 Let us continue with what we are doing in the previous lecture I am trying to prove this theorem that if X is an affine variety that the function field of X is just the quotient field of the affine coordinate ring of X which is the same as the ring of regular functions on X and if Y is a projective variety then the quotient the function field of Y is actually the degree 0 part of the homogenous localization at 0 ok and of course Sy localize at 0 itself will have a being homogenous localization will have a gradation and it is a degree 0 part of that if you take the degree 0 part you get field ok. Now I will explain more about this soon so but this is the point the point is that we define something in geometric ok and then we try to capture it using commutative algebra I mean that is the translation from algebraic geometry commutative algebra and that and what this is what this the theorem tells you what this translation means for a function field ok if it is an affine variety then the function field is the quotient field of the affine coordinate ring which is a projective variety then it is given by this which also produces a field ok. So let us try to prove that and then I can use this to go and show also that there are no non constant regular functions on a projective variety ok. So here is a proof so we start with x affine variety x is irreducible closed in some n and well ax is you have ax is just by definition this is affine coordinate ring of x is just the ring of polynomials in n variables namely the polynomial functions on this affine space restricted to x ok. So this is just this is by definition a of an divided by the ideal of x which is just and a of an is just can be identified polynomials in n variables over k the ring of polynomials in n variables over k these variables x i's are the coordinate functions on the affine space the n coordinate functions and here is your ax and by the way since x is affine this is the same as o x these two are one and the same ok. And what I am trying to show is I am trying to show that kx is the same as the quotient field of ax alright so yeah so you know we already have that o x sits inside kx this is something that we have already seen ok. In fact any regular function is also a rational function after all kx is supposed to be consist is supposed to be represented by regular functions on open sets and o x is regular functions defined on all of x and all of x is also an open subset of x ok therefore o x is a subring of kx ok this is something that we have already seen we have I have mentioned it here o x sits inside any local ring and that is further sits inside kx which is the function field of x and on the other hand o x is the same as ax ok and ax has its quotient field q of ax ok any integral domain sits inside its quotient field which is the quotient field is same as field of fractions any integral domain sits inside its quotient field ok and there is a property which says that you know the quotient field is smallest field in which the integral the quotient field of an integral domain is the smallest field in which the integral domain sits ok that is also called the universal property of the quotient field or field of fractions and therefore since the, since q ax is a quotient field of ax which is the same as o x it is a smallest field which contains o x and since kx is a field which contains o x ok this has to be contained in that therefore you know you get you get a unique ring homomorphism from here into this ok. So this ring homomorphism is a unique ring homomorphism which just expresses the fact that an integral domain sits inside its field of fractions or quotient field and the quotient field is smallest field which contains that integral domain. So if some other field contains that integral domain then this quotient there is an isomorphic copy of this quotient field inside that field ok and that is what is given by the image of this map which is an injective map. And in fact it is pretty easy to write out what this map is see the what is the quotient field of Ax you see Ax is given by polynomials f and g Ax is just O x, O x is regular functions but here the regular functions are just restrictions of polynomials ok. So you take an element of Ax it is a polynomial model of this ideal ok so you can write it as f bar ok. So if I take a polynomial in n variables namely f then there is this quotient which is going to model of the ideal of x and f goes to f bar which is an element here ok and of course f bar is the coset f plus i x ok. So you have this f bar so this quotient field will consist of things like this there will be things of the form f bar by g bar ok. So because of the field of fractions of an integral domain is consist of all fractions ok quotients of two elements of the integral domain with the denominator element not being 0 ok so the g bar cannot be 0. So this is how an element here will look like ok and what is the element to which you will send it after all f bar by g bar if you think of it as after all as a function this is just f by g ok after all f bar if what is f bar in Ax what does it mean it is simply restriction of the polynomial f which is which defines a function on this affine space to the subset x. So geometrically f bar is actually f restricted to x ok f bar is what you get algebraic it is an it is a coset it is an it is read of it is reading f modulo ix but geometrically what is it it is simply restricting the polynomial f to this closed subset x the subset x. So you know so if you take f bar by g bar you are just actually restricting f bar you are you are just evaluating f by g and where this f by g makes sense where g does not where g is does not vanish ok and where g does not vanish is an open set