 This is the continuation of the previous lecture where we were discussing projective and causing projective varieties. So the last thing that we saw was that if you have projective variety then there are no non-constant global regular functions and as we saw that the only morphisms from projective variety to an affine variety are the constant maps and putting the condition of projectiveness and affineness on a variety reduces it to a point okay. These are all just reflections of the fact that there are no non-constant global regular functions on a projective variety okay and therefore you know one thought of one line of thought from this is that if you still if one still goes by the philosophy of Felix Klein that the geometry of a space is controlled by the functions on it then it is very clear that geometry of a projective variety cannot be controlled by just looking at the global regular functions because there are not any non-constant of global regular functions and therefore you will have to concentrate on regular functions on open subsets of the projective variety and this leads to what is called as a birational geometry and that is studying geometry of open subsets okay and therefore the clue is that if you cannot keep track of the geometry of the projective variety by looking at its global regular functions which are only constants you can still keep track of the geometry of the projective variety by looking at regular functions on various open sets okay and this is covered in by studying sections of line bundles on a projective variety and this is something that you would see in a second course in algebraic geometry okay. So I am just trying to say that the statement of Felix Klein the philosophy of Felix Klein that the geometry of is controlled by the functions still applies in the also in the case of a projective variety only thing is that you it is not enough to just consider global functions because they are only constants you have to consider functions on open subsets okay and the device that attaches which keeps track attaches to every open set the regular functions on that open set is what is called the sheaf of regular functions okay so one needs to keep track of this using sheaf theory right which most probably you would see in a second course in algebraic geometry. So now what I want to do I want to I just want to continue saying the following that so basically you know I want to go back to our earlier argument where in I think couple of lectures ago where I showed that the affine space the projective n space was a union of n plus 1 copies of affine n space so you know so let us recall that. So you see we have so we have we have projective n space and we of course take the homogeneous coordinate ring for the projective n space as k x0 through xn and these xi's are they are the coordinates on the affine space above the projective space so you must remember that there is a there is an an n plus 1 minus the origin of which the projective space is a quotient by this quotient map which is quotient by the equivalence relation that identifies all points on a line passing through the origin in the affine space above and the coordinates here are the xi's and the corresponding coordinates here are called homogeneous coordinates because here only the ratios are the coordinates are ratios and now you look at the you look at the set ui where ui is the projective space minus the 0 set of xi in the projective space notice that each xi is homogeneous of degree 1 therefore it 0 set is a close subset of projective space it is called a hyperplane because it corresponds to the ith coordinate being 0 and it is complement is this opens at ui this consists of all points with homogeneous coordinates with ith with the subscript i coordinate non-zero ok and then I told you that there is a there is a map phi i from an to ui and I would ask one of you to check whether phi i was defined this direction in this direction or the other was this the map phi i or was it the other way around ok so it was the other way around so let me so that I stick to the notation I started with earlier and this map was very simple what you did was you took a point with coordinates lambda not etc lambda n and simply send it to the point lambda not by lambda 1 lambda i dot a dot lambda n by lambda i where of course omit lambda i by lambda i ok so this is the map we defined and we checked it is easy to check that this map is a bijective map ok and I asked you to check that this map is actually a homomorphism of topological spaces and so you know so what I want to say now is that suppose I start with the topology on the projective space given by the quotient topology for this map with the topology above being this one induced by the Zariski topology and affine space then this topology with P n with that topology will induce a topology on the ui and for that topology phi i is a homomorphism with a n with a n having the usual Zariski topology so the way this is interpreted is that if you want to give the topology on P n there are 3 ways of doing it one is you define the closed sets as 0 sets of a bunch of homogeneous polynomials then the other way is by giving the P n the quotient topology where this quotient map the third way is that you make each of the phi i's homomorphism namely you transport on the you give the topology on ui that makes this a homomorphism namely a set here is open or closed if you know leave its image here is open or closed and then these topologies on the individual ui's will agree well on the intersections and therefore they will give define a global topology on the projective space and that will be the same as the quotient topology or the topology for which closed sets are given by 0 sets of bunch of homogeneous polynomials okay so all these are the same so to you know now to make things more clearer in fact what I want to say is that these phi i are not they are not just homomorphism these phi i's are actually isomorphism of varieties okay so fact each phi i each phi i is an isomorphism of varieties okay and aya okay this is