 Ok, so you see uhhh let me tell you uhhh uhhh let me recall the notation that I used ok uhhh this is just to uhhh make sure that I do not mess up the notation. So, you know uhhh we have this situation that uhhh uhhh so so if you recall what we had was uhhh uhhh we had so you had x uhhh uhhh x and y varieties and you had small x in capital X small y in capital Y and uhhh uhhh you had uhhh psi uhhh from from the local ring of capital X at small x to uhhh the local ring of capital Y at small y uhhh you had uhhh an isomorphism k-algebra isomorphism ok we had this and then what we did was uhhh we wanted to show that uhhh if you have 2 varieties and there are points uhhh where the local rings are isomorphic we wanted to show that there are open neighbourhood surrounding those points which are actually isomorphic as varieties ok. So, in other words there is an open set we wanted to find an open set containing x and an open set containing y and an isomorphism between these 2 open sets ok and the point is that uhhh that isomorphism uhhh at local rings will induce this isomorphism ok that is what we wanted to show. So, uhhh what we did was uhhh so this was given to us ok and what we did was uhhh so uhhh of course because uhhh any variety is covered by uhhh finitely many open sets each of which is isomorphic to an affine variety what you did was we took an affine variety we took an open set containing small x which is an affine variety and an open set containing small y which is also an affine variety and therefore uhhh you know we had the diagram like this we had x here we had uhhh x1 inside uhhh x which is uhhh an affine open and uhhh we had uhhh small x belonging here then we had capital Y uhhh and we had this capital Y1 which is an affine open uhhh sub variety which contains the point small y ok. And uhhh if you look at it uhhh in in terms of uhhh uhhh uhhh the this is geometric part if you look at it commutative algebraically you got O x here that corresponds to a regular functions on x we have O y here and then you have this restriction of regular functions on x to regular functions on x1 so you have O x1 but that is the same as A x1 ok and uhhh and of course this is an this is an intuition because uhhh that is because uhhh uhhh if a regular function is uhhh 0 on a proper open set which is non-empty then it is 0 everywhere ok and another way of saying it is of course that if 2 regular functions coincide on a non-empty open set they coincide everywhere that is the reason why this is this map is injective it is uhhh uhhh and of course this map corresponds to restriction of regular functions to the open set uhhh x1 and similarly you have an inclusion uhhh with of this with O y1 and O y1 is the same as A y1 because uhhh uhhh and O x1 is the same as A x1 this is because x1 and y1 are affine varieties for affine varieties the ring of regular functions is the same as affine co-ordinate ring right and then of course then we go all the way to the point uhhh by going to the local ring at that point and uhhh the local ring will not change if you go to an open set if you go to a smaller open set which contains the point so this is the same as the local ring of uhhh uhhh x uhhh considered as a point of the open sub variety x1 right and similarly here we have uhhh the local ring of capital Y at small y which is uhhh the local also the local ring of uhhh small y at as a point of capital Y 1 alright and we know uhhh uhhh we know that you see the uhhh we had given uhhh names for these rings we call this ring as A we call this ring as uhhh B okay and uhhh uhhh the point uhhh small x as a point of x1 will correspond uniquely to a maximal ideal of A and we call that maximal ideal as uhhh uhhh M and the point y small y will be will correspond uniquely to a maximal ideal of B we call that as uhhh N or Vita and so you know x uhhh so this is so this local ring becomes uhhh can be it can be very infinite with A localised at M okay this is the expression for a local ring at a point you take the uhhh uhhh if you have a point of an affine variety then the local ring at that point is given by simply taking the affine co-ordinate ring which is the same as the ring of regular functions and then localising at the maximal ideal that corresponds to that point and similarly this is identified with B localised at N okay where M corresponds to uhhh so this x corresponds to M in the uhhh uhhh maximal spectrum of A where A is A of x1 and uhhh y corresponds to uhhh N which is uhhh in the maximal spectrum which is set of maximal ideals of B uhhh where B is of course uhhh the affine co-ordinate ring of y1 right and then uhhh of course we have uhhh uhhh then all this is happening inside the respective quotient fields all these things have the same quotient field uhhh I mean this uhhh uhhh uhhh uhhh the quotient field of A M is the same as the quotient field of this uhhh is the same as quotient field of this it is the quotient field of A and there is also the function field of X. So, uhhh what you must understand is that this is happening inside uhhh uhhh uhhh a quotient field of A which is the same as quotient field of A sub M. This is also equal to well function field of this is also equal to quotient field of that is also equal to the function field of X the field of rational