 What we need to now look ahead to in the course is to try to expand the category of varieties okay, so far we have been looking at affine varieties and quasi-affine varieties which are open subsets of affine varieties but then we need to also include more general affine varieties and the next in this list are the projective varieties and the quasi-projective varieties and the projective varieties are they have properties is very different from the properties of affine varieties okay. So you know what I wanted to start with is since we are looking at we have been looking at affine varieties okay, the first thing I wanted to say is about the so called Jacobian conjecture which is a very simply stated conjecture but which is open even the simplest case okay and the reason it makes sense to talk about that conjecture now is because you know what automorphisms you know what morphisms of varieties are you know what are isomorphisms of varieties and you know how to characterize isomorphisms of affine varieties okay. So I will first recall the following thing from the previous lecture so if you recall x any variety of course here for us any variety will mean either affine variety or quasi-affine variety okay and y an affine variety which means y is an irreducible close subset of some affine space over an algebraic closed field of course then we have a natural bijection so on on the one hand you have morphism of varieties from x to y that is bijective to homomorphisms of k algebras from Ay to O x okay we proved this bijection and in fact what was the what was the if you recall this map if I call this map as alpha then how does how is this map defined it is the map in this direction is just given by pull back of regular functions okay so you know in other words if f from x to y is an element on this side you need some morphism from the variety x to the affine variety y then you send it to alpha f this alpha f is going to be a k algebra homomorphism from the ring of polynomial functions on y to the regular functions on x and that is very very easy namely you give me a polynomial function let me put it as p on let me put it as capital P give me a polynomial function capital P on y okay you compose it with f to get a regular function on x note that a of y is the affine coordinate ring of y the coordinate ring of functions polynomial functions on y it is just polynomials restricted to y and these are the polynomial functions on the affine space in which y sits y is an affine variety so y sits inside some an the ambient the bigger affine space and on this bigger ambient affine space you have the polynomials in n variables and each of these polynomial functions by evaluation defines a map into k which can be thought of as a map into a1 and it is a regular function of course and you restrict such polynomials to any subset in particular to y and you get a polynomial function on y only the only thing is that this polynomial function on y is not represented by unique polynomial it is represented up to addition by a polynomial in the ideal of y okay which consists of polynomials which vanish on y okay so give me a polynomial restricted to y and I just compose it with f first apply P then apply f that is a regular function on x because a polynomial function restricted to y is of course a regular function on y and what is happening is I have a regular function on y and then by composing it with f I get the pullback of the regular function on x and the pullback of a regular function has to be a regular function because f is a morphism because that is already built into the definition of a morphism. So this is how this map is defined okay and there is also there is also the if you look at the map in the reverse direction alpha inverse how is that defined well give me a give me a home of some phi from k algebra home of some from Ay to OX okay and then of course you know if of course you know you assume that y is thought of a sitting inside affine space and Ay is just well kf is identified with kf x1 xn which is the polynomials on the affine space dxi being the coordinate functions divided by the ideal of y okay and so what you do is that you just take the xi you take the xi bars which are elements here they are the images of xi in this quotient okay. So xi bar just denotes the coset xi plus iy in the quotient ring okay and you simply send it to a certain function let us call it as hi it is a regular function on x and the fact is that you will what will happen is that from x to an you will have a map which will send any point x small x to this h1 x this n tuple defined by the hs hn of x because I have n of the xi's so I have their images here which are n of the xi bars. So I get n of the hi's okay and then I evaluate this point at each of these n functions I get an n tuple which is a point in an this is the map and the fact is that this map will factor through a morphism g through the close irreducible close of variety y of an this diagram will commute and this phi will be nothing but alpha of g this is the surjectivity okay. So start with a phi here then you get these hi's using the hi's you define