 So, you know we are what we are looking at in algebraic geometry is trying to look at things which are which have which have an intrinsic meaning ok. So, intrinsic means these are certain definitions and properties that you make or define which depend only on the object and not on other extraneous factors. So, for example you know what we are worried about is we are worried about varieties alright and then of course the varieties could be you know they could be affine or quasi affine or they could be projected or they could be quasi projective and when you say when you say these things you are trying to say that the variety is either sitting as an irreducible closed subset of some affine space or some projective space or it is sitting as an open subset of such a set ok. So, when we define variety we are already thinking of it as sitting inside somewhere either an affine space or a projective space. So, there is this and there is this fact that you are making it sit into inside something which gives a certain ambiguity ok because you can you for example if you take the line the line can sit inside any affine space in any way ok. So, you can think of the line just as a1 can also think of it as a one of the coordinate axis in a2 ok or even any other line in a2 and you can also think of it as a line in 3 space a3 ok but in any case a line is a line ok. So, what is interesting about the line is that it is a line alright and what is extrinsic about it is the way you are putting it inside you are thinking of the line just as a line or you are putting it thinking of it as a line inside the plane a2 or you are thinking of it as a line in 3 space which is a3 and so on. So, these are so you know what you must understand is that when we when we start working in algebraic geometry start building the theory of varieties we always start by embedding your object into some affine space or projective space even to define a variety you have to think of it as sitting inside some affine space or projective space as an irreducible closed subset or as an open subset non-empty open subset of such an irreducible closed subset. But then finally what we want to really analyze and define and analyze and study is not I mean we want to define and analyze and study only the properties which are intrinsic to the variety we are not interested in they should depend only on the variety and not the way in which it is embedded ok. So, the fact for example you know this saying that the line is one dimensional does not depend on whether the line is being caught of as a line a1 or whether it is being thought of as a line in a2 which is a plane or whether it is thought of a line in 3 space a3 whatever be the space in which you embedded it is a it is still one dimensional. So, you see we say that dimension is a very intrinsic property because and in fact our definition of dimension we have seen that we have given the definition of dimension in two ways one is as a topological space we have given we have defined the dimension of variety to be topological dimension and we also proved that you know this dimension is the same as if the if the variety is an affine variety then it is the same as the cruel dimension of the affine coordinate ring of the variety ok. So, the moral of the story is that you know the definition that the dimension is the topological dimension is a very intrinsic definition it just depends on the topology of the variety it does not depend on anything else ok whereas the definition that the dimension is the cruel dimension of the affine coordinate ring namely the coordinate ring of polynomials on the variety ok that ring the dimension cruel dimension of that ring is the dimension of the variety is another definition but that somehow seems to depend on that ring because that ring is it depends on the way in which the variety is embedded ok if the affine variety is the affine variety could be embedded in so many affine spaces in so many ways as an irreducible close-up variety and in each of those cases you will get a coordinate ring the rings of polynomials ok but thankfully for affine variety is we have seen that the coordinate ring itself is an invariant it is an intrinsic thing namely if you take two affine varieties then they are isomorphic if and only if they are affine coordinate rings are the same so which means that no matter you in which in what way you embed the affine variety the moment you say affine variety you are considering an affine variety is isomorphic to an irreducible close subset of some affine space it could be an irreducible close subset in an affine space of any dimension ok and the dimensions could be different the like the same line being embedded in a plane or being embedded in three space ok so the ambient the bigger affine space in which your variety is embedded as a close-up variety that could be different but still the ring of polynomials on that variety they are all the rings are all isomorphic the affine coordinate rings are isomorphic so we express this by saying that the affine coordinate ring is a is an intrinsic object that is defined connected with the variety ok now so you know this is what I am trying to say we are trying to it is very important that we begin by using extraneous or extrinsic things ok but then finally we try to find out what are the things that are intrinsic ok so in that list comes the affine coordinate ring of an affine variety ok which is intrinsic