 Ok, so I want to make a few remarks about this proof which is which says that there are no non-constant global regular functions on a projective variety uhhh the there is only one issue uhhh with this proof namely that I need that y intersection ui is non empty for every i ok. And what I want to say is that uhhh given uhhh a y uhhh a closed as a projective sub variety of projective space it could happen that y may not intersect a certain ui ok. And therefore uhhh in that case we reduce to a case we reduce this we can reduce this n ok to come to uhhh a situation where y is embedded in a projective space and it intersects every ui ok. And how we can do that is by realizing that if y does not uhhh hit a certain ui then y is completely continued the complement of that ui. But the complement of that ui is the locus where xi vanishes and the locus where xi vanishes is uhhh a projective space of one dimension less ok. So uhhh y is embedded in a projective space of one dimension less ok. And you can therefore go to you can work in that projective space. Now that projective space will again have an affine cover and you check whether y intersects each of those uhhh members in uhhh in an affine cover the moment it does not intersect one of the members in the affine cover it means that it is again contained in some hyper plane in that smaller projective space which is again a much more smaller projective space. So you can continue this process this process will have to stop at some stage giving rise to a y embedded in a suitably smaller dimensional projective space with the property that y intersection ui is never empty for every i ok. So that is the case that we have treated ok. So in with that in mind this proof covers all the cases ok that is something that you have to notice right ok. So uhhh so this so the the the point is that you know there are no global regular functions on a projective variety alright. And as you can see the the proof needs uhhh the notion of uhhh uhhh it needs the notion of the function field ok without uhhh the notion of a function field you cannot give this proof ok uhhh. So what I am going to do next is I am going to you know uhhh go back to uhhh uhhh I am going to go back to this issue about projective varieties uhhh that uhhh the uhhh uhhh that there is no uhhh uhhh proper analog of the affine coordinate ring for the projective case. So for the projective case uhhh the analog of the affine coordinate ring is the projective homogeneous coordinate ring. But the big deal is that while the affine coordinate ring is an invariant of the affine variety the projective coordinate ring the homogeneous coordinate ring of a projective variety is not an invariant ok. So to explain that what I am going to do is I am going to look at a very simple situation so I am going to do the following thing. So you know I am just going to take I am going to describe what is called the dupil embedding okay so the dupil embedding so the idea is the following so what we are going to do is we are going to take we are going to take projective n space okay and then so you fix d positive integer okay and well I will also fix n so let me write fix n and d integers okay and I am going to take pn and I am going to embed pn into p capital N okay where capital N is n plus d choose d minus 1 okay. So what is this number so you know the well the homogeneous co-ordinate ring of pn is polynomial ring in these n plus 1 variables okay and now what I am going to do is that I am going to look at monomials in these variables but all of degree d is given degree d okay so let m0 etc up to m sub n be the monomials of degree d in x0 through xn okay. So each mi looks like you know x0 to the power of m0 into x1 to the power of m1 and so on xn power m sub n with summation of all the mi is equal to d okay you are just you are just writing out monomials of total degree d okay I mean products of all these xi's powers of xi's such as the powers add up to d and how many such monomials you will get you can check that these many monomials is what you will get okay I mean n plus d choose d is what you will get and therefore you know if you look at the homogeneous co-ordinate ring of this bigger projective space it will be k of y0 etc up to y capital N okay and this n is just this n plus d choose d minus 1 right yeah so the number of monomials will be this much okay and if you will label them as m0 through mn you are actually getting n plus 1 that capital n plus 1 should be this. So this is the number of monomials of degree d in small n variables okay now what you are going to what we are going to do is we are going to define a very nice map so this is the so this is a map from pn to it is a monomial embedding so it is pn to pn p small n to p capital N and we call this as we use the word phi d I mean we use this let us use a symbol phi sub d and what we are going to do is the following thing see any point here is of the form lambda not etc lambda n this is how a point here looks like these are the homogeneous coordinates of a point in projective space okay and what you are going to do is it is very simple you are going