a morphism of x into an the morphism will land inside y and if you call that morphism is g then alpha of g is the phi that you started with okay that is the surjectivity that gives all the surjectivity and that also defines how that the alpha inverse of phi is this g okay g is just alpha inverse of t this is how you get the inverse map and that is how this is a bijection okay and as a corollary to this what happens is that if you know if x is also an affine variety then O x can be replaced by A x because for an affine variety the ring of regular functions it can be naturally identified with the polynomials restricted to the affine variety okay. So and alpha will take a bijection to a bijection okay and therefore what it will tell you is that if x is also affine then x and y are in fact I should say alpha will take an isomorphism to an isomorphism okay that is an invertible not just a bijection but an invertible morphism to an invertible morphism an invertible morphism varieties will go to an invertible ring home morphism okay that is a ring isomorphism and isomorphism of varieties will go to an isomorphism k-algebras okay. So alpha will carry isomorphism to isomorphisms provided you know x is affine alright and what that will tell you is that it will tell you that 2 affine varieties are isomorphic if and only if they are coordinate rings affine coordinate rings that is just the rings of polynomials on those varieties are isomorphic. So the corollary to this is the corollary to this is x affine and y affine are isomorphic if and only if if and only if Ax and Ay are isomorphic of course this the isomorphism when I say x and y are affine and isomorphic there I mean isomorphism is varieties and when I say Ax and Ay are isomorphic I mean isomorphism as k-algebras okay. So 2 affine varieties are isomorphic if and only if they are rings of polynomials the rings of polynomial functions on those affine varieties are isomorphic as k-algebras okay and how many such isomorphisms are there there are as many as there are as many isomorphisms here as there are isomorphisms here okay. So you know as a particular case what you can do is well you know I can take I can take for x and y just An itself. So you know take x equal to y is equal to An so what you will get is you will get morphisms of varieties from An to An is is bijective to the k-algebra homomorphisms from Af An to Af An to Af An and of course you know Af An is the polynomial ring. So you know if you want well both of these are equal to k if you want x1 etc xn okay if you take them to be the ring of polynomials if you take the ring of polynomials to be the ring of polynomials in n in determinants x i's okay and in particular if I look at the automorphisms so the word automorphism means a self isomorphism it is a morphism is from one object to another object an automorphism is a morphism from the object back into itself and in fact a morphism of an object back into itself in general is called an endomorphism okay and an invertible endomorphism is called an automorphism okay. So here of course when I say morphisms from An to An actually I am looking at the endomorphisms of An so this is all the endomorphisms these are maps from An back into itself they are endomorphisms so of course the you know the other notation is endomorphisms as varieties of An and of course here what I will have is endomorphisms as k-algebras of the polynomial ring Af An okay and what are the automorphisms are the invertible endomorphisms they are i's they are endomorphisms which are also isomorphisms okay so they are self maps they are morphisms of the object back into itself which can be inverted okay. So if you look at the automorphisms that is of course if you look at it carefully this is a group because the composition of two morphisms is again a morphism therefore the composition of two automorphisms is again an automorphism so this is a group and on the other hand you also have a group here this is automorphisms as k-algebras of k well let me write as A of An okay. So and then alpha carries automorphisms to automorphisms okay so this alpha will also give you a map like this okay you start with a morphism a morphism is an isomorphism which means it is here if and only if its image is here and conversely okay and now the you see the Jacobian conjecture is connected with automorphisms of the polynomial ring alright. So let me make a statement so what I want to look at is let us take any morphism from An to An okay how is it going to be given it is going to be given by N polynomials in the N variables okay any element of phi of R let me use f of morphisms of varieties from An to An is given by polynomials in fact so let me write this f1 etc fn in N variables so N polynomials in N variables that is just because of this bijection okay. So you know what are all these you know I start with so here is my f from An to An then you know I have this is mapped that is a point that is an element here it is mapped to an element here this is alpha of f and this alpha of f is a map from well k x1 etc xn to k x1 etc xn where I am identifying the ring of polynomials on An