not equal I am sorry this should be a subset of P n aq that is is that topology is generated by all these ui if you have a topology now yes because of these maps right now to give a topology on P n using these you mean that is a topology generated by aya yeah you can so it is a fact you can take the topology generated by this or so okay so thank you for pointing out this should have been subset not equal to of course so when I said that there is a topology on each ui that makes phi i a homomorphism and then you all these topologies put together give a topology on P n I mean the following thing a subset of P n is defined to be closed or open if it is intersection with each ui is respectively closed or open that is the definition and for this definition to work a subset of ui intersection uj is closed in ui only if and only if it is closed in uj and is open in ui if and only if it is open in uj that is the compatibility that you will have to check. So the topology on P n that I want to give by gluing the topologies on the ui is the following you define a subset of P n to be closed respectively open if it is intersection with this cover is closed respectively open in particular these ui is themselves will become open because if you take the set ui you will have to check that ui intersection uj is open in uj okay for every j different from i and ui intersection ui is ui and that is of course open in ui so each of the ui is by this definition will automatically become open subsets and then you are only requiring that subset of the ambient space is closed respectively open open if and only if that if it is intersection with respect to this cover is relatively closed or relatively open. So that is the topology that you can get by gluing the topologies on each ui okay now the fact I want to make is that each phi i is an isomorphism varieties alright and how does one do this you will have to show that phi you have to show that phi i is a morphism you have to show that phi i inverse is a morphism so let me do that properly. So what we will do is so on the so we will have to you know what I am trying to say is that the phi i is an isomorphism so I am trying to say that each ui is isomorphic to an but of course every each of these ui is a quasi affine variety because it is an open subset of a projective variety okay so sorry quasi projective variety because it is an open subset of a projective variety okay and I am just saying that these quasi projective varieties are actually affine because they are isomorphic to affine these affine varieties namely I am just saying that this natural open cover of course the projective varieties is actually isomorphic to so many copies of affine space okay and then so you know so the way to check it is that you do the following thing you use this lemma which I have stated earlier the lemma is a map zeta from x to y where x is any variety and y is an affine variety in say an in am with coordinates with coordinate functions x1 or let me put t1 etc tm is a morphism if and only if it pulls back each coordinate function ti to a regular function on x okay so this is the fact that we already seen I mean we saw this in the context of affine varieties or quasi affine varieties but you can go and check that the proof has got nothing to do with the source variety being affine or quasi affine it could have very well been projective or quasi projective. So the idea is if you just given a set theoretic map okay when do you check it is a morphism I mean how do you check it is a morphism when is it a morphism it is a very very important it is a very powerful lemma but it is very easy okay and it is very easy to use the proof is also easy but it is a very powerful lemma because if you want to check something is a morphism you have to check two things you have to check it is continuous number one then you have to check that it pulls back regular functions to regular functions so checking involves two steps but this lemma tells you that you can you know do away with that and do it in one go by just simply saying that if your target is an affine variety and your map pulls back coordinate functions to regular functions then it is a morphism okay. So if you use this lemma it is very clear that it is very easy to see that phi i and phi i inverse a morphisms and since they are inverse is set theoretic inverse of each other it will follow that phi i is an isomorphism okay. So you know so for example if you try to apply it so I try to apply this lemma phi i my source variety is ui which is a quasi projective variety and the target variety is an affine variety is just affine space okay and you know if I take so if I take a coordinate function on this and compose with this what I will just get so for example suppose I take the first coordinate function the first coordinate function will be if I take the if I take the coordinates here as t1 through tn okay then if I composed let us say t1 with phi i I will simply get this x0 through xn homogenous going to x0 by xi okay and if I compose it with tj it will be just this x0 through xn going to xj by xi and xj by xi is of course a regular function on the source projective space because it is a quotient of two homogenous polynomials of the same degree 1 and it is defined on the set I am dividing by xi is correct because I am on ui where xi is not 0 therefore it is very trivial to see that the pull back of the coordinate functions are regular functions so that makes phi i aomorphism alright and similarly if you go in this direction also okay. So you can go in one direction because this is affine and this is any variety okay so for showing that the map is a morphism in the other direction I will need to do something more so let me write this down let me write that down so who for fact let t1 etc tn be coordinate on coordinate functions on the target an then if I look at the pullback then their pullbacks by phi i the functions well t0 by I mean x0 by xi x1 by xi and xn by xi omitting xi by xi okay which are regular on ui therefore by the lemma each phi i is a morphism so phi i