 Okay so let us continue with discussion, so what we have just shown in the previous lecture is that if you start with any variety Y and you take an affine variety X then there is a bijection between the set of morphisms from Y to X and the K-algebra homomorphisms from the ring of affine functions, ring of polynomials on X to the regular functions on Y and this map is just the map induced by pull back of regular functions okay. So we are just saying that morphisms into an affine via the pull back of regular functions correspond to K-algebra homomorphisms from the ring of functions on the affine okay. So let me repeat that morphisms into an affine correspond via the pull back of regular functions to K-algebra homomorphisms of polynomials or regular functions on the affine okay fine. So now that we have this done let me tell you that there is a little bit more to this proof that can be extracted a nice little lemma, here is a lemma if Y is any variety in our case that means either affine or quasi-affine variety and X is an affine variety say X is in An affine okay then a set theoretic map psi from Y to X is a morphism, it is a morphism of varieties if and only if Y followed by Ti Ti is a regular function on Y for every coordinate function Ti on An okay. So actually this is a very beautiful statement it tells you how to quickly verify that a set theoretic map is a morphism okay and so all it is saying is that you know from any variety which may be affine or not affine if you have a set theoretic map into an affine variety to check that it is a morphism of varieties all you have to do is you compose it with the projections which are the coordinate projections the target variety is sitting inside some affine variety on that affine variety you have the projections giving you the various coordinate functions. You compose this map with those projections the resulting functions will be functions from Y to K and all you have to verify is that these are their regular functions on Y. So it is a very beautiful statement it says a map from any variety into an affine variety a set theoretic map is a morphism if and only if it pulls back every coordinate function to a regular function that is what it says okay and this is the proof is actually the proof is already contained in the earlier proof that I gave you okay in the proof of the statement that the morphisms into an affine variety correspond to K algebra homomorphisms from the ring of polynomial functions on that affine variety nevertheless let me repeat it okay. So you see if and mind you this is an if and only if condition one way the proof is very very simple if psi is a morphism then of course then since T i since T i restricted to X are regular functions on X it is clear it follows that T i circle Y I am sorry I think it should have been T i circle psi not T i circle Y okay that is a that is a that is wrong notation T i circle Y does not make any sense okay I should compose T i with psi okay please correct that okay. So T i circle psi will of course be a regular function on Y this is very simple that is just because of the definition of a morphism because a morphism is supposed to pull back regular functions to regular functions okay. So T the coordinate functions on A n if I restrict them to X they are going to be regular functions on X because after all the coordinate functions on A n are the polynomials these are the polynomial variables okay. So when I say coordinate function T i on A n it means that I am thinking of the functions on A n to be given by the polynomial ring in the n variables T 1 through T n okay so each T i is a polynomial okay and you know polynomial functions are of course regular functions and therefore if you restrict each of these polynomials to X they are in O X and mind you O X is the same as A X O X is the same as A X because X is a fine okay and psi is already given to be a morphism. So it will pull back regular functions to regular functions and what is the pull back of T i restricted to X under psi it is just T i circle psi okay it is just the composition and that is supposed to be a regular function on Y because of definition of a morphism okay. So one way is very easy it is the other way that we need to understand. Conversely let T i circle psi if you want of course here when I say T i circle psi I mean T i restricted to X circle psi does not matter but that is what it means because I am evaluating T i on the image of psi which is X because psi goes into X okay. Conversely assume that psi is just a set theoretic map with this property that it pulls back coordinate functions on the target affine variety into regular functions okay that is this T i circle psi is in O Y for all I okay suppose this holds then I will have to show that psi is a morphism okay now how do I show that psi is a morphism I have to take two properties of psi namely the first thing I have to check is that psi is I will have to check that it is a you know continuous then the other thing I have to check is that it pulls back regular functions to regular functions. So the idea is very very simple so let me say tell you the idea in a very qualitative way how do I check psi is continuous I check psi is continuous by checking that the inverse image of closed sets are closed but what is the closed set a closed set on X is just given by common zero locus of a bunch of polynomials okay and what is the inverse image of such a set under psi it is a common zero locus of the bunch of regular functions on Y that I got from these polynomials by composing with psi and a set of common zeroes of