 So, what we saw in the previous lecture was how to calculate the local ring of an affine variety at a point ok. So, the result was that if x is an affine variety and p is a point of x then the local ring of x at p is given by taking the localization of affine co-ordinate ring of x at the maximal ideal inside the affine co-ordinate ring of x Ax this maximal ideal mp has to be you have to localize Ax with respect to mp and the local ring that you get will be can be identified canonically with the local ring of x at p ok. Now what I am going to do now is I am going to try to do the same thing for a projective variety ok and just as in the case of affine variety the local ring is given in terms of localization of the affine co-ordinate ring in the case of projective variety it will be given by suitable localization of the projective or homogenous co-ordinate ring of the projective variety ok. So let me state that ok so here is the theorem let y be a projective variety let p be a point of y ok then o y p the local ring of y at p is isomorphic to the homogenous co-ordinate ring of y localized at mp ok is an isomorphism as k algebra where yeah I need to so I need to localize and also take degree zero part ok. So where sy is the homogenous co-ordinate ring mp is the ideal generated by functions in sy which vanish at p this notation means take the degree zero part of the graded ring ok. So let me explain this so you see the situation is you have y you know y sitting inside as an irreducible closed subset of projective space ok and then you know of course you have the homogenous co-ordinate ring of the projective space which is just the same as affine co-ordinate ring of the affine space above of which the projective space is a quotient of course a puncture affine space and this is going to be k x0 etc xn polynomial ring in n plus 1 variables ok and the homogenous co-ordinate ring of y is just the quotient of this by the ideal of y mind you y is a closed subset of projective space so it is given by a homogenous ideal ok and the homogenous ideal is going to live in this polynomial ring which is a homogenous co-ordinate ring of pn. So this is so sy is just sy of pn mod Iy Iy is simply all those Iy is just you know it is the ideal generated by all those homogenous polynomials which vanish on y ok. It is a homogenous ideal because it is generated by homogenous elements and therefore this quotient because you are taking a graded ring and you are going modulo homogenous ideal the quotient also becomes a graded ring ok so sy is also a graded ring ok and the gradation is simply given reading mod Iy right. Now the situation is how do we get this thing so you know we use the following facts which I said in the previous lecture the first thing is to calculate the local ring you can go to an open subset ok that is one fact. The second fact is of course that if your subset is affine then you have a formula for the local ring so what we are going to do is we are going to do a very simple thing we are going to take you are going to take the point p of y and of course the point p is going to line one of the ui's the ui's being the open subsets of pn there are n plus one of them each one of them isomorphic to an the finance space so p is going to line one of them then what you are going to do is we can just intersect so p will lie in y intersection ui but y intersection ui is an affine variety ok and therefore it is easy to we know by the previous theorem we know what the affine what the local ring is and we also know that the local ring that we get will be the same as the local ring of y at p. So this is what we are going to do ok so here is a so let me draw a diagram so there is a diagram like this so I have the following situation so I have pn I have ui ok. So let me write here p is in y which is contained in pn which is union i equal to 0 to n plus 1 of ui where ui is the set of all points lambda 0 with homogenous coordinates lambda 0 through lambda n such that lambda i is not 0 ok and so this how so p belongs to ui not for some i not as without loss of generality ok so let me so let it be like this so we do the following thing so I have this ui not which is isomorphic via the morphism phi i not with affine n space ok and we saw this in lecture before the last lecture oh no you are right I should start with 0 and end with n yes yes thank you yeah it should be u not through ui there are only n plus 1 of them thank you yeah so it should be i equal to 0 to n ok thank you for that so yeah so you see this we have already seen that this isomorphism is given in terms of rings in the following way so there is a phi i not so this phi i not induces phi i not star which is the which is given by pull back of regular functions that will go from the regular functions on an to so if you have a morphism regular function the target will by composition of this morphism give you a regular function the source ok so I will get regular functions here to regular functions here so I will get this to this but you know of course this is just because this is affine this is just a of an and that is identified with k t1 etc tn these are the coordinates on the an ok and o u is as we showed in a lecture before the last lecture it is just the coordinate ring of tn localized at xi and then you take the degree 0 part so this is so here it is i0 so here also I have to put i0 ok so this is something that we have seen and we have also seen what this map is we also have this we know what this what these maps in these two directions are so the map in this direction the map in this direction is given by well you give me a polynomial f of t1 etc tn then the map in this direction is given by homogenization it is given essentially by homogenization and divide and dividing