functions of X is also equal to the field of rational functions of X1 the field of rational functions does not change if you go to an open side ok and similarly here I am going to get this is all this is going to sit inside by the way I should also tell you that the when you go from the ring to its localization the ring is an integral domain this is also an injective map ok. From a domain to its local ring to any of its localizations the natural map given by localization is an injective map because the source is a domain when you localize a domain you get a bigger sub ring of the quotient field of the domain ok. So similarly B is a domain and if you localize B at the maximum ideal neta you are going to get a bigger sub ring of the quotient field of B this Q of B that is the same as a quotient field of any localization in this case it is also the quotient field of the field of this is a field of fractions of the local ring and this is also equal to the function field of y1 it is the function field of y they are all the same ok. And the point is that it is at this level so I have this isomorphism here so I have this psi here this psi is given to me alright somehow I have to use this psi to produce an isomorphism between an open subset of x1 containing the point x and with an iso with an open subset of y1 containing the point y so I have to somehow cook up those open sets ok. So well so what we had done is that we had assumed that see your A which is the affine co-ordinate ring of x1 if I look at the notation A was ks1 etc sn divided by the ideal of x1 this corresponds to you know this corresponds to putting because x1 is an affine variety ok x1 sits inside x1 sits inside some an ok and if you take the affine co-ordinate ring of an as a polynomial ring in the variables s1 through sn ok then the affine co-ordinate ring of x1 which is ax1 which we call as a is going to be just this polynomial ring divided by the ideal of x1 ok where of course here when I say x1 is being considered in an here it is being considered as an irreducible closed subset of an ok which identifies it as a co as an affine variety right and then we also have similarly b is also a polynomial ring in m variables divided by the ideal of y1 where similarly what we are doing is that we are we are thinking since y1 is an affine open subset of y it means that y1 is an open subset of capital Y but it is also abstractly isomorphic to an affine variety therefore you can identify it you can embed it as an irreducible closed subset of am and on this am if you take the t1 through tm to be the co-ordinate functions then the affine co-ordinate ring of that am is this polynomial ring in the t1 through tm and if you go model of the ideal of y1 you will get the affine co-ordinate ring of y1 right and now in terms of these rings we have this inclusion of a into am and this is b into bn ok and of course am is sitting inside the quotient field of a and bm is bn is sitting inside the quotient field of b and of course my the isomorphism that is given to me is between am and bn ok this is the k-algebra isomorphism that is given to me of course all ring homomorphisms we are considering are all k-algebra homomorphisms ok and am in essentially they are all as far as the same variety is concerned they all correspond to going to I mean going from here to here corresponds to taking germs of regular functions and then and then going from here all the way to the quotient field is trying to look at an equivalence class trying to look at a regular function as a rational function ok and across from one side to the other we always think of ring homomorphisms as coming because of morphisms by a pull back of regular functions but the point here is that we have to construct the morphism we have to construct an isomorphism between an open subset of x1 and an open subset of y1 ok. So what we did was if you remember so we have put we have put we had made all these assumptions we had put psi of si bar by 1 is equal to f i bar by g i bar ok. The si's are polynomials here and the si bars live in this quotient and you are considering them as because this is sub of this ok this is the subring of this so you are considering them as elements here ok. So si bar by 1 is an element here and then I apply psi I get a quotient I get an element here a quotient where g i bar is not in the where g i bar is certainly not in the maximum ideal n ok and then this has to be probably psi inverse 1 by g i bar lives here this has to be psi inverse ok and similarly I do for psi inverse I take the image of the coordinate the generating coordinate variables variable the generating functions the t j's and so this is what I have now what I want you want you to check is that see you if you take this set if you take if you take this localization a localize that product of g j bar product of a bar b i bar product of psi inverse of b j bar ok and you see and you take this you take this localization ok and then you take this localization on as far as so this is you know a is going to sit inside this localization and b is going to sit inside this localization and in this localization you take product of product of capital G I bar to product of a j bar b j bar and I think you will also have to take psi of b i bar product of psi of capital