to send it to this point where these small n small n plus 1 values are substituted for these Xi's okay in the right order in each of these monomials okay so that you get these this you get capital N plus 1 coordinates which will define a point in whose homogeneous coordinates will define a point in the bigger projective space. You know let me write it as m not of lambda not etc lambda n blah blah blah and you go on up to mn of lambda not etc so this is the map see you have lambda not to lambda n and then you give me a monomial like this you substitute for Xi lambda okay and you do this for each of these capital N plus 1 monomials you will get these capital N plus 1 coordinates take the point and of course these are the monomials in some order okay some order for example you can use lexicographic order on the powers if you want okay you can use the lexicographic order on the variables and their powers right you can have some suitable order after all this is a finite set so you can choose a decent order. So this is the map okay now the question is now the point is the following the point is that this map embeds the smaller projective space as a close sub variety of a bigger projective space okay this is a closed embedding okay that in other words this map is an isomorphism of its onto its image and the image is a irreducible close sub variety of higher bigger projective space okay and so well so what is that so how do you how does one see this so the first thing that one notices is that you know if you take so you know you define whatever is happening here in terms of coordinates can be reinterpreted in terms of commutative algebra in terms of the corresponding homogenous coordinates. So what we have is we have for this we have a map in this direction the opposite direction and mind you this is so I will call this as pd upper star okay because it is going to induce in some sense pull back of regular functions okay via pd okay and this is well k why not why n and this is k x not xn okay and what is this map this map is very simple you just send yi to mi the map is pretty simple after all there are exactly as many yi's as there are mi's okay. So send the corresponding yi to the corresponding mi so what this map does is that it takes mind you this yi is degree 1 alright whereas this mi is degree d it is a degree d monomial okay so what this map does is it takes every degree 1 thing to degree d alright and therefore this is called a graded homomorphism so what it will do is it will take the degree 1 piece to degree d piece okay and it will take the degree r piece to the degree r times d piece okay you know if I instead of yi if I put yi power r it will go to mi power r and the degree of mi power r will be rd. So this is what is called a graded homomorphism mind you these two are graded rings we are not thinking of them as affine coordinates rings of the affine spaces above no we are thinking of them the reason why we put this s is to remind keep reminding ourselves that there is a gradation that is going on and this gradation is very important whenever you are working in the context of projective varieties okay in this case they are projective spaces. So phi d, phi d upper star of you know if I take s r the degree r part p and k this will land inside s dr homogenous part of the small p and k this is what it does it takes so you know this is what is called a graded homomorphism so it is graded we say it is graded of degree d, phi d star is a graded k algebra homomorphism of degree d this is what it means it takes the any degree l piece to the corresponding l times degree l times d piece okay and it is a fact that if you take the kernel of a graded homomorphism it is always a homogenous ideal it will be a graded ideal it will be homogenous ideal. So the fact that these graded homomorphism we will tell you that its kernel of a homomorphism is always an ideal but the fact that it graded we tell you that the kernel is a it is a homogeneous ideal ok. So this will imply that kernel of pd star is a graded ideal is a homogeneous ideal I mean I am just saying that you know if a polynomial here goes to 0 ok I am just saying that if a polynomial here goes to 0 which is the same as saying that polynomial belongs to the kernel then every degree d I mean every homogeneous part of that polynomial also individually has to go to 0 that is because you know if you take the polynomial and take any degree l part that will go to a degree ld part ok and therefore if the image is 0 then it will you will get that each homogeneous part of the polynomial goes to 0 you know so it is very easy to verify that if f is in kernel pd star ok and f is sigma f sub l l equal to 0 to some m ok where f l belongs to sl ok this is the breaking up of f into its various homogeneous parts s sub l is the degree l homogeneous piece of this big pn ok. Then you know pd upper star of f 0 will be pd upper star of f that will be equal to just sigma l equal to 0 to m pd upper star of f l ok but you see this pd upper star of f sub l this will belong to s sub ld of this pn small pn ok and as l changes these are all in different pieces and