then of course the other thing that I introduce was the local ring at a point the local ring at a point is something that is that is defined in a very intrinsic way ok it is just defined using regular functions ok and the point is that if you change the variety up to isomorphism then the local ring will also change only up to isomorphism ok namely if you have a variety and a point with an isomorphism carrying this variety to another variety and this point going to another point then the local ring of the original variety at the given point is isomorphic to the local ring of the target variety isomorphic variety at the image of the point that is gotten by the image of this point ok so we say that the local ring is also something that is very intrinsic namely if you change the variety up to isomorphism the local ring will also change up to isomorphism ok. So what we are going to talk about so you know in that direction I want to introduce another important ring associated with the variety in fact this is a not just a ring it is a field it is called the function field of a variety ok so I want to introduce that and they want to tell you that I want to introduce it in a very intrinsic way and then show how to compute it for fine and objective varieties ok. So let me put the title of the lecture as the function field of a variety the function field of a variety and so what is it that we are going to do so you know the idea is very similar to looking at the way in which we got the local ring at a point ok so you know what we did is see if so we start with x a variety ok and of course which means you know it is to be a fine or quasi-fine or projective or quasi-projective and we have a point and then you know how do you define the local ring at a point of the variety. So normally what we did was if you have a point small x in capital X then what we did is we define O capital X small x this is a local ring of x at x to be O capital X small x tilda modulo and equivalence relation ok and how was this equivalence relation and what was this tilda O capital X small x tilda these were just you know pairs of functions which were regular in some open in some Zariski open neighbourhood of the point x. So you know this consists of pairs of the form u, f such that f is in O u and x belongs to u. So you are just looking at functions which are defined in an open neighbourhood of the point x and which are regular alright and you took such two such pairs and then you know what is the equivalence relation the equivalence relation was that two such pairs will be identified if they define the same function on the intersection ok. So you know u1, f1 is equivalent to u2, f2 if well f1 restricted to u1 is equal to f2 restricted I mean u1 intersection u2 is equal to f2 restricted to u1 intersection where of course we you know you have to remember that x is both in u1 and in u2 so x is in u1 intersection u2 so u1 intersection u2 is a non-empty open set so it is in fact dense ok and you are just saying that f1 restricted to u1 intersection u2 is equal to f2 restricted to u1 intersection u2. In other words these two functions can be glued together to give a bigger regular function on u1 union u2 ok. So in some sense in the sense of complex analysis we say that these are two function elements which are direct analytic continuations of each other ok. So now of course you can weaken this condition you could you could just say that f1 restricted to w is equal to f2 restricted to w but w is an open neighbourhood of the point x which is contained inside u1 intersection u2 that is also enough because you know we keep always we prove this and we keep using this all the time that two regular functions if they are equal on a non-empty open subset then they have to be equal everywhere ok this is true always right. So now you know so we got this local ring like this alright and of course we have seen we have seen that this how to compute this local ring ok we have seen how to compute this local ring and what it is in the affine case what it is in the what it is in the projective case we have seen it alright. But now what I am trying to say is if you want the function field it is very simple what you do is you just remove this restriction of concentrating attention at a point ok just do not worry about the point and the beautiful thing is that you if you do not concentrate attention at a point you go global ok you get something global. So what you do is put kx to be kx tilde to be a set of all pairs u,f such that f belongs to ou and now there is no restriction on u, u inside x as an open subset open non-empty ok. Of course here also whenever I say f belongs to ou I am I am it is implicit it is understood that u is an open subset of x ok u is a Zarski open neighbourhood ok fine. So now you see so what is the difference between this and this the difference is that here you are only taking open neighbourhoods of the point small x but here you are simply taking any any non-empty open set and you are taking a regular function of that ok. So you just drop attention to a point and lo behold what you get is you repeat the same procedure what you get is not a local ring you get a field ok. So now what you do is again you define the same thing you define same equivalence relation u1,f1 is equivalent to u2,f2 if and only if f1 restricted to u1 intersection u2 is equal to f2 restricted to u1 intersection u2. Of course for this I mean you need