b i bar ok. So, now the now the fact is that now the fact is that if you of course these localizations will sit inside am and here inside bn ok these localizations will sit here and what you can check is that the way we have defined psi inverse you will get an isomorphism like this ok you will get an isomorphism like this and that is just because of the universal property of localization you will get a you will get a you will get a map like this and you will get a map in the reverse direction ok you will get an isomorphism between these two alright and such that this so you will get this map here which I would like to call by some name maybe I will call it as I do not know what I had so it was probably phi. So, you know you have you have this map yeah so let me put the arrow in this direction call this phi inverse ok you will get a map like this k algebra isomorphism phi inverse phi inverse will just come because of the universal property of localization and the way because of the way in which these are defined and what will happen is that this phi inverse you know if you if you take this this phi inverse then I will also get phi inverse what this phi inverse will do is that it will also induce an isomorphism here ok and this isomorphism will be the same as the isomorphism psi you started with so you will get you will get something like this you will get this ok you see everything everything was constructed using psi alright. Therefore the the point about going inverting all these elements is to go to a proper I mean it is to go to a proper open subset to go to an open subset of the affine variety with co-ordinating a ok and the same way inverting all these things here is the point is to go to an affine open subset of the variety y1 whose co-ordinate ring is b ok. So this this is the affine co-ordinate ring of an affine open subset of y1 and this is the affine co-ordinate ring of the affine open subset of x1 ok and this isomorphism of affine co-ordinate rings gives an isomorphism between open subset so you know so you you therefore so hence you will get hence we get you see you have x and inside you have x1 and inside x1 you have this u ok and where what is this u this u has affine co-ordinate ring given by this whole expression so this is AU ok and similarly so this is this is this is affine open and this also again affine open ok and similarly at for y you have y1 which is affine open and here u we call it as b that will also be another affine open and what you will get is that you will get you will get a morphism so I think so notation wise let me let me make let me do something let I say that I will get an isomorphism phi ok such that you know if you take the isomorphism that is induced by that is induced by pullback of regular functions because an isomorphism between affine varieties corresponds to an isomorphism between the affine co-ordinate rings which is a K-algebra isomorphism so this will induce a phi hash so let me change this to phi hash ok and then the direction is correct so I will call this as phi let me call simply this as phi hash ok and then this isomorphism will take the point x to the point small x to the point small y so it will induce an isomorphism of the local ring of y with the local ring of x and that is exactly this so that is so that is just phi upper hash at y and mind you this is this is this is identified with o v y ok which is which is of course you know the same as o y1 y and is also the same as o yy right and this is identified with o u x ok and y is of course y is of course phi x ok so this phi hash at y this is isomorphism of the level of local rings this is the original psi that you started with ok. So so so the moral of the story is if you start with an isomorphism of an a K-algebra isomorphism between the local rings at points of two varieties then those two varieties are by rationale that means there are open subsets on each which are isomorphic to one another ok and I told you that the property of two varieties being isomorphic on open subsets is called birationality ok and so all you are saying is that if the local rings at two points of two varieties isomorphic then the two varieties are birational so this is just to tell you that the local ring carries a lot of information ok it carries a lot of global information also reasonably so global information in the sense that it does not carry information about the whole variety but it certainly carries information about a about an open set containing the given point where you are considering the local ring but that open set is of course you know dense open set it is a non-empty open set it is irreducible and dense because of the Zariski topology open sets are all huge ok therefore the local ring also contains information over a huge open set ok so that is a so you know that so as a matter of fact you know you must also notice that the dimension of the local ring is the same as the dimension of the variety so the moment the local rings at two given points of two varieties isomorphic it means that first of all it means that the varieties are having the same dimension ok so this would not happen if the varieties do not have the same dimension ok if the varieties are of different dimension then you cannot find an isomorphism even I mean you cannot