a bunch of regular functions is a closed subset of a variety therefore psi pulls back regular function therefore psi pulls back closed sets to closed sets so psi is continuous it is very simple so let me write that down. So you know it is actually this it is actually this argument it is actually this argument but let me write it here if z of j in X is a closed subset then psi inverse of z of j is I am just reproducing what I have written there it is psi inverse of z of j is the intersection of z of h where h belongs to j and this is the intersection h belongs to j of z of h circle psi which is closed in X okay the zero set of a single regular function is a closed subset okay and an intersection an arbitrary intersection of closed sets is again closed subset therefore this is a closed set so what I have proved is that the inverse image under psi of a closed set is a closed set so this implies this implies that psi is continuous okay. Now the only other property that I have to check that check in order to ensure that psi is a morphism is that it pulls back regular functions to regular functions so now there is a small there is a small h here in my argument I need to I have already used the fact that h circle psi is a regular function it is something that I will have to show okay and that is purely a matter of verification so let me do the following thing so the correct way to do it is to use the fact that these are regular functions yeah so basically it is a calculation so what I need to do is the following so let me wrap this off just to say it in sequence so I need to tell that start with psi I have to make a small calculation start with an element h in start with an element h in a x h is just a polynomial in t1 through tn restricted to x this is what it means because x is a is an affine variety in an and the ambient affine space an in which x sits is supposed to have these as coordinates so the polynomial ring generated by this is a set of all polynomial functions on the affine space in which x sits and any element in the polynomial function set in the ring of polynomial functions on x is the restriction of a polynomial to x okay but of course the function is uniquely defined but the polynomial is not uniquely defined it is defined only up to and up to addition by an element of the ideal of x okay so choose h instead of h choose a polynomial okay that represents h and call that also as h okay then say h is if you want let me call this as g bar where g is g of t1 etc tn g bar is the image of g in a x which is a quotient of the polynomial ring in n variables in these n variables okay which is the ring of all polynomial functions on a n okay now look at so I want to calculate what the pull back of h under psi is this is what I want to do okay so what is what is what is h circle psi what is let us calculate this so basically what is happening is that I have I have y I have psi I have x and here I have a polynomial h okay and this is my map h circle psi okay it is just pull back of h under psi it is just psi star of h okay it is just this what is what is it if I evaluate it so if you take a point small y in capital Y that goes under psi to psi of y and that will go under h and you know the psi of y will have some coordinates lambda 1 etc lambda n okay and that if I apply the function h of course the function h is going into k which is a1 if you want okay with the Zariski topology alright so if I apply h applying h is the same as applying g okay so what I will get here is g of lambda 1 etc lambda n this is what the map this is the this is the this is this map which is h circle psi okay and if you look at it now it will be clear that h circle psi is a regular function on certainly a regular function on capital Y because it is actually a polynomial in the ti circle size in exactly the way in which g is a polynomial in the lambdas in the in the ti's see g is g of t1 etc tn okay so if you calculate h circle psi it will be actually just g of t1 circle psi etc tn circle psi you will get this because you know evaluate it evaluate it and check since h circle psi if I evaluated at a point what will I get I will get h of psi of y and h of psi of y psi of y is lambda 1 through lambda n and then I evaluate h on that h is represented by a polynomial g so I will get g of lambda 1 etc lambda n okay and you will also see that g of t1 circle psi etc tn circle psi evaluated at a point at the point y is just evaluating g on t1 circle psi of y comma dot dot dot tn circle psi of y but then you see psi of y is the point lambda 1 through lambda n and if I apply ti to that I will get lambda i so I will simply get g of lambda 1 etc lambda n okay. So this calculation this calculation tells you that h circle psi is just a polynomial in the ti circle psi so what is h circle psi it is a polynomial combination it is a polynomial in these functions but what are these functions they are already given to be regular functions on y each ti circle psi is already a regular function on y and so if you write if you give me a bunch of regular functions on y and write a polynomial in those regular functions on y with coefficients in the field the result is again a regular function because sum and product of regular functions is again regular function okay so it is so all this will tell you that h circle psi is in O y okay since each ti circle psi is in O y you will get this. So what this will tell you in principle it will tell you that psi pulls back regular functions on x to regular functions on y okay so this is one of the conditions for the map psi to be a morphism