by the by xi to the power of degree f so it is going to be xi degree f times f of times f of x so this xi0 to the degree f xx0 by xi0 dot dot dot xn by xi0 and then divided by xi0 to the degree f this is how this map is given ok and of course when I write like this I mean forget forget xi0 by xi0 alright so that you have only n of them because there are only n variables that I can fill that I can substitute for ok so this how this map is given and we also know what the map in this direction is that is also something that we have seen if you give me any element here any element here will be of the form g of x0 etc xn this homogenous polynomial of certain degree divided by xi0 to the degree of g this is how this is how an element here will look like ok because you are localizing at xi0 ok which means you are dividing you are inverting powers of xi0 but then you want the degree 0 part which means that the numerator degree and denominator degree should be the same which means that numerator should be first of all homogenous polynomial of certain homogenous degree and the denominator should be the same degree as a numerator so the denominator should be xi to the power of to that power which is equal to the degree of the numerator polynomial. So it is going to be like this and what you are going to get here is just simply you know you substitute xi equal to so you just substitute so you divide you take x0 etc up to xn divided by xi0 ok and omit the xi0 by xi0 term you get n terms and you just no that was a map in this direction so what you do is you just put xi equal to 1 ok and then for the remaining x0 to xn leaving out xi you put t1 through tn that is what you do so the map in this direction is t1 blah blah blah 1 tn where this 1 is in the i0 place ok so that you now get you substitute for n plus 1 variables and of course you essentially you are substituting xi0 is equal to 1 ok so the denominator is going to just vanish I mean it is not going to vanish I mean it is going to be 1 so it is not going to show up here right. So this is how this isomorphism is this is the commutative algebra version of what is happening here in geometry in terms of regular functions everything here is in terms of regular functions. Now what I am having is I have inside pn I have this y which is a close subset of pn irreducible close subset of pn and then I get this yi0 which is just intersection of this is just intersection of y and ui0 ok. So what happens is that yi0 is an irreducible sub variety of ui0 because you see y is irreducible closed y is first of all closed in pn therefore y intersection ui0 is closed in ui0 so this is certainly a close subset of that and then this is yi0 on the other hand is also irreducible why because yi0 is y intersection ui0 which is an open subset of y and an open subset of something that is irreducible continues to be irreducible it is non-empty therefore mind you p the point p is in the point p is here the point p is in ui0 it is in y so the point p belongs here ok. So this is so this is a non-empty open subset so it is irreducible it is non-empty open in y so it is irreducible on the other hand it is also closed in ui0 so this becomes a irreducible close subset of ui0 which under this isomorphism is going to be carried to phi i0 of yi0 which is well I have to give this some name let me call it zi0 right. And so this isomorphism carries this irreducible closed subset into this irreducible closed subset but then this is a variety here ok and so what is going to happen to so this isomorphism of course is diagram commutes you are also a diagram there is no diagram here I am going to draw it in the next board. So this isomorphism can also be you can also write down what this isomorphism is ok so what this isomorphism is well let us write it down it is going to be the following so what is happening here so let me write this part alone there so I have O of an which is a of an which is a t1 tn and I have this I have this which is phi i0 star to OU0 which is spn localize at xi0 take the degree 0 part this is what it is so this is just translation of the of the of this arrow into completely algebra now I am going to do this so you see when you you go to an irreducible closed subset the homogenous coordinate ring is given by a quotient because sy is going to be given by sy is a quotient of s of pn ok and so you know so well maybe I will try to use this part of the board as well let me rub this so I can extend this diagram to the right so you know this pn here so there is ui not sitting as an open subset of pn so this is an open subset and that corresponds to this localization of s of pn so this is a this is a localization followed by well so in fact this is sitting inside the localization as the degree 0 part which comes out which comes as a localization of s of pn ok so this is the localization this sitting inside the 0 part and this part of the diagram corresponds to this ok and what is happening is that y is an irreducible closed subset of pn and it is homogenous coordinate ring is a quotient of pn so what is happening is that you have here you go mod iy you get s of y this homogenous coordinate ring of y so this you know corresponds to this here which is irreducible closed a closed subset corresponds to quotient ok and so this corresponds to this and then you know this thing which is the intersection of y with this open ui not ok that will correspond to something what is that it is going to be well it is going to be it is just you localize so you know this is a very important property in commutative algebra that localization commutes with taking quotients ok localization is what is called localization is exact ok the process of localization is exact ok so localization transforms exact sequences to exact