not require it on all of u1 intersection u2 it is enough to even require it on open non-empty open subset of u1 intersection u2 and you must always remember that this very important fact that by definition or varieties are all irreducible and therefore you take any non-empty open subset of a variety it will continue to be irreducible it will be dense and any two non-empty open subsets of a variety will always intersect ok. So this is the problem the Zariski topology the problem is that the open sets are huge ok they are dense their closure is the whole variety I mean of course non-empty open sets they are huge and any two of them will intersect right. And well so it is just enough to require that f1,f2 coincide in an open subset of u1 intersection u2 ok non-empty open subset of u1 intersection u2. So this definition this equivalence is same as equivalence here only thing is you are not focusing attention at a point alright and then you put kx to be kx tilde mod the equivalence ok. So of course when I write mod equivalence I mean equivalence classes ok. So this is the set of equivalence classes this is also set of equivalence classes. Now the beautiful thing is that just like in this case the set of equivalence classes gave you a local ring ok you got a commutative ring which is a local ring which had a unique maximal ideal which was represented by functions which are regular in a neighbourhood of x and which vanish at x ok and in fact the representatives the local ring namely the equivalence classes were called germs of functions ok. So this is the set of germs of regular functions at the point x and every germ of a regular function is represented like this it is represented by a regular function on an open neighbourhood of the point ok. Now here we call the elements here well they are kind of they are called rational functions ok they are called rational functions and basically they are to be thought of as functions are regular on an open set ok that is how you should think of rational functions. Rational function is nothing but a regular function on an open set it is not a regular function on the whole alright but it is a regular function it may be a regular function only on an open subset which may be a proper open subset ok so that which means that there could be points where it is not regular ok where it where you cannot think of where you might extend the function but it may not be regular right. So let me write that down elements equivalence classes of kx are called rational functions they are called rational functions so they represent they well they represent they are represented by a regular function on an open subset alright and further you see this kx this is actually this is actually a field kx is a field and why is that so because you see you see you take two elements in kx so you know so you know I have this I have this map kx tilde to kx this is a quotient map it is a surjection so that is why I am putting this double arrow head and this is a set of equivalence classes this is mod equivalence ok and what we are doing is you are taking if you give me a rational function u1, f1 you send it to its equivalence class which I will indicate by putting a square bracket ok and well if you give me another rational function u2, f2 I will send it to so rather let me use u, f and v, g so this will go to well v, g round bracket followed by a square bracket the square bracket indicating the equivalence class here and then how do you add these two guys it is very very simple you add them like this you give me two rational functions you just add them on their intersection because you know as I told you we keep always using the fact that any two non-empty open sets will intersect so you and we will intersect on the intersection both functions make sense so I could add them so I can do this I can take this equivalence class and define this to be the sum alright and then the same way I could define product I can put product here and I can put product here ok so depending on what your either sum or product can be defined like this and now I want you to check it is easy to check as you must have done in this case that this definition of addition and multiplication is independent of the choice of representatives because I have defined the sum when I define the sum of two rational functions I am actually taking the sum of the representatives and I am I am taking the sum of the representatives on the intersection on the intersection of the domains where the representative functions are defined ok. So and similarly when I do for the product I also use representatives but then whenever you know you use representatives you have to make sure that your definition is correct it is well defined so you have to check ok so check that check well defined as that is a simple exercise that the same kind of exercise as you would have undergone for this case of you know for the local ring. So this will make kx into a commutative ring in fact you will have that kx is commutative ring of course it will be commutative ring because addition and addition of functions and multiplication of functions is point wise and that is commutative because if they are the functions are taking values in the base field k which is commutative alright. So it is a commutative ring with 1 to be with 1 being the given by the pair x, constant function 1 and 0 element given by the pair x, constant function 0 ok. So it will be a