the only other condition now I have to check is the continuity of psi which will follow as I gave in the earlier proof. So only I have to check psi is continuous and that is because of the following thing psi is continuous because psi inverse of z of j is psi inverse of intersection of z of h let me put z of g, g belongs to j this is intersection g belongs to j z of g circle psi which is closed in y. So I am using the fact that if you give me a bunch of regular functions and look at the set of common zeros that is a close that gives a close to locus and the reason for that is every regular function is locally quotient of polynomials and looking at the zero set is actually the same as looking at the zero set of the numerator polynomial okay. So and we are done so you have verified that if psi pulls back coordinate functions to regular functions then not only is psi pulls back any regular function to a regular function but it is also continuous so psi is a morphism okay so it is a very beautiful segment it tells you very quickly how to verify that a map from in a given variety into a affine variety is a morphism okay it is very simple all you do is you just show that the pull backs of all the coordinate functions are regular functions that is all you have to check okay fine. So this is a fact that is kind of buried in the proof of the theorem that I gave in previous lecture so I just wanted to bring it out now what we will do is let us try to apply this and reconcile some of the results that we had that we have already talked about earlier. So if you recall so if you recall that you know if you have f in polynomial ring k1 etc to k of x1 through xn then and yeah then we have the bijection the following bijection. So you see you have you have you have an and this is just this is with this here is with coordinates x1 through xn okay and you had df the affine open the basic affine open set defined by f this is just the it is just the complement of the 0 set of f it is an-0 set of f okay so it is a locus where f does not vanish and that is an open subset and I told you that this is a basic open subset it is basic because every open subset can be written as union of such open subsets and because of Cauchy compactness in fact it can be written as finite union of such basic open sets and I told you that this the other beautiful thing about this subsets of this type is that they are actually themselves isomorphic to affine varieties and I told you that they are isomorphic to closed irreducible closed sub variety of affine space one dimension more okay. So how is how does that come about that comes about by taking an plus one of k with coordinates with coordinates the same x1 through xn but I add an extra coordinate y okay and what I do is well I look at the 0 set of yf-1 okay so f is a polynomial the first n variables and y times f becomes y-1 becomes a polynomial all the n plus 1 variables okay and this polynomial is irreducible and it is 0 set therefore gives an irreducible closed subset of affine space find you this is open and this is irreducible closed of course you must remember that this being an open non-empty open subset of affine space of course I am assuming this is a non-empty set okay f is a non constant polynomial okay and mind you that that is an open subset of affine space non-empty open subset means that it is it is also irreducible and it is also dense mind you this set here is irreducible and dense okay and this set here is irreducible and closed and we have we have this projection you project onto first n coordinates that is this projection map okay and under this projection map you have an isomorphism under the projection there is a bijective map from here to here this is something that I told you in and in fact I told you that this bijective map is actually an isomorphism of varieties okay in fact the map is very very simple to define you give me an element lambda 1 etc lambda n at which which is here which means f does not vanish at this element you simply associate to it the this this element lambda 1 through lambda n and then you add 1 by f of lambda 1 through lambda n then you know this satisfies the last coordinate y multiplied by f applied to the first n coordinates minus 1 equal to 0 which so it satisfies this equation and I can invert f of lambda 1 through lambda n because lambda 1 through lambda n is in the locus where f does not vanish okay. So I told this it is very easy to check that this is a bijective map but I told you actually to check as an exercise that is it is a homeomorphism of topological spaces I hope you have done that but now the time and then I told you that it is even an isomorphism of varieties I told you that it is an isomorphism of varieties and you know I told you that this is something that we will see later because when I told you at that time I had not defined what a morphism of varieties was but now that I have defined what a morphism of varieties is I can come go back and justify this statement okay. So if you look at this statement now okay what you can see immediately is that this is a this is a quasi-affine variety because this is an open subset one empty open subset of an affine variety and this is an affine variety okay and if you take this map I have a set theoretic map now it is even a bijective map I have a set theoretic map from here to here okay and let me take the map in this direction okay let me take the map in this direction it is a bijective map okay so you see I have a map from a variety to an affine variety this is an affine variety okay and it is a set theoretic