this so you know these are all pieces in various different degrees and if the sum is 0 then each piece has to be 0 that is what direct sum means so whenever a whenever a polynomial is 0 then every homogeneous piece has to be 0. So and these are the different homogeneous pieces of the image pd upper star of f ok so what this will tell you is that pd upper star of f l is 0 for every l this is what it will tell you and so what you are saying is that if pd upper star kills f then pd upper star kills every homogeneous component of f and that is precisely the same as saying that if f belongs to kernel pd upper star then every homogeneous component of f also belongs to kernel pd upper star that is another way of saying that kernel pd upper star is a homogeneous idea ok. So and you also notice that the kernel is also a prime ideal because the image is a domain after all. So tomorrow the story is that the kernel this kernel of pd upper star is a prime ideal it is a homogeneous prime ideal and therefore it is 0 set will define a projective sub variety a close sub variety of the bigger projective space and the claim is what is that what is that close sub variety of the bigger projective space it is actually the image of this ok. So further kernel pd upper star is prime since s is small and k is a domain target is a domain ok. So the 0 set of kernel pd upper star inside the big projective space is a sub variety. So projective variety it is an irreducible close subset ok and the claim is that image of pd is exactly this the claim is the image of pd pd is exactly z of kernel of pd upper star this is these are these are the claims and the second important claim is pd from this little pn to if you are restricted to that image is an isomorphism of varieties. So this is the claim the claim is that this pd the so called dupal embedding it maps this variety projective space isomorphically on to a close sub variety and that close sub variety is nothing but the 0 I mean it is the 0 set of the kernel of this homomorphism ok and so this is a claim alright this requires a little bit of computation ok. But you know before we try to settle these claims I just want you to look at the case when a small n is 1 ok and d is 2 ok it is a simplest case and the reason why we often look at that case is to tell you that you can have 2 projective varieties which are isomorphic but their homogenous coordinate rings are not isomorphic ok. So take n equal to 1 d equal to 2 ok if you take n equal to 1 d equal to 2 then you have p1 ok you have pd this is p2 into p well here I am going to get so I will get 3 choose 2 which is the same as 3 choose 1 this is 3 minus 1 this is 2 I am going to get 2 so it is p1 inside p2 ok and you know and I have this z of kernel of p2 of a star inside this. So you know I just want you to I just want to verify this for the case n equal I mean n equal to 1 d equal to 2 is the simplest case alright to verify for the general case we will need further calculation but there is a lesson to be learnt even in this simple case ok. So you know how does one ensure this so you know so what is the map the map is lambda lambda 1 goes to the monomials in you are looking at monomials of degree 2 in 2 variables ok. So you know so you have x0 and you have x1 so small n is 2 so you have x0 x1 and then you are looking at monomials of degree 2 in x0 and x1 so you get m0 which is x0 squared you get m1 which is x0 x1 and you get m2 which is x1 squared these are the 3 monomials you will get alright and therefore what is this map you are going to send lambda not lambda 1 to lambda not squared, lambda not lambda 1, lambda 1 squared that is what this simple map is ok and what is this phi 2 star this phi 2 star is the map that is going to go from S of p2 which is just identified with k of y0, y1, y2 to S of p1 which you know is k x0 x1 and you know what this map is you are just going to send y0 to well according to this definition you are going to send y0 to m0 which is x0 squared, y1 will go to x0 x1 and y2 will go to x1 squared ok so are you able to see that and what is the kernel of let us calculate kernel of phi 2 star what is the kernel of phi 2 star well kernel of phi 2 star you know you can see this y0 is going to x0 squared y1 is going to x1 squared I have crammed up everything here so y1 is going to x0 x1 sorry and y2 is going to x1 squared ok you can easily see that y1 squared minus y0 y2 goes to 0 ok. So this contains the ideal generated by y1 squared minus y0 y2 ok because this is going to go to 0 and if some element is seen in an ideal then the ideal generated by that element is also in that and mind you this is a degree 2 element so it is a homogeneous element so the ideal that it generates is a homogeneous ideal it contains in this and you know in fact if you use some if you use some commutative algebra you can show that this is exactly this kernel ok. So see the reason being that this is irreducible y1 squared minus y0 y2 is irreducible polynomial therefore the ideal it generates is a prime ideal ok and this is a homogeneous prime ideal right and the height of this ideal is going to be 1 that is because