is a morphism by the lemma now so what this will tell you is that phi i is a bijective bi continuous morphism but even that is not enough to ensure that phi i is an isomorphism because in the category of varieties the problem is that you can have a bijective map which is a morphism in one direction it could even be continuous in the other direction but it may fail to be a morphism in the other direction so you will have to do something to say that phi i inverse is also a morphism okay and for that of course to check that phi i inverse is a morphism I cannot apply this lemma because the target now is ui and I do not know for sure that ui is an affine variety to start with I do not know that ui is an affine variety it is only a quasi projective variety I cannot apply the lemma to show that phi i inverse is a morphism so I will have to prove phi i inverse is a morphism I will have to do it the hard way namely I have to check it is continuous and then I have to check that it is it pulls back regular functions regular functions now continuity is something that I have already asked you to check but it is probably easy to write down so phi i inverse is a morphism that phi i inverse is a morphism needs to be done needs to be shown cannot use the lemma because the target is ui and I do not know ui is affine but of course once I prove this fact it will follow that ui is isomorphic to an affine variety and therefore ui will become affine but I cannot assume it when I try to prove it okay. So phi i inverse is continuous because you know how will I show that it is continuous so you know I start with to show it is continuous I have to show that if I take a closed set here it is inverse image and phi i inverse which is the same as it is image and phi i is a closed set in an so I will just have to show that phi i is a closed map or an open map okay so because phi i is a closed map equivalently open and why is phi i a closed map the idea is very simple if you take a closed subset here okay that closed subset is the closed subset in the projective space intersect with ui because that is the induced topology. Now since it is a closed if you take its closure in if you take a closed subset here and take its closure in the projective space you will get a closure you will get a closed subset of projective space so it is the zero set of a homogeneous ideal okay now what you do is you take all those elements in the homogeneous ideal and then just de-homogenize them in terms of these coordinates okay you will get an ideal here and it is a zero set of this ideal which is the image of that closed set okay so let me write this down so what you must understand is okay so that gives me an opportunity to say something what is the algebraic translation of this morphism the algebraic translation is of course the algebraic part here is the ring of it is a polynomial ring it is the ring of regular functions on an it is the polynomials in t1 etc tn okay and what is happening here the fact is the regular functions here are just the resist this polynomial ring localized at xi okay that means you invert xi but then after you invert xi it is still degraded okay because you have inverted a homogeneous element and then you take this degree zero part of that that will give you a sub ring and that is the ring of regular functions in u1 okay so the fact is that this map induces an isomorphism of these rings and that is the geometric translation that is the algebraic translation of the fact that this map is an isomorphism of varieties okay so let me explain that so what you do is let f in u i be closed be a closed set so f bar in pn it is the ambient projective space in which u i sitting is closed which means f bar is the zero set in pn of i here homogeneous ideal okay and now what you do is that so you know take define a map from the fine coordinate ring of an which is k t1 etc tn to the set of homogeneous elements in h this is just union of sd d greater than equal to zero and what is this map so this map here is homogenization so this is the homogenization map and what is the homogenization map if you give me a polynomial in t1 etc tn okay then this polynomial in t1 etc tn has a certainly it has a maximal degree okay now what you but this has only n variables so what you do is that you homogenize it namely you make every monomial equal you make the degree of every monomial appearing in this polynomial to be equal to this highest degree by adding as many you know the required power of a new variable okay and that is a new homogenizing variable that you introduce okay and so the so the homogenization is I put xi power degree f of x0 by xi and so on xn by xi okay this is a homogenization process so what you do is that if you do it like this you must you must realize that when I write f of when I plug in for t1 through tn the x0 by xi x1 by xi and so on up to xn by xi of course omitting xi by xi okay then I will get a polynomial which will have the xi's in the denominator I will get a polynomial in the xi by xj I mean rather polynomial in the j by xi where j varies and i is fixed but then by multiplying it by xi to the degree f I cleared all the denominators okay so what you must understand is that this polynomial is an homogenous polynomial of degree equal to degree of f okay so this method this is called homogenization alright and there is a map there is a map like in this direction which is called de-homogenization so there is a map called de-homogenization and that is that is that is very simple you give me a polynomial g x0 etc xn and I do the obvious thing I simply send it to g x0 etc gt0 I mean gt1 etc I put one where I have where I have i then I put tn okay so this is called de-homogenization so what you do is there are these variables x0 through xn okay now for xi you put one okay if you put one for the xi then you get the remaining polynomial is only a polynomial in x0 through xn without xi so it is only n variables and these for these n variables you put t1 through tn in that order okay this is called de-homogenization and the fact is that let me call