map how do I check that it is a morphism I check that it is a morphism by checking that if I compose it with the coordinate projections okay then the resulting things give me regular functions on the source variety okay. So you see if I compose this map with the coordinate if I compose this map with the first coordinate projection I simply get the first coordinate projection here if I compose this map with the second coordinate projection and so on up to the nth coordinate projection I simply get the nth coordinate projection here which are of course regular functions okay because a polynomial is always a regular function and a regular function restricted to an open set is also a regular function there is no problem okay what about projection under the last coordinate okay if I project well I mean so that is it I mean I just have to I just check that if I project on to the last coordinate I get 1 by f okay if I take a point here I take the image there and I project on to the coordinate last coordinate what I get is a point here going to 1 by f of that point so I get the function 1 by f but 1 by f is also a regular function on df okay. So I have verified the condition that every projection of this map with the coordinate functions is a regular function therefore this becomes a morphism so it is a bijective morphism now okay now the only thing you have to worry about is that if it is a bijective morphism so you would be worried as I had warned in one of the earlier lectures that a bijective morphism need not be an isomorphism because the inverse map need not be a morphism okay but the fact is that what this induces at the level of regular functions okay is the following you will see that if I call this map as let me call this map by something let me call this as p no I should not call it this way is projection so I should not call it projection I should call it as I should call it something so let me let me give it some name well if you want okay let me call this as p sub f because this f is involved okay then you know this p sub f will induce alpha of p sub f which is a pullback map it is p sub f star it is a pullback map and what is this what is the pullback map it is it will go from the regular functions on the target so it will go from a of z of y f minus 1 to the regular functions on the source this is what you will get okay this is the pullback map this is the pullback map and see this is this coordinate ring the ring of functions is given by simply k of x 1 through x n, y divided by f y f minus 1 this is what it is this is the this is the ring of functions on that it is will close up it is just the it is just the ring of functions on the of ambient affine space which is the polynomial ring in these n plus 1 variables the first n being given by the x's and the n plus 1 variable given by y. Model of the ideal of this close subset the ideal of this close subset is the ideal generated by y f minus 1 because the y f minus 1 is an irreducible polynomial and it is an irreducible element in this polynomial ring which is a u f d unique factorization domain so the ideal it generates is prime so actually I should write here the radical of this ideal okay because if I take z of j then I will have to put I of z of j okay but I of z of j is rad j so I will have to put radical of y f minus 1 but radical of y f minus 1 is y f is ideal radical of the ideal generated by y f minus 1 is ideal generated by y f minus 1 because y f minus 1 is a prime ideal and that is because y f minus 1 is an irreducible polynomial okay and it is sitting inside this polynomial ring which is a u f d okay so this is the ring of functions and you can I want you to check that this is the same as polynomial ring in n variables localize that f okay so this is this is an exercise which I want you to do that to check that it is very clear that if you take an element here what is an element here in the localization it is of the form g by f power n where g is a polynomial in n these n variables these n axis divided by some power of f that is of course a regular function on df because it is a quotient of polynomials regular function is something that is locally a quotient of polynomials in this case it is globally a quotient polynomial so it is very clear that these guys are certainly here the fact is they are all you can check that the inclusion of this inside that is actually surjective map therefore this is actually equal to that okay this is this checking can be done and now after you do that checking you check that this map is the natural isomorphism you get in commutative algebra it is the map that sends y to it is a map that is given by sending xi to xi and send y to 1 by f okay this map comes from this direction because of the universal property of the polynomial ring in n plus 1 variables and the map comes in and this map is an isomorphism because the map in this direction comes from the universal property of the localization. So you can check that this pullback map is actually this isomorphism this isomorphism which is completely commutative algebraic okay so what this will tell you is that this map phi f which is a bijective morphism if you look at the pullback map the pullback map gives you an isomorphism okay and now you can if you believe that whenever you have this isomorphism of whenever you have an isomorphism then you know the object I mean this tells you that it is correct to define the it is correct to think of df of dff as an affine variety okay and to call this as a of dff which is what we started with we defined a of dff like this okay and I