that is because of you know that is just because of Krull's principle ideal theorem ok. So Krull's principle ideal theorem says that you take an oetherian ring and you take an element in the ring which is neither a 0 device or nor a unit then any minimal prime ideal that contains that element will have height 1. So if you take any minimal prime ideal that contains this it has to be equal to this because it is already prime and therefore its height is 1 and since its height is 1 the 0 set of this will define a hyper surface ok which is a 1 dimensional object ok whereas this will also have the same height ok. So what this should tell you is that the 0 set of y1 squared minus y0 y2 will contain the 0 set of kernel phi 2 star you will have this and this guy is 1 dimensional this is a 1 dimensional object ok this is a 1 dimensional object that is because you know if you just look at the 0 set of this in the affine space over this projective space you will get the affine space over this projective space is 3 dimensional ok and there I am having a single irreducible polynomial therefore its 0 set above and the affine space will give me 1 dimension less sub variety the co-dimension 1 sub variety so I will get a 2 dimensional sub variety but then when I remove the origin and come down to the projective space I will cut down 1 dimension therefore I will get only a 1 dimensional object. So therefore this is a 1 dimensional irreducible closed sub variety of projective space ok and that contains this ok but then mind you this is also an irreducible closed subset space but the point is that this contains phi 2 of p1 by definition it contains phi 2 of p1 because you know you take you see take any take any element see take any point take any point in phi 2 of p1 it is of this form ok and the fact that the way we have defined phi 2 star we will tell you that this is contained inside this ok I think that is probably pretty easy to see just a minute I think I just have to write it down so you know if g is in z of kernel phi 2 star g is say g of so g is a polynomial in 3 variables why not y1 y2 then phi 2 per star of g is 0 that is g of instead of y0 if I put x0 squared x0 x1 x1 squared is 0 this is what it means sorry g is just in kernel phi sorry suppose g is in the kernel ok not in not the 0 set suppose g is in the kernel write g is a polynomial of these 3 variables and then so if it is in the kernel phi 2 upper star of g is 0 so that means g of this is 0 mind you this is a polynomial this is happening in k of x0 x1 that is how this map is defined ok. So this implies that you know if you take a point for a point lambda not squared for a point lambda not comma lambda 1 of this p1 we have you know g of lambda not squared lambda not lambda 1 lambda 1 squared is 0 I mean if g of some polynomial is 0 then for the polynomial whatever variables are substitute that also should result in 0 so this implies so this calculation actually tells you that phi 2 of p1 the image of phi 2 image of p1 and phi 2 has to be in the 0 set of this kernel because everything in the kernel vanishes on this ok. So the so you know now what you must understand is that you know this is already one dimension ok this is also this is already one dimensional mind you phi 2 is topologically phi 2 is an injective in fact topologically you can check phi 2 is a homeomorphism ok phi 2 is injective actually phi 2 is injective it is topologically a homeomorphism ok these are all things that you can check. So since it is a since phi 2 is a homeomorphism phi 2 of p1 which is the image of phi 2 is topologically isomorphic to p1 and p1 is one dimensional therefore this is one dimensional ok so that means this is already one dimensional ok this will be one dimensional but this is also one dimensional and this is a closed subset ok and in a finite dimensional noetherian topological space ok if you have a closed subset of the same dimension then the closed subset has to be everything. In other words if you take if you go to a proper closed subset the dimension has to fall if you go to a closed subset and the dimension does not fall then the closed subset has to be everything ok this happens in a finite dimensional noetherian topological space which is the case with all our varieties ok. So this will tell you that this will tell you that this is equal to this equal to this ok. So the moral of the story is so you know if you use a little bit of topology then you will get that phi 2 of p1 is the same as the 0 set of kernel of phi 2 star and that is the same as 0 set of this ok and in fact what I want to say is that see this phi 2 see this phi 2 is actually a so therefore this phi 2 actually gives an isomorphism of p1 with this 0 set ok which is actually the 0 set of y1 square minus y0 y2 this is the same as the 0 set of y1 square minus y0 y2 these two are equal for dimension basis and this is an isomorphism of this with that ok. And in fact I am saying this is even an isomorphism of varieties this thing is even an isomorphism of varieties and you know the one way to check it is by you know to check that a morphism is has a certain property it is