you cannot find an isomorphism between local rings of from point from a point of one variety to a point of the other variety ok therefore the first thing you should notice is that once you have an isomorphism of local rings it already tells you that the two varieties are having the same dimension but the story is more the fact is that the isomorphism of local rings actually makes them makes them birational ok so that is to say about the power of local rings but then I also want to tell you about I also want to tell you in connection with this about the power of function fields ok so so the next result that I am going to talk about is about the fact that you know if you have two varieties such that their function fields are isomorphic as k-algebra ok then also they are birational ok. So isomorphism of function fields of two varieties is also you know it also has the same effect as having an isomorphism of local rings ok so let me give this notebook back to you so let me make so here is the theorem so this is kind of because it is connected with this I am continuing with this if two varieties so this tells you about the importance of function fields ok if two varieties have isomorphic function fields in their birational and conversely ok two varieties x and y with isomorphic as k-algebras function fields are birational ok. So this is you know now the now I am for a moment I am trying to tell you about the importance of function fields right so you know the function field of a variety is an invariant in the sense that it will only change up to isomorphism if you change the variety up to isomorphism ok so of course if two varieties are birational ok then there is an isomorphism on open sets ok and this isomorphism on open sets will lead to an isomorphism of the function fields ok and you know the function field will not change if you go to an open set ok so if two varieties are birational then certainly their function fields are isomorphic that is obvious ok but what is not obvious is the other way round if you just say that the field of rational functions on one variety isomorphic as a k-algebra ok to the field of rational functions on another variety that is good enough to say to saying that that is as good enough as saying that the varieties are themselves isomorphic on open subsets namely that they are birational ok. So so the so the proof for this also runs in a in a in more or less in the same I mean the ideas are nearly the same as in this proof ok so what you do is see you have see you have so I have so I have x I have y varieties alright and I have kx and I have ky they are the function fields and I am given and I am given an isomorphism let me call as a zeta ok this is a k-algebra isomorphism. So I have k-algebra isomorphism between the function fields of two varieties ok and what I want to say is that just as in this case when you have an isomorphism of local rigs as you are able to cook up two open sets where they are isomorphic you can also cook up open sets here and here where the two varieties are isomorphic ok just an isomorphism at the function field level is enough ok and how do you do that so you know again what we will do is we will we will go to so so just as we did there we went to affine open sets ok which is which helps us to translate everything into commutative algebra ok. So we can do the same thing here we can assume that x1 is an affine open in x and you know y1 is an affine open in y and well if you go to an affine open of course whenever I say open set it is a non-empty open set I am not worried about the empty open set ok. So if you go to an affine open set then well what happens is the function field does not change ok the function field does not change so this is the same as k of x1 and this is the same as k of y1 ok. So it is enough to show that x1 and y1 are birational ok because that will mean that x and y are also birational just because x1 is just an open subset of x and y1 is an open subset of y. So you know again I follow the same I follow with the same notation so what I do is I think of x1 as sitting inside an ok and I think of y1 as sitting inside am alright and I take on an I take the coordinates to be s1 through sn on am I take the coordinates to be t1 through tm ok and I write a to be the affine co-ordinating of x1 b to be the affine co-ordinating of y1 ok I use the same mutations but the only thing that is missing is I do not have this I do not have these two points and the local rings and isomorphism between local rings I do not have that but I have the isomorphism here I have the isomorphism at the quotient field level because after all the quotient field of the affine co-ordinating of variety is the same as the function field of variety. So what I have now is as before as before what we get we get the following diagram you get a diagram similar to this only that you do not get this ok this is not given to you there is no point here going to a point there such that the local rings are isomorphic there is only an isomorphism between the quotient field of a and the quotient field of b. So what happens we get this diagram so let me write it down here so a is the affine co-ordinating of x1 and you know what I have done is as in that case I have embedded x1 as an irreducible closed subset of