with k x1 etc xn okay and how is a from a polynomial ring a map is dictated by the images of the variables. So you know if I take alpha f of xi this is what alpha f takes xi2 and this N tuple dictates the k algebra of morphisms alpha of f because the universal property of the polynomial okay so what I want to tell you is that and it is these alpha f of xi that I am calling as fi's okay so let me write that here I need a better duster. So I have xi going to alpha of f of xi okay and I am calling this as fi okay so essentially what you are doing is that corresponding to this morphism An to An that corresponds to giving me N polynomials okay and that is what I have written in the previous line that a morphism that a morphism from N to N is simply given by N polynomials in N variables okay and when is this morphism an isomorphism when it has an inverse okay f is an automorphism that is it is a invertible morphism namely an isomorphism if it has has an inverse okay and so it has an inverse f inverse so you see what will happen is I will have an f inverse which again will go from An to An okay and that will be mapped under alpha to alpha of f inverse that will turn out to be again a map k algebra homomorphism from the polynomial in N variables to again to itself back into itself and that is again going to be dictated by the images of the xi's under this so alpha of f inverse of xi okay and well if I call this as you know if I call this as gi if I call this as gi which means I am just calling f inverse as gi okay by my previous notation if f goes to f1 etc fn then f inverse equal to gi will go to g1 etc gn alright and what you can see immediately is that see you will have if I start with the point x1 etc if I start with the point with coordinates lambda 1 etc lambda n that will go under if I apply the map f to it what I am going to get is f1 of lambda 1 lambda n and so on fn of lambda 1 etc lambda n this is the point is going to go to okay that is what it means to send the xi's to fi's under alpha of f okay and now this point is if I apply if I apply f inverse which is gi this point has to go back to this but under f inverse where will this point go see this point will go to g1 of f1 of the lambdas I will put a I will put it like this dot dot dot gn of f1 lambda fn lambda this is what it means so what where of course you know where lambda underline is just lambda 1 through lambda n okay. So what you get is gi of f1 lambda etc fn lambda is simply lambda i for every i this is what happens if you have an automorphism right and now what you can do is you know this holds for all lambdas okay so you can write this in a variable form as gi of f1 of x fn of x is equal to xi for every i you can write this in variable form okay which make sense alright and then you know for example if you want you can take the partial derivative on both sides with respect to any xj you will get of course if it is x if you take partial derivative with respect to xi you will get 1 on the right side if you take partial derivative with respect to xj of j not equal to i you will get 0 then what you can check is that you can check you can check that the Jacobian of dou fi by dou xj is a non-zero constant in k. So what you must understand is that you see I take each of these polynomials each of these fi's each fi is a polynomial in n variables and if I differentiate partially with respect to each variable I will again get a bunch of polynomials okay I have n polynomials I have the n fi's I differentiate each n of them with respect to the n variables. So I will get I get this Jacobian matrix okay that is in general going to be a matrix of polynomials okay because you take a polynomial in n variables and differentiate it partially with respect to one of the variables the resulting is the resulting thing is again a polynomial in n variables okay and then therefore if you look at the Jacobian determinant then in fact I think the correct maybe it is better to write it as course here x underline stands for x1 through xn and maybe it is better to write this as Jacobian of f and call it like this okay and the point I want to make is that you see if you calculate the Jacobian determinant it is going to only be you expect it only to be a polynomial because every entry is a polynomial gotten by taking a partial derivative with respect to certain one of the variables and the fact is f has an inverse g okay and therefore you know if you write down everything f followed by f inverse is identity okay and for the identity function if you take the Jacobian you will simply get the identity matrix okay so finally what will happen is that you will see that the Jacobian polynomial of f into Jacobian polynomial of g will if you take the product polynomial it will be equal to 1 which will be the Jacobian polynomial of the identity map which is just the identity matrix you have two product of two polynomials equal to 1 so each of them has to be a constant it has to be a non-zero constant that is why Jacobian of f will be a non-zero constant. So what is a simple argument shows so far is that you know if I start with an automorphism okay