had given you I had proved that if you take the whole polynomial ring then the O of that I mean if you take the whole affine space then the O of affine space is same as E of affine space I told you that holds not only for affine spaces it holds for affine varieties it holds for basic opens okay and that is this checking that you have to do and so all this tells you that what we originally started with namely with defining this as a of df is the definition is correct because we define a only for affine varieties and df is an affine variety because it is actually isomorphic to this the fact that this bijective morphism is actually an isomorphism namely the crucial fact that the inverse map is a morphism comes from the fact that this is an isomorphism okay it comes from this fact it is reflected in this fact that this is an isomorphism alright. So this is justification for defining a of df as this set as this ring okay that is one thing and then the final thing that I wanted to say is about the equivalence of categories of affine varieties and finitely generated k-algebras which are integral domains so for that I need a little bit of functoriality so here is a here is a lemma that you can easily check if y to x1 to x2 phi psi is a sequence of morphisms of varieties with x1 and x2 affine okay then the morphism then the k-algebra who morphism k sorry a of x2 to a of y given by the pullback of psi circle phi which is psi circle phi per star is the same is the same as the composition I should be careful here it is a x2 I should not write a y I should write o y because as I told you it is a convention that we write a only if it is an affine variety okay y is just a general variety need not be affine but only x1 and x2 are assumed to be affine yeah. So let me continue the statement given by alpha of psi, psi circle phi is the same as the composition a of x2 to a of x1 to o of y which is psi a per star this is the same as alpha psi and this phi a per star this is alpha phi so I should write it as first apply phi a per star then apply sorry first apply psi a per star then apply phi a per star which is the same as first applying alpha of phi then apply alpha of psi okay this is this is just functoriality this is functoriality I am just actually you know this looks a little long to write down but all I am saying is that the notion of pullback of maps is very functorial which is I mean if I if you have a composition of morphisms pulling back a function on the on the on x2 all the way to y is the same as first pulling it back to x1 via psi and then further pulling back the resulting function via to y via phi okay this is a very this is something that you can check very easily okay there is nothing much about it okay now this is a very easy lemma to check okay so I will leave it to you but then the important corollary to this lemma is the following two affine varieties x1 and x2 are isomorphic and when I say isomorphic it means as varieties if and only if ax1 and ax2 are isomorphic as k algebras so you know this is basically because morphisms so the proof is it follows from the fact that morphisms of varieties from x1 to x2 via this alpha map which is pullback is this set can be biologically identified with morphisms of k algebras from ax2 to ax1 mind you here I should have put I should have put o x1 but I can replace o x1 by ax1 because x1 is affine okay if x1 is not affine then I have to replace this ax1 by o x1 that is what we prove alright and of course this map is as usual you send a morphism to alpha phi which is phi upper star this is just a pullback alright and what I want you to understand is that if this morphism phi is has an inverse then there is a psi such that phi circle psi and psi circle phi are corresponding identity maps okay and because of this lemma it will follow that phi star and psi star will be inverses of each other and that will tell you that phi star phi is an isomorphism of varieties if and only if that induced map on pullback of regular functions phi star is an isomorphism of k algebras okay so you get this corollary as a result of this lemma and the earlier theorem that morphisms into an affine variety are in bijective correspondence with k algebra homophisms from the polynomials on the affine variety polynomial functions restricted on the affine variety okay so this completes the statement that we have an equivalence of categories so let me write that down that is the final statement which is as I mentioned couple of lecture at the end of the lecture before the last one that it is a grand you know statement of the Hilbert Nullscher and Schwarz namely it is an equivalence of categories so let me write that down here so you have corollary we have an equivalence of categories so here is an equivalence of categories on this side we have a category of affine varieties okay these are the objects and when I say category of affine varieties it means that the objects are affine varieties of morphisms are morphisms of varieties and on this side we have the category of affine coordinate rings over k which is which should be defined as category of finitely generated k algebras that are integral domains and of course the morphisms there are k algebra homophisms and you have an equivalence of categories namely you have a functor in this direction okay a functor is a generalization of function okay but since the source and target are not sets but the categories we do not use a word function we use a word functor so you have functor like this you have which has a kind of inverse functor in this direction okay and what is the