sequences and you know quotient fits into an exact sequence so you take a quotient and then localize is the same as localizing and then taking quotient and these are all standard from the first course in commutative algebra so if you go by that so here is so what I can do is I can take s y and I can localize it at x i not ok so this is the localization alright on the other hand what I can do is I can on the other hand first localize at x i not and then go mod iy localize at x i not ok so this is not only can you localize the ring you can also localize an ideal in the ring in fact you can localize a module ok so iy is an ideal there ok and you can therefore localize it ok this is the localization and then so you know then here what you will get is just s y localize at x i not and the degree 0 part which is which is again it will be a quotient here ok and it is this quotient and this diagram commutes and it is this quotient that corresponds to this irreducible closed set ok so so this irreducible closed subset corresponds to this quotient and this irreducible closed subset corresponds to this quotient ok and on the other hand what is going to happen on this side on this side z i not which is phi i not of y i not is an irreducible closed sub variety of the affine space so it is going to be given by its its coordinate ring is going to be given by the coordinate ring of affine space modulo the ideal of z i not ok so here it is going to be just here this this irreducible closed here this irreducible closed is going to correspond to quotient here and that quotient is just k it is just going to be k t so let me first write it as a a and a of z i not this is what it is going to be and this is just this is a this is just gotten by going mod mod ideal of z i not ok and so which is just so it is going to be identified with k t1 tn modulo ideal of z i not ok and you are going to get an isomorphism you are going to get an isomorphism like this this diagram is going to be commutant this isomorphism is precisely the isomorphism of it is this isomorphism phi i not that is carrying y i not to z i not ok so this is what is going to happen. Now so you know so you know the point you see if you take the point p the point p is here it goes to the point phi i not of p which is a point here ok and you know that because you have because you have an isomorphism of because you have an isomorphism of isomorphism of these two varieties the local ring of this at this point is going to be the same as the local ring of z i not at phi i not of p ok. So what you are going to get is you are going to get local ring of y i not y i not at p is going to be isomorphic via phi i not star to the local ring of z i not at phi p phi i not of p ok you are going to get this because you have seen that isomorphism of varieties is going to identify give rise to an isomorphism of local rings but then you see the local ring of y i not at p is the same as the local ring of y at p because y i not is an open subset of y ok because we have also seen that the local ring to get the local ring at a point you can restrict attention to an open subset which contains a point and y i not is an open subset of y which contains a point because it is y i not is just the intersection of y with this ambient open subset for the projective space. So this is well this is the same as this guy is the same as o y p ok this is what we want we want to calculate this and this is isomorphic to this right and what is this see this is this as we have seen this is the local ring at a point of an affine variety and therefore it is given by this is isomorphic to to you take the affine ring affine coordinate ring of z i not and localize at the maximal ideal corresponding to phi i not of p this is what we have already seen we have seen this is this theorem previous class that previous lecture that the local ring of at a point of an affine variety is just the affine coordinate ring localize at the maximal ideal corresponding to that point so this a so this local ring of z i not at phi i not of p is just the localization of the affine coordinate ring of z i not which is a z i not at the maximal ideal m of phi i not of p corresponding to the point phi i not of p ok and of course you know this map is the localization this is the localization map from the from the ring of coordinate functions to its localization and similarly this will also be the localization map it will be a localization map and the question is that you need to know what localization map this is so what is what is this going to be this is this is going to be this ring sy x i not ok and then you have to further localize it I mean this is take this degree 0 part ok and then further localize it at the at the maximal ideal that you get which is the image of this maximal ideal here ok so it is going to be localize it at phi i not star inverse so this there is going to be a map like this and that is going to pull back this maximal ideal phi i not p it is localizing you have to localize at this ideal ok this is what it is going to be alright if you just follow the commutative algebra this is what is going to be and you have to show that this thing you have to show that this thing is the same as that ok you have to show that this is this is the same as that right. So yeah so one has to really write this down so let me do the following thing so you have see you have this so you have this you have this projective space and you have this y and you have this point p okay suppose p of course p is uhhh p is p is in u i not okay so you know this p has uhhh p has homogenous coordinates lambda not blah blah blah lambda n okay and lambda you know lambda i not is not 0 okay because it is in u i not and now you look at the you look at this uhhh projective