I end up with I end up with you know you have bunch of functions I end up with n polynomial functions to start with an automorphism f I end up with n polynomial functions whose Jacobian is a non-zero constant okay that the converse of this is true is the Jacobian conjecture okay so the Jacobian conjecture is if f is not known to be f is not given to be to be an isomorphism initially does then the condition that the Jacobian of f is g is a non-zero constant implies f is an automorphism okay this is the Jacobian conjecture the Jacobian conjecture is that if I start with an f which I do not know is invertible I do not know is an isomorphism but suppose I have the condition that if you take the Jacobian of f namely you take the Jacobian of this n polynomials that specifies f okay under this correspondence if this Jacobian is a in generally you expect it only to be a polynomial in invariables but suppose it turns out to be a non-constant I mean a non-zero constant polynomial then the Jacobian conjecture says that f should be invertible that means you should be able to find another set of polynomials n polynomials which if you plug in to f the f's you will get back id that is the Jacobian conjecture and the point is that somehow so another way of stating that is that you know there is a map from here to the polynomial ring given by taking determinant of the Jacobian okay every endomorphism is given by n polynomials and you take the Jacobian determinant of those n polynomials you will get a polynomial. So you get a map from this endomorphism to your ring of polynomials and what you have seen is that the inverse image of k star that is non-zero elements of k that well that contains this okay namely every automorphism for every automorphism the Jacobian is a non-zero constant but the question is whether if you take the inverse image okay then it will be exactly this that is the question okay namely if you give me n polynomials okay for which the Jacobian determinant is non-zero constant do those n polynomials actually correspond to an automorphism is the question. So the question is whether the inverse image of the non-zero constants under the Jacobian determinant map from this endomorphism is exactly this we know it contains this but what is required to show is that this is exactly this okay that is the Jacobian conjecture and the beautiful thing is that even for k equal to complex numbers and even for n equal to 2 just polynomials in 2 variables this is open okay and it is a very difficult problem and the beautiful thing is that in the case of complex numbers we have also complex analysis holomorphic functions we have that theory also but it does not seem to have helped okay. So this is a very deep problem and working being able to solve this or being able to give a counter example to this is worth being awarded by a fields medal which is the equivalent of you know well priced in mathematics. So that is the depth of the problem it is a very hard problem called the Jacobian conjecture and the point is it can be stated now because you people know that there is a bijection between you know isomorphisms of affine varieties and k-algebra homomorphisms of their coordinate rings so that is the reason why I wanted to state it here. So maybe I hope that many or at least some of you will go ahead and try to tackle this problem in your future career okay alright. So now okay so now this is just to bring in this Jacobian conjecture what I want to do now is I want to go to the question of more general varieties okay so this is what I want to do next right. Now so let me give first an example so what are the varieties that we know so far we know affine varieties and we know quasi-affine varieties okay and quasi-affine varieties are open subsets of affine varieties and you know that there are quasi-affine varieties which are actually affine and for example basic open subsets they are all quasi-affine varieties but they are actually isomorphic to affine varieties in affine space of one higher dimension okay because of the so called Rabinovich trick okay. So the basic open set defined by non vanishing of a polynomial is a quasi-affine variety in that affine space but in an affine space of dimension one more it becomes a closed irreducible closed sub variety okay. So the question you can ask is are there quasi-affine varieties are not affine so I want to give an example of such a specimen okay and just to tell you that there are quasi-affine varieties are not affine varieties. So here is a fact so here is a claim or I can just put it as lemma a2k-origin is not affine is quasi-affine but not affine. So here is the lemma actually it is more than a lemma it is you can call it a theorem because you are going to use this result and you know that this result is in it is a grand version of the Nulstrand sets. So it is very bad to call it a lemma but anyway I will call it a lemma alright. So it is usually or maybe I will have second thoughts and at least call it proposition. So what are we going to do? See first let us understand the statement it is quasi-affine because it is an open subset of an affine