equivalence you give me any x you send it to A x so in this direction it is A okay and if you give me and a functor is not only supposed to map objects to objects it is also supposed to map morphisms to morphisms so if you give me another affine variety I get A y and if you give a morphism phi from x to y I get A of phi this A of phi is nothing but phi star pull back of functions which is just alpha of phi in all our previous notations okay so this is a functor for every object here you get an object there for every object here which is an affine variety I get an object there which is a finitely generated k algebras and that is an integral domain because it is a polynomial ring modulo a prime ideal alright so it is a finitely generated k algebras and given any morphism in this direction on this side I have pull back functions which is a k algebras homomorphism okay and this and that the set of all such psi is bijected to the set of all such k algebras homomorphism on this side is a statement that we have stated as a corollary there okay I mean it is part of the theorem that we proved and what is more what you should see is that the arrows are reversed an arrow in this direction use rise to an arrow in the other direction because you pull back functions from the target to the source so we say that A is a contra variant functor because it changes the direction of the arrows as it goes an arrow in this category is converted into arrow in the reverse direction in the target category so it is called a contra variant functor and what is the inverse functor in this direction the inverse functor in this direction is given by max spec okay it is given by max spec and namely if you start with the finitely generated k algebras if you start with the finitely generated k algebras which is k of x1 etc xn modulo some prime ideal okay then what you can do is that you can take max spec of that of that and this can be identified via the null still inserts to the 0 set of p as a as an affine variety in affine in an with coordinates with coordinates these xis okay and so if you go like this what happens is that you get the functor in this direction this functor is also contra variant okay and if you if you have bijection between sets what usually happens is that you start with an element here you go this way and then when you come back you should get the identity on this side and similarly for the other side but when you have a bijective equivalence of categories you will not get the identity what you will get is something up to isomorphism so if you start with something here you go you take ax and then if you take max spec of ax what you will get is something that isomorphic to x and why you get the isomorphism is because a of that and a of your original x will be isomorphic because of the choice of coordinates the isomorphism comes because you are choosing a bunch of coordinates okay the isomorphism comes because it depends when I write ax okay you are choosing coordinates. So here is the very very important subtle technical point the technical point is a following and affine space if you take an affine variety the affine variety can sit as a close subset in any affine in so many affine spaces for example take the plane the plane can simply sit inside a2 by the identity map it can sit as a plane it can sit as a 2 plane in a3 it can sit as a 2 plane in any an but you know if you but how is affine coordinate ring defined affine coordinate ring is defined based on the embedding based on the ambient affine space in which your affine variety is sitting. So if x is a plane sitting inside if x is the plane a2 then ax will become the polynomial ring in 2 variables but if x is say the xy plane sitting in 3 space then ax will become k xy z mod z which is again k xy okay. So the beautiful thing is this affine variety no matter in what affine space it is sitting in as a close sub variety the ax that you get will always be the same up to k algebra isomorphism okay that is a beautiful thing so the coordinate ring the ring of functions on the affine variety is a very nice object it does not depend in or it does not depend on the embedding of the affine variety in a certain affine space if it sits in some other affine space also the ring of functions will change only up to isomorphism which is a nice thing which goes on to tell you that the fact that something is an affine space is a very intrinsic thing okay that it tells you that trying to characterize an affine variety by its ring of functions is a very intrinsic thing because it does not depend on the extrinsic choice of embedding of putting that variety as a irreducible close subset of some affine space no matter in which affine space you put it as an irreducible close subset if you calculate the affine coordinate ring you will still get the same ring up to up to an isomorphism which means essentially a change of variables okay so you will get an isomorphic ring so what this tells you is that the affine coordinate ring or the ring of polynomial functions on an affine variety is defined up to isomorphism and depends only on the variety it does not depend on which affine space in which you are considering this as an affine variety okay so it is a very intrinsic object so this A is a very intrinsic object for an affine variety that is the subtle point. So we started with this definition then we are it is nice that we have come to this point where we are able to say that this does not depend on the affine space in which you are considering excess in irreducible close subset okay so with that I will stop this