it is an open set in affine space. So if it intersects x you will get an open subset of x ok so the moral of the story is that this will define this x intersection dg, f by g. So this is I have this element which makes sense here so what you must understand is that you see dg is the basic open set in affine space where the polynomial g does not vanish ok dg is just the complement of z of g which is a zero set of g ok the zero set of g is a zero set set of points where g vanishes ok and mind you that is you know that is a hypersurface ok of course g g need not be irreducible but even if you factor g into its irreducible factors then this z g will be union of those z of those g i's which where the g i's are irreducible factors of g and each z g i will be a hypersurface ok which means a co-dimension 1 sub variety ok that is n-1 dimensional close sub variety of an ok. So the z g will be a union of hypersurface and its complement is dg which is a basic open set mind you this is a basic open set in the affine space and this is itself isomorphic to an affine variety ok by the Rabinovich trick this is isomorphic to a close subset of an plus 1 ok. So this dg is an open subset and x intersection dg is an open subset of x and there f by g makes sense right and so this is how you define this map ok and so that is how this map is defined this is how you will define this map naturally you can check that this here I have defined this map in a very geometric way by thinking of everything as functions but then I told you this map also comes because of the universal property of the field of fractions of an integral domain ok you can check that there also the definition is the same ok. So this map that you get either geometrically or via commutative algebra it is one in the same map right and so all this tells you that the quotient field of Ax is certainly a sub field of kx ok what I am to prove is that they are equal ok. So what I will have to show is that I have to show this map is surjective alright and how do I show that this surjective well take so you know let me call this map as something ok let me call this alpha ok. So alpha is an injective k-algebra homomorphism k-algebra homomorphism well actually it is more actually you know the quotient field of Ax contains k quotient field of Ax is the field it contains k the the the scalars they are all thought of as constant functions mind you k is sitting inside here k is k every element of k defines a regular function namely constant functions. So it is the same k here so the constant functions are there and this is a field which contains this smaller field so this is the field extension and this is again another field extension ok kx is a bigger field extension in fact we want to show that these two fields are the same ok why is alpha surjective because you know how does an element in kx look like in any element of kx is of the form well is of the form u, f, u, phi where phi is in O u and u inside x is open non-empty I mean this is how we define kx it is equivalence classes of pairs consisting of an open set and a regular function on that open set ok and this open set has to be taken to be an open set inside x right and but then you see a little bit of fins part will tell you that this is the image of an element from the quotient field of Ax and what is that element is very easy to guess by just concentrating attention at a single point so you know if x belongs small x is a point of u inside x ok then phi looks like f by g right with x belonging to dg I mean this is how a regular function is defined a regular function is something that locally surrounding each point looks like a quotient of polynomials so phi looks like f by g means the function phi from u to k can be identified with the function f by g which is also defined on a open neighbourhood of the point x to k and of course that open neighbourhood should be a neighbourhood where g does not vanish so it has to be in dg ok of course x is in dg intersection x which is an open neighbourhood of small x in capital X right this is just the definition of regular function ok but then you have f bar by g bar this element if you take f bar by g bar that is certainly an element in the quotient field of Ax certainly and alpha of f bar by g bar will be you will get this as I have defined this map alpha it is just going to be x it is going to be x intersection dg, f by g this is what I am going to get ok but then notice that this is the same as well and of course you know but this is the same as w f by g these two are the same where w is the open neighbourhood of small x in x intersection dg where phi is equal to f by g see after all phi looks like f by g locally at every point so I am taking only one point it is enough to take just one point to call that point as small x so there for this point small x I have an open neighbourhood w ok which is in x intersection dg ok and there on the w phi is the same as f by g ok that means you know so f by g make sense on w phi also make sense on w and so this is the same as w, phi ok and that is the same as your u, phi because that is the equation this is w, phi restricted to w so I should write phi restricted to w the restrictions are obvious let me not write it ok so in other words so this implies alpha is this alpha is subject and you have done thus the quotient field of x is the same as well isomorphism is alpha as I have defined it it is the quotient field of the function field of x kx is the same as the quotient field of ax ok which is the same as quotient field of ox because ox is the same as ax ok so it is very simple to get the function field of a affine variety simply take it is co-ordinating and take the quotient field that is it ok that proves the first