commutative ring with unity and with this as 0 element alright and in fact it is in fact not only that it will in fact be a k algebra because you know a regular function multiplied a regular function multiplied by a scalar is also a regular function because scalars are of course regular function they are part of a constant regular function ok. So kx is a k algebra of course you know in our discussion we always fix small k to be an algebraically closed field where we are working that is where we are do studying our varieties. So and this kx is a k algebra ok all functions take values in small k all regular functions take values in small k and well now the point is the following the point is well in this case after you took if you took open neighbourhoods of a point and then you define the sequence and when mod this sequence you got a max you got a local ring you got a commutative ring which was also a small k algebra but it was a maximal it had only one maximal ideal and therefore it became a local ring but the point is in this case you do not get the only maximal ideal you will get a 0 and so you will get a field. So why is this a field to show that this is a field I will have to tell you that every non-zero element here is invertible ok. So if so you know if you take u,f not equal to 0 element the 0 element is x,0 in kx ok. So then you know what I want you to understand is that you see you take 0 take z of f ok see z of f inside x will be a closed subset ok. So what I want you to understand is that whenever you take a variety and you take a regular function on it ok and you take the set of the locus where it vanishes ok that is always that will always be a closed subset ok. So in fact you know I should say it is a closed subset of u in fact because f is defined only on u it should not write x here it should be careful right. So f, f is a regular function on u and mind you u itself is a variety, u is any open subset of a variety is again a variety ok. So u itself is a variety f is a regular function on u ok and if you take the 0 set of f namely set of points in u where f vanishes that is going to be a closed subset of u ok because actually you know roughly the idea is that the closed subsets are defined by vanishing of polynomials right in the affine case it is in the affine case they are common zeros of bunch of vanishing of bunch of polynomials and if you are the projective case then it is common zeros of a bunch of homogeneous polynomials ok in any case it is just vanishing of polynomials that gives the 0 the closed subsets for the risk it is a policy. And therefore if you take a regular function on a variety u is also a variety and f is a regular function on that then locally f is a quotient of polynomials and therefore looking at zeros of this regular function is like looking at locally the zeros of the numerator polynomial because locally f is a looks like a numerator polynomial divided by denominator polynomial and zeros of f will be just the zeros of the numerator polynomial ok and therefore essentially you are just looking at zeros of polynomials and therefore again you are going to get a closed set ok. So this is a closed subset and mind you z of f you know cannot be u ok z of f is not equal to u because if z of f is u that means f is identically 0 on u but then if f is identically 0 on u then f is equal to the 0 function which is identically 0 on x and that will contradict the fact that these two are different elements of the function field ok. So z of f is not equal to u because otherwise you will see that u, f is the same as u, 0 and this is the this is equivalent to x, z which is a contradiction ok. So I should say contradiction it is a contradiction to the fact that these two are different alright. So that means 0 z of f is a proper closed subset of u so it is complement in u is again an open subset of u. So you know u minus z of f inside u is non-empty open and of course you know where a regular function does not vanish its reciprocal is also a regular function ok. After all a regular function is a locally of the form quotient of polynomials where it will not vanish is where the numerator polynomial does not vanish. If the numerator polynomial does not vanish ok already when you are writing the quotient denominator polynomial cannot vanish, if the numerator also does not vanish then its reciprocal the reciprocal of these two numerator and denominator will give you also a rational we will again give you a regular function at that point locally ok. Therefore wherever a regular function does not vanish its reciprocal is also a regular function so what happens is that you will get this pair u minus z of f which is an open subset of u proper it is non-empty open subset of u mind you it is this is actually mind you it is dense in x itself ok it is a non-empty open subset of u and u is non-empty open subset of x so this is a non-empty open subset of x it is reducible it is dense ok it is a huge open subset ok and on this I have the function 1 by f 1 by f is certainly a regular function on that ok. So this this element if you multiply it with the original element u, f what you will get is just x, 1 you will get this ok by the definition of multiplication alright because after all 1 by f and f are going to multiply to give you 1 wherever f does not vanish ok and wherever f does not vanish is certainly is a huge open set alright therefore so what this tells you is that whenever u, f