this as f sub h the subscript h mean being homogenization and let me call this as g sub dh which means the de-homogenization okay instead of giving names to these maps so any f which is a polynomial in the ti's is homogenized to give an fh which is homogenous polynomial in the xi's conversely you start with the homogenous polynomial in the xi's you can de-homogenize it to give g sub dh is in homogenous polynomial not necessarily homogenous polynomial in the ti's okay so you have this map and the reason I want the reason I have this map is because you can now check it is a very simply set theoretic exercise to check that you know if you take if you take if you so I have now see I am trying to check that this map phi i is closed okay so I started with an f which is closed here and I took it is closure in the full space in this f closure is the 0 set of an ideal it is a homogenous ideal so now what I am going to do is I am simply going to take since it is a homogenous ideal is generated by homogenous elements alright so what I am going to do is I am going to take those homogenous elements and I am going to just de-homogenize it okay so essentially what I am doing is see the map is only from the homogenous elements to the homogenous elements so what you do is this ideal breaks up into direct sum of its homogenous pieces and so the ideal is equal to i intersection sd direct sum of i intersection sd and each i intersection sd is a subset here and I take its image there okay so i is direct sum i intersection sd d greater than or equal to 0 this is because i is homogenous okay and consider idh which is definition is summation of i intersection sd see mind you i intersection sd is this i intersection sd is sitting inside sd which is sub of sh so homogenous elements and on sh I have this de-homogenization map okay so what I do is I simply take i intersection sd and I de-homogenize it okay and then I take the sum okay let me for safety take the ideal generated by that alright and the fact is so this bracket is supposed to mean ideal generated by this alright and the fact now is that the image of the image of this f under this phii under this phii is I have to verify it is a closed set and what is a closed set it is just the closed set of this ideal mind you this ideal is an ideal in on the left side it is an ideal in k, t1 etc tn this is an ideal okay this is an ideal there and so it defines a closed set and the fact is what is a closed set that closed set is actually the image under f the image under phi of f so fact is z of idh is equal to phi of f which shows that phi is closed okay phi i of f showing phi i phi i is closed okay and since I have written down all this it is also easy to say why phi i inverse is closed and phi i inverse is also closed is also easy to check in the same way what I will have to do is that which I will have to start with a closed set here and show its image under phi i inverse is a closed set there and what will I do it is very simple a closed set here is given by an ideal it is given by an ideal in t1 etc tn and then I take this ideal in t1 etc tn and simply homogenize it I will get an ideal I will get a homogenous ideal and the 0 set of that homogenous ideal is going to give me a close subset of tn if I intersect it with ui that is going to be the image of this closed set under phi i inverse and that is how you show check that phi i inverse is closed okay so let me write that also so this is something that you can check okay so instead of saying fact I should say check similarly if z in so mind you this z is z this z is in an this z is a 0 set in an so similarly if z in an of j is a closed subset is a closed subset of an for an ideal j in k t1 etc tn then phi i inverse of z in kj is simply z of this z in pn of the homogenization of j which is jh okay is a homogenization of this ideal and it is that intersection with ui showing phi i inverse closed that is phi i is continuous okay. So so once you understand this homogenization de-homogenization you see clearly that is what is going on at the you know algebraic level in fact to be to be more precise what is happening is that you know once this fact has been proved once this fact has been proved what is going to say is that ui is isomorphic to each ui is isomorphic to an is an affine variety therefore you expect that the ring of regular functions on ui is the same as polynomial ring in n variables okay so that should give you some isomer ring isomorphism and you will have a nice ring description here okay that also can be checked so but before I do that let me complete the proof of this fact I needed to check that phi i inverse isomorphism I have already checked that phi i inverse is continuous I have to only check that phi i inverse pulls back regular functions regular functions so it remains to show that phi i inverse pulls back regular functions to regular functions to regular functions and that would make phi i inverse isomorphism and that together with the fact that phi i is a morphism we will tell you that phi i is an isomorphism so what should I do so I have to show that this map pulls back regular functions to regular functions so I should take a regular function here I should compose it with this map and show that the resulting function here is regular so that is also easily written in terms of the de-homogenization map okay. So start with regular function let me make sure that I am not messing up some notation start with a regular function g by h on an open subset of ui namely ui intersection projective space minus 0 set in projective space of h okay so you know a regular function on a quasi projective variety is just a quotient of homogeneous polynomials of the same degree so I start with a regular function g by h on an open subset of this okay where gh are homogeneous of the same degree say g, h are in some degree d part okay d of course I will assume now what you have is you look at so look at then phi i inverse of g by h okay so this is a rather confusing notation so you know I wanted to go back and look at this diagram phi i inverse