part ok now you are going to prove the second part which is about the which is the projective case right so how do you prove that so let you start with y the radius will be closed in Pn, Pm ok y is a projective variety and I am trying to now calculate the function field of y alright and well then you know of course the the the projective space is a quotient of the punctured affine space above and you know the homogenous co-ordinate ring of the whole projective space is the same as the affine co-ordinate ring of the affine space above and this is just polynomials in M plus 1 variables and the the homogenous co-ordinate ring of y is just this modulo ideal of y where ideal of y is the is all those homogenous polynomials which vanish on y so this i h I will put a subscript h to just remind you that it is a homogenous ideal it is ideal generated by homogenous elements namely it is generated by all those homogenous polynomials in these M plus 1 variables which vanish throughout the locus y ok. Now we have seen in earlier lectures that you know that the projective space has an affine cover consisting of M plus 1 open subsets which are each isomorphic to affine M space ok. So you know we have P M k is union of U i i equal to 0 to M U i is just the complement of the 0 set of X i ok this is 0 set in this is a projective 0 set this is 0 set in P M each X i is homogenous polynomial of degree 1 it is 0 set will define a close subset of projective space it is called the hyper plane projective hyper it is called the projective hyper hyper plane defined by X i vanishing of the X i and it is complement is this U i ok and these U i's and these U i's are all open sets and we have seen that there are isomorphisms phi i from U i to A M ok there isomorphisms like this and you know how these isomorphisms are written U i corresponds to all those points with projective homogenous coordinates with ith coordinate non-zero. So what you do is if you take a coordinate like this which is written as lambda not through lambda M with colons to indicate that it is a common that it can be scaled by a common ratio these are the homogenous coordinates you simply send it to lambda not by lambda i lambda M by lambda i where you omit lambda i by lambda i this is the map this is the map that is defined theoretically and we proved that this map is an isomorphism of varieties and under the isomorphism of varieties this translates to this is a geometric map and it translates to commutative algebra to an isomorphism of the affine coordinate rings namely A of A and K so this is my pullback is pullback regular functions and give me a regular function here you compose with this you get a regular function here that is a map like this the opposite direction and this is the K algebra isomorphism this is of course isomorphism varieties okay and here I will get O of U i and we proved that what this O of U i is actually it is the coordinate ring of Pn Pm localized at Xi and then you take degree 0 part okay so we proved this so we proved this in an earlier lecture right and now this is the picture with the whole projective space now you rewrite the picture that corresponds to this closed subset Y okay so what you will get is if you intersect with Y you will get Y also as a union of I equal to 0 to M Y intersection U i okay so Y is covered by this Y intersection U i and each Y intersection U i is an irreducible closed subset of U i and under the isomorphism of P i it will give you an irreducible closed subset of affine space so it will correspond to an affine variety. So this just expresses the projective variety as a union of at most M plus 1 affine varieties so you know we use this to show that any projective variety is a union of affine varieties and we later on proved that any quasi affine variety is also a union of affine varieties and therefore we proved that any variety is a finite union of affine varieties so the affine varieties are the building blocks of all varieties anyway so you have this and what happens is that what happens to this isomorphism this Y intersection U i you will have an isomorphism P i the P i will give you an isomorphism with P i of Y intersection U i which is a variety there and what will happen is that the and if you this if we will translate if you go to combative algebra this will translate to A of affine coordinate ring of P i of Y intersection U i which is an affine sub variety P i of Y intersection U i is a irreducible closed sub variety of A m ok and again this is P i star which is pull back of regular functions by P i ok this is also a K algebra isomorphism and what you are going to get here is the ring of regular functions of Y on Y intersection U i and you know we proved that this is this can in fact we identified with S of Y you localize at X i and then take the degree 0 part ok. So it is this thing when you intersect with Y you get this alright and now here is a point the point is well you know K Y is the same as K of Y intersection U i if Y intersection U i is not 0 is not empty ok because I told you that the function field is not going to change if you go to a non-empty open subset Y intersection U i is non-empty open subset of Y I mean of course I am choosing an i such that Y intersection U i is non-empty open subset of Y if some Y intersection U j is empty I do not choose that j ok that is certainly at least one i such that Y intersects U i ok and you take that Y intersection U i that is an open subset of Y non-empty open subset of Y it has the same function field but then K of Y intersection U i is the same as K of Phi i of Y intersection U i this is equal to this because Y intersection U i and Phi of Y intersection U i are isomorphic because Phi i is an isomorphic but then so this is the