variety namely it is an open subset of a2 and it is a non-empty open subset okay because I have only deleted the origin alright it is a complement to the origin. So it is a non-empty open subset so it is certainly a quasi-affine variety but I want to show it is not affine what do you mean by saying it is not affine what I mean by that is it cannot be isomorphic to any affine variety that means you cannot find an isomorphism of this punctured plane okay this is a punctured plane this punctured plane you cannot find an isomorphism of that with an irreducible closed subset of any affine space that is what it means okay that is what it means to say that it is not affine right. So it means that if I take any map from the punctured plane into any affine space certainly it is never ever going to be a closed embedding I can never expect it to be a closed embedding alright. So you see it is like if I if you state it in this generality it looks the proof looks very difficult to you know verify because you are trying to say that I will have to just try to think of look at all possible you know maps morphisms of this into various affine spaces and I will have to check that each one of these is not a embedding onto a closed subset it is not an isomorphism onto a closed subset that is how that is what it means okay but there is a but the way we prove it is rather we use all the techniques that we have developed so far that is some of them. So we prove by contradiction we go by contradiction so suppose a2-point a2-origin punctured plane is affine okay now again what that means is that I am assuming that is isomorphic to an affine variety okay. So see if it is affine then it has an affine co-ordinate ring alright. So then we have a bijection as we have seen there is this bijection alpha which is morphisms of varieties from a2-point-origin to a2 so I have this map from this into homomorphisms of k-algebras from A of a2 to O of a2-point okay. So I have this bijection that I have already written down here okay the set of morphisms from any variety into an affine variety is given by is in bijection with the set of all k-algebra homomorphisms from the co-ordinate ring of the target affine variety to the regular functions from the source variety right. So I am just applying that here but I am noting that well you know since I have assumed a2-point to be an affine variety this can be replaced by A of that okay and of course so this is this can be replaced by A of a2-0,0 and what this means is that what does this mean this means see because I have assumed a2-point as affine there is some embedding of it as a some embedding of it into some affine space some big affine space I do not know what dimension and certainly dimension greater than or equal to 2 alright and in that affine space since it is a realize as a closed sub variety I take the ring of polynomial functions there and that is what this means and these two are one and the same okay that is something that we have proved for an affine variety the regular functions and the ring of polynomials are one and the same okay of course by polynomials I mean polynomials are restricted to the affine variety right. So now the so in particular you know I will look at the canonical inclusion I will look at the inclusion map a2k-the origin this is sitting inside a2 and there is a natural inclusion there is this map sorry let me put the I here okay then I will get this will go to alpha of I okay and what is this alpha of I this alpha of I is going to be a map from k of let me write this like this k of x1 etc xn to you know O of a2k-the origin okay I will have this now the you know that under this bijective correspondence okay oh sorry I should know I should not write x1 etc xn it should be only just x1 x2 sorry of course yeah so alpha I is from k x1 x2 because it is just affine space two dimensional affine space is just polynomials in two variables so that n I wrote below before was equal to 2 right. So I have this map see okay so what I told you is that under this bijection you say I told you isomorphisms correspondence isomorphisms okay what I will prove what you can actually see is that alpha of I is actually an isomorphism therefore it will mean that I has to be an isomorphism but I cannot be an isomorphism because it is not even surjective because there is a point missing okay so that is how you will get the contradiction okay then this contradiction will prove that a2-a point cannot be identified with any close sub variety of any affine space so it is not affine okay. So we claim that alpha of I is an isomorphism this would imply that I is an isomorphism and that is the contradiction so this contradiction will prove that our assumption our original supposition that the point of plane is affine is wrong okay so this proof proceeds by contradiction right. So how to show that alpha of I is an isomorphism okay there are two things that need to be done it is a k-algebra homomorphism I have to show its injective then I have to show its surjective okay. So let me explain why alpha of I is injective so first of all let us understand what alpha of I is alpha of I