an so that it is affine co-ordinate ring becomes the affine co-ordinate ring of that an which is the polynomial ring in the co-ordinates s1 through sn more to the ideal of x1 in the affine space ok and I have on this side I have y1 also embedded as an irreducible closed subset of am and here again I will get if I take the co-ordinates on am to be t1 through tm then if I take k of t1 through tm this is going to be the this is the co-ordinate ring of the affine space am and then if I go modulo the ideal of y1 I end up with b which is the affine co-ordinate ring of y1 ok. And then what I now have is well this is sitting inside the quotient field of a and but the quotient field of a is just the function field of x1 and that is the same as the function field of x. And similarly this is sitting inside the quotient field of b which is the quotient the function field of y1 and which is also same the same as a function field of y and now I have this I have this isomorphism zeta which is an isomorphism between these two function fields ok. Now you know now I have to play nearly the same game as I did here and well the in fact the truth is that you know somehow in a way if I had proved this statement first ok then that statement would have been easier to prove because you know if I had given you this statement first then if you go back to this situation this isomorphism between local rings will induce an isomorphism between the the quotient fields. And once I have isomorphism between the quotient fields if I grant that I have already proved that it is birational I am already done. So in a way this could have been reduced from here but then there is no harm in doing it like this because you learn some techniques ok. So basically what we do in this situation is the following so you know you take you look at all these Si's ok which are in the polynomial ring they their images in A are the Si bars ok and they are all elements of this of this quotient field and you apply zeta to them ok. So what will happen is that you get all these Si each of these Si bars going to I mean you are thinking of Si bar as Si bar by 1 that is the way you think of elements of the domain as elements of the quotient field ok and you apply zeta to this. So I will get zeta of Si bar by 1 alright alright and the point is that all these what are these zeta Si bars by 1 well they are rational functions on they are rational functions on Y1 they are rational functions on Y1 and you know what is the so you know you should remember that K Y1 is rational functions on Y1 up to an equivalence ok. So rational function is given by a regular function on an open set and two such pairs consisting of a regular function and an open set on which it is defined are said to be equivalent if in the intersection of those two open sets the corresponding regular functions coincide ok and mind you any two non-empty open sets will always intersect because of irreducibility ok. So you know any rational function here is a regular function on an open set so you know each Si zeta I bar by 1 is regular on some open set it is a regular function on some open set and there are only n of them ok. So if you take the intersection of all those open sets then these fellows will become regular functions on that common open set ok. So you know so let me write that this is regular on U let me call this as these open sets Vi open inside Y1 ok these are all regular there and similarly I can do it the other way around. So what I can do is you know I can also look at the T's ok the Tj's are all coordinate functions here and their images in B will be the Tj bars and then you take their images here the quotient field then follow by zeta inverse ok and I am going to get rational functions on this side ok. So the same argument applies so if I take the Tj bar divided by 1 and then you know if I apply if I apply zeta inverse mind you then I will end up with zeta inverse of Tj bar Y1 and these fellows are going to be regular functions on UIs where UIs are open subsets of X1 ok. So these are going to be regular functions on UI which are open in X1 ok. So I am going to end up like this now you see if you so if you look at if you look at that carefully so let me rub this. So you know what you do is you do the following thing. So this zeta of Si bar by 1 is a regular function on Vi for each i equal to 1 to n ok. Now you take the intersection of all those Vi's ok put V equal to intersection or V is equal to intersection i equal to 1 to n Vi ok put this. Then you see then the zeta of S1 bar by 1 and so on zeta of Sn bar by 1 they are all regular functions on V alright because each one is regular on Vi and all of them put together is going to be regular on V alright. So you know the point is that you see we use this fact here. So if you recall if you recall the there is a there is a canonical bijection between the morphisms of varieties from any variety if I call it as so I am short of notation so let me use some script notation I am short of symbols so let me use some script notation you take any variety X and you take an affine variety Y ok and if you look at the set of morphisms of varieties from script X to script Y alright script X to script Y then this is in a natural bijection with the set of all homomorphisms of k algebras