is not the 0 element mind you this is 0 element in kx so whenever something is not 0 it has a reciprocal there is something with which you can multiply it to get 1 this is 1 this is the 0 in kx and this is the 1 in kx ok. So here also when I write 1 equal to x, 1 I mean this is the 1 in kx and this is 0 in kx ok. So what I proved is that every non-zero element is I have a commutative ring in which every non-zero element is invertible so it is a field. So the moral of the story is that kx is a field and is a field extension of k ok because it is a k algebra so you know it is an over ring of k it contains small k as a sub ring and whenever you have a larger ring which is a whenever you have a when the smaller when you have a ring extension and both of fields you it is actually a field extension ok. The smaller ring is k it is a field the larger one that we have defined and constructed is the is also a field extension right now and this is called the function field it is called the function field of x ok it is called the function field of x and what is so special about the function field of x. So the answer to that geometrically is the following that the function field of x will give you all the information that is true on a large open subset of x ok. So this is the philosophical importance this kx will contain all the information about x which you can find on a large open set of course any non-empty open set is large ok. So you take any general open set next and you take properties that are valid on a general open subset of x then all those properties they will be captured by kx ok. So kx geometrically captures what is happening on a large open set of x ok. So kx captures captures the geometry on a general geometry of ok. So the point is you see what I want you to understand is both in the local ring case and in the function field case you are just looking at pairs of functions ok I mean you are just looking at pairs which consist of a regular function defined on an open set ok. Here you are only look concentrating on open sets which contain a given point and you get the local ring and the local ring contains all the information in a neighbourhood of the point ok all information in a neighbourhood of that point how the variety behaves in a neighbourhood of that point all that information is controlled by this and captured by this ok. Whereas if you want to know what is happening on a general open subset of x that information is captured by kx ok. So this is a you know if you want to study a general open subset of x if you have a geometry question about a general open subset of x ok then what you have to study is this if you have question about what is happening at a point of x then the object you have to study is this the local ring ok. So these are two extremes ok and I will tell you how they are connected we are going to see how they are connected of course they are all connected by to each other and essentially they are all I mean everything is done by commutative algebra ok. So now comes the following question so you can ask you can ask how are these things connected. So the first thing I want to tell you is that you know we have maps so before that yeah we have maps we have natural maps oh oh x oh x x ok there are maps like so what is this map let f be a regular function on x you just send it to since f is a regular function of x then I have the pair u x, f and its equivalence class will give me a point it will give me an element of the local ring at this point x because the elements of the local ring at a point are simply germs of regular functions defined on the neighbourhood of the point and since f is a global regular function it is also a regular function at that point so I can take its germ so this is a germ of f at the point and well you what I can do is on the other hand you know this is this map alright mind you that I can also send a map from here to here namely if you take u, g let me write v, g what is this v, g? g is regular function on the open set v which contains a point x ok this is the germ of v this is the germ of g at the point small x ok and of course g this open set v may be a proper open set g may not extend to a regular function on the whole of x it may be restricted it might just be defined on this open set ok now but anything here I can further send it to the same thing here mind you I am using the same notation but the point is this equivalence is the equivalence in local ring this is the equivalence in the function field ok sorry but I am using the same square bracket. Now the point is that of course you know there is also a map which goes like this directly and that is simply that is simply going all the way to x, y which is just the composition of these two maps this will also go to x, y where again this square bracket is the equivalence class here and this square bracket is equivalence class here ok alright so I have maps like this now the fact is that these maps are well the fact is that these maps are injective k-algebra homomorphisms ok. So these are injective k-algebra homomorphisms ok these are injective k-algebra homomorphisms ok so of course they are k-algebra homomorphisms namely they are ring homomorphisms which are k linear ok so that is no problem it is very easy to see the point is the point is the injectivity. So the injectivity comes like this you see that is also something that uses again this fact that you know if a regular if two