space as a quotient of the affine space above I mean the punctured affine space above and then the inverse image of uhhh p uhhh uhhh will be just the line in affine space through the point lambda not lambda n uhhh minus the origin okay this is what you are going to get and you see what is the equation of this line what is the equation of this line uhhh in the in the in the affine space okay. So you see the picture is that you know you have something like this you you have the origin and then you have you have this line uhhh L okay which is this line L and uhhh of course there is a point on it which is given by lambda not etc lambda n and any other point on it on it is given by t lambda not etc t lambda n that is how uhhh uhhh any other point on this line is given by and therefore if you look at the uhhh uhhh if you look at the uhhh ideal of functions that vanish on this line uhhh it will be a homogenous ideal okay uhhh we have already seen that whenever a polynomial vanishes on a line it has to have no constant term and then every homogenous component of that polynomial also has to vanish on that line. So uhhh uhhh the the ideal of this line is simply uhhh gotten by eliminating t from the the equation from the uhhh equation of a general point on this line. So this line is given by you know xi is equal to t lambda i okay this how it is given where t is a parameter okay and if you you know if you if you if you eliminate t uhhh you will you will be writing xi by lambda i is equal to t is equal to xj by lambda j but then you cross multiplied because some lambda i or lambda j may be 0. So you will you will get that the ideal of this line uhhh in in affine in uhhh of of polynomials that vanish on this line is just ideal generated by xi lambda j minus xj lambda i it is just this ideal okay it is just this ideal and uhhh you know uhhh this ideal will also be the uhhh this will also be the ideal the homogenous ideal of the point uhhh of the the image of this ideal uhhh first of all this ideal will be a homogenous ideal of course because it is generated by all these elements which are all homogenous of degree 1 okay and uhhh and what is the 0 set of this homogenous ideal in projective space it is just this point okay so the so this this this will also be equal to the ideal of the point in in projective space okay and uhhh uhhh and therefore uhhh so this ideal corresponds to this point in projective space and uhhh and you know uhhh uhhh what will it correspond to as a point of y it will correspond to the image of this ideal uhhh in sy in sy okay. So you see this ideal is here alright it is homogenous ideal here and uhhh if you take its image here here it will correspond to uhhh the point uhhh p considered as a point of uhhh sy in other words I am saying if you take the image of this ideal here you will get exactly M of p namely the ideal of uhhh the ideal of uhhh the ideal generated by all homogenous functions the homogenous polynomials which vanish at p vanish at the point p okay. So I am just saying that this this is the same as M of p inside uhhh uhhh so uhhh it is quotient is M of p. So uhhh uhhh so this sits inside sy in in set S of pn uhhh and then you have a quotient which is S of y and its image there will be just M of p okay this is what you will get. So so this is what M of p is alright it is just it is just uhhh uhhh it is just the image it is just the quotient of this ideal okay mind you this ideal will contain uhhh uhhh the uhhh uhhh this ideal will contain the ideal of y okay this will contain is y okay and that that that is the reason why mod i y it also defines an ideal alright and uhhh it corresponds to the point p in y alright. Now uhhh so you know if you if you go and try to calculate what uhhh what this quantity is this quantity is just so you know sy uhhh localized at M p0 is going to be uhhh uhhh you know you see what I am doing is uhhh I am taking quotient okay uhhh I am taking quotient by an ideal and then I am uhhh and then I am localizing alright. But I already told you that uhhh localization commutes with quotients so what you can do is this is the same as you take the spn okay localize it with respect to that ideal ideal generated by xi lambda j minus xj lambda i okay and then what you do is uhhh uhhh you uhhh you localize at this and then go mod uhhh uhhh that is why you go i i of y localize at this ideal and then take the degree 0 part okay this is what this is what it is alright. But then the trick is that you know uhhh the point is the following the point is that you know since the point lies in u i not this lambda i not is non-zero lambda i not is non-zero that let me write it down. So, the point here is lambda not lambda n okay uhhh homogeneous coordinates uhhh that is going to the point here which is going to be uhhh lambda not by lambda i not uhhh uhhh lambda n by lambda i not that is going to be the point here okay. Therefore, if you look at this calculation you will see that the maximal ideal corresponding to this point is going to be the maximal ideal of phi i not of p is going to be just uhhh uhhh t uhhh t1 minus lambda not by lambda i not and so on tn minus lambda not by lambda i not uhhh uhhh of course I omit uhhh lambda i not by a lambda i not okay it is this ideal mod this is this ideal mod ideal of z not z i not this is what the maximal ideal here here is okay it is a it is a maximal ideal of this point. The maximal ideal of this point is just the maximal ideal of this point in the affine space modulo this ideal of uhhh z i not okay. So, this point you see this point uhhh corresponds to a point in a in affine space. So, it corresponds to a maximal ideal here that maximal ideal is this okay and that maximal ideal mod i of z i not will give me the maximal ideal in a z i