from Ay to OX and mind you Ay is the same as OY because Y is affine ok. See we have already seen this we have seen this bijection set the set of all morphisms from a variety into an affine variety is in bijective correspondence with the set of all k algebra homomorphisms from the fine co-ordinate ring of the target affine variety to the regular functions on the source variety ok. Now you know you see now you see what you do with these guys ok what you do with these guys is that you know I can use this to define I can use these regular functions and regular functions to define your morphism from V into X1 ok. So what you do is you see what you do is the following I mean it is exactly if you go back and look at it that is how we establish this bijection ok. So what you do is you see you have see zeta restricted to zeta restricted to A we will actually go from A into zeta of A which is sub of OV ok. You see zeta you see this zeta is from the quotient field of A to the quotient field of B alright and what is any element of A any element of A is a polynomial in the SI bars ok any see the SI generate the polynomial ring of affine space and their images will generate the SI bars will generate A any element of capital A is actually can be thought of as a polynomial in the SI bars ok and therefore you know but all the SI bars their images under zeta are all lying or they are all landing in the K algebra OV they are all landing in the K algebra OV and therefore you see zeta will land inside OV the image of zeta will land inside OV alright and as a result you see you get a K algebra homomorphism from A which is A of x1 to OV if you apply this bijection this will correspond to a morphism of varieties from V to V to x1 ok alright. So you know in this you apply with script x equal to V and script y equal to x1 ok and you will get a morphism from V to x1 so by this result this gives rise to a morphism of varieties from V to x1 ok you get a morphism like this and in fact you know it is even it is even easy to write what that morphism is it is even easy to write what the morphism is the morphism is see after all this V is see this V this V is an open subset of y1 and y1 is V is an open subset of y1 y1 is sitting in affine space so you know any point of V is given by m coordinates ok and you know you just take this point of V with m coordinates mu1 etc mu m and you know what you will have to send it to you will have to just send it to you know these are all regular functions on V ok these are regular functions on V alright and therefore they are given by quotients of polynomials in m variables on V and you substitute for those m variables these values ok. So that will give you an n tuple and that n tuple will precisely be an n tuple of an it will land inside x1 ok so if you call this as mu ok then this is just this is just the map psi s1 bar by 1 of mu and so on sorry zeta s1 bar by 1 acting on I mean evaluated at mu and so on up to zeta sn bar by 1 evaluated at mu. So you will get this you will get this n tuple and this n tuple will be a point of n in fact it will be a point of x1 that is how this bijection was established if you go back to that lecture and try to recall ok. So you get this morphism like this ok and you know if you call this so you know if you call this morphism as again let me use some notation so now let me use psi here or let me use some other notation so I will use capital psi ok I will use a block psi like that so I have morphism like this right and what is the meaning that this comes from this it means that this is actually psi hash ok. The map from here to here is just neta going to neta hash because you know given a morphism of varieties you have the morphism at the regular functions level given by pull back of regular functions ok and what does it mean to say that I started zeta restricted to a here and I ended up with this psi there it means that this psi is zeta sorry I started with the zeta restricted to a here and I got a psi here it means that psi hash is that zeta ok so this zeta restricted to a psi hash alright and moreover and this psi you know what you must remember is that this psi induces psi hash is zeta and mind you this zeta is part of an isomorphism at the quotient field level this zeta from a to ov if I go to the quotient field ok if I go to the quotient field level it is an isomorphism zeta is an isomorphism quotient field level that is something that you should not forget alright but anyway you know you can so what we have done is you know if you look at the so let me rub this diagram lest it cause any confusion so you know the situation is now at what we have done is we have we have this x and we have this x1 which is affine open inside x and well of course we embedded x1 inside affine space but that is not important now and then there is y1 here which is also affine open in y and you see what I have done is I have got this v which is an which is an open set in y1 and I have got a morphism like this I have got a morphism like this alright and in the same and for that I use the fact that all the zeta si bar they are all regular functions on this open set now I play the same game with zeta inverse of the tj bars okay so so similarly if we put u equal to intersection i equal to 1 to I mean j equal to 1 to