is you know this alpha is just the map that is induced by pull back of regular functions okay so what is the meaning of alpha of I it means you give me an element here namely it is a polynomial in two variables so it is a function on this a2 okay and if you compose it with I which amounts to just restricting the polynomial to the punctured plane that is what it goes to. So it is just a polynomial going to polynomial restricted to a2-a point a2-the origin this is what the this is what the map is because pull back means you take a regular function the target you compose it with the map in this case the morphism is I but composing a morphism with I is the same as restricting that morphism because I is just the inclusion of this subset okay so pulling back a back a map under an inclusion is just restriction to the subset corresponding to the inclusion. So what is this alpha of I it is just take a polynomial and restricted to the punctured plane okay now how would I show that alpha of I is injective home morphism by showing its kernel is 0 because after all it is a K algebra home morphism to check it is injective I have to just show its kernel is 0. So if I have a polynomial function okay if I have a polynomial which if I restrict to a2-a point vanishes okay then it vanishes everywhere because you see if a polynomial vanishes on a set it will also vanish on the closure of the set because of the continuity of the polynomial for the Zariski topology okay. Therefore if the polynomial vanishes if this is 0 then you are saying that the polynomial restricted to the punctured plane is 0 but the punctured plane is a dense open it is an open subset it is a dense open subset any non-empty open subset is irreducible and dense okay. Therefore this polynomial is going to vanish on a dense open subset therefore it will vanish everywhere by continuity. So what will tell you is that this polynomial as a function vanishes everywhere and in this case this because you are working with an infinite field is an algebraically closed field so it is infinite. So if a polynomial vanishes as a function then it has to vanish as a polynomial. So see so that tells you that alpha of I is injective okay so alpha of I of P is equal to P restricted to a2- the origin is equal to 0 implies P equal to 0 so alpha of I is injective this tells you that alpha of I is injective the only thing that I will have to now prove is that alpha of I is surject if I prove that then I would have then I am done then it will then I would have proved that alpha of I is an isomorphism and we would get the contradiction that we want alright. So how do I prove it is surjective so we do it like this so well alpha of I is surjective how do I prove this well so what I do is I start with I start with the regular function on the punctured plane and I have to show that it is given by the restriction of a polynomial okay. So what is the statement I start with the regular function on the punctured plane and I have to prove that this regular function is nothing but restriction of a polynomial in two variables okay. So start with a regular function phi in on the punctured plane to show phi is equal to restriction of a polynomial in kx1x2 this is what I have to prove alright this is what I have to prove. Now so you know so let me draw a diagram so you have something like this you have the origin this is a2 and I have thrown out the origin so I will put a circuit here okay so this is the punctured plane and I have a regular function on this namely well I have a phi which takes values in a1 so this is my phi and I have to show this phi is actually coming from a polynomial right and what is the meaning of saying that this is a regular function let us go back into that give me any point x in a2 okay then you know I can find an affine neighborhood I can find a neighborhood of x ux okay so you know so this is a neighborhood of x okay and this is I must admit that this diagram is not accurate because the neighborhood does not look like that any neighborhood of a point is going to be a non-empty open set so it will be dense okay so more ideally I should think of the neighborhood well as a compliment of some curves okay right this is how it should look like but then you know so you know if I take a neighborhood ux of the point x it is a compliment of points of curves alright because only closed subset the closed subsets here are curves and these are the one dimensional closed subsets and zero dimensional closed subsets will be points so it will be it will be the compliment of some curves and some may be finitely many points okay that is how an open set here will look like so given a point x I have this neighborhood ux and then what I have is the fact that it is a regular function means that on this ux it is given by a quotient of polynomials with the denominator polynomial not vanishing on ux okay so let me write that there exists an open set open of course the Zariski topology ux containing x and polynomials g i gx hx in kx x1 x2 such that phi restricted to ux is the same as gx by hx restricted to ux and hx does not vanish on ux okay this is the definition