function field of Phi i of Y intersection U i Phi i of Y intersection U i is Phi i of Y intersection U i is a sub variety of A m ok and so it is affine it is an affine variety and we have just now proved we have just now proved here the first part that if you have an affine variety then the function field is given by taking the quotient field of the field of a quotient field of the ring of regular functions in the affine coordinate ring so this is the same as quotient field of the affine coordinate ring of Phi i of Y intersection U i ok and of course here you know when I put equality maybe I should put here isomorphism because I am using the fact that if two varieties isomorphic then the function fields are isomorphic so I have this isomorphism between Y intersection U i and Phi i of Y intersection U i so these two are isomorphic varieties isomorphic varieties have isomorphic function fields ok so that is the fact that is pretty easy to understand I did not state it but let me try to write that down here suppose you have X so let me put you some other notation so if X1 is a variety and X2 is a variety and Phi is an isomorphism of varieties then I have Phi star the pull back of functions will induce isomorphism of K X2 with K X1 ok so if two varieties are isomorphic then their function fields are of course isomorphic the idea is very simple because you see if you give me an element of K X2 it is going to be a rational it is going to be it is a rational function so it is a regular function and open subset of X2 and if you pull it back you will get a regular function on the inverse image of that open subset in X1 and that is going to define an element of K X1 and you can check that this is an isomorphism it is a trivial check that if two varieties are isomorphic then their function fields are isomorphic ok so an isomorphism of varieties induces an isomorphism of function fields in the opposite direction always the isomorphism at the level of regular functions or at the level of local rings or at the level of function fields ok all isomorphisms that involve functions either locally or globally they are all induced by pull back that is something that you should not forget ok. So isomorphic varieties have isomorphic function fields and since y intersection ui is isomorphic to phi i of y intersection ui they have isomorphic function fields but the quotient field of phi i of y intersection ui is the quotient field of its affine co-ordinate ring ok that is what we have seen in the first part and then but this is the same as well you know so this is the same as quotient field of this isomorphic to the quotient field of o of y intersection ui because again if two rings are isomorphic if two integral domains are isomorphic then their quotient field isomorphic so the quotient field of this isisomorphic to the quotient field of this because these two integral domain is isomorphic but I already but I have already identified o of y intersection ui with sy localize it, so this is just the portion field of sy localize at xi take this degree 0 part. And now I want you to check it is a very simple check in combative algebra that this is the same as simply sy localize at 0 and take the degree 0 part ok. So it is the last equality that has to be checked and it is kind of it is very easy to check ok because you know so this is the only thing that needs to be checked, so you must understand what I am doing here, here I am using the fact that if two varieties isomorphic their function field isomorphic and here I am using the fact that if two rings are isomorphic two integral domains are isomorphic then the quotient fields are isomorphic and then I come to this ok. And yeah so how does one prove this last equality that is also little bit of combative algebra, so let me rub this part so that I can continue. So what is sy localize at xi is just elements of the form f by xi to the power of r equivalence classes like this where f is in f is a polynomial in these m plus 1 variables and r is greater than equal to 0. I mean this is what it is and this square bracket is equivalence classes ok this is what the localization is sy localize at xy is just invert xi ok invert xi means you are allowing denominators with positive integral powers of xi right and what is sy xy xy xi localize at 0 is well it is such things which with homogeneous degree 0 which means that you know the numerator should also be homogenous denominator should also be homogenous and they should be homogenous of the same degree so that when you take the difference of degrees it is 0 ok. So this is this will consist of things of the form well sums of finite sums finite sums and of course also finite products for that matter well I could simply write it instead of saying all that I could simply write it as f by xi to the power of r such that f is homogenous of degree r this is a degree 0 part right and of course the degree 0 part sits inside this ok and what about this so that is and you know if I take the quotient field of this ok this is an integral domain alright and I take its quotient field so if I take quotient field of sy xy at 0 this is what this will sit inside and this will consist of quotients of things like this so it will be something like f by xi power r divided by some g by xi power s it will be quotients like this where f, g homogeneous of degree r and s respectively this is what it will be alright if I take the quotient field of that right. And well and what you can understand is that you know I can identify I can identify an element like this with the element f times xi power s by g times xi power r ok this if you just divide these two you know very nively this is what you will get ok and the numerator will have degree r plus s