regular functions are equal on an open set then they are equal everywhere ok of course that open set has to be a non-empty open set of course because you need a point to test you need points to test whether two functions are equal actually ok. So well you see you can see why this is why this is injective is very very clear because you know if x, f is 0 ok then it means that f vanishes in a neighbourhood of small x ok but that means f is equal to the 0 function in an open set containing small x ok but then if you use the fact that two regular functions if they are equal on a non-empty open set they are equal throughout. So if f is if small f vanishes in a neighbourhood of small x then it vanishes everywhere that means f itself is 0 ok that will tell you why this is injective and why is this injective that is also literally the same argument if you take a v, g germ of a function g which is regular in this open neighbourhood v about the point small x if it goes to 0 here it means that it is equal to 0 on some non-empty open subset of v ok but in any case that will also force it to be 0 everywhere in fact if this goes to 0 here it will tell you that g is the 0 function on v ok and the 0 function on v extends to the 0 function on x ok and therefore this will itself be 0 alright. So the moral of the story is both these ring homomorphisms have 0 kernel so they are injective ring homomorphisms and what is the moral of the story the moral of the story is the beautiful thing that the regular functions sit inside the local rings the local rings sit inside the function field this is how the you know so the idea is that somehow when you are looking at it topologically you have the whole your you have the whole variety x you concentrate attention to a point you get a larger ring ok. If you have the whole variety x you have o x which is a regular functions on x ok now you concentrate attention to a point you get a larger ring ok that is a local ring at that point ok which is larger than the ring of regular functions and then on the other hand if you concentrate instead of concentrating at a point if you concentrate on an open set you will get a much larger ring which is in fact a field and that is the field of rational functions. Now what I want you to understand is that you know in all these things nothing will go nothing will change if I replaced x by an open subset u ok. So in fact what will happen is that in fact if u is an open subset of x is open and small x is in u then you have we have a commutative diagram we have equalities in fact o x to o x x I mean at the level of local rings and at the level of function fields you are not going to have any difference so you know so I will put a hook arrow saying that you are considering this as sub rings what will happen is this will be the same as o u x is o u and this will be the same as k u and this will sit inside ok. So if you the function field depends only on an open subset of x so the function field of variety will not change if you replace x by an open subset so k x the function field of x if you replace x by u non-empty open set then k x will be the same as k u by going to a non-empty open set you are not changing the function field and that is just reflection of the fact the function field captures informational on a general open set so if you go to a general open set the function field does not change ok and the local ring at a point also depends only on a neighbourhood of the point it does not depend on the ambient variety. So whether you are considering x is a point of the open set u of capital X and calculating the local ring of small x at the of the variety u or whether you are calculating this local ring of small x with respect to the bigger variety ambient variety capital X you will get the same result ok. So the local ring also depends only on a neighbourhood of the point ok and of course the fact what is what is what is happening here what is happening here is that you also get an inclusion like this you also get an inclusion like this because every regular function on x can be restricted to get a give you a regular function on u and the restriction maps restrictions of functions they are all injective because of the same old reason that if a regular function if two regular functions coincide with an open subset non-empty open subset then they have to coincide everywhere or other way of saying it is if a regular function vanishes on open subset then it is identically 0 it will vanish on the whole variety. So you see this is what happens if you if you go to an open subset and why these two are equal is something that you can easily you can easily you can get this equality in a very very easy way namely what you do is you know you and any element of kx is of form u,f where f is a regular function on u and you take the equivalence class and what do you have to do it is very very simple I should not use the same u let me use v,g or even better let me use w,h w is an open subset of x h is a regular function on w then I have this pair w,h and its equivalence class is the rational function here ok and what will I send it is very simple you see w is a non-empty open subset of x, u is a non-empty open subset of x so they intersect so I can make sense of h also on the intersection that is what and that is the element I am going to send it to. So I am simply going to send it to w intersection u,h restricted to w intersection u I take this class