of what a regular function is a regular function is something that is locally given by quotients of polynomials and to make sense of the function defined by quotient of polynomials the denominator polynomial should not vanish because you cannot divide by 0 okay so this is what it means alright but now notice that the you know the what I need to prove is that phi comes from a polynomial alright and if phi came from a polynomial then that polynomial restricted to ux will be equal to this quotient of polynomials restricted to ux okay and you know if I cross multiply it what I will get is that I will get hx times that polynomial equal to gx everywhere okay because the fact is if two polynomial functions agree on an open set they agree everywhere if two regular functions agree on an open set they agree everywhere okay the reason is because open sets are dense and polynomial functions regular functions they are all continuous and if a continuous function is 0 on a non-empty open set it is identically 0 so if two continuous functions are equal on a non-empty on a dense set okay then they have to be equal everywhere if two continuous functions are equal on a dense set then they have to be equal everywhere by continuity alright so what this will tell you is that you know finally I have to prove phi is a polynomial okay so it will tell you that hx has to divide gx okay it will tell you that the polynomial will be equal to gx by hx on ux okay but if I cross multiply it will tell you that polynomial into hx is equal to gx on ux but then that will mean polynomial into hx equal to gx on whole affine space because if two polynomials coincide on an non-empty open set they are the same okay. So what it will finally tell you is that hx divides gx okay so to obtain a contradiction I will assume hx does not divide gx and I will try to obtain a contradiction so assume hx does not divide gx for every x I assume that of course you know I am in the polynomial ring which is a unique factorization domain and it makes sense to talk about when one divides the other because you have unique factorization any polynomial can be uniquely factor into irreducible polynomials so this assumption is this assumption will be true only if phi does not come from a polynomial function mind you please understand I have to prove phi comes from a polynomial function okay in other words I have to show phi is just restriction of a polynomial alright but if phi is a restriction of polynomial it will mean that h divides g okay conversely if h divides g alright then phi will be restriction of a polynomial alright so if you assume phi does not come from a polynomial which means if you assume subjectivity is false then you are actually assuming that hx does not divide gx that is what you are assuming okay so let us assume that and come to a contradiction. Now note that on ux intersection ux prime we have well gx by hx is equal to gx prime by hx prime okay you will have this alright and that will tell you that you know gx hx prime is equal to gx prime hx alright and now you see hx divides the right side so hx divides this but hx does not divide gx so it has to divide hx prime okay and you will similarly get hx prime divides hx so the moral of the story is that hx and hx prime will differ by a nonzero concept okay so this will tell you that hx prime is equal to some lambda x x prime into hx this is what you will get okay that is because you see hx let me again repeat the argument hx divides the left side so hx divides this product so it has to divide it does not divide gx that is our assumption so it has to divide hx prime but this argument is symmetric in x and x prime so you will also get hx prime divides hx okay and therefore if two polynomials divide each other they have to just be constant multiples of one another and that constant of course would be a nonzero constant okay so this lambda xx prime is not nonzero and well if you put that back into this what you will get is that you will get the gx into lambda xx prime is equal to gx you will get this okay so it also tells you that the g's are they differ by a nonzero constant multiple of one another alright now what I am going to do is I am going to do the following thing you know a2k is noetherian this is noetherian and you know any subspace of a topological space that is noetherian is also noetherian so this will tell you that a2- the function plane a2-0 0 a2- origin is also noetherian and you know noetherian topological space is quasi compact okay therefore what will happen is that so what you will get is that this is quasi compact so you will get a2- is quasi compact in fact one characterization of noetherian topological space is that every subspace topological space is noetherian if and only if every subset is quasi compact okay so this is quasi compact but you know if I take all the ux's as x varies in a2-0 point I get an open cover for a2-0 point and that is quasi compact so finitely many of these should be enough to cover a2-0 point so this implies that the open cover ux x belonging to a2-0 point a2 puncture plane has a finite sub cover say ux1 union uxn uxm is