denominator will have degree r plus s so this is the quotient of homogeneous polynomials of the same degree ok so what you are going to get is you are going to get an element here ok so see this belongs to sy localize at 0 degree 0 part I will tell you what this means see sy localize at 0 is equal to is gotten by inverting all nonzero homogeneous polynomials ok this is the notation is ring localize at a prime ideal ok so the point what I want you to notice is that this is homogenous localization see normally when you localize a ring at a prime ideal you are supposed to invert everything outside that prime ideal ok but here it is homogenous localization what you are doing is you are not you are inverting every nonzero polynomial but you are inverting only nonzero homogeneous polynomials the reason why you are inverting only homogenous polynomials is because only then this localization will get a gradation ok if I mind you this sits inside sy localized at 0 so you know this when I write like this this is the usual localization 0 is a prime ideal and ring localize at a prime ideal is invert everything outside that prime ideal so this is actually the quotient field of sy this is quotient field of sy this is very big this is very big because this consists of quotients where in the denominator you have put any polynomial which is not 0 in fact you are inverting even nonzero homogeneous polynomials even now I mean sorry you are in fact you are inverting even non homogeneous polynomials alright but this is smaller you are inverting only homogeneous nonzero polynomials only those homogeneous polynomials which are nonzero you are inverting them so this gives you a subring of this which is the quotient field ok but this is graded because the gradation can be defined as degree of the numerator minus degree of the denominator because denominator consists only a homogeneous polynomial which is nonzero because you are here you are inverting only the nonzero homogeneous polynomials and in this if you take the degree 0 part that is what this is so this consists of exactly elements like this ok this consists of exactly elements which are given by numerator homogeneous polynomial denominator also homogeneous polynomial they are both of the same degree ok and therefore you can see that this is certainly sitting inside this homogeneous localization degree 0 part to which this belongs and that is a map you send this guy to this guy alright and you can check that this map is injective alright what we are trying to say is that this is also surjective that is what we want to prove we want to show that this there is an isomorphism between this and this ok I have written equality but in most cases the equality is just isomorphism with proper identification an isomorphism can be treated as an equality I mean if you think of this function as the same as this ok as a function then you get equality but if you prefer to write it like this with all these weird brackets here and another weird bracket here if you want ok then you can you can think of it as a formal isomorphism ok but if you just think of them as functions they is just equality right. So this is sitting inside this and conversely you know you can check that if you give me any element here it comes from an element here ok so the only fact that needs to be proved is that this is surjective alright and how does one show that it is very very simple conversely you know see if you want let me call this map as alpha sub i let me call this map as alpha sub i so I just want to say that alpha sub i is well an injective kalgebra homomorphism of course you know see the fact is that this is a non-zero homomorphism from a field and you know a non-zero homomorphism from a field is always injective. So you do not have to worry about verifying the injectivity because the sources are field alright being the quotient field of this integral domain alright I have to only worry about the surjectivity so how do you get surjectivity surjectivity is also pretty easy well take an element here how does an element here look like it will look like a quotient of 2 homogenous polynomials of the same degree alright. So take f by g take a quotient in this which means that f and g are homogeneous of the same degree and of course g is non-zero because whatever you put in the denominator whatever you invert is non-zero so any element here looks like this and I want to tell you that this element actually comes from an element here ok and that is perhaps pretty easy you can easily see that I just have to divide by the same power of xi as the degree of f which is the same as degree of g so it is obvious so you can see that so it is surjective obviously so clearly this element f by xi degree f degree of f divided by g by xi degree g this element this fellow is in sy localize at xi degree 0 and its image under alpha i is f by g ok. So that implies that alpha i is an isomorphism surjective hence an isomorphism ok as functions if you consider everything as functions then these two are not just isomorphic they can even treat them as equal ok so this proves the other fact that ky is nothing but take the homogenous coordinate ring of y you localize at you take homogenous localization at 0 and then take the degree 0 part this is an isomorphism ok so that proves the theorem ok. So it is very clear that you can you know how to write commutative algebraically what the function field is for an affine variety for alpha projective variety it helps you with calculations ok and there is another fact that that comes with there is another fact that comes with that comes with in connection with these two diagrams the fact is the following the fact is that of course the function field is an extension of the base