enough to check it on a cover a cover suitable cover of the target and then you pull back that cover to get a cover of the source and restrict the morphism to each of these members of the cover and if this morphism has a particular property for each member of the cover then it has that property throughout for example so you know if I want to show that this map from p1 to the image if I want to show that it is an isomorphism it is enough to show it on a cover and what cover will I use I will use the usual cover of p2 which p2 has cover consisting of u1, u2, u3 I mean u0, u1, u2 which are the three a2's which cover p2 so you know I can make a computation involving that ok. So well you know let us try to do that computation for a moment so you know if you look at so you know I have this so I have this p2 here and I have this suppose I take u0 then I have this so I have this z of kernel p2 star and then I will get z of kernel p2 star intersection u0 so I get this diagram this is this intersecting with u0 so and you know and you know this is well u0 is identified with a2 by phi0 you know and u0 corresponds to the place where y0 is non-zero alright and therefore under this isomorphism this will be identified with a close sub variety of a2 ok and if I take so I have so I have p1 to p2 I have this phi2 if I take phi2 inverse of this z of kernel or I simply take phi2 inverse of u0 it will land inside so you know it will land inside this ok. So I have u0 u1 and u2 the 3 a2's that cover p2 and I am working with u0 and I am taking the inverse image of u0 taking the inverse image of u0 under phi2 they are taking the same as taking inverse image of u0 intersection z kernel phi2 because z kernel phi2 is actually the image of this that we already seen it goes like this the map factors like this ok and mind you said theoretically this map is you must understand that said theoretically this map is injective this map is said theoretically injective and it is said theoretically surjective ok why is it said theoretically injective because you know if you have lambda0 lambda1 you have lambda0 prime lambda1 prime suppose they go to the same thing ok the fact that you have this mixed product here first of all you will get lambda0 squared is equal to lambda0 prime squared and you will get lambda1 squared is lambda1 prime squared so the only problem is that you might get it when you take square roots you could have taken different square roots but then the fact that this lambda0 lambda1 is also equal to lambda0 prime lambda1 prime will ensure that lambda0 lambda1 you started with should be equal to lambda0 prime lambda1 prime you can easily check injectivity and you can get you can check surjectivity because you know any point here is actually here any point here is actually here so the square of the middle coordinate is equal to power of the first and the last coordinate ok therefore you can take the square root of the square root of the first coordinate and you can take a square root of the last coordinate that will give you a point here which will go to that ok if you take the correct square roots therefore it is both injective and surjective is a very easy set theoretic checking only things that you can take square roots because you are in algebraically closed field you always will have square roots of elements ok that is where there you need algebraic closeness of course otherwise for all our arguments about varieties we are we are at the back always assuming algebraic closeness ok so this is very easy to check this is map is bijective it is very easy to check that this map is a homeomorphism ok that is very easy to check so but you know the big deal is I am looking at this map ok what is this map going to do see it is the same map it is lambda0 lambda1 going to after all let me write the map above it is lambda0 colon lambda1 going to lambda0 squared colon lambda0 lambda1 colon lambda1 squared this is the map ok and the moment I go into you know I mean that means that my lambda0 squared is not 0 ok and lambda0 squared is not 0 means that lambda0 itself is not 0 the point is that this is an affine variety ok the point is that this is an affine variety here I need a larger diagram let me draw it again so you know so I have this p1 so I have this p1 I have p2 p2 and I have u0 here which is identified by p0 with a2 and you know what this map is this is you know this map is just something that sends y t0 t1 t2 is mapped to t0 is not 0 on u0 so I will send t1 by t0 t2 by t0 this is what this map is ok this is what this map is and what is this map this map is lambda0 lambda1 going on to lambda0 squared lambda0 lambda1 lambda1 squared that is what p2 is ok and I am taking and mind you this map actually factors through this 0 set of kernel of p2 star inside this ok which I mean let me just call it z so that is easy to write down so and if I intersect with u0 I will get I will get z intersection u0 and that under this isomorphism phi0 will go into z0 ok and if you follow it up so you know if I take phi2 inverse of u0 that is an open set here in p1 alright and if I now write out this map all the way to z0 ok