equal to a2-0 point and mind you each open set is being taken in a2-0 point okay I am for every point in this I am looking at an open set there if you want the only problem is that this open subset might contain the origin but then you can throw it out you can simply throw it out from an open set and replace it with you can puncture it with the origin if it contains the origin so you see this therefore this union will be this alright now well you see what this will tell you is that if you look at it the it will tell you see the hx does not vanish on ux okay so it will tell you that ux is contained in dhx okay so it will tell you that ux is contained in dhx alright and therefore what it will tell you is that all the dhx corresponding dhx's will certainly contain this alright. So this will tell you that dhx1 union dhxn xm will be a2-0 point or it may even be a2 itself so in any case every dhx contains ux so you know if you take the union of all these dhx it has to contain this punctured plane it might even contain the origin alright. Now what I want to tell you is that each of these has a meaning if you if you exhaust each case you will get the contradiction if dhx1 union dhxn is equal to a2 that is equal to the punctured plane what this will tell you is that if you take compliment it will tell you z of hx1 hxn is the point because you know if I take compliment of dh I will get zh and intersection of all the zh is just z of this okay but you see for a point this is actually z of the maximal ideal x1 x2 because the point 0, 0 corresponds to the 0 set of the maximal ideal x1, x2 alright. But you see and mind you that you know all the h's they are all multiples of constant multiples of one another so if you look at the 0 set of all the h's it is essentially 0 set of a single polynomial and you are seeing the 0 set of a single polynomial is a point which cannot happen because 0 set of a single polynomial has to be a curve okay because you know that the you know there are these we have seen this in one of the earlier lectures that there is a notion of geometric hyper surface and there is a notion of commutative algebraic hyper surface it is a locus which is given by vanishing of single polynomial and it has a dimension one less okay. So if all the h's are just multiples of one polynomial then this is just the 0 set of one polynomial and you are seeing a 0 set of one polynomial in two variables is just a point that cannot happen the 0 set of a polynomial in two variables has to be a union of hyper surface. So in this case it should be a union of curves it cannot be a single point so that is a contradiction a contradiction okay the 0 set of a single polynomial in two variables cannot be a single point. On the other hand so the other possibility is that the union of all this is A2 okay if the union of all these things is A2 it will tell you that all the this is something that we have already seen it means that all the h's will generate the unit ideal the ideal generated by the h's will be the whole polynomial okay if dh hx1 dhxm is A2 what it means it will mean is that z of hx1 hxm is a null set okay and this means that the ideal generated by hx1 hxm is the whole polynomial because one version of the notion says that if you take a non if you take a proper ideal a non trivial ideal then the 0 set of the ideal cannot be the null set okay. So the 0 set of an ideal is a null set if and only if that ideal contains is the unit ideal right. So but then you know but all the h's are all multiples of one another so you are saying that this polynomial in two variables is generated by a single element and that is again a contradiction okay. So again a contradiction okay it is a contradiction because you know all the hx's are multiples of one another so this is the ideal generated by single polynomial and you are saying the ideal generated by single polynomial contains one. So it will mean that that polynomial multiplied by some other polynomial is equal to 1 which would mean that that polynomial itself is a non-zero constant but then of course I have assumed all the polynomials the h's so it will finally reduce to assuming that all the h's are constants but then that will tell you that phi's are all polynomials but I already assumed that phi does not come from a polynomial so again get a contradiction okay. So both these contradictions demonstrate that you know if you assume that this phi does not come from a polynomial then you get a contradiction so it means phi comes from a polynomial every regular function and the punctured plane is the restriction of a polynomial in two variables therefore the map is surjective and we are done okay. So that finishes the proof and therefore the moral of the story is the punctured plane is an example of a Cauchy affine variety which is not affine okay. So we do have Cauchy affine varieties which are not affine right so I will stop here and what I am going to do in the next lecture is I am going to tell about projective varieties and Cauchy projective varieties which are the more general varieties than affine and Cauchy affine varieties okay. So let me stop here.