m of uj okay then then zeta inverse of t1 bar by 1 etc to zeta inverse of tm bar by 1 they are all going to be in o u okay where u is an open subset of u is an open subset of x1 okay and therefore if you take zeta inverse and restrict it to b it is going to give you a map from b to o u okay and so I should say from b to zeta b which is sitting inside o u mind you this is an isomorphism this is also an isomorphism because they are images under an injective ring home k algebra homomorphism k so zeta b is so zeta inverse b goes into o u alright and so you know again by this yoga what will happen is that you are going to get a morphism of varieties from u to y1 given by some let me call it as phi capital phi like this okay such that well when you take this morphism and you take the pullback by regular functions it gives you get phi hash and this phi hash is actually equal to zeta inverse alright so this means that you know we have got one like this so you have you have u open inside x1 and you have got a you got a so if this is if this is this is psi then this is phi okay so u is here I have u is sub of this so I have psi inverse if u this is an open subset of this okay and then here I take phi inverse of c inverse of u that is that here is an open subset okay and the fact is that phi will carry this into this and psi will carry this into this the only thing that you will have to worry about is that the image of psi the image of psi has to intersect u okay that is an important thing because if the image of psi does not intersect u then psi inverse u will become empty alright and similarly the image of phi should intersects psi inverse u okay and so these non-emptiness statements come because the both psi and phi at the at the pullback of regular functions level they induce in fact injective maps okay and it is this injectivity that ensures that first that if you take u will intersect the image of psi and that similarly psi inverse u will intersect the image of phi okay that is because of the injectivity of that follows from the way we have constructed them right and therefore you get this you get this you get these maps and they will be inverses to each other so the fact that you are using is that you know if you have two open sets and you have if you have two open sets I mean if you have two varieties and you have two morphisms going in opposite directions such that you know the composite morphisms so phi followed by c here and then psi followed by phi there if they induce identity at the quotient field level okay if they induce identity at the quotient if they induce the identity map at the quotient field level identity of the quotient field then they have to be inverse isomorphisms okay so you know if you go to the quotient field level psi followed by phi is just going to be applying you know this psi is going to induce you see this psi psi induces zeta at the quotient field level phi induces zeta inverse at the quotient field level so you know psi composition phi and phi composition psi they will induce identity maps at the quotient field level if you have two morphisms alright which induce upon composition identity at the quotient field level then those morphism then those morphisms have to be inverses of each other okay so that will tell you that you have so this gives open subsets of x1 and y1 which are isomorphic as well as what I wanted to remark is that you know this when we the way we got the isomorphism the by ration in the case of isomorphism local rings so the same thing will also work in this case okay so what you can do is that you can you can you can do something like this okay which was in the local ring case right so in fact you can write you can get these things more explicitly by doing like this okay so you know you have so if I so if I have this off and try to write that out explicitly okay so you just have to take see if you take zeta of si bar by 1 this is a this is an element here in the quotient field of B so it is a it is a fraction so it is a so it is some fi bar by gi bar okay and then so you know and then if you take the if you apply zeta inverse to 1 by gi bar alright I will get some ai I will get yeah I will get an ai bar by bi bar okay so 1 by so if it takes zeta of si bar by 1 I will get fi bar by gi bar where fi bar and gi bar are you know polynomials in the t bars okay they are they are elements of B and because this is after all the fraction field of fractions of B every element here looks like a quotient of elements of B okay and if I take the 1 by gi bar okay which also makes sense in the quotient field of B okay and I apply zeta inverse I will again get ai bar by bi bar alright and zeta inverse of if I apply zeta inverse to tj bar by 1 I will end up with some capital ai bar by bi bar I mean aj bar by bj bar where aj capital bj are polynomials here okay in the s's right and if I take I apply zeta to 1 by bj bar okay then that is going to go to some small fj bar by small gj bar okay and now what one has to do is that so whenever I have applied zeta I am on the I am on the you know I am on the side of the B I am working with the t bars and whenever I applied zeta inverse I am working with the s bars okay so these two these two are connected with the t bars and these two are connected with the s bars.