the map will be lambda0 colon lambda1 going to well you see first of all lambda0, lambda1 will go to this ok and this will go to division by lambda0 squared so this is so this map will finally land in if I go all the way to z0 you know the point I will get this I divide this by lambda0 so I will get lambda1 by lambda0, lambda1 squared by lambda0 this is the point I picked ok when I follow it all the way up and mind you since I am taking inverse image of u0 lambda0 squared is not 0 which implies lambda0 is also not 0 which means actually this thing is actually in the u0 corresponding to this projective space this p1 also has a cover by 2 a1 namely u0 and u1 so you know if so I will I will I will I will just call them v0 and v1 so that I do not confuse with the u0 here ok so in fact this is this fellow is inside u so this is inside v0 ok where this v0 in this projective space is opens this is the open which is identified with a1 by a phi0 here is identified to a1 ok and this do not confuse this phi0 with that phi0 ok maybe I can so that you know I let me not call this phi0 I will call this psi0 this is standard identification of v0 with a1 v0 is a locus where the first coordinate x0 does not vanish ok and you see and under this identification this point here will go to what is the point it will go to the point lambda1 by lambda0 ok and therefore finally if you take this and phi2 inverse u0 is an open subset of v0 so its image here will be an open subset of a1 ok and if you finally write out the morphism the morphism is this and mind you is not I should not put square brackets it is round brackets here also it is so these are coordinates in affine space ok they are round brackets and they are coordinates separated by commas ok not you use square brackets only for homogenous coordinates. So you know finally when I write out this for phi2 if I take the inverse image of u0 from phi2 inverse u0 to u0 if I write out the morphism after going to the affine piece here and the affine piece the corresponding affine piece there the morphism is lambda1 by lambda0 going to lambda1 by lambda0, lambda1 squared by lambda0 squared that is the map. So the map is simply t going to t, t squared that is what the that is the map the map is just t going to t, t squared and it is a map from a1 to a2 so when I compute this map phi2 from p1 to p2 ok in on this affine piece on this affine a1 where the first coordinate in this p1 is first homogenous coordinate in this p1 is not 0 and on the affine piece here where the first coordinate piece the first coordinate in this p2 is not 0 which is an a2 then and when I translate this map finally I am getting the map t going to t, t squared is not that a morphism that is a morphism because projection on each coordinate gives me a regular function. So projection on the first coordinate gives me identity projection on the second coordinate gives me t squared ok. So this is a regular function so the moral of the story is that phi2 inverse u0 see if you take so phi2 restricted to phi2 inverse u0 from phi2 inverse u0 to its image which is z intersection u0 is a morphism ok this is a morphism alright. And actually the funny thing is it is actually an isomorphism this is actually an isomorphism because you know if I take this map t going to t, t squared that is an isomorphism because you know if there is the inverse morphism which is given by projection on the first coordinate if I take the map t going to t, t squared ok and if I project on the first coordinate I will get the projection on the first coordinate will send t, t squared back to t ok therefore it will projection on every coordinate is certainly a morphism alright. So the moral of the story is that I have an inverse morphism such that this followed by this is identity ok and where is the inverse morphism define I am looking at the graph of I am actually looking at points of this form and what are these points these are the points of the parabola y equal to x squared this is this is lying see these are all points on z of y minus x squared after all t, t squared is a parametric representation of the parabola and z of y minus x squared is just it is a conic it is a conic it is a plane conic it is a conic in a2 it is a parabola in a2 and all the points t, t squared is if it if you vary t you are simply going to get all the points on this conic only thing is t is not 0 ok no I can have lambda 1 0 so lambda not is not 0 but lambda 1 can be 0 so this t can be 0 as well. So the moral of the story is the moral of the story is that this map is not just a morphism it is actually an isomorphism and in fact an isomorphism ok so what I have proved is phi2 if you take if you restrict to phi2 to phi2 inverse u0 to its image that is an isomorphism I did this for u0 you try it and write it down you do it for u1 and u2 also it will be an isomorphism. So what you have done is you have checked on a cover that phi2 is an isomorphism and therefore it is an isomorphism so phi2 actually use an isomorphism of p1 on to z ok. So I repeat what I have proved is phi2 from phi2 inverse u0 to z intersection u0 is an isomorphism ok similarly I want