not to which the point corresponds to in z i not okay. So, this is what it is alright and you know if I take each of this elements like t1 minus lambda not by lambda i not okay if I take any such element and if I take its image under this phi i not star a map induced by phi i not star what I will get is you know what this map is this map is just homogenize with respect to x i not okay. So, what I will get is I will get uhhh uhhh uhhh you know uhhh I will simply get x not by x i not uhhh minus lambda not by lambda i not okay uhhh this is what I will get when I when I when I uhhh when I when I apply this map I am supposed to homogenize it with respect to x i not and then divide out by the same power of x i not okay. So, this is what I will get and this is clearly an element of uhhh uhhh and and similarly you know if you take the element tn minus lambda not by uhhh uhhh sorry this is uhhh this should be lambda n, tn minus lambda n by lambda i not that will go to uhhh under this phi i not star it is going to go to x n by x i not minus lambda n by lambda i not this is what it is going to go to because after all the map from here to here is you just send f to homogenization of f divided by uhhh uhhh and then you you remove this power of x i that you put to homogenize it okay. So, if you apply this this is what you will get therefore what this calculation tells you is that it tells you that this uhhh this uhhh ideal okay is actually it is no x i not uhhh uhhh x i not by x i x not by x i not is degree 0 and lambda not by lambda i not is also degree 0 this is a difference of 2 degree 0 elements. So, it is degree 0 it it certainly x i x not by x i not is certainly in the degree 0 part of uhhh this okay. So, you know if you do this what you will get is uhhh uhhh you see all these see all these elements are here these elements they are all here okay and these elements generate a maximal ideal and it is locally and that maximal ideal is precisely this maximal ideal okay at which you have to localize this to get o y i not p. So, you know so phi i not star inverse of m of phi i not of phi is actually the ideal generated by all these guys x not by x i not uhhh minus lambda not by lambda i not dot dot dot x n by x i not minus lambda n by lambda i not of course you know I am able to write all this because lambda i not is non zero that is why I can write all this and therefore this uhhh this ring what you what you get is just so so o y i not comma p is just s y localize at x i not take this degree 0 part and further localize at this guy x i x not by x i not minus lambda not by lambda i not and so on x n minus by x i not minus lambda and by lambda i not this is what it is. So, let me let me write this uhhh so let me indeed write this what I am getting here is this is uhhh uhhh s y localize at m p m p is just the uhhh is this ideal mod i y okay. So, it is just s y uhhh this ideal ideal generated by x i lambda j minus x j lambda i uhhh mod uhhh uhhh i y uhhh then localize at 0 okay. But you see uhhh what I want you to understand is that these two are the same you see what I have done here is I have taken s y I have inverted x i not okay then I have taken the degree 0 part and then I am inverting everything outside this okay. But then you know if you if you uhhh think of it naively think of this x i not as going out and getting multiplied with this okay see if you if you already inverted x i not okay and you want to uhhh uhhh and you want to do this okay it is the same as multiplying this this thing throughout by x i not. So, what I want to say is that this is the same as s y localize at this and then take the degree 0 part which is exactly that okay. So, I am claiming that these two are the same okay you have to just uhhh you have to just satisfy yourself that is true purely it is a purely commutative algebraic thing see what you understand is in s y localize at x i not you know x i not is a unit mind you x i not is invertible therefore you know to localize this here it is just enough to localize this at x i not x not minus you can multiply everything by x i not okay because x i not is after all a unit. So, it is enough to localize this at x instead of localizing it at this ideal it is enough to localize it at x not minus x i not times lambda not by lambda i not that is you can multiply by x i not throughout okay but then localizing uhhh but you know x not minus x i not lambda not by lambda i not is the same as up to a multiple of lambda i not it is just x not lambda i not minus lambda not x i not which is of this form. So, these two are one and the same okay. So, uhhh so, so, so, so let me maybe continue here and say that that you know. So, let me let me erase this line and write that this guy here is you know it is s y x i not uhhh localize at 0 and then further localize that see now I am going to multiply throughout by x i not because x i not is a unit alright. So, it it will become x not. So, in fact I can also multiply by lambda i not because lambda i not is also uhhh uhhh uhhh a nonzero element. So, I multiply throughout by lambda i not x i not so, what I will get is I will get x not lambda i not minus uhhh lambda not x i not blah blah blah and then I will get x n uhhh lambda i not minus uhhh uhhh lambda n x n I will get this. So, the way to do it is you remove the 0 here then you take the 0 okay you can do this. I will uhhh I will leave it to you to uhhh I will leave it to you to check that you know this guy here and this guy here are one and the same okay. It is just a matter of uhhh some commutative algebra okay to check that this is the same as this. So, it is an easy check you will have to check whether this is also equal to this alright. But I leave it to you to check that this is equal to this okay and I will stop here.