field small k ok and the fact is that the transcendence degree of the function field O of a variety over small k will give you the dimension of the variety ok so that is also true. So you know so here is a so that is a very well corollary corollary is if x is affine dimension of x this is topological dimension of x is a dimension cruel of ax this is also the dimension cruel of ax at mx by x of the local ring this is also equal to a transcendence degree over small k of the quotient field of ax which is the same as transcendence degree over small k of the function field of x and mind you mind you that this guy is the same as this isomorphic to O x small x and this guy is isomorphic to O x so this is one corollary so you know you have so many formulas for the dimension of an affine variety it is either the topological dimension of x or it is the cruel dimension of ax if the ring of polynomials recited to x ok or it is also the cruel dimension of the ring O x which is a k algebra ok it is also the cruel dimension of ax localized at mx because you know ax localized at mx is the local ring of capital X at small x ok and this is the same as the transcendence degree of the quotient field of this and that is the same as the transcendence degree over k of 4 small k of capital Kx so this is any this is one important corollary and if you have and you know the point is that if it is projective also all these statements are correct except that you do not bring in the global regular functions because as I told you we are going to prove that for a projective variety all the global regular functions are all the only global regular functions are all constants ok so if if if y is projective then you know I will get I will have to forget this ok this is the only thing that I have to throw out but everything else everything else can be written so let me write that out if y is projective then also you will see that dimension y topological dimension of y is the same as the cruel dimension of the the local ring of y at this point small y and there is also the transcendence degree over small k of the function field of this is what will happen ok and so why is this a corollary of all that we have seen I have to explain that and the see we have already proved the topological dimension is the same as cruel dimension of the affine coordinate ring for an affine variety and but you know but then you know if you take the cruel dimension of the local ring ok then you see the cruel dimension of the local ring is just the cruel dimension of this ok but the cruel dimension of this is actually the same as the height of this maximal ideal ok see because when you localize at a prime ideal in the local ring your only the only prime ideals that survive are the prime ideals which are contained in the in the localized prime ideal in the in the prime ideal that you are localizing ok. So what is one thing in community algebra that you should remember about localization if you take a ring if you take a commutative ring A and if P is a prime ideal if you go to AP what are the prime ideals in AP the prime ideals in AP are precisely the prime ideals in A which are contained in P by going to AP you are you are throwing out you are removing all the prime ideals which are outside P which have elements outside P. So by going to AP you are actually only focusing attention on prime ideals which contain P so you know the cruel dimension of this thing this is a local ring so the cruel dimension will be it has only one maximal ideal and you know cruel dimension of a ring is just supremum of the heights of all its prime ideals so it will be the height it will be here the biggest ideal is this maximal ideal so the cruel dimension of this ring is the same as the cruel dimension of the height of this maximal ideal. But then this local ring model of this maximal ideal is the is just the base field K ok this Ax localize at Mx modulo Mx is just K ok and dimension formula will tell you the dimension of K is 0 therefore the dimension of Ax mod therefore the dimension of Ax localize at Mx is the same as the height of Mx ok and but but then that height will be equal to the cruel dimension of Ax ok and that is the reason why you get this equality ok and this equality comes because any integral domain and it is that if you have a so this equality comes because of the fact that you know so this also comes from a formula that I have stated earlier if you have a finitely generated K algebra which is an integral domain then the transcendence degree of the quotient field of that integral domain is the same as the cruel dimension of that of that algebra ok so so so if you apply that to Ax you will get that this is equal to this ok because Ax is a finitely generated K algebra which is an integral domain ok and it is so it is transcendence degree over K has to be equal to the cruel dimension of Ax ok so that gives this equality alright so you do not get this is equal to this rather you get this is equal to this and therefore this is equal to this because these two are equal ok and of course we have proved Q of Ax is same as Kx so this is also equal to the transcendence degree over K of Kx ok and the same kind of argument will give you will yield these equalities in the affine case ok and you have to only remember that when you if you want to do this for y it is enough to do it for yi where yi is an affine piece of y which is gotten by intersecting with ui ok and if you do it with a instead of if to do this computation it is enough to do this computation with y replaced by yi intersection ui which is yi if you want ok and then that case is already covered here therefore this follows ok therefore that is the proof of this statement ok so I will stop with that.