you to check that phi2 from phi2 inverse u1 to z intersection u1 is an isomorphism you can write it out and phi2 from phi2 inverse u2 to z intersection u2 is also an isomorphism ok and z intersection u1 z intersection u2 z intersection u0 form a cover of z because u0 u1 u2 form a cover of p2 and what you have done is you have checked that morphism is an isomorphism on a cover and therefore it is an because of property of being an isomorphism with a local property ok. So what this will tell you is that phi2 is actually an isomorphism of p1 on to z ok so let me write that here and now I want to state the important point similarly you can check phi2 in restricted to phi2 inverse of ui from phi2 inverse of ui to z intersection ui is an isomorphism for i equal to 1, 2 thus phi2 from p1 to z is an isomorphism because z is the union of z intersection u0 z intersection u1 and z intersection u2 ok. So this is why you get that the image of p1 here is this and it is an isomorphism on to the image but now comes the big deal this is also a p1 ok and that is actually a twisted p1 you know if you want to think of p1 as a projective line ok then the image here you have seen locally it is a conic it is a it has been twisted into a parallel parabola. So what has happened is this line p1 has been mapped isomorphically on to a conic locally ok but the beautiful thing is even though these two are isomorphic projective varieties if you calculate their coordinate rings the coordinate rings are the homogenous coordinate rings are not isomorphic ok that gives us the lesson that the homogenous coordinate ring of a projective variety is not as well behaved as the affine coordinate ring of an affine variety it is not an invariant ok. So you know if you calculate the homogenous coordinate ring of p1 what you will get is you know this is just k x0 x1 ok and what is the homogenous coordinate ring of z it is the homogenous coordinate ring of the target p2 model of the ideal of z and what is that that is just k y0 y1 y2 model of what is the ideal of this z it is just the ideal generated by y2 minus y0 y2 ok and these two are not isomorphic even though so the see this is a this is a polynomial ring in two variables alright and this is a polynomial ring in three variables and you are going model over degree 2 homogenous polynomial these two these two rings are not isomorphic you can well you can try to prove that as an exercise that this ring cannot be isomorphic to this ring because actually you know this ring will have the full universal property for a polynomial ring in two variables this would not that is how you check that this is not isomorphic this cannot be isomorphic to this because a polynomial ring has universal property this is a polynomial ring in two free variables so it has universal property whereas that property will not hold you cannot find an analog of that property for this ring and that will show you that this cannot be isomorphic to that ok. So I want you to check as a as an exercise that these two rings are non isomorphic ok as k algebras they are not isomorphic and therefore the homogenous coordinate rings of p1 and z are different but yet p1 and z are isomorphic as projective varieties so that gives us the lesson that the homogenous coordinate ring of a projective variety is highly dependent on the embedding ok if you take the if you so the embedding the embedding decides how your homogenous coordinate ring is going to look like and the homogenous coordinate ring can change even though your variety up to isomorphism does not change. So this is the problem that for projective varieties you cannot keep track of them just by looking at their homogenous coordinate ring. So to sum up this and the previous lecture what I want to tell you is that unfortunately with projective varieties neither can you use global regular functions the ring of global regular functions for the simple reason it is just constants nor can you use the homogenous coordinate ring because it is not invariant and this is really in sharp contrast with affine varieties where the ring of global regular functions is same as affine coordinate ring and that is an invariant you can completely recover the affine variety from its ring of regular functions is same as its affine coordinate ring ok but you cannot do that with projective variety. So this necessitates that you this problem that you have in which is of interest in higher algebraic geometry that you take a projective variety to study it you have to study all its embeddings into various projective spaces and look at the geometric properties of that embedding of those embeddings and try to extract more information from those embeddings just looking at the homogenous coordinate rings with respect to the embeddings will not help ok you need to extract some more information ok. So that is and usually this kind of study is accomplished by studying so called line bundles and linear systems on which are connected with embeddings namely morphisms of a variety into projective space and this is usually done in a second course in algebraic geometry ok. Thanks for watching and stop.