is in this direction okay my regular function g by h is here it is defined on this I composite phi i inverse okay then I get the pull back so this means this is simply means that you first apply phi i inverse and then apply g by h and if you write it out is the function if you give me a point with coordinates t1, t1 etc tn what I am supposed to do is I am supposed to apply phi i I am supposed to apply phi i inverse to it so the point to which it will go is it is going to be t1 dot dot dot 1 tn this is these are the homogeneous coordinates where this one is in the ith position this is the map which is phi i inverse okay and then what I am going to do is I am going to apply g by h to this okay and so what I will get is I will get g of t1 etc 1 and tn by h of t1 1 tn okay this is what I will get but then what is g of t1 but g of t1 if I put g of t1 etc tn with 1 in the ith t ith position then this is the de-homogenization of g high of g so this is actually so what this tells you is that phi i inverse of g mod h is nothing but the de-homogenization of g by the de-homogenization of h which is of course regular on d of h dh so I am done so I am just saying that I mean this is if you write it down it seems a little complicated but it is not it is quite straight forward I am just saying that if you give me a regular function here which is a quotient of polynomials okay if I pull it back the function that I get here is simply the same quotient of not the same polynomials but the corresponding de-homogenized polynomials. But since it is again a quotient of polynomials it is a regular function on affine space because regular function on affine space is supposed to be defined by locally by a quotient of polynomials so it is pulling back regular functions to phi i inverse is certainly pulling back regular functions regular functions and what is helping is this language of homogenization and de-homogenization okay so that completes the proof of the fact that phi i is an isomorphism okay so that completes the proof of this fact and then let me also tell you once this is done so let me add one line to this okay the pull back the pull back via phi i induces an isomorphism induces a k-algebra isomorphism from the regular functions on ui which is the following you take you take this polynomial ring in n person variables the homogenous coordinate ring of projective space you localize it you localize it at xi, S sub xi is the localization of xi okay and then you take its degree 0 path okay see localizing xi means you are inverting xi so any element in the localization will look like a polynomial by a power of xi now for such a polynomial for such a quotient namely a polynomial by a power of xi you can define degree to be the degree of the numerator polynomial minus the degree of the power minus the power of the xi that is occurring in the denominator with this degree definition you can check that the localization S sub xi is also a graded ring and you take its degree 0 path okay that will be a k-algebra that k-algebra is exactly this that is exactly the ring of regular functions on ui okay and so in other words I am just saying the regular functions on this ui are simply quotients of they are globally given by quotients of polynomials with the numerator polynomial being a homogenous polynomial the denominator polynomial being a power of xi whose power is equal to the degree homogenous degree of the numerator polynomial so the only regular functions on this ui global regular functions are the form g by xi power n g by xi power t okay where t is the degree of g they are the only regular global regular functions on this and that is what you get when you take the this homogenous coordinate ring localize at xi namely invert xi and then take the degree 0 path for the induced gradation okay and this will you will get an isomorphism of this k-algebra with the affine coordinate ring of affine space this is what pullback pullback of regular functions will induce an isomorphism from the ring of regular functions here to the ring of regular functions here okay and that is described like this okay and you can check this you can check this state this equality okay it involves a little bit of diligent checking that the global regular functions on ui is given by this ring and you can also check that this map this map is given by the pulling back of pullback of regular functions from the okay I think my directions are wrong if I want to use phi i then I must pull back regular functions from here to here so my directions are wrong so maybe what I should say is I should rather change this to phi i inverse or if I want to use phi i should change this so let me change it and phi i induces a k-algebra isomorphism from a of an to the regular functions on ui which is s localized at xi and you take the degree 0 path so this is the so the fact is so let me repeat it not only is each phi i an isomorphism but once you know phi i is an isomorphism I am just saying that the variety ui which originally was a quasi projective variety is actually isomorphic to an affine variety because it is isomorphic to this an it is in fact an affine space and you know that whenever you have an isomorphism affine varieties it has to induce an isomorphism of the corresponding regular functions and therefore this isomorphism by pull back should give me an isomorphism regular functions here to the regular functions here the regular functions here are just the polynomials in n variables and the regular functions on ui are you can check exactly the homogeneous localization of this infinity ring with respect to xi which means you localize with respect to xi and take the degree 0 path which means you are taking quotients of quotients of the form homogeneous polynomial of certain degree by xi to the to that degree as a raised as a power okay and you will get an isomorphism of this right and this map is also described in terms of this map and it is inverse as you can check is also described in terms of homogenization and de-homogenization you can write it down right so with that I will stop.