and then I take its equivalence class and that is a element I am going to send it and in principle I should write this as a map and you can see that this map will be a k-algebraism of some but then I put equality because actually as far as the functions are concerned I am simply restricting the same function so that is why I put equality there ok if you think of them as functions then you can you can put equality here and you can put equality here ok. So if you think of them really as functions ok of course you have to remember that here you should think of representative as functions on a non-empty open set ok and here you must think of representative as functions on a regular functions on a non-empty open set which contains the points molex that is the only difference ok. So you have this nice diagram alright and now what I have to do is to tell you what this kx will be if x is an affine variety what this kx will be if x is a projective variety ok we need to understand what that is ok we have already understood what the local rings are when x is an affine variety or a projective variety ok they are given by suitable localizations ok we have to repeat this extend this kind of argument to the function fields ok. So let me just state the result then we will probably prove it in the next lecture here is theorem if x is an affine variety then kx is isomorphic to quotient field field of ax which is equal to o x ok. So if x is an affine variety then you know we have we have defined ax the affine coordinate ring of x and we have proved that that affine coordinate ring is equal to the ring of regular functions ok and this equality being thought of as functions ok we put this equality because we think of a function here also as defining the function there and conversely ok and what if x is projective if x is projective then kx is let me put y whether I do not confuse if y is projective then ky is just you take the homogenous coordinate ring of y ok the homogenous coordinate ring is defined the same for a projective variety is defined the same way as affine coordinate ring is defined namely you take the ring of polynomials of the ambient space that you are considering and then go modulo the ideal of functions that vanish on the variety in this case of course you will have to take all whom you have to take polynomial ring in of the affine space over the projective space in which this variety y is embedded and go modulo the ideal of y which will be homogenous ideal and then you take this and then you know what you do is you take the 0 ideal ok that is a prime ideal and this is localizing at 0 and then I will put another 0 here saying that you are taking the degree 0 part of the homogenous localization of sy ok so this is degree 0 part of the homogenous localization sy so here is the formula for the function field of affine variety and a nice formula for the function field of a projective variety ok and you know if you try to write out if you try to write out this kind of diagram including the you know the local rings and the quotient fields then what will happen is that you know if you take O x small x that is going to sit inside kx and you know this kx is just ax quotient field of ax ok q represents a quotient field of ax that is kx that is what this first part of the theorem says this is isomorphism and we have already seen that this is nothing but ax localize at mx where mx is the maximal ideal in ax corresponding to functions that vanish at small x so I should put small x here so mx is equal to ideal of functions vanishing at x ok we have already proved this and we have this inclusion ok and you must recall a fact in from commutative algebraics very easy if you have an integral domain ok then it is quotient field will be the same as the quotient field of any of its localizations ok the quotient field of quotient field or field of fractions of an integral domain is the same as the quotient field or field of fractions of any of its localizations. In fact the quotient field is the largest localization ok it is a localization where you have inverted everything except the only element which you cannot invert which is 0 and 0 is a prime ideal ok and so you are localizing at the prime ideal ok so this q ax is just ax localized at the prime ideal 0 ok this is what it is you are inverting everything outside 0 right because localization at a prime ideal means invert everything outside the prime ideal in this case invert everything that is not 0 that will give you the quotient field. So this is the picture you get for the affine case ok and you will get a similar picture for the for the projective case if you take o y y small y point of capital Y where capital Y is a projective variety this is going to sit in side ky and what is going to happen is that the function field of ky will be just sy localized at the prime 0 degree 0 part and this will contain sy localization at my and degree 0 part where here my so this is of course this diagram commutes this diagram commutes this isomorphism of the local rings has already been proved ok and this isomorphism of the function fields that I have to prove and here again my is equal to all those ideal of ideal of homogeneous elements of sy that vanish at small y ok. So this is how you get it you get the local ring the quotient field the affine case and this is the local ring and the quotient field in the projective case ok so we will prove this in the next talk.