also an isomorphism for the same reasons ok. The only thing the way I have done it is I have used I have just used the pull back I am just saying that this morphism which corresponds to the natural quotient of punctured affine space to projective space if you take the pull back the pull back induces a K-algebra homomorphism but that K-algebra homomorphism from the source to the target it is injective it is image is precisely the degree 0 part. In other words it is an isomorphism on to the degree 0 part thus proving that the regular functions on ui is exactly this the degree 0 part of the localization at xi it is called the which is called the homogenous localization at xi ok. So that is the proof of that is the proof of this statement that these two are one and the same and you know to show that this pull back via phi this map this is an isomorphism in terms of rings you can actually give the ring isomorphism from K T1 Tn to this homogenous localization. So what is that so what is the what is the other way what is the other statement so you see I have O of an and I have this isomorphism K-algebra isomorphism which is given by pull back via phi i pull back of regular functions via phi i this is supposed to give as we proved last time an isomorphism with O ui but now you know this is K of T1 etc Tn and this is as we have seen just above it is just s localize at xi so let me let me use a let me expand what s is k x0 etc to xn localize at xi and then degree 0 part ok this is this is s localize at xi degree 0 part and what is this what is this isomorphism what is a what is the K-algebra map in this direction what is the K-algebra map in this direction this map is given the map in this direction is given by homogenization the map in this direction is given by de homogenization which is what we used last in the last lecture to prove that this is an isomorphism ok. So what is the map in this direction so you take f of let me take p of T1 etc Tn what am I going to send it to I am going to send it to the following thing I homogenize it ok. So which means I take xi to the degree p degree p of times p of x0 by xi and so on xn by xi and of course when I write this I omit omit xi by xi alright and I take this and you know and then I divide by xi to the degree p ok see if I take p is a polynomial in n variables ok and if I just take the numerator this is the homogenization of p it is you add the new variable xi ok and then you get the homogenized polynomial. Now that is that will continue to have degree p that will be homogenize of degree p ok here degree by degree p I mean the degree of the highest monomial here ok the degree of the polynomial like this will consist of monomials product of powers of Ti ok and multiplied with some coefficients and some of such finitely many such and among them you take the monomials of the highest degree and take the highest degree and call that as a degree ok that could be of course several monomials of the same highest degree ok but you take the highest degree monomial ok and call that as a degree of p and if you take this expression on the numerator what I will get is the homogenization of p. So this homogenization is achieved by adding a new variable xi ok and raising it to the degree p and multiplying it with this ok. Now once I do this numerator becomes a homogenous polynomial in x0 through xn of degree p alright and if I divide it by xi to the degree p what I get is a regular function on ui because a regular function on ui is of this form it is some homogenous polynomial in the x0 through xn divided by a power of xi which is equal to the degree of that homogenous polynomial. So this is certainly an element of ui so this is how the map goes in this direction you can explicitly write this map ok and what is the map in this direction the map in the other direction is pretty simple that is also very easy to write. So let us write that down that is just the homogenization so what is the map in this direction so this is pull back via ti so I have here let me write that k x0 etc xn localize at xi take the degree 0 part and here this is k t1 etc tn we have this so what is the map in this direction the inverse map is you start with any g by xi to the degree g ok and what you will do it do is simply send it to you know g of t1 dot dot dot 1 tn this is what you will do so what you are doing is you are simply sending so this is the homogenization this is just the homogenization alright. So this g is a polynomial in x0 etc xi xn g is a homogenous polynomial in this n plus 1 variables alright and know what you do is in wherever xi comes you put 1 ok wherever xi comes you put 1 ok and for the remaining x0 through xn with xi left out which are remaining n you simply substitute t1 through tn that is what this means so this is the homogenization so this map is this is homogenization divided by the right power of xi so this is this map in this direction is given by homogenization and this map is given by the homogenization and therefore so the model of the story is that and you can see that if I now start with this g and homogenize it and divide by degree g I will simply get back this so it will you can very easily see that this is the inverse of this map ok. So it is so what I am trying to say is that you see that this isomorphism of this ui with an geometrical isomorphism if you translate it to commutative algebra it is just this isomorphism between a polynomial ring in n variables ok and polynomial ring in n plus 1 variables localized at one of those variables and taking the degree 0 part ok the significance of this isomorphism which you can write down just from commutative algebra you do not need any geometry for that ok you can write this you can write these maps down just using commutative algebra right. The fact that this isomorphism is an algebraic fact and geometric manifestation of that is that it is giving you an isomorphism of corresponding open subset of projective space with an affine space that is what it means. So this is the commutative algebraic translation of this isomorphism commutative algebraic translation of this geometric isomorphism ok and this is a very very important theme in algebraic geometry whatever you see in terms of geometry you should translate it to commutative algebra and whatever you see in terms of commutative algebra you should try to translate it to geometry ok and I am saying that in this case it is completely you know you can really write it down there is nothing complicated about it one is able to write down all the maps ok. So what I need to say is next which I will do in the next lecture is to tell you that I will use this to tell you that you know affine varieties are the building blocks of all varieties. So I will tell you that any variety can be covered by finitely many open sets which are each themselves isomorphic to you know affine varieties ok. So affine varieties are the building blocks of all varieties ok. So I will do that in the next lecture.